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https://www.unitsconverters.com/en/Angstrom-To-Centimeter/Unittounit-8115-4134
Angstrom to Centimeter Units Converters AccelerationAngleAreaEnergyForceLengthPowerPressureSpeedTemperatureTimeVolumeWeight Percentage error Subtract fraction LCM of three numbers Discover more Unit Converter Angstrom to Centimeter (A to cm) -10%Copy+10%-10%Copy+10% = ⇄ Angstrom to Electron Compton Wavelength | Angstrom to Micrometer | Angstrom to Centimeter | Angstrom to Meter | Angstrom to Nanometer | Angstrom to Kilometer | Angstrom to Megameter More 👎) 👍 Converting 2X of 1 ▶1/2X of 1 ▶ 5X of 1 ▶1/5X of 1 ▶ 8X of 1 ▶1/8X of 1 ▶ Exameter Smallest Meter Base Terameter Biggest Result 1 Angstrom is equivalent to 1E-8 Centimeter Discover more Unit Converter Home » Wavelength » Angstrom to Centimeter Formula Used 1 Meter = 10000000000 Angstrom 1 Meter = 100 Centimeter ∴ 1 Angstrom = 1E-08 Centimeter Other Angstrom Conversions Angstrom to Terameter⇄ [A to Tm⇄] (Biggest) Angstrom to Gigameter⇄ [A to Gm⇄] Angstrom to Megameter⇄ [A to Mm⇄] Angstrom to Kilometer⇄ [A to km⇄] Angstrom to Hectometer⇄ [A to hm⇄] Angstrom to Decameter⇄ [A to dam⇄] Angstrom to Meter⇄ [A to m⇄] (Base Unit) Angstrom to Nanometer⇄ [A to nm⇄] Angstrom to Decimeter⇄ [A to dm⇄] Angstrom to Centimeter⇄ [A to cm⇄] (You are Here) Angstrom to Millimeter⇄ [A to mm⇄] Angstrom to Micrometer⇄ [A to μm⇄] Angstrom to Electron Compton Wavelength⇄ [A to Compton Wavelength⇄] Angstrom to Proton Compton Wavelength⇄ [A to P Compton Wavelength⇄] Angstrom to Neutron Compton Wavelength⇄ [A to N Compton Wavelength⇄] Angstrom to Petameter⇄ [A to Pm⇄] Angstrom to Exameter⇄ [A to Em⇄] (Smallest) Discover more Unit Converter Angstroms to Centimeters Conversion A stands for angstroms and cm stands for centimeters. The formula used in angstroms to centimeters conversion is 1 Angstrom = 1E-08 Centimeter. In other words, 1 angstrom is 100000000 times smaller than a centimeter. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale. Convert Angstrom to Centimeter How to convert angstrom to centimeter? In the wavelength measurement, first choose angstrom from the left dropdown and centimeter from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from centimeter to angstrom? You can check our centimeter to angstrom converter. Discover more Unit Converter FAQ about converter How to convert Angstrom to Centimeter? The formula to convert Angstrom to Centimeter is 1 Angstrom = 1E-08 Centimeter. Angstrom is 100000000 times Smaller than Centimeter. Enter the value of Angstrom and hit Convert to get value in Centimeter. Check our Angstrom to Centimeter converter. Need a reverse calculation from Centimeter to Angstrom? You can check our Centimeter to Angstrom Converter. How many Meter is 1 Angstrom? 1 Angstrom is equal to 1E-08 Meter. 1 Angstrom is 100000000 times Smaller than 1 Meter. How many Megameter is 1 Angstrom? 1 Angstrom is equal to 1E-08 Megameter. 1 Angstrom is 100000000 times Smaller than 1 Megameter. How many Kilometer is 1 Angstrom? 1 Angstrom is equal to 1E-08 Kilometer. 1 Angstrom is 100000000 times Smaller than 1 Kilometer. How many Centimeter is 1 Angstrom? 1 Angstrom is equal to 1E-08 Centimeter. 1 Angstrom is 100000000 times Smaller than 1 Centimeter. Angstroms to Centimeters Converter Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like wavelength finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like A to cm through multiplicative conversion factors. When you are converting wavelength, you need a Angstroms to Centimeters converter that is elaborate and still easy to use. Converting Angstrom to Centimeter is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Angstrom to Centimeter conversion along with a table representing the entire conversion. Discover more Unit Converter HindiFrenchSpanishMarathiPortugueseGermanPolishDutchItalianRussianGujaratiPunjabiTurkishKorean Percentage calculatorFraction calculatorLCM HCF Calculator About Contact Disclaimer Terms of Use Privacy Policy © 2016-2025. A softUsvista venture! Let Others Know ✖ Facebook Twitter Reddit LinkedIn Email WhatsApp Copied!
901
https://artofproblemsolving.com/wiki/index.php/Subset?srsltid=AfmBOooWN96W6fwiAe1FBkC4LOYgjlD8-0X7T12Rluj_OwfxXhyye0vv
Art of Problem Solving Subset - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Subset Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Subset We say a set is a subset of another set if every element of is also an element of , and we denote this by . The empty set is a subset of every set, and every set is a subset of itself. The notation emphasizes that may be equal to , while says that is any subset of other than itself. In the latter case, is called a proper subset. The following is a true statement: The set of all subsets of a given set is called the power set of and is denoted or . The number of subsets of is . Example Problems Introductory Counting the number of subsets in a set Intermediate 1992 AIME Problem 2 Retrieved from " Category: Set theory Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
902
https://sites.google.com/online.island.edu.hk/maths/year-7/unit-5-shapes/symmetry-squares
IS Mathematics - Symmetry Squares Search this site Embedded Files Skip to main content Skip to navigation IS Mathematics MYP Mathematics Year 7 Unit 1: Numbers Number Bracelets Diffy 1089 Multiplication Methods Max Product Four 4s Partitions Unit 2: Algebra I Think of a Number Addition Strips Pyramids Matchstick Patterns Tiles Multilink Staircases Climbing Stairs Earrings Frogs Using Spreadsheets Unit 3: Measuring Uncle Russ's Measuring Task Draw a Sports Field Borrett Rulers Mixing Paint Unit 4: Chance What are the Chances? Recur or Terminate Beads in a Bag Paper Fractions Multilink Cubes Spreadsheet Simulations Capture-Recapture Unit 5: Shapes Circle Patterns P=12 Triangles in Circles Folding Paper A = P Dotty Shapes The 9-Pin Geoboard Symmetry Squares Viral Spirals Rangolis Unit 6: Statistics Mean, Median, Mode Two Poems Penalty Shoot-out Dishonest Teacher Hong Kong Observatory Student Survey End of Year Activities Rangolis Tangrams Pentominoes Year 8 Unit 1: Numbers Divisors The Sieve of Eratosthenes The Jailer Problem Happy Numbers Shake Hands Divisibility Tests Loopy Snooker Unit 2: Algebra Grid Staircases Number Squares Odds & Evens Triangle Corners Diagonal Differences Unit 3: Chance The River Game Spinners Odds & Evens Sarah's Game Designing a Game Coloured Cubes Unit 4: Shapes Adding It All Up P-Numbers Pizza Slices Coordinate Codes What's My Line Mango Shop Graphing Stories Transformations Unit 5: Number Skills Fermi Questions The World as a Village Is That a Big Number? Unit 6: Statistics Matchmaking Are You Square? Just a Minute Mayfield High School End of Year Activities Secret Shapes V = 12 Deconstructed Cubes Exploring Solid Shapes Partially Painted Cubes Year 9 Unit 1: Numbers Fractions, Decimals and Percentages Equivalence Recurring Decimals (Extended Curriculum) Percentages Financial Maths Prime Factorisation, HCF, LCM (Extended Curriculum) Ratio and Proportion (Extended Curriculum) Rounding, Estimation and Bounds (Extended Curriculum) Unit 2: Algebra Negative Numbers (Prior Knowledge) Forming Expressions and Function Machine Simplifying Expressions Rearranging Formulae Substitution Expanding Brackets Factorising Equations Inequalities Simultaneous Equations Standard Form (Extended Curriculum) Indices (Extended Curriculum) Surds (Extended Curriculum) Unit 3: Shapes Angles Basics (Prior Knowledge) Angles in Polygons (Prior Knowledge) Quadrilaterals and Polygons (Prior Knowledge) Constructions Pythagoras' Theorem Right-Angled Trigonometry Circle Theorems (Extended Curriculum) Unit 4: Chance Sets Probability Tree Diagrams (Extended Curriculum) Unit 5: Graphs Straight Line Graphs Real Life Graphs Functions Unit 6: Three Dimensions Arcs and Sectors Volume and Surface Area Similar Shapes 3D Pythagoras and Trigonometry (Extended Curriculum) IGCSE Unit 1: Number Negative Numbers (Prior Knowledge) Decimals Fractions Fractions, Decimals and Percentages Equivalence Recurring Decimals Percentages Financial Maths Ratio and Proportion Direct and Inverse Proportion (Extended Curriculum) Rounding, Estimation and Bounds Unit 2: Graphs and Powers Prime Factorisation, HCF, LCM Indices Surds Standard Form Straight Line Graphs Real Life Graphs Simultaneous Equations Unit 3: Algebra 1 Rearranging Formulae Expanding Brackets Factorising Equations Inequalities Unit 4: Statistics and Transformations Transformations Vectors Averages and Spread Unit 5: Algebra 2 - Quadratics Equations Inequalities Simultaneous Equations Algebraic Fractions Algebraic Proofs Unit 6: Geometry 2 - Shapes Angles Basics (Prior Knowledge) Angles in Polygons (Prior Knowledge) Constructions Pythagoras' Theorem Similar Shapes Right-Angled Trigonometry 3D Pythagoras and Trigonometry (Extended Curriculum) Circle Theorems Arcs and Sectors Volume and Surface Area Non-Right-Angled Trigonometry IS Mathematics MYP Mathematics Year 7 Unit 1: Numbers Number Bracelets Diffy 1089 Multiplication Methods Max Product Four 4s Partitions Unit 2: Algebra I Think of a Number Addition Strips Pyramids Matchstick Patterns Tiles Multilink Staircases Climbing Stairs Earrings Frogs Using Spreadsheets Unit 3: Measuring Uncle Russ's Measuring Task Draw a Sports Field Borrett Rulers Mixing Paint Unit 4: Chance What are the Chances? Recur or Terminate Beads in a Bag Paper Fractions Multilink Cubes Spreadsheet Simulations Capture-Recapture Unit 5: Shapes Circle Patterns P=12 Triangles in Circles Folding Paper A = P Dotty Shapes The 9-Pin Geoboard Symmetry Squares Viral Spirals Rangolis Unit 6: Statistics Mean, Median, Mode Two Poems Penalty Shoot-out Dishonest Teacher Hong Kong Observatory Student Survey End of Year Activities Rangolis Tangrams Pentominoes Year 8 Unit 1: Numbers Divisors The Sieve of Eratosthenes The Jailer Problem Happy Numbers Shake Hands Divisibility Tests Loopy Snooker Unit 2: Algebra Grid Staircases Number Squares Odds & Evens Triangle Corners Diagonal Differences Unit 3: Chance The River Game Spinners Odds & Evens Sarah's Game Designing a Game Coloured Cubes Unit 4: Shapes Adding It All Up P-Numbers Pizza Slices Coordinate Codes What's My Line Mango Shop Graphing Stories Transformations Unit 5: Number Skills Fermi Questions The World as a Village Is That a Big Number? Unit 6: Statistics Matchmaking Are You Square? Just a Minute Mayfield High School End of Year Activities Secret Shapes V = 12 Deconstructed Cubes Exploring Solid Shapes Partially Painted Cubes Year 9 Unit 1: Numbers Fractions, Decimals and Percentages Equivalence Recurring Decimals (Extended Curriculum) Percentages Financial Maths Prime Factorisation, HCF, LCM (Extended Curriculum) Ratio and Proportion (Extended Curriculum) Rounding, Estimation and Bounds (Extended Curriculum) Unit 2: Algebra Negative Numbers (Prior Knowledge) Forming Expressions and Function Machine Simplifying Expressions Rearranging Formulae Substitution Expanding Brackets Factorising Equations Inequalities Simultaneous Equations Standard Form (Extended Curriculum) Indices (Extended Curriculum) Surds (Extended Curriculum) Unit 3: Shapes Angles Basics (Prior Knowledge) Angles in Polygons (Prior Knowledge) Quadrilaterals and Polygons (Prior Knowledge) Constructions Pythagoras' Theorem Right-Angled Trigonometry Circle Theorems (Extended Curriculum) Unit 4: Chance Sets Probability Tree Diagrams (Extended Curriculum) Unit 5: Graphs Straight Line Graphs Real Life Graphs Functions Unit 6: Three Dimensions Arcs and Sectors Volume and Surface Area Similar Shapes 3D Pythagoras and Trigonometry (Extended Curriculum) IGCSE Unit 1: Number Negative Numbers (Prior Knowledge) Decimals Fractions Fractions, Decimals and Percentages Equivalence Recurring Decimals Percentages Financial Maths Ratio and Proportion Direct and Inverse Proportion (Extended Curriculum) Rounding, Estimation and Bounds Unit 2: Graphs and Powers Prime Factorisation, HCF, LCM Indices Surds Standard Form Straight Line Graphs Real Life Graphs Simultaneous Equations Unit 3: Algebra 1 Rearranging Formulae Expanding Brackets Factorising Equations Inequalities Unit 4: Statistics and Transformations Transformations Vectors Averages and Spread Unit 5: Algebra 2 - Quadratics Equations Inequalities Simultaneous Equations Algebraic Fractions Algebraic Proofs Unit 6: Geometry 2 - Shapes Angles Basics (Prior Knowledge) Angles in Polygons (Prior Knowledge) Constructions Pythagoras' Theorem Similar Shapes Right-Angled Trigonometry 3D Pythagoras and Trigonometry (Extended Curriculum) Circle Theorems Arcs and Sectors Volume and Surface Area Non-Right-Angled Trigonometry More MYP Mathematics Year 7 Unit 1: Numbers Number Bracelets Diffy 1089 Multiplication Methods Max Product Four 4s Partitions Unit 2: Algebra I Think of a Number Addition Strips Pyramids Matchstick Patterns Tiles Multilink Staircases Climbing Stairs Earrings Frogs Using Spreadsheets Unit 3: Measuring Uncle Russ's Measuring Task Draw a Sports Field Borrett Rulers Mixing Paint Unit 4: Chance What are the Chances? Recur or Terminate Beads in a Bag Paper Fractions Multilink Cubes Spreadsheet Simulations Capture-Recapture Unit 5: Shapes Circle Patterns P=12 Triangles in Circles Folding Paper A = P Dotty Shapes The 9-Pin Geoboard Symmetry Squares Viral Spirals Rangolis Unit 6: Statistics Mean, Median, Mode Two Poems Penalty Shoot-out Dishonest Teacher Hong Kong Observatory Student Survey End of Year Activities Rangolis Tangrams Pentominoes Year 8 Unit 1: Numbers Divisors The Sieve of Eratosthenes The Jailer Problem Happy Numbers Shake Hands Divisibility Tests Loopy Snooker Unit 2: Algebra Grid Staircases Number Squares Odds & Evens Triangle Corners Diagonal Differences Unit 3: Chance The River Game Spinners Odds & Evens Sarah's Game Designing a Game Coloured Cubes Unit 4: Shapes Adding It All Up P-Numbers Pizza Slices Coordinate Codes What's My Line Mango Shop Graphing Stories Transformations Unit 5: Number Skills Fermi Questions The World as a Village Is That a Big Number? Unit 6: Statistics Matchmaking Are You Square? Just a Minute Mayfield High School End of Year Activities Secret Shapes V = 12 Deconstructed Cubes Exploring Solid Shapes Partially Painted Cubes Year 9 Unit 1: Numbers Fractions, Decimals and Percentages Equivalence Recurring Decimals (Extended Curriculum) Percentages Financial Maths Prime Factorisation, HCF, LCM (Extended Curriculum) Ratio and Proportion (Extended Curriculum) Rounding, Estimation and Bounds (Extended Curriculum) Unit 2: Algebra Negative Numbers (Prior Knowledge) Forming Expressions and Function Machine Simplifying Expressions Rearranging Formulae Substitution Expanding Brackets Factorising Equations Inequalities Simultaneous Equations Standard Form (Extended Curriculum) Indices (Extended Curriculum) Surds (Extended Curriculum) Unit 3: Shapes Angles Basics (Prior Knowledge) Angles in Polygons (Prior Knowledge) Quadrilaterals and Polygons (Prior Knowledge) Constructions Pythagoras' Theorem Right-Angled Trigonometry Circle Theorems (Extended Curriculum) Unit 4: Chance Sets Probability Tree Diagrams (Extended Curriculum) Unit 5: Graphs Straight Line Graphs Real Life Graphs Functions Unit 6: Three Dimensions Arcs and Sectors Volume and Surface Area Similar Shapes 3D Pythagoras and Trigonometry (Extended Curriculum) IGCSE Unit 1: Number Negative Numbers (Prior Knowledge) Decimals Fractions Fractions, Decimals and Percentages Equivalence Recurring Decimals Percentages Financial Maths Ratio and Proportion Direct and Inverse Proportion (Extended Curriculum) Rounding, Estimation and Bounds Unit 2: Graphs and Powers Prime Factorisation, HCF, LCM Indices Surds Standard Form Straight Line Graphs Real Life Graphs Simultaneous Equations Unit 3: Algebra 1 Rearranging Formulae Expanding Brackets Factorising Equations Inequalities Unit 4: Statistics and Transformations Transformations Vectors Averages and Spread Unit 5: Algebra 2 - Quadratics Equations Inequalities Simultaneous Equations Algebraic Fractions Algebraic Proofs Unit 6: Geometry 2 - Shapes Angles Basics (Prior Knowledge) Angles in Polygons (Prior Knowledge) Constructions Pythagoras' Theorem Similar Shapes Right-Angled Trigonometry 3D Pythagoras and Trigonometry (Extended Curriculum) Circle Theorems Arcs and Sectors Volume and Surface Area Non-Right-Angled Trigonometry Symmetry Squares Introduction In this lesson, you will be learning about reflection and rotational symmetries. (Click on the hyperlinks if you are unsure what they are.) 3 x 3 squares How many reflection and rotational symmetries does the diagram on the right have? By shading in different boxes of the 3x3 square (PRINT ME), how many other ways could exactly 2 squares be shaded so that the shape has exactly one line of symmetry? How many different “types” of solution are there? How many other ways could the 2 squares have been shaded so that the shape has exactly two lines of symmetry? Repeat points 2 and 3 above if you can now shade in 3 squares. What about 4 squares? Can you write down the answer for 5, 6, 7 squares without doing any work? Why/ why not? 5 x 5 squares Now look at the 5x5 square on the right, Shade one more square so that the array has reflection symmetry. How many solutions are there? What are they? Where are the mirror lines? Take a look at this worksheet (PRINT ME). Find the two different solutions for each puzzle. Include the mirror line (line of symmetry) on your solution. For each question there are two grids so draw one solution on each grid. When you have finished, for each solution write down the number of dots that there are on the mirror line. What can you say about the number of dots on the mirror line? Why? 4x4 squares Look at the 4x4 squares on the right, describe the symmetries in each of these patterns. Now, use these 4x4 squares to create shapes that match these rules: Use 8 dots and make four different patterns that only have rotational symmetry (not line symmetry). Produce four patterns that have reflection (lines of) symmetry but no rotational symmetry. Use an odd number of dots to produce four patterns, each one with four lines of symmetry. Come up with your own interesting rules to investigate. Further Questions and Challenges In the 5x5 square task above, all the problems have exactly two solutions. That is, there are two different places to put the last dot if you want to leave a symmetrical pattern. Can you develop a problem that has: 0 solution 1 solution 3 solutions 4 solutions What is the greatest number of possible solutions for a single problem? Use these 5x5 squares (PRINT ME). Further Practice In this lesson, we learnt about rotational and line symmetries. Practise the relevant skills on DrFrostMaths, CorbettMaths,MyiMaths and Eedi. Watch any video and/or go through any online lesson as you see fit. Transum Look at this video if you need further support. Starters: Dice Reflections: A dice is reflected in two mirrors. What numbers can be seen? Freemason's Cipher: Find symmetric words in this ancient cipher. Match Fish: A classic matchstick puzzle designed to challenge your spacial awareness. Mirror Maths: The bottom half of some symmetrical calculations are shown above. Can you work out the answers? Reflective Cat: On squared paper copy the drawing of the face then reflect it in three different lines. Rotational Symmetry: Draw a pattern with rotational symmetry of order 6 but no line symmetry. Wrapping Paper: Find the order of rotational symmetry of the repeating pattern. Activities: Pattern Clues: An interactive activity challenging you to reproduce a pattern of coloured squares according to given clues. Symmetry Table Challenge: In how many cells can you draw symmetrical shapes with the given row and column headings? Xmas Symmetry Pairs: Match the pictures with the description of their symmetry. Snowflake Generator: See how the hexagon can be transformed into a snowflake with some basic translations. Polygons: Name the polygons and show the number of lines and order of rotational symmetry. Rotational Symmetry Pairs: The traditional pairs or pelmanism game adapted to test knowledge of rotational symmetry. For more goodies on Symmetry on Transum, click on the hyperlink. Extension Take a look at these two interesting videos: Symmetry, reality's riddle: Marcus du Sautoy' TED talk illustrating some aspects of symmetry. The science of symmetry: An illustrated talk about the notion of symmetry and how it applies to nature. Report abuse Page details Page updated Report abuse
903
https://www.publichealth.columbia.edu/research/population-health-methods/missing-data-and-multiple-imputation
Skip to content Columbia University Irving Medical Center Columbia University Mailman School of Public Health Missing Data and Multiple Imputation | Overview | Software | | Description | Websites | | Readings | Courses | Overview Data that we plan to analyze are often incomplete. Study design strategies should ideally be set up to obtain complete data in the first place through questionnaire design, interviewer training, study protocol development, real-time data checking, or re-contacting participants to obtain complete data. When obtaining complete data is not feasible, proxy reports or the collection of characteristics associated with the missing values can help. Missing data can be categorized in multiple ways. Perhaps the most troubling are the data missing on entire observations (e.g., due to selection bias) or on entire variables that have been omitted from the study design. Somewhat more tractable, but still potentially problematic, are data missing on a subset of variables that are missing for a subset of the observations. In this case, it can be useful to label those observations without missing data as “complete cases” and those with some missing data as “partial cases.” Ideally, we hope that the amount of missing data is limited, in which case we will rely less heavily on our assumptions about the pattern of missing data. Missing data can bias study results because they distort the effect estimate of interest (e.g. β). Missing data are also problematic if they decrease the statistical power by effectively decreasing the sample size, or if they complicate comparisons across models that differ in both the analysis strategy and the number of included observations. Description The amount of bias potentially introduced by missing data depends on the type of missing data. What you hope for: Missing completely at random (MCAR). By stating that data are MCAR, we assume that the missing values are not systematically different from the values we did observe. For example, imagine a standardized test which randomly assigns a subset of questions to each student. We could reasonably assume that the characteristics of students receiving different versions of the test would be similar, given large enough sample sizes. Even though some of the questions will have missing data, we have a clear understanding of the random process leading to these missing data patterns. Second best: Missing at random (MAR). When data are MAR, the missing values are systematically different from the observed values, but the systematic differences are fully accounted for by measured covariates. In this situation we can use what we know about partial cases to compensate for bias due to missing data. For example, imagine a pop quiz administered on a single day to all students, with complete data among those present and missing data for all who were absent. By linking to the full enrollment and attendance records, we see that quiz scores were lower on average among students with a poor attendance record, and there was more missing data for this group. Yet if we assume that being absent on quiz day was random after you account for the prior attendance record, we can use the available data to extend what we know about observed scores to the missing scores. The worst: Non-ignorable (NI) missing data, also sometimes labeled not missing at random (NMAR) or informative missing data. Concerns about NI data may be raised when missing values are thought to systematically differ from observed values. This can happen if (1) the missing value itself influences the probability of missingness or (2) some unmeasured quantity predicts both the value of the missing variable and the probability of missingness. Building on the example given above, let’s consider an optional quiz for which scores will be displayed publicly. Students who are apprehensive about their quiz score may avoid participating. They may have an unobserved history of low scores on practice quizzes, or the high-level of anxiety itself may hinder their performance. In either case, the characteristics of those abstaining from the quiz would make it difficult to identify a comparable group of students who completed the quiz. Other examples could include loss to follow-up as a direct result of illness in a prospective health study, or study assessments that were incomplete due to participant symptoms during the procedure. How can we distinguish MCAR, MAR, and NI missing data? In reality, we often have to rely on prior knowledge and assumptions. Showing that observed characteristics are similar among those with and without missing data can help to support a MCAR assumption. However, we cannot usually rule out NI missing data, since these are defined by a systematic difference across unmeasured quantities. Often, the best we can do is to investigate how sensitive our results are to different missing data assumptions. Another way to categorize missing data patterns is as monotone or arbitrary, a distinction that has practical implications in planning your strategy to address missing data. The most concise definition of monotone missing data that I’ve seen is that the data can be arranged such to make the following true: if Variable J is missing then Variable K is also missing for all K>J. This is often depicted visually is an array with observations as rows, and variables as columns, as a triangular or square block of data missing from the lower right corner. I can most easily imagine a monotonic missing data pattern occurring from loss to follow up: everyone with missing values at a particular study visit has dropped out and is also missing those values at all subsequent visits. Monotone missing data are in some ways simpler to work with, but this pattern is often suggestive of NI missing data if not by design. Options for analysis Options for dealing with missing data are relatively easy to implement in standard software. Comparisons across multiple methods may reveal that results are robust to the assumptions made about missing data, or they may provide extreme cases that likely surround the truth. Complete case (aka listwise deletion) is often the default, provided that missing data are coded in a way that the software recognizes (e.g., “.”). This approach discards partial cases, and is asymptotically unbiased if data are MCAR. Missing values can be treated as a separate category. Using this approach for confounders may allow for residual confounding if the missing category is not homogenous. (Note: if you decide to use this approach with continuous variables by replacing missing values with the mean, consider adding an interaction term between the predictor of interest and the indicator of missingness to minimize bias.) Censoring is a strategy commonly used for longitudinal data in a proportional hazards model when the outcome is missing. When the outcome can no longer be observed for certain individuals, those individuals are simply removed from the comparisons going forward. Another type of censoring may take the form of a “floor” or “ceiling” beyond which data are missing. Censoring-related strategies use the available information and may be appropriate for extreme NI missing data. Single imputation essentially consists of filling in the missing data with plausible values. The range of single imputation strategies differ in their strengths and weaknesses: Impute to mean or median (simply filling in a typical value for all missing data may be biased, but it limits the leverage of missing data) Impute based on regression analysis (accounts for MAR data, but is optimistic because the regression error term is not carried forward) Stochastic regression imputation (like above but appropriately adds uncertainty) Hot deck imputation (non-parametric approach based on matching partial and complete cases) Cold deck (like above, but matched to external data) Carry forward/carry backward (for longitudinal data with relatively stable characteristics) interpolation/extrapolation (for longitudinal trends, usually assumes linearity) Worst-case analysis (commonly used for outcomes, e.g. missing data are replaced with the “worst” value under NI assumption) Multiple imputation relies on regression models to predict the missingness and missing values, and incorporates uncertainty through an iterative approach. Key advantages over a complete case analysis are that it preserves N without introducing bias if data are MAR, and provides corrects SEs for uncertainty due to missing values. Tips for implementing multiple imputation Input variables to include: any that predict whether data are missing as well as variables that are correlated with the value of the missing data. Often this includes exposure, covariates, outcome, and other available data on study administration or on proxies for the variable with missing data Consider transformations to improve normality of variables with missing data or to enforce restrictions (e.g. log-transformation to force positive values only) Include interactions or nonlinear forms if they improve the models predicting missingness or missing values Diminishing returns make 5-10 imputed datasets sufficient in most situations (but some recommend as few as 3 or as many as 20) Set a seed number in order to get reproducible results (otherwise, results will vary slightly from one run to the next) Make sure data are logically consistent after MI (avoid “impossible” combinations e.g. never-smokers with a non-zero value for pack-years) Readings Textbooks & Chapters Allison, P.D. (2002) Missing Data. Sage Publications.A nice brief text that builds up to multiple imputation and includes strategies for maximum likelihood approaches and for working with informative missing data Little, R.J.A. and Rubin, D.B. (1987) Statistical Analysis with Missing Data. J. Wiley & Sons, New York. Rubin, D.B. (1987) Multiple Imputation for Nonresponse in Surveys. J. Wiley & Sons, New York. Schafer, J.L. (1997) Analysis of Incomplete Multivariate Data. Chapman & Hall, London. Gelman, A. and Hill, J. (2007) Ch 25: Missing-data imputation in Data Analysis Using Regression and Multilevel/Hierarchical Models. Cambridge University Press, New York. Methodological Articles Use of multiple imputation in the epidemiologic literature Author(s): MA Klebanoff, SR Cole Journal: American journal of epidemiology Year published: 2008 What do we do with missing data? Some options for analysis of incomplete data Author(s): TE RaghunathanJournal: Annu Rev Public HealthYear published: 2004 Imputation of missing values is superior to complete case analysis and the missing-indicator method in multivariable diagnostic research: a clinical example Author(s): GJ van der Heijden, AR Donders, T Stijnen, KG MoonsJournal: J Clin EpidemiolYear published: 2006 Multiple imputation for missing data in epidemiological and clinical research: potential and pitfalls Author(s): JA Sterne, IR White, JB Carlin, M Spratt, P Royston, MG Kenward, AM Wood, JR CarpenterJournal: BMJYear published: 2009 Multiple imputation versus data enhancement for dealing with missing data in observational health care outcome analyses Author(s): PD Faris, WA Ghali, R Brant, CM Norris, PD Galbraith, ML KnudtsonJournal: J Clin EpidemiolYear published: 2002 Software/Programming Articles State of the Multiple Imputation Software Author(s): RM YucelJournal: J Stat SoftwareYear published: 2011 Much ado about nothing: A comparison of missing data methods and software to fit incomplete data regression models Author(s): NJ Horton, K KleinmanJournal: Am StatYear published: 2007 Application Articles Association of black carbon with cognition among children in a prospective birth cohort study Author(s): SF Suglia, A Gryparis, RO Wright, J Schwartz, RJ Wright Journal: Am J Epidemiol Year published: 2008 Survival associated with two sets of diagnostic criteria for congestive heart failure Author(s): GD Schellenbaum, TD Rea, SR Heckbert, NL Smith, T Lumley, VL Roger, et al.Journal: Am J EpidemiolYear published: 2004 Early-life and adult socioeconomic status and inflammatory risk markers in adulthood Author(s): RA Pollitt, JS Kaufman, KM Rose, AV Diez-Roux, D Zeng, G HeissJournal: Eur J EpidemiolYear published: 2007 Race/ethnicity and breast cancer estrogen receptor status: impact of class, missing data, and modeling assumptions Author(s): N Krieger, JT Chen, JH Ware, A KaddourJournal: Cancer Causes ControlYear published: 2008 Chlorpyrifos exposure and urban residential environment characteristics as determinants of early childhood neurodevelopment. Author(s): GS Lovasi, JW Quinn, VA Rauh, FP Perera, HF Andrews, R Garfinkel, L Hoepner, R Whyatt, A RundleJournal: Am J Public HealthYear published: 2011 Software Description: All standard statistical programs can be used to implement missing data techniques, though some may allow for more sophisticated techniques over others. We recommend using this webpage by Stef Van Buuren which provides an annotated list of the software and packages that can be used to implement missing data techniques. Websites Statistical Computing Seminars: Multiple Imputation in Stata, Part 1 Website overview: This webpage is hosted by UCLA’s Institute for Digital Research and Education. This particular page is the first of a two part series on implementing multiple imputation techniques in Stata. Missingdata.org Website overview: This website is solely devoted to missing data. It has information on courses, books and workshops, as well as discussion groups and other helpful tips on how to address missing data. Website overview: This website is a companion to the book “Flexible Imputation of Missing Data” by Stef Van Buuren. This website contains an overview, course materials as well as helpful information for implementing missing data techniques in numerous software packages such as R, Stata, S-Plus, SAS and SPSS. Courses Missing Data and Multiple Imputation Host/program: The Epidemiology and Population Health Summer Institute at Columbia University (EPIC) Software used: SAS and Stata Join the Conversation Have a question about methods? Join us on Facebook JOIN JOIN Back to top
904
https://www.shmoop.com/common-core-standards/ccss-hs-g-c-3.html
More on Common Core Standards The Standards About High School: Number and Quantity High School: Algebra High School: Functions High School: Modeling High School: Geometry High School: Statistics and Probability Grade 8 Grade 7 Grade 6 Grade 5 Grade 4 Grade 3 Grade 2 Grade 1 Kindergarten High School: Geometry High School: Geometry Circles HSG-C.A.3 3. Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle. At the heart of it, the study of mathematics is the study of relationships. Before jumping into Romeo and Juliet, though, let's be clear that the relationships of interest in mathematics are those between mathematical objects. Not between two star-cross'd lovers. But in terms of depth, Shakespeare's got nothing compared to the beauty with which Euclid explored the deep and intrinsic connections between circles and triangles. Yeah, we're talking about a triangle's inscribed and circumscribed circles. Students should know that an inscribed circle is the largest circle that can fit on the inside of a triangle, with the three sides of the triangle tangent to the circle. A circumscribed circle is one that contains the three vertices of the triangle. Students should know the difference between and be able to construct both of these circles. Students should understand that while circles have one defined center, triangles try to outdo them by having four different centers: the incenter, the circumcenter, the centroid, and the orthocenter. Imitation is the sincerest form of flattery and all, but do you think maybe they're compensating for something? Students should know the definitions of these centers and how each one differs from the others. But how do we get to the true center of the triangle? The world may never know. We can draw any triangle by plotting three points on a circle. Students can take this one point further, plot four points on a circle, connect them, and make a cyclic quadrilateral. (They're called that because the vertices are all on the circle, in kind of a, well, cycle.) Students be able to prove theorems about cyclic quadrilaterals, the most important of which is that opposite angles in a cyclic quadrilateral add to 180°. Or, if students are wearing their fancy pants today, they can say that opposite angles in a cyclic quadrilateral are supplementary. We would say that too, but our fancy pants are at the dry cleaners. This only touches on the deep and fascinating love triangle between circles, triangles, and quadrilaterals. So maybe these shapes are less Romeo and Juliet and more A Midsummer Night's Dream. Well, the Lysander-Helena-Demetrius bit, anyway. Links Aligned Resources Video More standards from High School: Geometry - Circles Aristotle Albert Einstein Tired of ads? Logging out… Logging out... You've been inactive for a while, logging you out in a few seconds... Why's This Funny?
905
https://www.youtube.com/watch?v=_omfCpLSphc
B.Sc. maths, Envelope & Evolute (Q. find the evolute of the parabola y^2 = 4ax) H.E mathematics 1470 subscribers 412 likes Description 24257 views Posted: 1 Apr 2022 Hello friends students in this video we will solve a problem based on evolute 43 comments Transcript: हेलो हेलो फ्रेंड्स एंड स्टूडेंट्स स्पीड कम यूज करेंगे एंड ऑफ द वर्ल्ड के बारे में भी केवल यह समझते थे वर्ल्ड हमारा क्या होता है कि CBSE कब का समय वाउल्ड फाइंड करना है तो हम उसके घर पर पहले फाइनल करेंगे इक्वेशन ऑफ नॉर्मल और नार्मल का एंड फ्लॉप ही मुश्किल आता है लाइक नॉरमल कार्यक्रम के नार्मल का एनवेलप नार्मल काइंड ऑफ वर्ल्ड है को पहन कर दो हम रिमाइंड करना है फाइंड ए व्होल लोट आफ पैराबोला वॉइस के रिक्वेस्ट फ्लैट तो पहले हम आएंगे इस पर इक्वेशन ऑफ नार्मल को पर नार्मल का जो हमारा चैनल अब होगा वह हमारे ब्लड आता है तो हमें वर्ल्ड निकालने के लिए पहले नार्मल निकाला गंध नार्मल सा नमक ऐड करना हुआ नॉर्मल ही हमें निकालना वह क्या एंड फुल आफ लाइफ विकेट पहले इसको नार्मल 1 से निकाल रहा हूं और यह मानकर चल लव लैटर MB झाल अंधेरे का अगर हम इस तरह का मजाक प्ले स्टोर कर सकते हैं में हिमालयन एप्स है यहां पर यह हमारा वायक्षरा इट और यह हमारा बराबर होता है यह यह वाला आर्म होल होता है अगर वोल्टेज हम इस पॉइंट पर एक पॉइंट पीली हमने इस पर हमें Android गया तो नॉर्मल आ रही हो जाएगा लाइफ हमारा त्यौहार है तो यह मालूम और इस नॉर्मल है कि इस नार्मल कहां इस नॉर्मल ही एडाप्टिव बॉलीवुड के जाता है अब हम पहले नार्मल एक्शन फाइट करने जा रहे हैं और साथ हम इस पॉइंट्स पेमेंट कर लेंगे स्लॉट की फॉर्म में लाइफ इसकी हमने स्लो स्लो पुलिस प्रवक्ता कि हम यह मान लें जा रहा हूं मैं मान लूंगा लेफ्ट एवं विदा इस मोर MB दाणा ए स्लो हो आफ नॉरमल स्लोप आफ नॉरमल लाइफ इन मान लिया कि हमारा एमजेड मास लोग है नॉर्मल का लाइफ एंड कैमरा स्लोगन नार्मल का मतलब नार्मल के प्रवक्ता है नॉर्मल के प्रवक्ता ए नॉरमल डिलीवरी के प्रवक्ता राइट अध्ययन के जॉइंट करेंगे हमें पैराग्राफ को जिम्मेदार ठहरा बोला है हमारा वाइफ इज इक्वल टू फॉरेक्स लाइक तो पहले में फंसे फर्स्ट डेरिवेटिव निकालूंगी यह फोन डीएक्टिवेट डिफरेंसेस विद रिस्पेक्ट मुझे मिल जाए टू माई डिवाइस अपांडेक्स इज इक्वल टू फ्लाइट तो मुझे यहां से मिलाकर डिवाइस अपांडेक्स इस इक्वल टू 14.21 तो मुझे यह वाले पॉइंट इस पॉइंट टू यह देखिए यह हमारा नार्मल है तो इसका वैसे प्रॉब्लम होगा और - बन टाइम्स लाइक तो हम जैसे निकाल दे तो यह क्या स्लॉट पेंडेंट है कि हमारा स्लो कर दो ए स्ट्रेंज Android और होने का लवली स्लॉट नार्मल स्लॉट नार्मल हमारा माइनस वन अपॉन लोकरंजन आपने पढ़ा हुआ है कि 1005 अब देखिए हमारा सिलेक्ट का यह स्लो हमने मान लिया है मै यह मैंने हुआ एम तो हमारा एवं बराबर हो जाएगा माइनस वन अपॉन 0.2 यह फोन बाइक तो हम यहां से मिल जाएगा देखिए हुआ है को आवास उपलब्ध का यह मिल जाएगा एवं इस इक्वल टू माइनस 1.2 क्लिक कर देंगे इसका मीनिंग आफ यॉजिक पोस्ट में अमन-चैन कायम - 2015 में ऑनलाइन कर लिया वाइट है वह हमको यह पॉइंट का वह रिकॉर्ड नहीं मिल रहा है हमको पॉइंट निकाल हैं हम पहला पॉइंट पी निकाले थे गाइड फिल्म इजीली हम जो 2199 जॉइंट कर लेंगे अब माइक्रोवेव या फिर कर देगी पर बोला मैंने तो हमें एक बार मिल जाएगा फुल हम कि इक्वेशन वन है तो हम यहां पर मिल जाएगा फुर पेश किए टेनिस प्लेयर इज इक्वल टू फॉर नोएडा एक्सप्रेस वे कैंसर हो गया तो हमें मिले ₹50 हुआ था और एक ही से ही तो है एम्स पीर विजिट्स पॉइंट टू बैक टो अप्वॉइंट अगर यह हो गया और यह हमारा एक है यह मेरा भाई है तो पॉइंट हमको मिल गया पॉइंट रिकॉर्ड में हमको मिल गए एक्सप्लेनेशन पीएम ओं कि यह शहर को - बैक टू बैक हमने पॉइंट निकाले पॉइंट हमारा यह इसके - 281 फाइनलाइजेशन आफ नॉरमल एंड पॉइंट्स फीस तो गई प्वाइंटों पर गिरा और स्लो हमारे पास हीरा स्लो अपने मन हुआ तो हमारा एक सो नार्मल का यह मुश्किल रूप में एवं कृषि फॉर्म में हमारा स्लो एक सो नार्मल हमको मिल जाएगा और हुआ है कर दो मेरा टैब पॉइंट p.m. कि इस बाई - बाय वन इज इक्वल टू एडमिन टू एक्स माइनस 10 फॉल होता है वह यौवन कितना प्लस किट्टू एवं है हु इज इक्वल टू नेम एक्स - एक्स वन हमारा एग्जाम इस वेबसाइट तो यह बजाओ हवाई प्लस टू एवं इज इक्वल टू एक्स माइनस सफेद का ट्यूबलाइट इसको मैं ले जाऊंगा तो मुझे इसी प्लस टू एक्स माइनस टू हैव - यह आम का फूल की वाइफ यही है हमारा एक तरफ नार्मल परिस्थिति का हम रिमाइंड करना है एंड विल टॉक अब हमें इसका LMB फाइंड करना है यहां तक पर लंबा करना होगा फिर क्वेश्चन आता है माघ मास में जयपुर सावन मास में और एंड कि हमारा व्हेन इज द पैरामीटर है कि अ है मैम के नंबर पर मित्रों अब हम इस पर एंड डिफरेंट करेंगे हम यहां से हम कोई लिमिट कर देंगे विद थे हेल्प आफ ठेर डिफरेंसेस इन पाटिल डिप्रेशन का यूज करना होगा और मैं यहां से हम कोई लिमिट कर दूंगी कि हमें करना क्या हुआ आपका तो मैं इसको 15 वाइट यह सब्सक्राइब कर देंगे अब हम यह इस पर याद रखना है टाइम इसको पढ़ रहा हूं पहले मैं अ है लेकिन फ्रेंड्स अब हमें इस का एल्बम को साइन करना है तो मैं इसको निमिया एक्शन टू राइट इन उबंटू मैंने इसको नाम दे दिया है तो एक तो अब मैं इसका एंड्रोफाइंस दूंगा तो मैं इसका डिफरेंस कर दूंगा पार्स्ली विद रिस्पेक्ट टो हिम तो मुझे क्या मिलेगा का यहां पर मैं लिखूंगा डिफरेंस एंड पार्शियली है है इधर एक्सप्रेस टू एंड तो मुझे वाकई विटामिन जामुन अधिक उजागर केवल इलेक्शन - 2m का उपयोग नस काट हुए और - ए गेम क्योंकि जब अतिशय 10 कर दो 15 अब यहां से मैं हम केवल उस पॉइंट करूंगा तो इसको मैं ले जाऊंगा फ्री एडमिट किया इज इक्वल टू एक्स माइनस टू हैव यू फॉर बॉयज वेयर इज इक्वल टू जरा एक्स माइनस 12.38 तो एवं हमको मिल जाएगा Samsung Galaxy हो गया और कबहूं न एक्ट - 28130 का रूट रूट ला सकते हैं आप और वाइट वाइट इसे हम का बल यहां पर हम कट कर देंगे लेकिन थोड़ा सा सिस्टेमेटिक इनवेस्टमेंट करेंगे कि फ्रॉम एक्शन टू स्ट्रांग रिलेशन टू थोड़ा देश में चेंजेस करेंगे हम इसको लॉग इन बॉयज इक्वल टू हम यहां से को मिलूंगा तो यह जरा एक्स माइनस टू बैक यौतूबे और यहां आपको मिलेगा - एग स्पीयर लाइक एवं हमने अशोक मनेरिया ऐप को मजबूत कितना कंडीशन में हमको वापिस रिक्वेस्ट है मिल जाएगा का यहां हम केवल एक्स माइनस टू हुई है कि अपऑन थ्य का रूट इंदु एक्स माइनस तुम्हें यहां पर हमारे - नमस्कार हमारा एक्स माइनस 2013 ए ए कि एक्स माइनस टू अपऑन ठेर राइट तो हमारे ऐसे कई लोग हैं और कोलेस्ट्रॉल को माना गया तो हमको मिल जाओ बाय रिक्वेस्ट ऊ कि एक्सेस - 12511 2.38 ड्राई यहां से 16 मार्च 2018 का या और ब्रैकेट में हमको मिल जाए वन माइनस 1.3 लख कि मंत्रियों को मिलेगा व्यक्ति - टू वे को मारा गया यहां से बंद का व्यापक असर - का 193 लाइक टो वेयर इज इक्वल टू जावरा एक्स माइनस टू बैक टू थ्री फोर वन व टू प्लस का वन और यह हो गया टू वाइफ इन तो यह 2.3 रूट्स चाहिए इस मौके पर हम अगर देंगे तो यह धारा 323 लिए इंप्रूव आप इज इक्वल टू टू इन टू एक्स माइनस झुक पावर आफ डिवाइडिंग अब देखिए स्कैन हम यहां पर कर देंगे इसके रिंग कर देंगे दोनों साइड पर स्कैन करेंगे तो यह में क्या मेरे यहां पर स्क्रीन कर देंगे दोनों साइड का तो जरा ध्यान से देखिए अमेरिका या इधर प्रजा फ्री विनोद 36 यौवन टू 4 इंच की स्क्रीन करेंगे तो यह तो कैंसर हो जाएगा योजना में एक्स माइनस टू बैक का पोर्शन क्योंकि हमारा फॉर एडवांस रिसर्च में उगे थैंक्यू मिलते हैं नेक्स्ट वीडियो में है और स्ट्रेस देखिए इ तो यही है हमारा यह वॉल्यूम यह हम उस महिला का यह है कोई नार्मल का जो हमारा एंड अप होगा नार्मल का जो आर्म बड़ा होगा वह हमारा एवं करके आता है यही हमारा दीपांशु रिक्वेस्ट एंड एग्रो फूड ओं हुआ है हुआ है हुआ है हुआ है में रिक्वायर्ड कि बॉलीवुड अ कि आप The पैराबोला है की वाइफ इज इक्वल टू फॉरेक्स जब हमको यह वाला पॉइंट करना होगा हम पहले उसका लोकल नार्मल जॉइंट करेंगे और उस नार्मल का एंड ब्लैक टी आउटपुट कैसा आएगा कर उसका योर व्होल डे मील कुक नॉर्मल और नार्मल का एंड ऑफ द वर्ल्ड के ओके थैंक यू इन नेक्स्ट वीडियो में
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https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-circle-basics/v/informal-argument-for-the-area-of-circle-formula
Area of a circle intuition (video) | Circles | Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. We've updated our Terms of Service. Please review it now. 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We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content High school geometry Course: High school geometry > Unit 8 Lesson 1: Circle basics Getting ready for circles Circles glossary Area of a circle intuition Proof: all circles are similar Math> High school geometry> Circles> Circle basics © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Area of a circle intuition Google Classroom Microsoft Teams 0 energy points AboutAbout this videoTranscript Using triangles to create an informal argument for the area of a circle formula. Skip to end of discussions QuestionsTips & Thanks Want to join the conversation? Log in Sort by: Top Voted Michael Adikaibe 8 years ago Posted 8 years ago. Direct link to Michael Adikaibe's post “I know this is off topic,...” more I know this is off topic, but what is a seven sided polygon called? AnswerButton navigates to signup page•CommentButton navigates to signup page (20 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer Ahmad 4 years ago Posted 4 years ago. Direct link to Ahmad's post “7 sides is a heptagon.” more 7 sides is a heptagon. 1 commentComment on Ahmad's post “7 sides is a heptagon.” (20 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Satvik 8 years ago Posted 8 years ago. Direct link to Satvik's post “Even if n approaches infi...” more Even if n approaches infinity the area of the circle won't be extremely precise. Wouldn't it be off by a little AnswerButton navigates to signup page•CommentButton navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer andrewp18 8 years ago Posted 8 years ago. Direct link to andrewp18's post “In the sense that we can ...” more In the sense that we can only evaluate the area for finite 𝑛, the answer is yes, we will always be a little off even for very large 𝑛. However, the concept Sal is getting at here is the notion of a limit. That is, what does some function approach as the input approaches some value? In this case, even though the area function only approximates the area of a circle for all finite 𝑛, we can take the limit as 𝑛 → ∞ which will give us the area of the shape that this process is approaching as we increase 𝑛 (namely a circle). The concept of the limit is easily misunderstood when it is first introduced and there are rigorous definitions that make it even harder to understand, but hopefully you see the intuition behind Sal's logic. We are not looking at subcases for finite 𝑛, rather we are seeing what the area approaches as 𝑛 gets arbitrarily large (𝑛 → ∞). Also, I highly recommend to anyone who asks about calculus that they focus on algebra, trigonometry, and geometry first before studying calculus. So don't worry about limits now if you haven't made yourself familiar with everything that precedes it. Comment if you have questions! 1 commentComment on andrewp18's post “In the sense that we can ...” (16 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more captainamericawhyso 3 years ago Posted 3 years ago. Direct link to captainamericawhyso's post “But saying that the area ...” more But saying that the area of a circle is approaching πr^2 is different than saying: the area IS πr^2...right? AnswerButton navigates to signup page•CommentButton navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer David Severin 3 years ago Posted 3 years ago. Direct link to David Severin's post “This is better understood...” more This is better understood when you take precalculus because Sal is talking about limits. Sal states that as n approaches infinity, the area approaches πr^2 which is a true statement, as you add more and more sides to the polygon, the area gets closer and closer to the formula. If there were such a thing as an infinite number of sides (a circle does not have sides) which would mean each point on the circle were considered a side, then it would reach the area formula. The theory might be beyond what you have seen in math before. CommentButton navigates to signup page (12 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Arbaaz Ibrahim 6 years ago Posted 6 years ago. Direct link to Arbaaz Ibrahim's post “At about 7:30 minutes int...” more At about 7:30 minutes into the video, Sal said, nb is approaching the circumference, but how is this possible? AnswerButton navigates to signup page•2 commentsComment on Arbaaz Ibrahim's post “At about 7:30 minutes int...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer Jason Wu | 编者masterwu 5 years ago Posted 5 years ago. Direct link to Jason Wu | 编者masterwu's post “hi, I'm here to help. whe...” more hi, I'm here to help. when sal said nb is approaching the circumference he also mentioned when n approaches infinity and multiplied by b it will approach the circumference since the space will become 0.I hope that I have helped you.If not ask Sal. CommentButton navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Arthur, Hi😁 10 months ago Posted 10 months ago. Direct link to Arthur, Hi😁's post “help I don't know how to ...” more help I don't know how to do this AnswerButton navigates to signup page•1 commentComment on Arthur, Hi😁's post “help I don't know how to ...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer Jerry Nilsson 10 months ago Posted 10 months ago. Direct link to Jerry Nilsson's post “Don't worry, this is not ...” more Don't worry, this is not the type of problem you'll see on a test or anything. The purpose of this video is just to give an intuition for why the area of a circle is equal to 𝜋𝑟², so that we can actually use that formula with a bit more rigor behind it instead of having to resort to blind faith. As is often the case with proofs (although this is not a formal proof) it takes quite a few steps to get to the end result and it's easy to lose track along the way of what we're actually trying to achieve, which is to show that the area of the polygon gets closer to 𝜋𝑟² as the number of sides increases, and because the area of the polygon also gets closer to that of the circle it's fair to assume that the area of the circle is equal to 𝜋𝑟². The actual math involved is really nothing fancy, we're just using the formula for the circumference of a circle, 𝐶 = 2𝜋𝑟, and the formula for the area of a triangle, 𝐴 = 𝑎𝑏∕2, together with some rudimentary algebra, so if you just take it slowly you should be able to follow along. The hardest part is really the thought experiment towards the end – what happens to 𝑎 and 𝑛𝑏 as 𝑛 gets larger (i.e. approaches infinity)? – but I think Sal explains it pretty well in that 𝑛𝑏 is the perimeter of the polygon, which by observation seems to approach the circumference of the circle, and thereby 𝑎, which is the distance between the perimeter and the center, should approach the radius of the circle. 1 commentComment on Jerry Nilsson's post “Don't worry, this is not ...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more dlopez5125 a month ago Posted a month ago. Direct link to dlopez5125's post “this is very sigma” more this is very sigma AnswerButton navigates to signup page•CommentButton navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer Julianngthowhingplays 5 years ago Posted 5 years ago. Direct link to Julianngthowhingplays's post “When all those properties...” more When all those properties approach the properties of a circle, does it ever become it? AnswerButton navigates to signup page•CommentButton navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer J15 a year ago Posted a year ago. Direct link to J15's post “No. For a circle to be pe...” more No. For a circle to be perfect, we would need to measure an infinite number of points around the circle's circumference to know for sure. A perfect circle is impossible to achieve in real life. From subatomic particles to carefully built structures, nothing in the physical world can be a perfect circle. CommentButton navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Ahmed Rahmi 9 months ago Posted 9 months ago. Direct link to Ahmed Rahmi's post “I assume if you throw a l...” more I assume if you throw a limit in there we get a firm proof instead of an intuition AnswerButton navigates to signup page•CommentButton navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer mistudubey 5 years ago Posted 5 years ago. Direct link to mistudubey's post “Hi! Do you guys have a vi...” more Hi! Do you guys have a video on the various parts of a circle like the chord,secant,tangent,etc etc. AnswerButton navigates to signup page•CommentButton navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer J15 a year ago Posted a year ago. Direct link to J15's post “ more the previous video, on Circles Glossary. Hope that helps! CommentButton navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Asia Quispe 3 years ago Posted 3 years ago. Direct link to Asia Quispe's post “Fun fact: this video was ...” more Fun fact: this video was published December 22, 2015. That was 7 years ago as of March 29, 2022. AnswerButton navigates to signup page•CommentButton navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show previewShow formatting options Post answer Jayanthika a year ago Posted a year ago. Direct link to Jayanthika's post “How do you find out when ...” more How do you find out when the video was published? CommentButton navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript - [Voiceover] What I'd like to do in this video is make an informal argument for why the formula for the area of a circle is Pi r squared. And we're gonna start just with the most traditional definition of the number Pi and that's that and I'll just do it here in a corner some place that Pi is equal to the ratio of the circumference and the diameter of the circle or the ratio of the circumference to the diameter of a circle or, of course, we could write this as ratio of the circumference to, instead of the diameter, I could write two times the radius or I can multiply both sides times two times the radius and we get our traditional formula for the circumference, for the circumference of a circle, but once again, this literally just comes straight out of the definition for the number Pi. The number Pi is defined as the ratio between the circumference and the diameter, so you just multiply both sides out of it times the diameter, you get the circumference is equal to Pi times the diameter. Now, we have the circumference formula right here, once again, this comes out of the definition of Pi, but from this, I'd like to at least get an intuitive feel for why the area formula is given by Pi r squared and to think about that, we're going to approximate the area of polygons, the areas of polygons, that are inscribed in a circle. So over here, I have this, what's this? There's a five-sided polygon right over here and its area is going to be equal to, so the area of this polygon, it's going to be five times the area of each of those triangles and the area of each of those triangles, the height is a, the base is b, so it's going to be base times height or height times base times 1/2. So this, once again, five times ab over two, it's not that good of an approximation. This would be the area of just the inscribed polygon, so we're definitely underestimating the area of the entire circle or leaving out all of these little, these little chunks outside of the polygon, but still inside of the circle. But as we add more size to the polygon, we see that we're leaving out less. We see, when we have now, this is a one, two, three, four, five, six, seven-sided polygon, we're leaving a little bit less. We're underestimating still, but we're underestimating by less. This area that we're giving up isn't as large as this area right over here. So in this approximation, we have, what did I say? Seven triangles? One, two, three, four, five, six, seven triangles. And the area of each of those triangles is once again ab over two. Now, a and b here are different than a and b over here, and notice what's happening. As we increase, as we increase the number of triangles, not only is it approximating the area of the circle better, but a is getting longer. And you can see, you could imagine, as we increase many, many, many more triangles, a is going to approach r. Now another thing to think about is what is seven, what is seven times b approaching? So we're saying that a is approaching r as we add more of these sides of the polygon, as we add more triangles, now what is the number of triangles times the base of the triangle, what is that approaching? Well this is going to approach the perimeter of the, or this is going to be the perimeter of the polygon, so seven times b is that plus, let me actually, let me draw this, it's that plus that plus that, I think you get the point, plus that, plus that, plus that, plus that. So once again, seven, let me write this at seven, times b, that is the perimeter of the polygon, perimeter, perimeter of the polygon. So think about what's happening. As we have more and more sides of the polygon, our a, our height of each of our triangles, is going to approach our radius, is going to approach the radius. It's going to get the height of each of the triangles. It's gonna get longer and longer, it's gonna approach the radius as we have, as we approach an infinite number of triangles, and then the number of polygons we have, not the number of polygons, the number of sides we have times the bases, that's going to be the perimeter of the polygon and as we add more and more sides, as we add more and more sides, the perimeter of the polygon is going to approach, is going to approach, is going to approach the circumference of the circle. I'll write it out, circumference, circumference, and you'll see that even more clearly right over here. So, once again, how many sides do I have here? I have one, two, three, four, five, six, seven, eight, nine, 10 sides, so this, I can write the perimeter of the polygon as 10 times b and then if I multiply that times a over two, if I multiply that times, we'll use another color. a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle 'cause a times b over two, that's the area of each of these triangles and then I have 10 of these triangles, but now let's think about this more generally. Let's think about it if I were to have n, if I were to have an n-sided polygon, so I have n-sided polygon, then I'd be approximating the area as n times b, n times b, we see this right over here. When n is equal to 10, you have 10 times b. So it's n times b times a over two. Times a over two. I just wrote, this isn't something mysterious. The base times the height divided by two, this right over here, that's the area of each triangle and then I'm gonna have n of these triangles, so this is our approximation for the area, so let me write this, the area is going to be approximately that right over there. It's going to be n, the number of triangles I have, times the area of each triangle. Now what's going to happen, as n approaches infinity, as I approach having an infinite sided polygon, as I have an infinite number of triangles, so let's just think this through a little bit. 'cause this is where it gets interesting. Now this is the informal argument. To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence. So let's just think about what happens as n approaches infinity. So, as n approaches infinity, we've already said, as we have more and more sides and we have more and more triangles, a approaches r, so let's write that down. So, a is going to approach r, the height of the triangles is going to approach the radius and what else is going to happen? Well, n times b, the perimeter of the polygon, the perimeter of the polygon is going to approach the circumference. So, a is going to approach r and n times b is going to approach the circumference, is going to approach the circumference, or another way of thinking about it, if it's approaching the circumference, we could say that n times b is going to approach two Pi times the radius 'cause that's what the circumference is going to be equal to. So, if a is approaching the radius and nb is approaching two Pi r, well then, what is the entire, what is the area of, what is the area of our polygon or what is the area of our polygon going to, or, the area of our circle going to be? Well, it is going to approach, it is going, or I should say, the area of our polygon is going to approach, nb is going to approach two Pi r, nb is going to approach two Pi r. Instead of nb, I'm writing two Pi r there. a is going to approach r. a is going to approach r, and then I'm dividing it, and then I am dividing it by two. So as n approaches infinity, as we have infinite number of sides of our polygon, an infinite number of triangles, the area of our polygon will approach this, which is equal to what? Well you have two divided two and then Pi r times r is equal to Pi r squared. So as we approach having infinite number of triangles and infinite, I wanna keep that there, infinite number of sides, we see that we approach the area of the circle and as we approach the area of the circle, we are approaching Pi r squared. So hopefully this gives you an intuitive sense why this right over here is the formula for the area of a circle. You could think about it as the area of an infinite sided polygon that is inscribed in the circle, which will be equal to the area of the circle. 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907
https://www.whatisbiotechnology.org/index.php/exhibitions/antimicrobial/index/early_observations
Home Education Exhibitions People Places Sciences Interviews Timeline Menu Short presentation of the key points in this exhibition Introduction What is antimicrobial resistance and how big is the problem Early warnings about AMR René Dubos - antibiotics pioneer and AMR campaigner The history of penicillin Pencillin before Fleming Early therapy with Fleming's penicillin Mounting evidence of AMR Salvarsan - the first antimicrobial drug Sulphonamides - a new weapon Mary Barber - AMR surveillance champion Uncovering AMR's biological mechanisms Discovery of bacteria genetic transfer Esther Lederberg - a pioneer of bacterial genetics Plasmid-mediated resistance Microbial virulence Emergence of microbial population genetics Early methods for identifying bacteria Overview of AMR diagnostics Testing for antibiotic susceptibility Resistance testing - molecular diagnostics begins Value of rapid diagnostic - case study of pneumonia Andrew Conway-Morris and Vilas Navapurkar discuss developing their pneumonia diagnostic Affordable portable AMR diagnostic at the point of care Ankur Mutreja discusses developing an affordable rapid AMR diagnostic Archana Madhav discusses developing dipstick for affordable diagnostic Alternative treatment avenues to antibiotics Stephen Baker discusses the potential of vaccines and monoclonal antibody drugs to overcome AMR problem Timeline Glossary Acknowledgements and credits First observations of bacterial genetic transfer Figure 4.1.1: Frederick Griffith, n.d. (Credit: Maclyn McCarty, Steven Lehrer, Wikipedia). Griffith (1879-1941) was born in Hale, Lancashire county, England, and completed a medical degree at University College in Liverpool in 1901. After graduating Griffiths served as a house-physician and house surgeon at Liverpool Royal Infirmary alongside working as a pathologist at the Thompson Yates Laboratory. He also was involved in scientific investigations for the Royal Commission on tuberculosis. Griffith moved to London in 1910 to work as a medical officer for the Local Government Board and in 1911 joined the Pathology Laboratory run by the British Ministry of Health. His career was cut short in 1941 when he was killed in an air raid during the London blitz. The first person to observe that bacteria might have the capacity to transfer genetic material was Frederick Griffith, who from 1918 became involved in efforts to help create a vaccine to stop the spread of bacterial pneumonia which had contributed to a disproportionately large number of deaths during the influenza pandemic of 1918-19. Before he could create a vaccine he needed to understand the epidemiological patterns of pneumonia and its pathology. In order to do this Griffith collected a large number of pneumococci samples isolated from patients around the country which he classified into different groups. Overall Griffith found there to be two forms of pneumococci bacteria. Each of them differed in their appearance. One was smooth (S) and had an outer coat (capsule) of polysaccharides. By contrast the other (R) was rough and lacked a capsule. Out of the two strains the S one was discovered by Griffith to be more lethal when injected into mice. While mice who received the S strain died within a few days, those given the R strain survived. Griffith subsequently discovered that he could destroy the virulence of the S strain by heating the bacteria. Soon after his initial observations, Griffith noticed another strange phenomenon - all the mice died when injected with a mixture of heat-killed S bacteria and live R bacteria. This was particularly striking because no mice had died previously when given the bacteria by themselves. Furthermore, when Griffith isolated the live bacteria from the hearts of the dead animals injected with the mixed strains they had smooth capsules characteristic of the S strain. Additional experiments indicated that this new characteristic was passed on to succeeding generations of bacteria. Puzzled by what he had found Griffith repeated the same experiments several times, each time getting the same result. Publishing his findings in 1928, Griffith hypothesised that a chemical substance was somehow being transferred from the heat-killed S strain to transform the previously harmless R strain. What the chemical nature of the transforming factor was remained a mystery. All that was known was that it could survive heat (Griffith). Figure 4.1.2: Griffith's experiments with different pneumococcal strains (Credit: Madeleine Price Ball). Griffith's discovery was largely ignored until 1934 when Oswald Avery launched a series of experiments (Figure 4.1.2) to identify the specific molecule that seemed to transform the R strain into the more virulent S strain. This he did with Maclyn McCarty and Colin MacLeod. They hypothesised that either a protein or a nucleic acid was responsible for the genetic change in the bacteria. Taking careful measures to purify the pneumococcal extracts, the three scientists established that DNA was the key chemical substance that lay behind the change. It was the first time DNA was directly observed to transform an organism. Despite the results produced by the Avery group, not everyone was willing to accept that DNA, rather than proteins, stored and transmitted genetic information. Figure 4.1.3: Oswald Avery, 1937 (Credit: Rockefeller Institute). Avery (1877-1955) was born and grew up in Halifax, Canada, before moving with his family to New York when he was 10. After completing a medical degree at Columbia University College of Physicians and Surgeons, Avery spent some time as a practising physician. In 1907 he switched his career to conduct laboratory research on bacteria. By 1913 Avery had moved to the Rockefeller Institute where he spent the next thirty-five years studying Diplococcus pneumoniae, a species of bacteria that causes pneumonia. Avery believed that the development of an effective treatment for bacterial pneumonia rested on a detailed biochemical understanding of the pneumococcal cell. In 1923 he demonstrated that the sugar coating of pneumococci helped trigger the immune response. Based on this discovery he began looking for an agent to destroy the sugar coating of the bacteria. One of the people who helped him in this pursuit was Rene Dubos. Click here for more information about Rene Dubos. Figure 4.1.4: Avery, McCarty and MacLeod's experiment (Adapted from Avery-MacLeod-McCarty). Figure 4.1.5: Photograph of Maclyn McCarty (left) shaking hands with Francis Crick and James Watson by Marjorie McCarty (Credit: Lederberg, Gotschlich). McCarty (1911-2005) was born in Indiana and completed his first degree at Stanford University and then did his medical training at Johns Hopkins. By 1940 he had moved to New York University where he began working with Avery at the Rockefeller Institute. Figure 4.1.6: Colin McLeod, n.d. (Credit: Wikipedia). McLeod was born in Port Hastings, Nova Scotia, and completed a medical degree at McGill University. In 1941 he became the chair of the Department of Microbiology at New York University and soon after joined Avery's laboratory. Figure 4.1.7: from Griffith, credit: Rockefeller University Press. It shows the dramatic change linked to the transformation from 'rough' (and avirulent) cells (left) to 'smooth' (virulent) cells (right). The question as to whether proteins or DNA carried genetic information remained contentious for nearly a decade. It was only resolved following a series of experiments carried out at Cold Spring Harbor Laboratory by the American geneticists Alfred Hershey and Martha Chase between 1951 and 1952. Their experiment involved tracking the transfer of proteins and DNA between a virus and its host. For their research they chose to work with the T2 bacteriophage, a type of virus that infects bacteria. One of the advantages of the bacteriophage (phage) was that, like all bacterial viruses, it had a simple structure which consisted of a protein-based outer wall and a DNA core. Moreover, it was known to reproduce by injecting its genetic material into a bacterium while leaving its protein shell outside the bacterium. Once inside the bacterium, the phage replicates by hijacking the bacterium’s reproductive machinery. This process eventually destroyed the host (Oregon State University SCARC). While researchers knew that phages left their protein shell on the outside of the bacterium, they could not be sure whether genetic material was carried by certain proteins into the bacterium itself. Hershey and Chase proposed to settle this question by conducting two different types of experiment. In the first they tagged the phage’s DNA with a radioactive tracer called Phosphorus-32 and in the second they tagged its proteins with sulfur. Once tagged the phages were introduced into E. coli bacteria and the infected bacteria were then broken up with a blender and the phage’s protein shells were separated from their host with the help of a centrifuge. Following this measurements were made of the radiation levels in both the E. coli cells and protein shells. Contrary to the expectations of the two scientists, the experiments revealed that phage only transferred DNA, rather than protein, into the interior of bacterium. This provided conclusive evidence that DNA, not protein, was the source of genetic material. (Hershey Chase). Figure 4.1.8: Martha Chase and Alfred Hershey, 1953. Photographer: Karl Maramorosh (Credit: Oregon State University Special Collections & Archives Research Center). Born in Owosso, Michigan, Hershey (1908-1997) gained a degree in chemistry and then did a doctorate in bacteriology at Michigan State University. Chase (1927-2003) was born in Cleveland, Ohio and studied the genetics of fruit flies as part of her undergraduate degree at the College of Wooster, New York. Following her degree, Chase joined Hershey at Cold Spring Harbour Laboratory (CSHL) as a research assistant in 1950 where she helped design and carry out the seminal experiments demonstrating DNA to be the transmitter of genetic information. Despite Chase’s major role in the research Hershey failed to acknowledge her contribution when he was awarded the Nobel Prize in Physiology or Medicine in 1962 on the back of the work. Chase left CSHL in 1953 to join Oak Ridge National Laboratory as a research assistant and then went to the University of Southern California where she completed a doctorate in microbiology in 1964. Sadly her scientific career was ended by a series of setbacks. References Avery, OT, MacLeod, CM, McCarty, M (1 Feb 1944) Studies on the chemical nature of the substance including transformation of pneumococcal types', The Journal of Experimental Medicine, 79, 137-58Back Griffith, F (1928) ‘The significance of pneumococcal types’, Journal of Hygiene, 28/2: 113-59.Back Hershey, AD, Chase, M (1958) ‘Independent functions of viral protein and nucleic acid in growth of bacteriophage’, Journal of General Physiology, 36/1: 39-56. Back Lederberg, J, Gotschlich, EC (2005) 'A Path to Discovery: The Career of Maclyn McCarty', PLoS Biol 3/10, e341. Back Oregon State University Special Collections & Archives Research Center (2008) ‘The Hershey-Chase Blender Experiments’, The Pauling Blog Back << Unravelling the biological mechanism behind AMR Respond to or comment on this page on our feeds on Facebook, Instagram, Mastodon or Twitter. 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908
https://www.khanacademy.org/math/ncert-class-11/xea4762213f311c5e:introduction-to-three-dimensional-geometry-ncert-new/xea4762213f311c5e:untitled-1055/v/distance-between-2-points-in-3d
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909
https://wisc.pb.unizin.org/chem109fall2020ver03/chapter/day-31/
Day 31: Le Châtelier’s Principle, Equilibrium and Gibbs Free Energy – Chemistry 109, Fall 2020 Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Book Contents Navigation Contents Introduction Welcome to Chemistry 109! Learning Requires Your Effort How Will This Course Help Me Learn? What Will I Learn in Chemistry 109? Unit One Day 1: Chemistry, Matter, Energy, Models D1.1 Substances and Chemical Reactions D1.2 Atoms, Molecules, and Ions D1.3 Chemical Symbols, Formulas, and Equations D1.4 The Periodic Table D1.5 Matter, Energy, Models D 1.6 Structure, Energy, and States of Matter D1.7 What’s Ahead? Day 2: Atomic Spectra and Atomic Orbitals D2.1 Electromagnetic Radiation D2.2 Atomic Spectra D2.3 Atomic Energy Levels D2.4 The Quantum Mechanical Model of the Hydrogen Atom D2.5 Wave-Particle Duality D2.6 Atomic Orbitals and Quantum Numbers Day 3: Orbital Energy and Electron Configuration D3.1 Atoms with More than a Single Electron D3.2 Orbital Energy Level Diagrams D3.3 Electron Configurations D3.4 Valence Electrons D3.5 Effective Nuclear Charge D3.6 Periodic Variation in Atomic Radius Day 4: Periodic Trends; Forces between Atoms D4.1 Periodic Variation in Ionization Energies D4.2 Periodic Variation in Electron Affinities D4.3 Electron Configurations of Monoatomic Ions D4.4 Unpaired Electrons and Magnetism D4.5 Ionic Radii D4.6 Forces Between Atoms D4.7 Metals Day 5: Ionic Compounds; Covalent Bonding D5.1 Ionic Compounds D5.2 Lattice Energy D5.3 Properties of Ionic Compounds D5.4 Covalent Bonding: Molecular Orbitals D5.5 Molecular Orbital (MO) Diagram D5.6 Electron Configurations and Bond Order Day 6: Molecular Orbitals; Lewis Structures D6.1 Second-Row Diatomic Molecules D6.2 Bond Length and Bond Enthalpy D6.3 Bonding in Molecules with More Than Two Atoms D6.4 Lewis Structures for Covalent Molecules D6.5 General Guidance for Drawing Lewis Structures D6.6 Exceptions to the Octet Rule Day 7: Covalent Molecular Substances; Hydrocarbons D7.1 Covalent Molecular Substances D7.2 Hydrocarbons D7.3 Alkanes D7.4 Alkenes D7.5 Alkynes D7.6 Petroleum Chemistry D7.7 Attractions Between Atomic-scale Particles Unit Two Day 9: Bond Properties; Valence Bond Theory D9.1 Bonds, Molecules, and Structures D9.2 Bond Polarity D9.3 Electronegativity D9.4 Formal Charge D9.5 Resonance Structures D9.6 Aromatic Molecules D9.7 Valence Bond Theory Day 10: Hybrid Orbitals; Molecular Geometry D10.1 Types of Hybrid Orbitals D10.2 Predicting the Geometry of Bonds Around an Atom D10.3 Three-dimensional Bond Geometry D10.4 Molecules with More Than One Central Atom D10.5 Hybridization and Bond Angles D10.6 Hybridization in Resonance Hybrids Day 11: Molecular Structure: Isomers D11.1 Line Structures D11.2 Isomeric Structures D11.3 Conformations D11.4 Constitutional Isomers D11.5 Stereoisomers: Geometric Isomers D11.6 Stereoisomers: Enantiomers D11.7 Intermolecular Forces Day 12: Intermolecular Forces; Functional Groups D12.1 Intermolecular Forces: Dipole-Dipole Attractions D12.2 Functional Groups D12.3 Aldehydes and Ketones D12.4 Ethers D12.5 Esters Day 13: Alcohols, Carboxylic Acids, Amines, Amides; Hydrogen Bonding D13.1 Alcohols D13.2 Carboxylic Acids D13.3 Amines D13.4 Amides D13.5 Reactions of Alcohols, Amines, and Carboxylic Acids D13.6 Intermolecular Forces: Hydrogen Bonding D13.7 Intermolecular Forces: Water Solubility Day 14: Macromolecules D14.1 Addition Polymers D14.2 Polymer Structure and Properties D14.3 Various Addition Polymers D14.4 Conjugated Diene Polymers D14.5 Copolymers Day 15: Condensation Polymers, Proteins D15.1 Condensation Polymers D15.2 Polyesters D15.3 Polyamides D15.4 Proteins 15.5 Amino Acids D15.6 Protein Structure D15.7 Protein Folding and Denaturation Day 16: DNA and Lipids D16.1 DNA D16.2 Base Pairing D16.3 Lipids D16.4 Glycerolipids D16.5 Phospholipids 16.6 Proteins, Lipids, and Fatty Acids Unit Three Day 18: Reaction Rate D18.1 Reaction Rate D18.2 Relative Rates of Reaction D18.3 Factors Affecting Reaction Rates D18.4 Effect of Concentration: Rate Laws D18.5 Method of Initial Rates D18.6 Reaction Order and Rate Constant Units Day 19: Integrated Rate Law D19.1 Integrated Rate Laws D19.2 First-Order Reaction D19.3 Second-Order Reaction D19.4 Zeroth-Order Reaction D19.5 Flooding Method: Pseudo-Order Reaction Day 20: Rate of Radioactive Decay D20.1 Radioactive Decay D20.2 Types of Radioactive Decay D20.3 Half-Life of a Reaction D20.4 Radioactive Half-Lives D20.5 Radiometric Dating Day 21: Reaction Energy Diagram and Arrhenius Equation D21.1 Factors that Affect the Rate Constant D21.2 Reaction Energy Diagrams D21.3 Temperature and Maxwell-Boltzmann Distribution D21.4 Activation Energy and Temperature D21.5 Steric Factor D21.6 Arrhenius Equation and Arrhenius Plot Day 22: Elementary Reactions D22.1 Elementary Reactions D22.2 Unimolecular Elementary Reactions D22.3 Bimolecular Elementary Reactions D22.4 Trimolecular Elementary Reactions Day 23: Reaction Mechanisms D23.1 Multi-step Reactions and Rate-Determining Step D23.2 First Step is Rate-Determining D23.3 Equilibrium Approximation D23.4 Catalysts and Reaction Mechanisms Day 24: Enzymes and Enzyme Catalysis D24.1 Enzymes D24.2 Enzyme Kinetics: Michaelis-Menten Mechanism D24.3 Enzyme Denaturation and Inhibitors Day 25: Homogeneous and Heterogeneous Catalysis D25.1 Homogeneous Catalysis D25.2 Heterogeneous Catalysts Unit Four Day 27: Thermochemistry and Enthalpy D27.1 Energy, Temperature, and Heat D27.2 Calorimetry D27.3 Enthalpy D27.4 Standard-state Reaction Enthalpy Change, ΔrH° D27.5 Bond Enthalpy and Reaction Enthalpy Change D27.6 Hess’s Law D27.7 Standard Enthalpy of Formation Day 28: Entropy, Gibbs Free Energy D28.1 Entropy and Microstates D28.2 Predicting the Sign of ΔS D28.3 Second Law of Thermodynamics D28.4 Third Law of Thermodynamics D28.5 Gibbs Free Energy D28.6 Calculating ΔG° Day 29: Gibbs Free Energy, Chemical Equilibrium D29.1 Temperature Dependence of Gibbs Free Energy D29.2 Chemical Equilibrium D29.3 Concentration Equilibrium Constants D29.4 Equilibrium Constant and Partial Pressure D29.5 Calculations Involving Equilibrium Constants D29.6 Equilibrium Constants and Product-favored Reactions Day 30: ICE Table, Reaction Quotient, Le Châtelier’s Principle D30.1 ICE Table D30.2 “All-Reactant” or “All-Product” Starting Point D30.3 Reaction Quotient D30.4 Le Châtelier’s principle: Change in Concentration D30.5 Le Châtelier’s Principle: Change in Pressure or Volume Day 31: Le Châtelier’s Principle, Equilibrium and Gibbs Free Energy D31.1 Le Châtelier’s Principle: Change in Temperature D31.2 Catalysts and Equilibrium D31.3 Gibbs Free Energy and Equilibrium D31.4 Effect of Temperature Day 32: Gibbs Free Energy and Work, Kinetic Metastability D32.1 Gibbs Free Energy and Work D32.2 Gibbs Free Energy in Biological Systems D32.3 Kinetic Metastability D32.4 Haber-Bosch Process Day 33: Acids and Bases D33.1 Definition of Acids and Bases D33.2 Autoionization of Water D33.3 pH and pOH D33.4 Acid Constant Ka and Base Constant Kb D33.5 Acid Strength and Molecular Structure Day 34: Acid-Base Reactions D34.1 Polyprotic Acids D34.2 Acid-Base Reactions D34.3 Reaction Between Amphiprotic Species D34.4 Amino Acids Unit Five Day 36: Buffer Solutions D36.1 Buffer Solutions D36.2 Henderson-Hasselbalch Equation D36.3 Selection of a Suitable Buffer D36.4 Buffer Capacity Day 37: Acid-Base Titration D37.1 Titration D37.2 Titration Curves D37.3 Acid-Base Indicators D37.4 Titration of Polyprotic Acids and Bases Day 38: Oxidation-Reduction Reactions, Voltaic Cells D38.1 Oxidation-reduction Reactions and Electrochemistry D38.2 Redox Reactions and Oxidation Number D38.3 Balancing Redox Reactions D38.4 Introduction to Voltaic Cells Day 39: Voltaic Cells, Half-Cell Potentials D39.1 Voltaic Cell Potential D39.2 Cell Notation D39.3 Standard Half-Cell Potentials D39.4 Using Standard Half-Cell Potentials Day 40: Thermodynamic Properties of Voltaic and Electrolytic Cells D40.1 Relationships among ΔrG°, K°, and E°cell D40.2 Nernst Equation D40.3 Concentration Cells Day 41: Electrolysis; Commercial Batteries D41.1 Electrolysis D41.2 Commercial Batteries D41.3 Primary Batteries D41.4 Secondary Batteries D41.5 Fuel Cells Appendix Atomic Weights Elemental Abundances Electron Configuration Radius Ionization Energies Electron Affinity Common Polyatomic Ions Bond Enthalpy and Length Electronegativity Amino Acids Solubility Product Constant Acid-Base Ionization Constant Thermodynamic Standard Potential Basics of Organic Nomenclature Alkanes Alcohols Carboxylic Acids Esters Amines Amides Review Section Review: Waves D1.2 Wave Properties Review: Photons D1.3 Blackbody Radiation D1.4 Planck’s Quantum Theory D2.1 The Photoelectric Effect In Depth: Atomic-scale Particles and Waves D3.1 Atomic-Sized Particles Have Unusual Properties In Depth: The Heisenberg Uncertainty Principle D3.2 The Heisenberg Uncertainty Principle In Depth: In-Phase and Out-of-Phase D6.2 Constructive and Destructive Interference Chemistry 109, Fall 2020 Unit Four Day 31: Le Châtelier’s Principle, Equilibrium and Gibbs Free Energy D31.1 Le Châtelier’s Principle: Change in Temperature When a chemical reaction is at equilibrium and the temperature changes, the reaction’s equilibrium constant is different at the new temperature. Le Chatelier’s principle can be used to predict which direction an equilibrium shifts and hence whether increasing temperature increases or decreases K. Remember that, according to Le Chatelier’s principle, an equilibrium shifts in a direction that partially counteracts the change in conditions. Consider the reaction N 2(g) + O 2(g) ⇌ 2 NO(g) Δ r H° = 180.5 kJ/mol at 25 °C As shown by the enthalpy change, this reaction is endothermic: when the reaction takes place in the forward direction, the temperature is lowered. Because enthalpy change does not vary significantly with temperature, the forward reaction is endothermic at all temperatures where all reactants and products are in the gas phase. The reverse reaction is always exothermic. Suppose that the reaction is at equilibrium at a particular temperature and the temperature is suddenly increased. To partially compensate for the temperature increase, the reaction shifts toward products (the endothermic direction), which lowers the temperature a bit. Thus, when equilibrium is reached at the higher temperature, the concentration of NO is larger and the concentrations of N 2 and O 2 are lower. This results in a larger value for K c at the higher temperature. If the temperature of the reaction is suddenly lowered, the reaction shifts to partially raise the temperature—in the exothermic direction. In this case, the shift is from products to reactants. Thus, at a lower temperature the concentrations of reactants are larger, the concentrations of products are smaller, and the equilibrium constant is smaller. Summarizing, An increase in temperature shifts an equilibrium in the endothermic direction (the direction with positive Δ r H°) because the endothermic reaction partially counteracts the increase in temperature A decrease in temperature shifts an equilibrium in the exothermic direction (the direction with negative Δ r H°) because the exothermic reaction partially counteracts the decrease in temperature. The different concentrations in the new equilibrium system (after the shift resulting from the temperature change) correspond to a different value for the equilibrium constant. The larger the magnitude of Δ r H° is the larger the shift in the equilibrium is and the greater the change in the equilibrium constant is. Exercise 1: Applying Le Chatelier’s Principle to Temperature Change D31.2 Catalysts and Equilibrium A catalyst speeds up the rate of a reaction, allowing the equilibrium to be reached more quickly (by speeding up both forward and reverse reactions). Hence, catalysts influence the kinetics of a reaction. However, a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations. D31.3 Gibbs Free Energy and Equilibrium The standard Gibbs free energy change for a reaction indicates whether a reaction is product-favored at equilibrium (Δ r G° < 0) or reactant-favored at equilibrium (Δ r G° > 0). A strongly product-favored reaction (large negative Δ r G°) has a large equilibrium constant (K>> 1) and a strongly reactant-favored reaction (large positive Δ r G°) has a very small equilibrium constant (K<<1, a very small fraction because K cannot be negative). These qualitative statements suggest that there may be a quantitative relationship between the equilibrium constant and Δ r G° for a given reaction. The reaction quotient, Q, was introduced as a convenient measure of the status of a reaction. When Q<K, the reaction proceeds spontaneously in the forward direction until equilibrium is reached (Q = K). Conversely, if Q>K, the reaction proceeds spontaneously in the reverse direction until equilibrium is achieved. The relationship between Δ r G°, Q, and K is illustrated graphically in Figure 1 in graphs of G vs. reaction progress. In each graph, on the far left, the system is all reactants and Q = 0. On the far right, the system is all products and Q = ∞. The slope at any point on each graph is Δ r G/Δ(reaction progress). Because Δ(reaction progress) is always positive, the sign of the slope is the sign of Δ r G. In the light cyan region where Q<K, the slope of the plot is negative, corresponding to negative Δ r G, which predicts spontaneous forward reaction. In the light pink region where Q>K, the slope is positive, corresponding to positive Δ r G which predicts spontaneous reverse reaction. Where the slope is zero (bottom of the curve), Δ r G = 0, and the system is at equilibrium with Q = K. Hence, we can think of reaction progress as rolling down the sides of a Gibbs free energy valley, with equilibrium at the bottom (minimum G). Figure 1. These plots show the Gibbs free energy (G) versus reaction progress for systems whose standard Gibbs free energy changes (Δ G°) are negative (left), and positive (right). One end of the x-axis represents all reactants, the other end all products. Non-equilibrium systems proceed spontaneously in whatever direction is necessary to minimize Gibbs free energy and establish equilibrium. Where equilibrium lies along the reaction progress depends on the sign of Δ r G°. When Δ r G° < 0, the equilibrium (minimum in the curve) is further to the right, indicating that there are more products than reactants when equilibrium is reached. When Δ r G° > 0, the equilibrium is further to the left, indicating that reactants predominate. The Gibbs free energy change at any point along the reaction progress involves adjusting Δ r G° by the factor RT(ln Q): Δ r G = Δ r G° + RT( ln Q) At equilibrium, Q = K and Δ r G = 0, therefore: 0 = Δ r G° + RT( ln K°) Δ r G° = −RT( ln K°) or K° = e-Δ r G°/RT Note that in these last equations the equilibrium constant is represented by K°. The standard equilibrium constant, K º, is either the concentration equilibrium constant (K c) with each concentration divided by the standard-state concentration of 1 M or the pressure equilibrium constant (K p) with each pressure divided by the standard-state pressure of 1 bar. Hence, K º is truly unitless. Dividing by the standard-state concentration or pressure means that if concentrations in K c are expressed in M (mol/L) the numerical values of K º and K c are the same. Similarly, if partial pressures in K p are expressed in bar, the numerical values of K º and K p are the same. | Kº | Δ r G° | | --- | > 1 | < 0 | Product-favored at equilibrium. | | < 1 | > 0 | Reactant-favored at equilibrium. | | = 1 | = 0 | Reactants and products are equally abundant at equilibrium. | Exercise 2: Gibbs Free Energy and Equilibrium Exercise 3: Gibbs Free Energy—Nonstandard Conditions D31.4 Effect of Temperature Recall that: Δ r G° = Δ r H° – T Δ r S° Therefore: −RT( ln K°) = Δ r H° – T Δ r S° This equation can be used to calculate K° at different temperatures, if we assume that Δ r H° and Δ r S° for a reaction have the same values at all temperatures. This is a good, but not perfect, assumption and we will use it in this course unless specified otherwise. It is not a good assumption if there is a phase change for a reactant or a product within the temperature range of interest. Dividing both sides of the equation by -RT gives: ln K° =−Δ r H∘R(1 T)+Δ r S∘R y = mx + b A plot of ln K° vs. 1 T is called a van’t Hoff plot. The graph has slope = −Δ r H∘R and intercept = Δ r S∘R. If the concentrations of reactants and products are measured at various temperatures so that K° can be calculated at each temperature, both the reaction entropy change and enthalpy change can be obtained from a van’t Hoff plot. Figure 2. Van’t Hoff plots. The reaction N 2 + O 2 = 2 NO has Δ r H° = 180.5 kJ/mol; for this endothermic reaction, as T increases (smaller 1/T) the equilibrium constant increases. The reaction N 2 + 3 H 2 = 2 NH 3 has Δ r H° = −92.2 kJ/mol; for this exothermic reaction, as T increases (smaller 1/T) the equilibrium constant decreases. Based on the equation for the van’t Hoff plot, an exothermic reaction (Δ r H° < 0) has K° decreasing with increasing temperature, and an endothermic reaction (Δ r H° > 0) has K° increasing with increasing temperature. This quantitative result agrees with the qualitative predictions made by applying Le Chatelier’s principle. It also shows that the magnitude of Δ r H° dictates how rapidly K° changes as a function of temperature. In contrast, Δ r S° affects the magnitude of K° but not its temperature dependence. For example, suppose that K°1 and K°2 are the equilibrium constants for a reaction at temperatures T 1 and T 2, respectively: ln K 1∘=−Δ r H∘R(1 T 1)+Δ r S∘R ln K 2∘=−Δ r H∘R(1 T 2)+Δ r S∘R Subtracting the two equations yields: ln K 2∘−ln K 1∘=(−Δ r H∘R(1 T 2)+Δ r S∘R)−(−Δ r H∘R(1 T 1)+Δ r S∘R)ln K 2∘K 1∘=−Δ r H∘R(1 T 2−1 T 1)=Δ r H∘R(1 T 1−1 T 2) Thus calculating Δ r H° from tabulated Δ f H° and measuring the equilibrium constant at one temperature allows us to calculate the equilibrium constant at any other temperature (assuming that Δ r H° and Δ r S° are independent of temperature). Exercise 4: Temperature Dependence of the Equilibrium Constant Podia Question Ammonium nitrate is an important fertilizer that supplies nitrogen to crops. It also can be used as an explosive to remove tree stumps from farm fields or in terrorist bombings. There are two reactions that can occur to produce the explosive effect. At lower temperatures ammonium nitrate decomposes to form dinitrogen monoxide and water vapor, but at higher temperatures it decomposes explosively to form nitrogen, water vapor, and oxygen. The reaction equations and thermodynamic parameters at 298 K are (1) NH 4 NO 3(s) → N 2 O(g) + 2 H 2 O(g) Δ r H °= −36.026 kJ/mol Δ r S° = 446.42 J K−1 mol−1 (2) NH 4 NO 3(s) → N 2(g) + ½ O 2(g) + 2 H 2 O(g) Δ r H °= −118.076 kJ/mol Δ r S° = 520.749 J K−1 mol−1 a) Which of the reactions is(are) product-favored at room temperature? Which of the reactions is(are) product-favored at 1000 K? Explain briefly. b) Is there a temperature below which one of the reactions is more product-favored and above which the other reaction is more product favored? If there is such a temperature for these two reactions, determine what that temperature is. c) Given your results, write an appropriate scientific explanation of the fact that nitrogen and oxygen are produced explosively at higher temperatures but dinitrogen monoxide is produced at lower temperatures. Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer. Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. Previous/next navigation Previous: Day 30: ICE Table, Reaction Quotient, Le Châtelier’s Principle Next: Day 32: Gibbs Free Energy and Work, Kinetic Metastability Back to top License Chemistry 109, Fall 2020 Copyright © by John Moore; Jia Zhou; and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. Share This Book Pressbooks Powered by Pressbooks Pressbooks User Guide |Pressbooks Directory |Contact Pressbooks on YouTubePressbooks on LinkedIn
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https://www.ck12.org/flexi/physical-science/momentum/show-how-momentum-and-impulse-have-the-same-units-through-derivation/
Flexi answers - Show how momentum and impulse have the same units through derivation. | CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Physical Science Momentum Question Show how momentum and impulse have the same units through derivation. Flexi Says: Momentum (p) is defined as the product of an object's mass (m) and its velocity (v). So, p = m x v. The units of momentum are kilogram meters per second (kg.m/s). Impulse (J) is defined as the force (F) applied to an object over a time interval (Δt). So, J = F x Δt. Force is mass times acceleration (F = m x a), and acceleration is change in velocity over time (a = Δv/Δt). Substituting these into the impulse equation gives J = m x (Δv/Δt) x Δt. The Δt in the numerator and denominator cancel out, leaving J = m x Δv. This shows that impulse has the same units as momentum: kilogram meters per second (kg.m/s). Analogy / Example Try Asking: What would happen if an asteroid were strong enough to blast off half of Ganymede's mass?Avery hurls the 0.482 kg football to Tyrone. The ball reaches Tyrone with a speed of 19.3 m/s. Tyrone catches the pass, bringing it to a dead stop in 0.0211 seconds before heading up field. Determine the impulse delivered to the ball by Tyrone. (Magnitude only; no + or -.) Impulse N•sWhat is the formula for calculating momentum using mass and velocity? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
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https://www.vedantu.com/rs-aggarwal-solutions/class-10-chapter-4-triangles
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Sign In All Courses NCERT, book solutions, revision notes, sample papers & more Find courses by class Starting @ ₹1,350 Find courses by target Starting @ ₹1,350 Long Term Courses Full Year Courses Starting @ just Rs 9000 One-to-one LIVE classes Learn one-to-one with a teacher for a personalised experience Courses for Kids Courses for Kids Confidence-building & personalised learning courses for Class LKG-8 students English Superstar Age 4 - 8 Level based holistic English program Summer Camp For Lkg - Grade 10 Limited-time summer learning experience Spoken English Class 3 - 5 See your child speak fluently Learn Maths Class 1 - 5 Turn your child into a Math wizard Coding Classes Class 1 - 8 Learn to build apps and games, be future ready Free study material Get class-wise, author-wise, & board-wise free study material for exam preparation NCERT SolutionsCBSEJEE MainJEE AdvancedNEETQuestion and AnswersPopular Book Solutions Subject wise Concepts ICSE & State Boards Kids Concept Online TuitionCompetative Exams and Others Offline Centres Online Tuition Get class-wise, subject-wise, & location-wise online tuition for exam preparation Online Tuition By Class Online Tuition By Subject Online Tuition By Location More Know about our results, initiatives, resources, events, and much more Our results A celebration of all our success stories Child safety Creating a safe learning environment for every child Help India Learn Helps in learning for Children affected by the Pandemic WAVE Highly-interactive classroom that makes learning fun Vedantu Improvement Promise (VIP) We guarantee improvement in school and competitive exams Master talks Heartfelt and insightful conversations with super achievers Our initiatives Resources About us Know more about our passion to revolutionise online education Careers Check out the roles we're currently hiring for Our Culture Dive into Vedantu's Essence - Living by Values, Guided by Principles Become a teacher Apply now to join the team of passionate teachers Contact us Got questions? Please get in touch with us Vedantu Store RS Aggarwal Solutions RS Aggarwal Class 10 Solutions - Triangles RS Aggarwal Class 10 Solutions - Triangles Download PDF Study Materials NCERT Solutions for Class 10 Important Questions for Class 10 Revision Notes for Class 10 Maths MCQ NCERT Books Worksheets Maths Formula For Class 10 Sample Papers Sample Paper for Class 10 Maths Sample Paper for Class 10 Science Sample Paper for Class 10 English Sample Paper for Class 10 Hindi Sample Paper for Class 10 Social Science Practice Paper Class 10 Test Series Previous Year Question Papers Class 10 Maths Question Paper Class 10 Science Question Paper Class 10 English Question Paper Class 10 Hindi Question Paper Class 10 Social Science Question Paper Class 10 Sanskrit Question Paper Exam Info Date Sheet Marking Scheme Result Syllabus Maths Syllabus Science Syllabus English Syllabus Hindi Syllabus Social Science Syllabus Textbook Solutions RD Sharma Solutions RS Aggarwal Solutions Lakhmir Singh Solutions NCERT Examplar Triangles Solutions for RS Aggarwal Class 10 Chapter 4 Unable to formulate the answers to all the questions of RS Aggarwal textbook? Do not panic! We're here for you, bringing to you detailed solutions not only for the questions to Chapter 4 Triangles but all the Chapters and every single Question. The RS Aggarwal Class 10 Triangles Solutions are formulated by our mathematics experts at Vedantu who have been in the profession for years now. By following these solutions PDF, you shall receive the right guidance. Following these RS Aggarwal Solutions to Chapter 4 Triangles, you will learn how to approach the problems stepwise and solve them perfectly! You can download NCERT Solutions for Class 10 Maths and NCERT Solution for Class 10 Science to help you to revise the complete syllabus and score more marks in your Examinations. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Stuck at a question? Don't worry, Vedantu has got your back! Do you need help with your Homework? Are you preparing for Exams? Study without internet (offline) Download full PDF Download PDF of RS Aggarwal Class 10 Solutions - Triangles Courses Competitive Exams after 12th Science CBSE JEE Main JEE Advanced NEET Olympiad CUET RS Aggarwal Solutions for Class 10 Maths Chapter 4 We have provided step by step solutions for all Exercise questions given in the PDF of Class 10 RS Aggarwal Chapter 4 - Triangles. All the Exercise questions with solutions in Chapter 4 - Triangles are given below: Exercise (Ex 4A) 4.1 Exercise (Ex 4B) 4.2 Exercise (Ex 4C) 4.3 Class 10 Chapter 4: Triangles As kids, we see numerous shapes in our surroundings in day to day life. However, we never really cared to discover more about the properties of these shapes. Hence, we learn about all these properties as we progress through higher school. Triangles are one of the most common shapes we've been seeing in our surroundings. In our portions, we study Triangles in-depth and understand every little aspect of them. Important Points RS Aggarwal Solutions Class 10 Chapter 4 Triangles There are three different types of Triangles when segregated based on angles within a Triangle, Acute Triangle: An acute Triangle is a Triangle where each of the angles within a Triangle is less than 90 degrees, implies that the measure of each angle within the Triangle is less than 90 degrees Obtuse Triangle: An obtuse Triangle is a Triangle wherein the measure of at least one angle of the Triangle exceeds 90 degrees Right Triangle: A right-angled Triangle is a Triangle in which at least one of the angles of the Triangle is equal to 90 degrees. Based on the lengths of the sides of a Triangle, there are three kinds as well: An isosceles Triangle is a special variety of Triangle in which two sides of the Triangle are equal and two angles are equal as well. A right Isosceles Triangle is a Triangle in which two angles of the Triangle are equal to 45 degrees and one angle is equal to 90 degrees. An equilateral Triangle is a Triangle in which each of the sides is equal and all the angles are equal to 60 degrees. Pythagoras Theorem: The Pythagoras theorem is one of the most important theorems in right-angled Triangles. It is a very useful theorem to approach problems of Class 10 Chapter 4 Triangles. The theorem states that the square of the length of the longest side, that is the hypotenuse, will be equal to the sum of squares of the lengths of the other two sides. Median of a Triangle: The medians of a Triangle are the line segments drawn from one vertex to the midpoint of the opposite side. Centroid: The point of intersection of all the medians is called the centroid of the Triangle. Interestingly the centroid of the Triangle also divides each median in a ratio of 2:1. Altitude: Altitudes of a Triangle are the line segments drawn from the vertices of the Triangles to the side opposite to the Triangle such that the line segment makes a right angle with the side it intersects. Orthocentre: The orthocenter is the point of intersection of all the altitudes within a Triangle. Angular Bisectors: The line segment bisecting each angle of a Triangle and meeting the opposite side is called an angular bisector of the Triangle. Incentre: The incentre of a Triangle is the point of intersection of all the angular bisectors of the Triangle. Interestingly, in an equilateral Triangle, the median, the altitude, and the angular bisectors are the same, and hence all the points, incentre, orthocentre and centroid are concurrent to each other. The sum of all the internal angles of a Triangle is always equal to 180 degrees. The sum of lengths of two sides is always greater than the length of the third side of the Triangle. Heron's Formula: A formula to find the area of any Triangle by just knowing the length of each side of a Triangle. s(semiperimeter):(a+b+c)/2 Where a,b,c are the lengths of sides of the Triangle respectively. Area = √s(s–a)(s–b)(s–c) Area of Triangle: ½bh where b is the base of the Triangle and h is the height or the altitude of the Triangle. Area of an Isosceles Triangle: A = (1/4) × b × √(4a 2 – b 2) Where b is the length of the unequal sides of the Triangle and a is the length of sides that are equal. Area of an Equilateral Triangle Area of Equilateral Triangle = √3a 2/4 Where a is the length of a side of the Triangle Preparation Tips While preparing for the topic Triangles, you must ensure to be thorough with all the important theorems and formulae. While solving the problems you need to implement the right property at the right place. This will make your problem solving much simpler. Following Vedantu solutions to RS Aggarwal will make your problem solving more systematic and simple! Importance of learning CBSE Class 10 Maths Chapter 4- Triangles Triangles is an important Chapter as we learn about its different properties which will come in handy in the future Classes Triangles form the very basis of some of the important theorems Trigonometry becomes easier to understand Complex calculations become much easier if one properly learns about Triangles Areas such as Engineering, Astronomy, Physics and Navigation become smoother to understand Unknown quantities in geometry can be easily determined Concepts of Euclid’s Geometry also becomes easier Best Seller - Grade 10 View More> ### Vedantu Challenge 100 Science & Mathematics Class 10 Books Set Of 2 | CBSE Chapter-wise PYQ with Solutions | NCERT Based | 2026 Exam Prep Guide | Competency-Based | Objective & Subjective Questions | Practice Book by Shimon Joseph | Score Booster Series ₹1398.00 Sale ₹999.00 + ADD TO CART ### Vedantu Class 10 QR Flashcards & Booklets Combo Of 5 | Quick Revision Cards for Science & Social Studies with 3 QR Booklets for Science Activity Maths & Geography | CBSE 2026 Exam Preparation Guide ₹1819.00 Sale ₹1399.00 + ADD TO CART ### Vedantu Tatva Pro Question Bank Book CBSE Class 10 Set of 3 Books |Physics, Chemistry, Biology | Chapterwise - Topicwise Theory and Previous Year Questions | Free CBSE Class 10 Year Long Recorded Course ₹1299.00 Sale ₹999.00 + ADD TO CART ### Vedantu Tatva Pro Question Bank Book CBSE Class 10 Mathematics Chapterwise - Topicwise Theory and Previous Year Questions | Free CBSE Class 10 Year Long Recorded Course ₹1099.00 Sale ₹699.00 + ADD TO CART ### Vedantu Tatva Pro CBSE Class 10 book set- 4 subjects ₹1559.00 Sale ₹1199.00 + ADD TO CART ### Vedantu Challenge 100 Science Class 10 Book | CBSE Chapter-wise PYQ with Solutions | NCERT Based | 2026 Exam Prep Guide | Competency-Based | Objective & Subjective Questions | Practice Book by Shimon Joseph | Score Booster Series ₹699.00 Sale ₹499.00 + ADD TO CART ### Vedantu Challenge 100 Mathematics Class 10 Book | CBSE Chapter-wise PYQ with Solutions | NCERT Based | 2026 Exam Prep Guide | Competency-Based | Objective & Subjective Questions | Practice Book by Shimon Joseph | Score Booster Series ₹699.00 Sale ₹499.00 + ADD TO CART ### Vedantu's Instasolve - 1 Month - 24 hours Unlimited Instant Doubt Solving ₹2998.00 Sale ₹1999.00 + ADD TO CART ### Vedantu's Instasolve - 12 Months - 24 hours Unlimited Instant Doubt Solving ₹17998.00 Sale ₹12000.00 + ADD TO CART ### Vedantu's Instasolve - 3 Months - 24 hours Unlimited Instant Doubt Solving ₹9998.00 Sale ₹5499.00 + ADD TO CART ### Biology - Vedantu - Round Neck T-Shirt ₹998.00 Sale ₹499.00 VIEW DETAILS ### Doctor in the House - Women's Round Neck T-Shirt ₹998.00 Sale ₹499.00 VIEW DETAILS ### Doctor in the House - Men's Round Neck T-Shirt ₹998.00 Sale ₹499.00 VIEW DETAILS ### Dream Hustle Achieve - Men's Hooded Sweatshirt ₹1598.00 Sale ₹799.00 VIEW DETAILS ### Vedantu - (Bag + Bottle + Coffee Mug) & (Set of 6 Notebooks, Highlighter Set, Set of 4 Pens) ₹1799.00 Sale ₹1499.00 VIEW DETAILS FAQs on RS Aggarwal Class 10 Solutions - Triangles What is the correct method for solving problems based on the Basic Proportionality Theorem (BPT) in RS Aggarwal Class 10, Chapter 4? The solutions for RS Aggarwal Class 10 Chapter 4 demonstrate the correct step-by-step method for applying the Basic Proportionality Theorem (BPT). The key steps include: Clearly identifying the triangle and the line parallel to one of its sides. Stating the theorem as the reason for setting up the ratio of sides. Substituting the known values into the proportion, such as AD/DB = AE/EC. Solving the resulting equation to find the unknown side length. Following this structured approach, as shown in the solutions, ensures you do not miss any steps and helps in scoring full marks. How can I find the solutions for specific exercises like Exercise 4A or 4B in RS Aggarwal Class 10 Triangles? Vedantu provides comprehensive, exercise-wise solutions for Chapter 4, Triangles, from the RS Aggarwal textbook. You can navigate through the chapter page to find detailed, step-by-step answers for every question in each exercise, such as Exercise 4A, 4B, and so on. This makes it easy to check your work or understand the method for a specific problem you are stuck on. When solving a similarity problem, why is stating the criterion (e.g., AA, SAS, SSS) so important in the RS Aggarwal solutions? Stating the similarity criterion (like AA, SAS, or SSS) is a crucial step that carries marks in an exam. The RS Aggarwal solutions emphasise this because: It acts as the logical foundation for your proof, showing the examiner you understand why the triangles are similar. It justifies the subsequent step of writing the corresponding sides in proportion. Skipping this step can lead to a loss of marks, as it demonstrates an incomplete understanding of the geometric proof process as per the CBSE pattern. How do the RS Aggarwal solutions help with questions on different types of triangles, like isosceles or right-angled triangles? The RS Aggarwal solutions for Chapter 4 cover a wide variety of problems involving special triangles. For example, when solving a problem on an isosceles triangle, the solutions will guide you on how to correctly apply properties like equal sides and equal opposite angles along with similarity or Pythagoras theorems. For right-angled triangles, the solutions clearly demonstrate the application of the Pythagoras theorem and trigonometric ratios where applicable. What are some common mistakes to avoid when applying the theorem on 'Areas of Similar Triangles' as per the solutions? The RS Aggarwal solutions help clarify common pitfalls. When using the theorem on Areas of Similar Triangles, students often make these mistakes: Forgetting to square the ratio of the corresponding sides. The theorem states that the ratio of areas is equal to the square of the ratio of sides. Incorrectly matching corresponding sides, altitudes, or medians. Applying the theorem to triangles that are not proven to be similar first. The solutions guide you to first prove similarity and then correctly apply the area theorem. What is the typical structure for solving a proof-based question in the Triangles chapter, according to RS Aggarwal solutions? The RS Aggarwal solutions follow the standard CBSE format for proofs, which is essential for scoring well. The structure includes: Given: Stating all the information provided in the question. To Prove: Clearly mentioning what needs to be proven. Construction: Describing any additional lines or points needed for the proof (if any). Proof: A logical, step-by-step deduction with a valid reason (e.g., theorem, property, axiom) for each step. This methodical approach helps in building a clear and convincing argument. I'm confused between the Basic Proportionality Theorem (BPT) and the Mid-point Theorem. How do the solutions help differentiate their application? The RS Aggarwal solutions clarify this common confusion. The Basic Proportionality Theorem (BPT) applies when a line is parallel to any one side of a triangle and intersects the other two sides at distinct points, establishing a ratio between the divided segments. The Mid-point Theorem (from Class 9) is a special case of BPT where the line joins the mid-points of two sides. The solutions demonstrate that while BPT gives a general ratio, the Mid-point Theorem specifically proves the line is parallel to the third side and is half its length. How should I use the RS Aggarwal solutions for Class 10 Triangles for self-study and exam preparation? For effective preparation for the 2025-26 exams using the RS Aggarwal solutions for Triangles, follow this method: First, attempt to solve the exercise problems on your own. If you get stuck or want to verify your answer, refer to the step-by-step solution provided. Pay close attention to the theorems and properties cited in each step of the solution. Do not just copy the answer; focus on understanding the 'how' and 'why' of the method to master the concepts for the board exams. 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912
https://ui.adsabs.harvard.edu/abs/arXiv:2205.00870
The Shape of Central Quadrilaterals - Astrophysics Data System Skip to main content Now on article abstract page Toggle navigation ads === Feedback Submit Updates General Feedback ORCID Sign in to ORCID to claim papers in the ADS. About About ADS What's New ADS Blog ADS Help Pages ADS Legacy Services Careers@ADS Sign Up Log In Search Bar to Enter New Query quick field:Author First Author Abstract Year Fulltext Select a field or operator All Search Terms Your search returned 0 results Your search returned 0 results Show Menu Full Text Sources view AbstractCitations (1)ReferencesCo-ReadsSimilar PapersVolume ContentGraphicsMetricsExport Citation The Shape of Central Quadrilaterals Show affiliations Loading affiliations affiliations loading Rabinowitz, Stanley (0); Suppa, Ercole (1) Abstract The diagonals of a quadrilateral form four component triangles (in two ways). For each of various shaped quadrilaterals, we examine 1000 triangle centers located in these four component triangles. Using a computer, we determine when the four centers form a special quadrilateral, such as a rhombus or a cyclic quadrilateral. A typical result is the following. The diagonals of an equidiagonal quadrilateral divide the quadrilateral into four nonoverlapping triangles. Then the Nagel points of these four triangles form an orthodiagonal quadrilateral. Publication: eprint arXiv:2205.00870 Pub Date:April 2022 DOI: 10.48550/arXiv.2205.00870 arXiv:arXiv:2205.00870Bibcode: 2022arXiv220500870R Copied!Keywords: Mathematics - History and Overview; Mathematics - Metric Geometry; 51M04 (Primary) 51-08 (Secondary) Feedback/Corrections? full text sources Preprint Preprint PDF | Preprint article © The SAO Astrophysics Data System adshelp[at]cfa.harvard.edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement 80NSSC25M7105 The material contained in this document is based upon work supported by a National Aeronautics and Space Administration (NASA) grant or cooperative agreement. Any opinions, findings, conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the views of NASA. Resources About ADS ADS Help System Status What's New Careers@ADS Web Accessibility Policy Social @adsabs ADS Blog Project Switch to basic HTML Privacy Policy Terms of Use Smithsonian Astrophysical Observatory Smithsonian Institution NASA 🌓 × Back How may we help you? #### Missing/Incorrect Record Submit a missing record or correct an existing record.#### Missing References Submit missing references to an existing ADS record.#### Associated Articles Submit associated articles to an existing record (e.g. arXiv / published paper).#### General Feedback Send your comments and suggestions for improvements. Name Email Feedback Submit You can also reach us at adshelp [at] cfa.harvard.edu This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply. Close without submitting
913
https://www.youtube.com/watch?v=Q8BbZxkZSSA
Conic Sections - Circles, Semicircles, Ellipses, Hyperbolas, and Parabolas The Organic Chemistry Tutor 9680000 subscribers 843 likes Description 52385 views Posted: 13 Nov 2024 This precalculus video tutorial provides a basic introduction into circles, semicircles, ellipses, hyperbolas, and parabolas. Conic Sections - Free Formula Sheet: Video Lessons by Chapter: Full-Length Final Exam Videos & Worksheets: Finding The Radius and Center of a Circle: Writing Equations and Graphing Circles: Writing Equations of Ellipses: Circumference of an Ellipse: Area of an Ellipse: Eccentricity of an Ellipse: Hyperbolas - Conic Sections: Parabolas - Focus and Directrix: How To Find the Vertex of a Parabola: Conic Sections Quiz: Arithmetic Sequences: Geometric Sequences: Probability - Basic Intro: Intro to Statistics: Final Exams and Videos Playlists: 82 comments Transcript: in this video I want to go over some formulas associated with conic sections so if you have a pen and paper feel free to take down some notes so let's start with the first one the circle so let's say we have a circle with radius R and the center has the coordinates H comma K the standard form of the circle the equation is x - h2+ y - k^ 2 is = to R 2 so in order to write the equation of a circle you need the coordinates of the center and the radius R now some other things you may want to know is that the eccentricity of a circle is zero the area of a circle is p piun r^ 2 the circumference is 2 pi r and the diameter is 2 the radius now in other less common situations you may need to know the domain and range of a circle given its equation so the domain for a circle is going to be H - R comma H + r so all you need is the coordinates of the center and the radius and you could find the domain of the circle now the range is going to be Kus R and it's going to go up to k + r now from the equation of a circle you can get the equation of a semicircle so let's say if you were to solve for y in this equation solving for y you'll get this Y is equal to the square < TK of R 2 minus x - h^2 + K now when you take the square root you can get the positive answer and you can get the negative answer if you just have the positive sign you'll get a semicircle that is going to be the upper half of the circle the negative sign if you put that in front you'll get the other part of the circle the lower half now of course the semicircle like the circle can be shifted up down left right by the values of H and K so it can be in various quadrants now if you solve for x you'll get X I may need to get more space here so let's clear this away from the circle equation if you solve for x you'll get this x is equal to the square < TK of r^ 2 minus y - k^ 2 + H and like the other one it can be positive or negative when X is positive you're going to get the semicircle on the right side when X is negative you'll get the semicircle which is the left side of the circle so you can get four different semicircles with those four equations and for those of you who want to print out of these equations including the domain and range of the semicircle uh feel free to check out the link links in the description section below I'm going to have a print out with the graphs and the corresponding equations as well so now let's move on to the next conic section and that is the ellipse so with the ellipse there's two versions you need to be familiar with so there's this one where the major axis is horiz horizontal and here is the other form where the major axis is vertical so this is a this is negative a this is going to be B and negative B the length of the major axis you could see in this case which is the width it's 2 a the length of the minor axis is going to be 2B and that's the height for that ellipse now for the one on the right this is negative a this is a a is bigger than b for the ellipse by the way the length of the major axis is still 2 a but it's no longer the width this time for this ellipse is the height the length of the minor axis which is the width now is 2b so the length of the major axis is always 2 a and the length of the minor axis is always 2B it's just the width and the height might be different so this is the width in this case is 2 a here the width is 2b here the height is 2b here the height is 2 a but the length of the major axis is always 2 a now the equation for the ellipse on the left is going to be x - h^ 2 over a^ 2 + y - k^2 over b^ 2 is equal to 1 so notice that a is associated with the x axis for the ellipse on the left so you want a squ to be under x b is associated with the Y AIS so b^ s will be under y so knowing that will kind of help you to know which equation applies for which shape for the one on the right it's going to be x - h^2 but notice that b is associated with the xaxis so we're going to have B ^2 underneath this and then plus y - k^ 2 over A2 and this is going to equal one now for both equations here we have the Pythagorean relationship which is c^2 is equal a 2us b^ 2 as opposed = to a 2 + b 2 so Meson that equation that tells you that a is larger than b and a is larger than C so when you're trying to determine which one is a a is going to be the bigger one the bigger number the center of the ellipse is going to be the same as the center of the circle it's h comma K now to calculate the eccentricity of an ellipse it's C over a for a circle we said that the eccentricity is zero because the circle is perfect it's like completely even and symmetrical the ellipse is not completely even as you could see A and B are different if a and b were the same then the ellipse would basically be a circle now for an ellipse the eccentricity is less than one but it's greater than zero so it's between 0 and 1 for a circle it's zero for a parabola the eccentricity is one for a hyperbola it's greater than one so think of hyper hyper means higher than or above now C is going to be along the same axis as a so this is going to be our C values actually let me use the same color so this is positive C negative C here C is going to be on on the vertical axis with a now c will help you to find the coordinates of the fosi so we have two focal points for an ellipse for a circle you really don't need to worry about that but here's how you can find the coordinates it's going to be h plus or minus C comma K so because C is along the x axis it's it's going to be here with h in the coordinates for the the folky now on the right side the coordinates for the folai is H comma K plus or minus C now the major vertices is h plus or minus a comma K that's these two vertices here those are the major vertices on the right it's going to be H comma K plus or minus a so that's these no that is these two here now if you need to find the coordinates of the minor vertices let's put VM for minor vertices it's going to be H comma K plus or minus B so using a different color that would indicate these two and for the graph on the right it's h plus or minus B comma K which is the vertices here now if you want the other formulas like the domain the range the X and Y intercepts uh you could find that in the formula sheet now let's talk about the area of an ellipse so here we have a circle and here we have an ellipse for the circle we have the radius R and R for the ellipse this can be a and this is B now the area of a circle it's p pi r 2 basically it's Pi R R the area of an ellipse is very similar instead of R R it's Pi a B so you can see the similarities there the circumference of a circle is basically 2 pi r you could think of it as Pi r + r r + r will give you 2 R you can approximate the circumference using this formula for ellipse Pi A+ B very similar to r+ R but a plus b this is actually like the least accurate of the circumference equations for an ellipse but it's still pretty close you may get like 98 or 97% of the true answer now there's some other formulas where you can get a more accurate answer for calculating the circumference this one is longer but it'll give you a more accurate result it's 2 pi Square < TK a 2 + b^ 2/ 2 + Pi ab/ 2 so this equation basically it averages this form which is what we have here and basically the root means Square form and it does lead to a more accurate answer of the true circumference value if you want to find the exact answer you need to use calculus but this equation will give you an answer that's a very close to the exact answer but I'm going to put all those formulas in the formula sheet including the calculus version now let's move on to the next conic section the hyperbola so on the left we're going to have a hyperbola with a horizontal transverse axis so it's going to open left and right and on the right we're going to have a hyperbola with a vertical transverse axis so it's going to open up and down so these points here represent the vertices we have two vertex and the coordinates of the vertices well we'll talk about how to get that soon but the center of the hyperbola is H comma K which in this case is going to be right in the middle now the equation for the hyperbola on the left it's x - h^2 over a 2 - y - k^ 2 over b^ 2 is equal to 1 so the difference between the ellipse and the hyperbola is that the ellipse will have a negative sign I mean I take that back the ellipse will have a positive sign but the hyperbola will have a negative sign so that is a big distinction between them the second distinction is the Pythagorean relationship for a hyperbola we have the equation c^2 is equal to a^ 2 + B2 for an ellipse this is minus so for an ellipse a was greater than C but for hyperbola C is going to be greater than a so this here is a the distance between the center and one of the vertex and the focal point is going to be where the hyperbola opens up towards the distance between the center and the focal point uh that distance is c as you can see C is larger than a now we know that the eccentricity of a hyperbola you can calculate it the same way as an ellipse it's C over a so because C is larger than a the in the eccentricity of a hyperbola is going to be larger than one for an ellipse it's less than one because C is less than a but for hyperbola C is greater than a so e is greater than 1 now for the hyperbola on the right we're going to have the formula y - k^ 2 over a^ 2 minus x - h^2 over b^2 is equal to 1 now for an ellipse a is always larger than b and that's how you would distinguish a from B for an ellipse for an hyperbola a is not always larger than b sometimes a is larger sometimes it's smaller but how you distinguish how you identify a is a is always going to be associated with the Positive term B is associated with the negative term that's how you distinguish a from B for a hyperbola now if you want to find the coordinates of the two focal points here it is it's going to be h plus orus C comma K notice that c is associated with the x- axis so it's going to be associated with h you got to add C and subtract C from H for the hyperbola on the right we need to add and subtract C from K so this is going to be a and this will be C and here will be the focal point that c will lead us to so starting from the center we need to add plus or minus C de K to get the two focal points now to get the vertices which would be these two points we need to add and subtract a from H or from the the center here we need to add and subtract a from the center but along the Y AIS so from K so this is going to be H comma K plus or minus a and that will give us this vertex and the other vertex as well so that's how we can find the vertices of the hyperbola now the equation of the asmp tootes it's going to be y - K is equal to plus or minus b/ a x - H for the one on the right the slope is going to be inverted so instead of plusus b a it's plus- a b but everything else will be the same the plus minus tells us that we have two ASM tootes this will be the let me draw that again so this here will be the positive ASM toote that is the one with the Positive slope and this one will be the one with a negative slope and the same is true for uh this hyperbola now for those of you who want example of how to graph hyperbolas ellipses and parabas I'm going to be posting a list of videos in the description section below so once you look at those videos you could see how to put uh some of these formulas into practice and how to use them if you want to find the x intercept for this ellipse it's going to be plus or minus A over B the plus or minus tells you that you can have potentially up to two inter STS so plus orus a b s < TK k^ 2 + b s + H for the one on the right if you want to find the X intercepts it's plus or minus B over a square < TK k^2 - a^ 2 + H and this is all in the formula sheet including the Y intercepts and the domain and range but real quick for this hyperbola on the left the range is always all row numbers the domain for the hyperbola on the right it's our R numbers now let's move on to the parabola so here we have a parabola that opens to the right and over here we'll have a vertical Parabola that opens upward so for this equation it's going to be Yus k squar is equal to 4 P x - H now the vertex of the parabola instead of the center the coordinates is going to be H comma K and that's true for the other one as well the eccentricity of a parabola is one now on the right the equation is going to be x - h² is equal to 4 P y - K now what helps me to remember these two formulas I think of this one as Y is equal to x^2 and we know that's a probability that opens either up or down here this is X is equal to y^2 and that opens left or right what determines the direction in which they open is the sign of P when p is positive for this equation it will open to the right when p is negative it will open to the left for the equation on the right when p is positive it opens upward if p is negative it will open downward the focal point or the focus it's always going to be in the direction where the parabola opens towards so here is so this is the focus and this will be the focus as well now for Parabola the focus is going to be P units away from the vertex so to find the coordinates of the focus is going to be for this one it's h + P comma K so we traveled P units from the vertex to get to the focus but we moved along the x axis so we add P to H for the one on the right we're going to add P to K so it's h comma k + P now this distance between the focus and the curve that's equal to 2p and this part here is 2p so there's something called the lattice rectum which is this distance here and it's equal to 4 P if you need to find the coordinates of the lattice rectum uh that's going to be in the formula sheet so you could look that up for those of you who might be interested in that but let's just call the lattice rectum LR just know that it's equal to 4p so whatever this value is that is the ltis rectum and the same is true here this is 2 p and this part is 2p so if you know the value of P you can determine the focal point and you could determine two points on a curve which will help you to graph it now the next thing you need to know is the directrix the directrix is p units away on the other side of the parabola it's p units away from the vertex so because this one is vertical the equation for the derric will be xal to H minus P so starting from the vertex we traveled P units to the left now for a parabola that opens up or down the directrix will be horizontal the equation for the directrix is y is equal to K minus p now let's talk about the axis of symmetry so if you have a parabola that opens to the right or to the left the axis of symmetry it passes through the vertex and we know the vertex is H comma K so because it's horizontal it's going to equal the y coordinate of the vertex so it's so the a s axis of symmetry is y is equal to K now for a parabola that opens up or down the axis of symmetry is going to be vertical it still passes through the vertex and so it's going to be the x coordinate of that vertex so it's X is equal to H now the x intercept for a parabola that opens to the right or to the left is equal to k^2 over 4 P plus h for for the one that opens up or down it's X is equal to plus or minus so there's potentially two of them Square < TK -4 P K plus h by the way notice that when it opens to the right or to the left you can only have uh One X intercept so here will be an example let's say if it look like this it's only going to touch the x-axis one once so that's why we we don't have plus or minus we just have one x intercept whereas if it opens up or down you can have a parabola that looks like like this and notice that potentially it can touch the xaxis at two points so thus you have plus or minus and the same is true for the Y intercepts but you could find those equations in the formula sheet so that's it for this video uh hopefully gave you a good introduction into conic sections and some of the formulas that you need in order to graph them and answer questions associate with them so thanks for watching
914
https://mathoverflow.net/questions/302917/for-a-proof-of-the-three-square-theorem-without-using-dirichlets-theorem-on-pri
nt.number theory - For a proof of the three-square theorem without using Dirichlet's theorem on primes in arithmetic progressions - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more For a proof of the three-square theorem without using Dirichlet's theorem on primes in arithmetic progressions Ask Question Asked 7 years, 3 months ago Modified7 years, 3 months ago Viewed 2k times This question shows research effort; it is useful and clear 13 Save this question. Show activity on this post. The three-square theorem states that n∈N={0,1,2,…}n∈N={0,1,2,…} is the sum of three squares if and only if it is not of the form 4 k(8 m+7)4 k(8 m+7) (k,m∈N k,m∈N). This was first proved by Legendre during 1797-1798. In 1837 Dirichlet proved his famous theorem on primes in arithmetic progressions. Now I can only find proofs of Legendre's three-square theorem using Dirichlet's theorem, see, e.g., M. B. Nathanson's book "Additive Number Theory-The Classical Bases" (GTM 164, Springer, 1996). Legendre's original proof did not involve Dirichlet's theorem which was proved later. A proof without using Dirichlet's theorem might be self-contained and hence suitable for the course of elementary number theory. QUESTION: Where can I find a proof of the three-square theorem without using Dirichlet's theorem? nt.number-theory reference-request sums-of-squares Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Jun 16, 2018 at 7:51 Zhi-Wei SunZhi-Wei Sun 17.1k 1 1 gold badge 23 23 silver badges 76 76 bronze badges 1 I think the question is addressed in Vicky Neale's outreaching book "closing the gap" about Zhang's 2013 breakthrough and subsequent Polymath8b project. If I remember correctly, she uses quadratic forms as GH suggests in his answer below.Sylvain JULIEN –Sylvain JULIEN 2018-06-16 22:10:30 +00:00 Commented Jun 16, 2018 at 22:10 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 13 Save this answer. Show activity on this post. I am not sure if Legendre's proof was complete (see this related MO entry), but surely Gauss gave a proof without Dirichlet's theorem (which came decades later). A nice transparent proof (without Dirichlet's theorem) can be found in Serre's book "A course in arithmetic". Very roughly, the proof goes as follows. Assume that n n satisfies the necessary local conditions. Step 1: There exists a positive integer t t such that t 2 n t 2 n is the sum of three squares. This relies on the theory of quadratic forms over Q Q. Step 2: The minimal t t that works in Step 1 is t=1 t=1. This relies on the following simple but key property: for any (x 1,x 2,x 3)∈Q 3(x 1,x 2,x 3)∈Q 3 there exists (y 1,y 2,y 3)∈Z 3(y 1,y 2,y 3)∈Z 3 such that ∑(x i−y i)2<1∑(x i−y i)2<1. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 16, 2018 at 21:48 answered Jun 16, 2018 at 21:35 GH from MOGH from MO 112k 8 8 gold badges 314 314 silver badges 426 426 bronze badges 9 3 @Zhi-WeiSun: I trust Serre and Weil more than Wikipedia. Serre's book proves Dirichlet's theorem, but later in the book (besides, I have read that book). Weil's historical book discusses issues with Legendre's proof of the three square theorem (not Gauss's refined version which relates the problem to class numbers).GH from MO –GH from MO 2018-06-17 00:56:54 +00:00 Commented Jun 17, 2018 at 0:56 5 @Zhi-WeiSun: Your question was "Where can I find a proof of the three-square theorem without using Dirichlet's theorem?". I just answered that. Serre's book has such a proof. BTW I don't think Legendre or Gauss had a simpler proof than Serre's book.GH from MO –GH from MO 2018-06-17 04:46:47 +00:00 Commented Jun 17, 2018 at 4:46 1 @GHfromMO this is the other question, but I am still curious how did Legendre or/and Gauss prove it (but I am lazy to ask a separate question, since a similar question already exists).Fedor Petrov –Fedor Petrov 2020-10-25 22:34:01 +00:00 Commented Oct 25, 2020 at 22:34 2 @FedorPetrov: Gauss, using his genus theory, could give a formula for the number of primitive representations of n n as a sum of three squares. It equals (24/w)h(−4 n)(24/w)h(−4 n) when n≡1,2,5,6(mod 8)n≡1,2,5,6(mod 8), and (48/w)h(−n)(48/w)h(−n) when n≡3(mod 8)n≡3(mod 8). Here h(D)h(D) is the class number of primitive binary quadratic forms of discriminant D D, and w w is the number of automorphs of such a form: w=2 w=2 for D<−4 D<−4, w=4 w=4 for D=−4 D=−4, w=6 w=6 for D=−3 D=−3. This is in Gauss' Disquisitiones arithmeticae, §291§291, but I have not read the original source, nor do I know a good modern course. Continued in next remark.GH from MO –GH from MO 2020-10-26 04:17:41 +00:00 Commented Oct 26, 2020 at 4:17 2 Gauss's formula is a special case of Siegel's mass formula, which expresses the (weighted) average representation number over a genus of classes of primitive positive quadratic forms (in any number of variables) as a product of local densities. As the genus of x 2+y 2+z 2 x 2+y 2+z 2 consists of a single class, the averaging really equals the representation number sought, while the product of local densities returns (24/π)n−−√L(1,χ−4 n)(24/π)n L(1,χ−4 n) or (24/π)n−−√L(1,χ−n)(24/π)n L(1,χ−n) depending on the residue of n n modulo 8 8. Here χ D χ D denotes the unique primitive quadratic character modulo D D.GH from MO –GH from MO 2020-10-26 04:22:51 +00:00 Commented Oct 26, 2020 at 4:22 |Show 4 more comments You must log in to answer this question. 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https://mathworld.wolfram.com/SteinerConstruction.html
Steiner Construction -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Geometry Geometric Construction Steiner Construction A construction done using only a straightedge. The Poncelet-Steiner theorem proves that all constructions possible using a compass and straightedge are possible using a straightedge alone, as long as a fixed circle and its center, two intersectingcircles without their centers, or three nonintersecting circles are drawn beforehand. For example, the centers of two intersecting circles can be found using a straightedge alone (Steinhaus 1999, p.142). See also Circle-Circle Intersection, Geometric Construction, Mascheroni Construction, Matchstick Construction, Neusis Construction, Poncelet-Steiner Theorem, Straightedge Explore with Wolfram|Alpha More things to try: angle trisection circle squaring 1 vigintillion References Dörrie, H. "Steiner's Straight-Edge Problem." §34 in 100 Great Problems of Elementary Mathematics: Their History and Solutions. New York: Dover, pp.165-170, 1965.Rademacher, H. and Toeplitz, O. The Enjoyment of Mathematics: Selections from Mathematics for the Amateur. Princeton, NJ: Princeton University Press, p.204, 1957.Steiner, J. Geometric Constructions with a Ruler, Given a Fixed Circle with Its Center. Translated from the first German ed. (1833). New York: Scripta Mathematica, 1950.Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, 1999. Referenced on Wolfram|Alpha Steiner Construction Cite this as: Weisstein, Eric W. "Steiner Construction." From MathWorld--A Wolfram Resource. Subject classifications Geometry Geometric Construction About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
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https://byjus.com/maths/linear-equations/
Linear equations are equations of the first order. The linear equations are defined for lines in the coordinate system.When the equation has a homogeneous variable of degree 1 (i.e. only one variable), then it is known as a linear equation in one variable. A linear equation can have more than one variable. If the linear equation has two variables, then it is called linear equations in two variables and so on. Some of the examples of linear equations are 2x – 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, 3x – y + z = 3. In this article, we are going to discuss the definition of linear equations, standard form for linear equation in one variable, two variables, three variables and their examples with complete explanation. Table of Contents: Definition Forms of Linear Equation Standard Form Slope-Intercept Form Point Slope Form How to Solve Linear Equations Solution of Linear Equation in One variable Solution of Linear Equation in Two variables Solution of Linear Equations in Three Variables Solving Linear Equations Practice Questions FAQs Linear Equation Definition An equation is a mathematical statement, which has an equal sign (=) between the algebraic expression. Linear equations are the equations of degree 1. It is the equation for the straight line. The solutions of linear equations will generate values, which when substituted for the unknown values, make the equation true. In the case of one variable, there is only one solution. For example, the equation x + 2 = 0 has only one solution as x = -2. But in the case of the two-variable linear equation, the solutions are calculated as the Cartesian coordinates of a point of the Euclidean plane. Below are some examples of linear equations in one variable, two variables and three variables: Linear Equation in One variableLinear Equation in Two variablesLinear Equation in Three variables 3x+5=0 (3/2)x +7 = 0 98x = 49y+7x=3 3a+2b = 5 6x+9y-12=0x + y + z = 0 a – 3b = c 3x + 12 y = ½ z Forms of Linear Equation The three forms of linear equations are Standard Form Slope Intercept Form Point Slope Form Now, let us discuss these three major forms of linear equations in detail. Standard Form of Linear Equation Linear equations are a combination of constants and variables.The standard form of a linear equation in one variable is represented as ax + b = 0, where, a ≠ 0 and x is the variable. The standard form of a linear equation in two variables is represented as ax + by + c = 0, where, a ≠ 0, b ≠ 0 , x and y are the variables. The standard form of a linear equation in three variables is represented as ax + by + cz + d = 0, where a ≠ 0, b ≠ 0, c ≠ 0, x, y, z are the variables. Slope Intercept Form The most common form of linear equations is in slope-intercept form, which is represented as; y = mx + b Where, m is the slope of the line, b is the y-intercept x and y are the coordinates of the x-axis and y-axis, respectively. For example, y = 3x + 7: slope, m = 3 and intercept = 7 If a straight line is parallel to the x-axis, then the x-coordinate will be equal to zero. Therefore, y=b If the line is parallel to the y-axis then the y-coordinate will be zero. mx+b = 0 x=-b/m Slope:The slope of the line is equal to the ratio of the change in y-coordinates to the change in x-coordinates. It can be evaluated by: m = (y 2-y 1)/(x 2-x 1) So basically the slope shows the rise of line in the plane along with the distance covered in the x-axis. The slope of the line is also called a gradient. Point Slope Form In this form of linear equation, a straight line equation is formed by considering the points in the x-y plane, such that: y – y 1 = m(x – x 1 ) where (x 1, y 1) are the coordinates of the point. We can also express it as: y = mx +y 1 – m x 1 Summary: There are different forms to write linear equations. Some of them are: Linear EquationGeneral FormExample Slope intercept form y = mx + b y + 2x = 3 Point–slope form y – y 1 = m(x – x 1 )y – 3 = 6(x – 2) General Form Ax + By + C = 0 2x + 3y – 6 = 0 Intercept form x/a+ y/b= 1 x/2 + y/3 = 1 As a Function f(x) instead of y f(x) = x + C f(x) = x + 3 The Identity Function f(x) = x f(x) = 3x Constant Functions f(x) = C f(x) = 6 Where m = slope of a line; (a, b) intercept of x-axis and y-axis. Also, read: Application of linear equations Linear Equations One Variable Worksheet Graphing Of Linear Equations Pair Of Linear Equations In Two Variables Linear Equations In Two Variables Class 9 How to Solve Linear Equations? By now you have got an idea of linear equations and their different forms. Now let us learn how to solve linear equations or line equations in one variable, in two variables and in three variables with examples. Solving these equations with step by step procedures are given here. Solution of Linear Equations in One Variable Both sides of the equation are supposed to be balanced for solving a linear equation. The equality sign denotes that the expressions on either side of the ‘equal to’ sign are equal. Since the equation is balanced, for solving it, certain mathematical operations are performed on both sides of the equation in a manner that does not affect the balance of the equation. Here is the example related to the linear equation in one variable. Example: Solve(2x – 10)/2 = 3(x – 1) Step 1: Clear the fraction x – 5 = 3(x – 1) Step 2: Simplify Both sides equations x – 5 = 3x – 3 x = 3x + 2 Step 3: Isolate x x – 3x = 2 -2x = 2 x = -1 Solution of Linear Equations in Two Variables To solve linear equations in 2 variables, there are different methods. Following are some of them: Method of substitution Cross multiplication method Method of elimination We must choose a set of 2 equations to find the values of 2 variables. Such as ax + by + c = 0 and dx + ey + f = 0, also called a system of equations with two variables, where x and y are two variables and a, b, c, d, e, f are constants, and a, b, d and e are not zero. Else, the single equation has an infinite number of solutions. Solution of Linear Equations in Three Variables To solve linear equations in 3 variables, we need a set of 3 equations as given below to find the values of unknowns. Matrix method is one of the popular methods to solve system of linear equations with 3 variables. a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 and a 3 x + b 3 y + c 3 z + d 3 = 0 Also check:Solve The Linear Equation In Two Or Three Variables 8,43,763 Solving Linear Equations Example 1: Solve x = 12(x +2) Solution: x = 12(x + 2) x = 12x + 24 Subtract 24 on both sides of equation x – 24 = 12x + 24 – 24 x – 24 = 12x Simplify 11x = -24 Isolate x: x = -24/11. Example 2: Solve x – y = 12 and 2x + y = 22 Solution: Name the equations x – y = 12 …(1) 2x + y = 22 …(2) Isolate Equation (1) for x, x = y + 12 Substitute x =y + 12 in equation (2) 2(y+12) + y = 22 3y + 24 = 22 3y = -2 or y = -2/3 Substitute the value of y in x = y + 12 x = y + 12 x = -2/3 + 12 x = 34/3 Answer:x = 34/3 and y = -2/3 Practice Questions Solve the following linear equations: 5y-11=3y+9 3x + 4 = 7 – 2x 9 – 2(y – 5) = y + 10 5(x – 1) = 3(2x – 5) – (1 – 3x) 2(y – 1) – 6y = 10 – 2(y – 4) y/3 – (y – 2)/2 = 7/3 (y – 3)/4 + (y – 1)/5 – (y – 2)/3 = 1 (3x – 2)/3 + (2x + 3)/3 = (x + 7)/6 (8y – 5)/(7y + 1) = -4/5 (5 – 7y)/(2 + 4y) = -8/7 Stay tuned with BYJU’S – The Learning App and download the app to learn all Maths concepts and also get the video lesson to learn the concepts quickly. Frequently Asked Questions on Linear Equations Q1 What is a Linear equation? Linear equations are the equations of degree 1. It is the equation for the straight line. The standard form of linear equation is ax+by+c =0, where a ≠ 0 and b ≠ 0. Q2 What are the three forms of linear equations? The three forms of linear equations are standard form, slope-intercept form and point-slope form. Q3 How do we express the standard form of a linear equation? The standard form of linear equations is given by: Ax + By + C = 0 Here, A, B and C are constants, x and y are variables. Also, A ≠ 0, B ≠ 0 Q4 What is the slope-intercept form of linear equations? The slope-intercept form of linear equations is given by: y=mx+b Where m denotes the steepness of line and b is the y-intercept. Q5 What is the difference between linear and non-linear equations? A linear equation is meant for straight lines. A non-linear equation does not form a straight line. It can be a curve that has a variable slope value. Test your knowledge on Linear Equations Q 5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Start Quiz Congrats! Visit BYJU’S for all Maths related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted View Quiz Answers and Analysis X Login To View Results Mobile Number Send OTP Did not receive OTP? Request OTP on Voice Call Login To View Results Name Email ID Grade City View Result Comments Leave a Comment Cancel reply Your Mobile number and Email id will not be published.Required fields are marked Send OTP Did not receive OTP? Request OTP on Voice Call Website Post My Comment JOY ABHISHEKOctober 31, 2020 at 12:32 pm Thanks for the explaining Reply Register with BYJU'S & Download Free PDFs Send OTP Download Now Register with BYJU'S & Watch Live Videos Send OTP Watch Now × To continue watching the video please share the details. Send OTP Did not receive OTP? Request OTP on Voice Call Play Now
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https://www.facebook.com/frugalitaph/posts/frugality-defined-according-to-merriam-webster-frugality-is-a-careful-management/168926208797352/
Frugalita PH - FRUGALITY defined According to... | Facebook Log In Log In Forgot Account? Frugalita PH's Post Frugalita PH February 8, 2022 · FRUGALITY defined According to Merriam-Webster, frugality is a careful management of material resources such as time and money. Being frugal, is doing things and spending money that matters to you the most. Living a frugal lifestyle teaches us to live within means and also to maximize our potential to save for the future. Here are some of practical ways to live a frugal life without sacrificing the quality of a good life. Hope these tips help you to kick-start your frugal living. Keep tuned to more tips on how to live a frugal life. #Frugality #LivingLifeWithinMeans +7 Like Comment
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https://artofproblemsolving.com/wiki/index.php/System_of_equations?srsltid=AfmBOopDI1eeqCzvC_gNtwStf49rTpcg7n6cLv_pinsQ1E1wya0BB5W1
Art of Problem Solving System of equations - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki System of equations Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search System of equations A system of equations is a set of equations which share the same variables. Below is an example of a system of equations. Contents [hide] 1 Solve 2 variable equations in less than 5 seconds!!! 2 Solving Linear Systems 2.1 Gaussian Elimination 2.1.1 Problem 2.1.2 Solution 2.2 Substitution 2.2.1 Problem 2.2.2 Solution 2.3 Graphing 2.3.1 Problem 2.3.2 Solution 2.4 Advanced Methods 3 Convenient Systems 3.1 Symmetry 3.2 Clever Substitution 4 Problems 4.1 Introductory 4.2 Intermediate 5 See Also Solve 2 variable equations in less than 5 seconds!!! Video Link: Solving Linear Systems A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations. Gaussian Elimination Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example: Problem Find the ordered pair for which Solution We can eliminate by adding twice the second equation to the first: Thus . We can then plug in for in either of the equations: Thus, the solution to the system is . Substitution Main article: Substitution The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution. Problem Find the ordered pair for which Solution The first equation can be solved for : Plugging this into the second equation yields Thus . Plugging this into either of the equations and solving for yields . Graphing The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect. We'll go back to our same example to illustrate this. Problem Find the ordered pair for which Solution We graph the two lines as follows: From the graph, we can see that the solution to the system is . Advanced Methods Matrices can also be used to solve systems of linear equations. In fact, they provide a way to make much broader statements about systems of linear equations. There is a whole field of mathematics devoted to the study of linear equations called linear algebra. Convenient Systems Some systems can be solved by taking advantage of specific forms. Such systems can often seem tough to solve at first, however. Symmetry Consider the below system. The key here is to take advantage of the symmetry. If we add up all 5 equations we will have a total of 4 of each variable on the LHS. On the RHS we will have . Thus So then subtracting the first equation from this leaves on the LHS and on the RHS. Subtracting this equation from the second equation leaves on the LHS and on the RHS. And thus we continue on in this way to find that Clever Substitution Consider the below system. We can let and to get the two-variable linear system below. Solving the system results in and . Substituting that back results in and . We can do another substitution by letting and substituting to get . Rearranging results in , so . Finally, by substituting back in, we get . Plugging back satisfies the system. Problems Introductory 2002 AMC 8 Problems/Problem 17 2007 iTest Problems/Problem 2 Intermediate 1989 AIME Problems/Problem 8 1993 AIME Problems/Problem 3 See Also Algebra Substitution Retrieved from " Category: Algebra Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
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https://www.zhihu.com/question/324734890
为什么二次函数横向平移是左加右减呢？ - 知乎 关注推荐热榜专栏圈子 New付费咨询知学堂 ​ 直答 切换模式 登录/注册 为什么二次函数横向平移是左加右减呢？ 关注问题​写回答 登录/注册 数学 函数 解析几何 二次函数 函数图像 为什么二次函数横向平移是左加右减呢？ 而不是左减右加？显示全部 ​ 关注者 133 被浏览 140,975 关注问题​写回答 ​邀请回答 ​好问题 15 ​3 条评论 ​分享 ​ 54 个回答 默认排序 帽谱mop ​​ 浙江大学 软件工程硕士 ​ 关注 更新： 修改了一下二重积分的符号，之前是连用两个不定积分符号，现在改成了二重积分 我一直觉得高中教导的左加右减上加下减就是误人子弟的。这分明就是用换元法能一步写明白，一步想明白的东西，非要整个似是而非的口诀。而最过分的是，如果你对其中的原理掌握不明白的话，乱用口诀反而会得到错误的结果。 比如直线方程 x+y+6=0 x+y+6=0 请问这个直线向上（或者说沿y轴正向）平移六个单位后的方程是下面哪个？ A：x+(y+6)+6=0 A：x+(y+6)+6=0 B：x+(y-6)+6=0 习惯上加下减是不是很容易就选A了？但是仔细想想，原本的直线过点 （0，-6） ，向上平移后直线过点 （0，0） ，因此答案是B才对。 实际上对图像平移，可以直接用换元来解决。 还是以上面的直线方程举例，将直线向左平移2个单位，向上平移3个单位。 设直线上某点坐标 (x,y) ，该点平移后的坐标为 (x',y') 则很容易知道 x'=x-2 y'=y+3 即 x=x'+2 y=y'-3 用 x' 和 y'来替换 x 和 y 得 (x'+2)+(y'-3)+6=0 即 x'+y'+5=0 因此平移后的直线方程为 x+y+5=0 这里换元法的优势在于，对于三维甚至更高维的直线方程、曲面方程，都是可以适用的，三维坐标你怎么用左加右减上加下减呢？ 实际上，我在备战考研时，对于类似下面这样的习题，也会用换元法对积分区域进行平移来简化计算。 计算二重积分 I=\iint_{}^{}xdS ,积分区域为 D={(x,y)|x^2+y^2\leq2x } 很容易知道，区域D是一个以 (1,0) 为圆心，半径 r=1 的圆，我们把这个区域向左平移1个单位，把他的圆心平移到原点上，这样就可以很方便的用极坐标进行替换计算了。 将区域向左平移一个单位，对于平移后的横坐标 x' 有 x'=x-1 即 x=x'+1 用 x' 替换 x ，因此原二重积分 I=\iint_{}^{}(x+1)dS=\iint_{}^{}xdS+\iint_{}^{}dS ,积分区域 D'={(x,y)|x^2+y^2\leq1 } 显然区域关于x对称，因此 \iint_{}^{}xdS=0 ，故 I=\iint_{}^{}dS=\pi 或者这里我们计算相同积分区域的另一个二重积分 I=\iint_{}^{}x^2dS 平移后 I=\iint_{}^{}(x+1)^2dS=\iint_{}^{}(x^2+2x+1)dS 其中我们知道 \iint_{}^{}2xdS=0 , \iint_{}^{}dS=\pi 而\iint_{}^{}x^2dS=\int_{0}^{2\pi}cos^2\theta d\theta\int_{0}^{1}r^3dr=4I(2)\frac{1}{4}=\frac{\pi}{4} 得 I=\frac{\pi}{4}+\pi=\frac{5\pi}{4} 展开阅读全文​ ​赞同 367​​84 条评论 ​分享 ​收藏​喜欢 Algebra ​ 欢迎进群唠嗑：339820728 ​ 关注 不得不说，题主提了一个好问题。 什么叫好问题呢？ 就是那些可以让人思考，且在解决完该问题后如果按照其思路，会得到一些更有趣的结论的问题。 左加右减这玩意确实很有意思，和照镜子很像 镜子里的你的左手，其实是你现实中的右手。 同样的，假如一个函数的方程为 y=f(x)[y = f\left( x \right)] ，那么其向右平移一格后的方程为 y=f(x−1)[y = f\left( {x - 1} \right)] 。 乍一看，这玩意确实与我们的直觉不符，毕竟在坐标轴上，往右走是加，向左走是减 所以第一件事，我们得尝试证明一下这玩意，然后再说其他的。 还是先从简单的例子入手，从特殊到一般是研究数学问题基本的套路 首先，下图是一个普通的不能再普通的二次函数，它的方程为 y=x^2 。 我们把它往右平移一格（变成蓝色的那条曲线） 那这条蓝色曲线的方程怎么求呢？这就是我们接下来需要研究的。 我们先回忆一个平凡的事实哈，一条曲线，你往左平移也好，往右平移也罢，它的竖直方向是不会有变化的。 就像你照镜子的时候，镜子里的你的左手实际上是你的右手，但你的头并没有变成你的脚（比喻可能不恰当，但大致就是这么个意思） 回到刚刚的问题上来，现在我们在蓝色曲线上随便找一点，不妨就叫它点 A 吧，它的坐标是 [\left( {{x_0},{y_0}} \right)] ，如图所示： 然后呢，我们过 A 点，作一条平行于 x 轴的直线，像这样： 再把这条直线与最初的那条曲线，也就是方程为 y=x^2 的曲线的交点设为 B 可以看出， B 点只不过是 A 点向左平移一格所得到的点 什么意思？ 就是它的竖直方向的坐标是不会有变化的。 意思就是，如果我们设 A 点的坐标是 [\left( {{x_0},{y_0}} \right)]，则 B 点的坐标显然就是 [\left( {{x_0} - 1,{y_0}} \right)]。 接下来就到了最重点的地方了，注意！！！ 不管怎么说， B 点也是曲线 y=x^2 上的一点，既然是 y=x^2 上的点，那就必然满足其横坐标的平方等于其纵坐标。 而 B 点的坐标是 [\left( {{x_0} - 1,{y_0}} \right)]，这也就意味着它的横坐标是 {x_0} - 1 ，纵坐标是 {y_0} ，所以我们得到了： [{y_0} = {\left( {{x_0} - 1} \right)^2}] 。 打住！ 在这打住！ 回忆一下，因为 A 点是在蓝色曲线上任意取的，而它的坐标[\left( {{x_0},{y_0}} \right)]满足关系式[{y_0} = {\left( {{x_0} - 1} \right)^2}] 这也就意味着蓝色曲线上任意一点都满足关系式[{y_0} = {\left( {{x_0} - 1} \right)^2}] 好了，问题已经解决了，所以蓝色曲线的方程就是 [y = {\left( {x - 1} \right)^2}] 。 那除了这种方法外，还有没有其他的理解方式呢？ 答案是有的，不过我们得先引入坐标系变换的概念。 首先，咱们得对什么叫“一个点的坐标”有一个共识，举个例子，在下面这个坐标系中，点 A 的坐标为 [\left( {1,1} \right)] 。 这个 [\left( {1,1} \right)] 是什么意思？ 换句话说，点的坐标究竟指的是什么？ 这是百度百科上的解释（说句题外话，能把“轴”写成“周”，现在的百度百科真是越来越拉了） 百度上的解释没有错，初高中时你可以这么理解，但现在我并不建议 我建议用向量与坐标系的基来理解，不妨假设这个坐标系的原点为 O ，基为 [\overrightarrow {{e_1}} ] ， [\overrightarrow {{e_2}} ] 。 从原点 O 出发，到 A 点，作一个向量 [\overrightarrow {OA} ] 那么[\left( {1,1} \right)]指的是 [1 \cdot \overrightarrow {{e_1}} + 1 \cdot \overrightarrow {{e_2}} ] 也就是说，点 A 在某坐标系下的坐标是 [\left( {1,1} \right)] 等价于[\overrightarrow {OA} {\rm{ = }}1 \cdot \overrightarrow {{e_1}} + 1 \cdot \overrightarrow {{e_2}} ] 虽然书写的时候我们写作 [A\left( {1,1} \right)] ，但请不要在写的时候忘记它最初最想表达的意思。 好了，现在问题来了，同样一个点，在不同坐标系下的坐标一样吗？ 显然是不一样的，还是拿刚刚的点 A 举例，不过这一次换了一个新的坐标系，它的原点为 O'，基为 [\overrightarrow {{e_1}} '] ， [\overrightarrow {{e_2}} '] ，此时就有： [\overrightarrow {O'A} = 0 \cdot \overrightarrow {{e_1}} ' + 1 \cdot \overrightarrow {{e_2}} '] 。 所以在这个新的坐标系中，点 A 的坐标为 [\left( {0,1} \right)] 。 好，接下来，我们还是取蓝色曲线上的任意一点 A 不过这一次我们先研究 A 点在原点为 O'，基为 [\overrightarrow {{e_1}} '] ， [\overrightarrow {{e_2}} ']的坐标系上的坐标 我们不妨假设在此坐标系下 A 点的坐标为 [\left( {x',y'} \right)] ，即 [\overrightarrow {O'A} = x' \cdot \overrightarrow {{e_1}} ' + y' \cdot \overrightarrow {{e_2}} '] 又因为在此坐标系下，蓝色二次曲线为标准二次曲线，在此坐标系下曲线上任意一点的纵坐标都是横坐标的平方，所以我们有： [y' = {\left( {x'} \right)^2}] 接下来我们研究 A 点在原本坐标系下的坐标： 还是先设 [\overrightarrow {OA} = x \cdot \overrightarrow {{e_1}} + y \cdot \overrightarrow {{e_2}} ] ，即： 最后，我们显然有 [\overrightarrow {OO'} = 1 \cdot \overrightarrow {{e_1}} + 0 \cdot \overrightarrow {{e_2}} ] \overrightarrow {{e_1}} =\overrightarrow {{e_1}} ' , \overrightarrow {{e_2}} =\overrightarrow {{e_2}} ' 注意到： [\overrightarrow {OA} ] [ = \overrightarrow {OO'} + \overrightarrow {O'A} ] [ = 1 \cdot \overrightarrow {{e_1}} + 0 \cdot \overrightarrow {{e_2}} + x' \cdot {\overrightarrow {{e_1}} ^\prime } + y' \cdot {\overrightarrow {{e_2}} ^\prime }] [ = 1 \cdot \overrightarrow {{e_1}} + 0 \cdot \overrightarrow {{e_2}} + x' \cdot \overrightarrow {{e_1}} + y' \cdot \overrightarrow {{e_2}} ] [ = \left( {x' + 1} \right) \cdot \overrightarrow {{e_1}} + y' \cdot \overrightarrow {{e_2}} ] 而之前我们设 [\overrightarrow {OA} = x \cdot \overrightarrow {{e_1}} + y \cdot \overrightarrow {{e_2}} ] 所以此时有： [\left{ {\begin{array}{{20}{c}} {x = x' + 1}\ {y = y' \ \ \ \ \ \ \ } \end{array}} \right.] ，即 [\left{ {\begin{array}{{20}{c}} {x' = x - 1}\ {y' = y \ \ \ \ \ \ \ \ } \end{array}} \right.] 回忆起 [y' = {\left( {x'} \right)^2}] ，代入，有： [y = {\left( {x - 1} \right)^2}] 也就是说 A 点在坐标系 [\left[ {O;\overrightarrow {{e_1}} ,\overrightarrow {{e_2}} } \right]] 上有关系式 [y = {\left( {x - 1} \right)^2}] ，而 A 点是任意的，故将二次曲线向右平移一格后的方程为 [y = {\left( {x - 1} \right)^2}] 。 严格意义上来说，前面我只是运用坐标轴变换的思路简单实验了一下 那有没有一般形式的变换呢？ 有的！ 同样的思路 假设 [\overrightarrow {OA} = x \cdot \overrightarrow {{e_1}} + y \cdot \overrightarrow {{e_2}} ]， [\overrightarrow {O'A} = x' \cdot {\overrightarrow {{e_1}} ^\prime } + y' \cdot {\overrightarrow {{e_2}} ^\prime }] 那么 x , y , x' , y' 有什么关系呢？ 假设 [\left{ {\begin{array}{{20}{c}} {\overrightarrow {{e_1}} ' = {c_{11}}\overrightarrow {{e_1}} + {c_{21}}\overrightarrow {{e_2}} }\ {\overrightarrow {{e_2}} ' = {c_{12}}\overrightarrow {{e_1}} + {c_{22}}\overrightarrow {{e_2}} } \end{array}} \right.] ， [\overrightarrow {OO'} = {d_1} \cdot \overrightarrow {{e_1}} + {d_2} \cdot \overrightarrow {{e_2}} ] 其中 [\left| {\begin{array}{{20}{c}} {{c_{11}}}&{{c_{12}}}\ {{c_{21}}}&{{c_{22}}} \end{array}} \right| \ne 0] （这是为了保证这步变换是可逆的，不用太在意） 故： [\overrightarrow {OA} ] [ = \overrightarrow {OO'} + \overrightarrow {O'A} ] [={d_1} \cdot \overrightarrow {{e_1}} + {d_2} \cdot \overrightarrow {{e_2}} + x' \cdot \overrightarrow {{e_1}} ' + y' \cdot \overrightarrow {{e_2}} '] [ = {d_1} \cdot \overrightarrow {{e_1}} + {d_2} \cdot \overrightarrow {{e_2}} + {c_{11}}x'\overrightarrow {{e_1}} + {c_{21}}x'\overrightarrow {{e_2}} + {c_{12}}y'\overrightarrow {{e_1}} + {c_{22}}y'\overrightarrow {{e_2}} ] [ = \left( {{c_{11}}x' + {c_{12}}y' + {d_1}} \right) \cdot \overrightarrow {{e_1}} + \left( {{c_{21}}x' + {c_{22}}y' + {d_2}} \right) \cdot \overrightarrow {{e_2}} ] 又因为 [\overrightarrow {OA} = x \cdot \overrightarrow {{e_1}} + y \cdot \overrightarrow {{e_2}} ] 所以我们得到关系式： [\left{ {\begin{array}{{20}{c}} {x = {c_{11}}x' + {c_{12}}y' + {d_1}}\ {y = {c_{21}}x' + {c_{22}}y' + {d_2}} \end{array}} \right.] 如果觉得不美观，可以将其写成矩阵形式： [\left( {\begin{array}{{20}{c}} x\ y \end{array}} \right) = \left( {\begin{array}{{20}{c}} {{c_{11}}}&{{c_{12}}}\ {{c_{21}}}&{{c_{22}}} \end{array}} \right)\left( {\begin{array}{{20}{c}} {x'}\ {y'} \end{array}} \right) + \left( {\begin{array}{{20}{c}} {{d_1}}\ {{d_2}} \end{array}} \right)] 说实话，第一次学到这时，怎么看怎么别扭 因为按照正常思路，我们是希望知道 x , y ，然后求出 x' , y' ,但这个公式却恰好反过来了 不过从某种意义上，这样造成了所谓“左加右减”的反直觉经验 因为如果我们将那个式子整理一下，就有： [\left( {\begin{array}{{20}{c}} {x'}\ {y'} \end{array}} \right) = \left[ {\left( {\begin{array}{{20}{c}} x\ y \end{array}} \right) - \left( {\begin{array}{{20}{c}} {{d_1}}\ {{d_2}} \end{array}} \right)} \right]{\left( {\begin{array}{{20}{c}} {{c_{11}}}&{{c_{12}}}\ {{c_{21}}}&{{c_{22}}} \end{array}} \right)^{ - 1}}] 而平移的时候， [\left( {\begin{array}{{20}{c}} {{c_{11}}}&{{c_{12}}}\ {{c_{21}}}&{{c_{22}}} \end{array}} \right)] 为单位矩阵，所以 [\left( {\begin{array}{{20}{c}} {x'}\ {y'} \end{array}} \right) = \left( {\begin{array}{{20}{c}} x\ y \end{array}} \right) - \left( {\begin{array}{{20}{c}} {{d_1}}\ {{d_2}} \end{array}} \right)] 这就是“左加右减”的根源所在！ 当时大二刚学完高等代数就打算写一下这篇文章，结果拖到现在才弄完…… 只能说，人类的鸽能力是无止境的！ 展开阅读全文​ ​赞同 111​​9 条评论 ​分享 ​收藏​喜欢 假装做国王 有天才的病，没有天才的命 ​ 关注 问得好，首先需要说明的是，这并不只是对于二次函数。对于任意的函数都存在这样的现象。以下不从定义角度解释，给题主从逻辑的角度讲讲看。 首先，我们理解一下函数的本质，是指一个映射的关系。从图像的角度看，一个函数y=f(x)，y是图像上的纵坐标，f是映射关系，x是图像上的横坐标。 那么我们来看横向平移的情况， 假设移动前图像表达式是y=f(x)，移动后图像表达式是m=h(n)。平移距离为d，d>0则是向右平移。 那么我们要求m=f(n)是y=f(x)横向平移得到的。横向平移的特点是什么，用通俗的语言来讲，我一米八，往左走了一步还是一米八，这才叫平移。你不能说我往左走就是一米九，往右走变一米七了。那不叫平移。 所以，对于图像中每一个单独的点，平移后的纵坐标是不能变的。 也就是说对于任意特定的点（x',y'）在y=f(x)的图像上，自然有y'=f(x')。由平移后纵坐标不变，必然有h(x'+d) = y'，即（x'+d，y'）在平移后的图像上 例如（1，3）在平移前的图像上，d=2，向右平移2个单位，则（3，3）必然在平移后的图像上 用数学语言描述，那就是f(x) = h(n+d)。翻译回人话就是“平移前函数上所有的点，高度都跟他平移d距离后的高度一样” 但是你是已经知道了f(x)和d，h是未知的。所以这个式子变换为f(x-d) = h(n)。也就是我们说的左加右减 所以你的疑惑解决了，为什么左加右减这么反人类？ 从坐标上来看，平移确实是左减右加，（1，3）平移+2变成（3，3）。 坐标转换为表达式就是f(x) = h(n+d)。这里也还是左减右加。但是在这个式子中，f(x)、d已知，你想求的是h(n)，已知量扔到一边这是很正常的思维，所以我们有一个类似移项的过程，就是这个过程导致了d的符号跟我们直觉上理解的恰好相反。 再更进一步，为什么纵向平移y就是直接加减，横向就会有这个问题呢？ 说得通俗一点，对于y=f(x)、m=h(n)这两个平移前后的式子来说，纵向平移建立的是y、m的对应关系，他们不套在f、h里面，所以直接加减没有任何问题，向上平移3，m = y+3。此时你知道y和d求m。 但是横向平移建立的是x、n的对应关系，使得f(x) = h(n+d)，而此时你知道f、d想求h，势必需要把d跟f统一起来，简化h的形式。这就导致了上面说的需要"移项"，从而导致了符号跟直觉相反。 展开阅读全文​ ​赞同 131​​5 条评论 ​分享 ​收藏​喜欢 查看剩余 51 条回答 ​写回答 2 个回答被折叠（为什么？） 下载知乎客户端 与世界分享知识、经验和见解 知乎 AI 产品榜上线啦 相关问题 一次函数为什么左右平移是左加右减？ 9 个回答 一次函数为什么左右平移是左加右减呐？ 4 个回答 已知一个二次函数解析式，将其向上平移，它与x轴的交点和向上平移的关系是什么？ 1 个回答 为什么函数的平移是“左加右减”？ 4 个回答 函数的平移法则：上加下减 左加右减究竟怎么个操作法？求解答 靴靴? 2 个回答 帮助中心 知乎隐私保护指引申请开通机构号联系我们 举报中心 涉未成年举报网络谣言举报涉企侵权举报更多 关于知乎 下载知乎知乎招聘知乎指南知乎协议更多 京 ICP 证 110745 号 · 京 ICP 备 13052560 号 - 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https://aga-fileuploader-bucket.s3.us-east-2.amazonaws.com/AGA%20Gastroparesis%20Guideline%202025-04-15-6.pdf
AGA Clinical Practice Guideline on Management of Gastroparesis Authors: Kyle Staller1, Henry P. Parkman2, Katarina B. Greer3, David A. Leiman4, Margaret Zhou5, Shailandra Singh6, Michael Camilleri7, Osama Altayar8 on behalf of the AGA Clinical Guidelines Committee First author(s) Senior author(s) Institutions: 1Division of Gastroenterology, Massachusetts General Hospital and Harvard Medical School, Boston, MA 02114; 2 Gastroenterology Section, Department of Medicine, Temple University, Philadelphia, PA 19140; 3 Department of Medicine, Louis Stokes Cleveland VA Medical Center, Case Western Reserve University School of Medicine, Cleveland, OH 44106; 4Division of Gastroenterology and Duke Clinical Research Institute, Duke University School of Medicine, Durham, NC 27710; 5Division of Gastroenterology and Hepatology, Department of Medicine, Stanford University School of Medicine, Stanford, CA 94305;6 Section of Gastroenterology and Hepatology, Department of Medicine, West Virginia University, Morgantown, WV 26506; 7 Division of Gastroenterology and Hepatology, Mayo Clinic, Rochester, MN 55905; 8 Division of Gastroenterology, Washington University School of Medicine, St Louis, MO 63110 Corresponding Author Address: Chair, Clinical Guidelines Committee, American Gastroenterological Association, National Office, 4930 Del Ray Avenue, Bethesda, Maryland 20814. Email: clinicalpractice@gastro.org. Manuscript Information: • Word Count: 14,600 • Number of References: 102 • Number of Tables/Figures: 3 tables and 1 figure • Number of Supplemental Tables/Figures: 53 tables and 57 figures Acknowledgements The authors would like to thank Ms. Caitlin Bakker, Archer Library at University of Regina, Regina, Saskatchewan for their expertise in designing and conducting the literature searches. The Guideline Panel also acknowledges the support of AGA guideline staff. ABSTRACT Background & Aims Gastroparesis is a complex gastric motility disorder characterized by symptoms of nausea, vomiting, and other symptoms associated with a delay in gastric emptying in the absence of mechanical obstruction. Variations in diagnostic testing and limited effective treatments make caring for this patient population challenging. The American Gastroenterological Association developed this guideline to provide recommendations on 1) ensuring an accurate diagnosis and 2) identifying evidence-based, effective treatments among the available pharmacological and procedural interventions for patients with idiopathic or diabetic gastroparesis. Methods The Grading of Recommendations Assessment, Development and Evaluation (GRADE) framework was used to assess evidence and develop this guideline. The Guideline Panel prioritized clinical questions and outcomes, conducted an evidence review, and used the Evidence-to-Decision Framework to develop recommendations. Results The Guideline Panel agreed on 12 recommendations. A conditional recommendation was issued against using 2-hour gastric emptying testing and in favor of 4-hour testing in patients with suspected gastroparesis. There are conditional recommendations for the use of metoclopramide and erythromycin in patients with gastroparesis. A conditional recommendation was issued against the use of gastric per-oral endoscopic myotomy except in select patients with disease refractory to medical therapies. Moreover, conditional recommendations were issued against the use of domperidone, prucalopride, aprepitant, nortriptyline, buspirone, cannabidiol, pyloric botulinum toxin injection, and gastric electrical stimulation in patients with gastroparesis. The role of surgical pyloromyotomy and surgical pyloroplasty is uncertain in patients with gastroparesis because of identified knowledge gaps. Conclusions Other than 4-hour testing for diagnosis, several recommendations require shared decision-making based on patient preferences and supportive evidence. There is still considerable unmet need in the treatment of gastroparesis. (Note: Include only if needed, while complying with the word limit.) Key Words: prokinetic; anti-emetic; neuromodulator; endoscopic myotomy; pyloroplasty EXECUTIVE SUMMARY Description of the Health Problem Gastroparesis is a disorder of gastric motility defined by a measurable delay in gastric emptying associated with a constellation of upper gastrointestinal (GI) symptoms in the absence of mechanical obstruction. Variation in diagnostic testing strategies have made diagnosis and effective treatment challenging, and differentiating gastroparesis as the source of upper GI symptoms can be difficult. Despite diagnostic challenges, symptoms suggestive of gastroparesis are highly prevalent and patients with gastroparesis incur significant direct and indirect healthcare costs as well as a reduction in quality of life.1 To date, there is only one treatment approved by the Food and Drug Administration (FDA) for gastroparesis (metoclopramide), but newer medical and procedural approaches to this condition have been tested or are emerging. An objective assessment of the available evidence of tested treatments is to be provided by this Guideline. Aims and Objectives of the Guideline The purpose of these recommendations is to provide evidence-based guidance for patients with suspected and confirmed gastroparesis including gastroparesis refractory to medical therapies based on a systematic and comprehensive synthesis of the literature. The guideline document recommendations are intended to appraise the benefits of the available therapies including the improvement in the cardinal symptoms of nausea, vomiting, early satiety, and postprandial fullness in addition to quality of life while minimizing the potential adverse effects. The recommendations also account for the feasibility of the available interventions. Target Audience While the Guideline Panel is aware that this guideline and its recommendations will be read by a wide audience, including policymakers and industry, that is not its primary audience. This guideline, like all AGA guidelines, is primarily targeted towards healthcare providers in the field of gastroenterology and primary care who depend on such evidence-based, expert recommendations to inform their clinical practice and shared decision-making with patients. We intend these recommendations to be used by clinicians to guide the management of patients and inform considerations of benefits and harms of treatments in each individual case, rather than providing a specific standard to adhere to. In this way, the guideline and its recommendations are intended to contribute to the improved delivery and management of GI health care services. Recommendations o Table 1: List of Guideline Recommendations Supplementary Material Refer to the guideline’s Supplementary Material, Clinical Decision Support Tool, and Spotlight, while including space for the journal to incorporate weblinks to each for easy access by the reader. 1. GUIDELINE DEVELOPMENT PROCEDURES 1.1 Overview This document represents the official recommendations of the AGA and was developed using the Grading of Recommendations Assessment, Development and Evaluation (GRADE) framework. It adheres to best practices in guideline development as outlined by the National Academy of Medicine (formerly Institute of Medicine), using a process outlined previously,2 1.2 Organization and Panel Composition The Guideline Panel included 8 members (Supplemental Table 1), all selected based on their specific expertise. There were 6 Clinical/Content Experts with clinical and/or research expertise in the clinical topic and 2 Methodologists with specialized GRADE guideline development skills. A librarian assisted the Panel members with designing and executing the required literature searches. All Panel members participated in evidence review and the Senior Methodologist oversaw the evidence synthesis. All members of the guideline committee helped review the synthesized evidence, contributed to discussion, and helped develop the recommendations and clinical decision support tool (Figure 1). 1.3 Conflict of Interest Each Guideline Panel nominee underwent a vetting process that required disclosure of all conflicts of interest. Reported conflicts were reviewed by the Chair of the AGA Clinical Guidelines Committee (CGC) and adjudicated against rules and criteria in the CGC COI Policy. Only nominees whose COI status (Supplemental Table 2) complied with this policy were appointed. 1.4 Guideline Funding AGA provided all financial support for the development of this guideline. No funding from industry was offered or accepted to support the writing effort. 1.5 Document Review The guideline underwent several levels of review including an open, 30-day Public Comment period, external Peer Review by at least three known topic experts, and Patient Review. Organizational-level review was carried out by AGA Institute’s Clinical Guidelines Committee and Governing Board. At each stage, the Guideline Panel considered all reviewer comments and feedback and revised the guideline manuscript to address feedback as needed. 1.6 Guideline Updates Guidelines are living products. To remain useful, they need to be updated as new information accumulates. This document will be updated when major new research is published. The need for update will be determined by the authors and AGA Clinical Guidelines Committee. 2. METHODS 2.1 Scope Gastroparesis is a disorder of gastric motility defined by a measurable delay in gastric emptying in the absence of mechanical obstruction, associated with a constellation of upper gastrointestinal (GI) symptoms. Cardinal symptoms include nausea and vomiting, which may be accompanied by other symptoms such as early satiety, postprandial fullness, abdominal pain, and bloating.3 Because of differences in testing methodology in combination with the pervasiveness of upper GI symptoms among patients with other conditions such as functional dyspepsia, making an accurate diagnosis of gastroparesis and identifying successful treatment strategies has been heretofore challenging.4 To date, there is only one FDA-approved agent for the treatment of gastroparesis (metoclopramide) and many agents in the pipeline have failed to show meaningful benefit over placebo.5 New procedural approaches—including gastric per oral endoscopic myotomy (G-POEM)—are nonetheless emerging alongside the existing pharmacologic treatments. Despite diagnostic challenges, gastroparesis has a high prevalence, estimated to be between 21.5 to 24.2 cases per 100,000 persons, with a disproportionate impact on women (75% of cases).6 There appears to be an increased risk of mortality in gastroparesis as well, seemingly driven by the subset of patients with diabetic gastroparesis and related to complications of diabetes.6 Patients with gastroparesis represent a significant burden on the healthcare system, with more than 140,000 outpatient visits yearly in the US and rising number of hospitalizations.7, 8 Furthermore, patients with gastroparesis also have decreased quality of life and higher rates of comorbid anxiety, depression, and somatization.9 People with gastroparesis are more likely to identify as disabled, suffer from unemployment, or have reduced working hours.3 Amidst diagnostic uncertainty and significant societal burden, the Guidelines Panel focused its review and research questions on 1) ensuring an accurate diagnosis of gastroparesis and 2) identifying evidence-based, effective treatments among a vast landscape of pharmacologic and procedural interventions spanning more than 40 years of research. 2.2 Formulation of Clinical Questions The clinical questions which underpin this guideline were developed by the Guideline Panel, with methodologists and clinical content experts working together to develop specific questions to address current knowledge gaps in this area. The PICO format was used to outline the specific patient population (P), intervention (I), comparator (C), and outcome(s) for each clinical question. The list of PICO questions is presented in Supplemental Table 3. The panel selected desirable and undesirable patient-important outcomes (benefits and harms). Critical and important outcomes for decision-making for all the interventions included in this review are summarized in the evidence profiles (see 2.5). The panel used the rating of the Gastroparesis Cardinal Symptom Index-Daily Diary (GCSI-DD) composite score and its subscales when evaluating the effect of treatment on symptoms. The scale rates the symptoms as none (0), mild (1), moderate (2), severe (3), very severe (4) scale on the worst severity of the symptom over the last 24 hours. 2.2.1 Definition of minimum clinically important difference The panel defined the thresholds for minimum clinically important difference (MCID) a priori to guide the certainty of evidence (CoE) assessment. The thresholds reported by Revicki et al. were utilized after re-scaling the original 0 to 5 range to standardized scores of 0 to 4. The MCID utilized for nausea and vomiting is 0.44, early satiety and postprandial fullness is 0.62, abdominal pain is 0.24, and for the GCSI-DD total score 0.58.10 After the review was completed and the results were completed, a new validation study proposed a 1 point change on a 0-4 scale as a meaningful change threshold of symptom change when using the GCSI-DD.11 However, the latter patient response outcome had not been used in the vast majority of prior studies, and when it was used, it was not the primary endpoint anyway (e.g. in tradipitant trials).12 2.3 Search Strategy The systematic review process was guided by a search protocol developed a priori by the Guideline Panel members in collaboration with the medical librarian. The librarian conducted a comprehensive search of the following databases: MEDLINE, Embase, and Cochrane Central Register of Controlled Trials, and Cochrane Database of Systematic Reviews from inception to April 8, 2025 using a combination of controlled vocabulary terms supplemented with keywords. The search was limited to English language and human adults. The final strategy is available in Supplement Table 4. The bibliography of prior guidelines and the included references were searched to identify relevant studies that may have been missed. Additionally, content experts helped identify any ongoing studies. To ensure that the guideline document is up-to-date with recent studies, the literature was searched periodically up to the date listed above. 2.4 Study Selection, Data Collection, and Analysis The inclusion and exclusion criteria were based on the PICO questions developed by the Guideline Panel (Supplement Table 3). Searches from all the databases were combined in bibliographic software, duplicates were removed, then the references were uploaded to Covidence software for systematic reviews.13 For the diagnostic question comparing 4-hour versus 2-hour gastric emptying tests, we aimed to identify randomized controlled trials (RCTs) that compared the outcomes of interventions based on those tests. As we were unable to identify any such trial, we then relied on studies reporting the diagnostic test characteristics of gastric emptying studies as measured by sensitivity, specificity, and positive/negative predictive values. For the therapeutic interventions, we aimed to identify RCTs that compared the interventions of interest to placebo, sham intervention or no intervention. For the procedural and surgical interventions, we anticipated absence of RCTs for some of the PICO questions and aimed to identify comparative observational studies for those interventions. One clinical content expert and one methodologist screened each reference and conducted a full-text review of the eligible studies (Supplement Figure 1). Any conflicts were resolved with adjudication by consensus. Data were extracted from each study, including study characteristics such as year of publication, study site, study population, intervention, comparison group, outcomes, and methods for risk-of bias assessment. Data from the included studies was extracted by two panel members independently. If an RCT had multiple arms, the panel combined groups so that the only difference between the intervention and control group was the intervention of interest. For cross-over trials, we used data that combined both periods and considered the impact of the washout period in the risk of bias assessment. When the numeric values were not reported in the text, we extracted the data presented in plots (e.g. box and whisker plots) using a plot digitizer.14 If an abstract of an unpublished clinical trial was identified, we contacted the authors for additional data. We also contacted the corresponding authors of published studies as needed to request additional data (e.g. data on individual symptoms). Risk of bias was assessed using version 2 of the Cochrane risk-of-bias tool for randomized trials including for parallel and cross-over trials, the Risk Of Bias In Non-Randomized Studies of Interventions (ROBINS-I) tool for non-randomized studies, and QUADAS-2 tool for diagnostic accuracy studies.15-17 The risk of bias assessment for each study was assessed by both the senior and junior methodologists separately and in a blinded manner, and disagreements were resolved by discussion. The robvis visualization tool was used to produce the traffic light plots.18 Meta-analyses were conducted when more than one study contributed data for the same intervention and outcome. Dichotomous outcomes were pooled to obtain a relative risk (RR) and continuous outcomes were pooled to obtain a mean difference (MD) with 95% confidence interval (95% CI). If different scales were used by the different studies, we scaled them to be consistent with GCSI-DD (0 to 4 scale). For the meta-analyses, the generic inverse-variance method of weighting was used and the random-effects model applied using the restricted maximum-likelihood estimator to estimate between study variance. However, if number of studies was too small (less than 4) to estimate the between-study variance, a fixed-effect model was used. The Guideline Panel assessed the statistical heterogeneity using the I2 statistic. When a sufficient number of studies was presented with no substantial heterogeneity (I2 < 50%), we planned to assess for publication bias using funnel plot asymmetry tests.19 When we did not identify trials comparing an intervention to placebo or no intervention, but we did identify trials that compared an intervention to another intervention, we conducted a network meta-analysis using a random-effects model to estimate the effect of the intervention relative to placebo via the indirect comparison. For the diagnostic accuracy studies, the accuracy estimates from the studies were pooled using the logit transformation and the bivariate random-effects model. The Panel used the packages meta (version 7.0-0), netmeta (version 2.9-0), and mada (version 0.5.11) in R version 4.4.1 to conduct the statistical analyses.20-23 2.5 Certainty of the Evidence The Guideline Panel used the GRADE approach to assess the certainty of evidence for the effect of the intervention on each outcome using the GradePro Guideline Development Tool software ( The GRADE approach considers factors such as study design, population studied, risk of bias, inconsistency, indirectness, imprecision, and risk of publication bias to rate the certainty of evidence as high, moderate, low, or very low (Supplemental Table 5 - Interpretation of the Certainty of Effects Using the GRADE Framework). The results of certainty assessment are reported with each recommendation. 2.6 Development of Recommendations The process of translation of evidence into guideline recommendations followed the GRADE Evidence to Decision framework and was achieved by means of discussion during virtual meetings of the guideline committee. The Evidence to Decision framework considers the certainty of evidence, balance of benefits and harms, patient values and preferences, feasibility, acceptability, and equity. Evidence to Decision tables with each recommendation. The recommendations were developed by the Panel using the evidence to decision framework. The panel did not explicitly incorporate cost or cost-effectiveness. The panel agreed on the recommendations (including direction and strength), remarks, and qualifications. The interpretation of strength of recommendations is summarized in Supplemental Table 6. The certainty of evidence and the strength of recommendation are provided for each clinical question. As per GRADE methodology, recommendations are labeled as “strong” or “conditional”. The words “we recommend” indicate strong recommendations and “we suggest” indicate conditional recommendations. A systematic search for studies on health equity and disparities was performed using the MEDLINE/PubMed Health Disparities and Minority Health Search Strategy filter, and the identified references were reviewed and summarized when identified. 3. Recommendations 3.1 Should 2-hour versus 4-hour gastric emptying studies be used to diagnose gastroparesis? 3.1.1 Summary of the Evidence We did not identify any randomized controlled trial or comparative study that compared the impact of testing protocol on outcomes. Evidence informing the recommendation for the diagnosis of delayed gastric emptying was derived from cohort studies that reported the diagnostic test performance of a 2-hour gastric emptying study compared to a 4-hour gastric emptying study. We identified 8 studies that reported the results of gastric emptying studies evaluated at 2 hours and 4 hours (Supplemental Table 7).24-31 The studies included 4,309 patients who underwent gastric emptying studies using solid meals radiolabeled with technetium-99m. The indications for studies varied and were not limited to patients with cardinal gastroparesis symptoms. The pooled sensitivity and specificity of 2-hour gastric emptying studies were 0.57 (95% CI 0.51-0.62) and 0.94 (95% CI 0.91-0.96), compared to 4-hour gastric emptying studies as reference standard. The studies were at high risk for bias mainly due to patient selection (Supplemental Table 8). We also identified two prior systematic reviews that evaluated the association between gastric emptying with symptoms, and response to therapy supporting the 4-hour gastric emptying test over shorter test durations.32, 33 3.1.2 Benefits and Harms Although a 2-hour test may be more convenient and could potentially decrease the cost of testing, it is associated with large rate of false negatives and false positives. The average prevalence of gastroparesis in the included studies was around 30% and the pooled sensitivity of a 2-hour test was 0.57 (95% CI 0.51 to 0.58; Supplemental Figure 2); thus a 2-hour test may be associated with a 12% false negative rate. Those patients will be assumed not to have Recommendation 1: In individuals with suspected gastroparesis, the AGA suggests against the use of a 2-hour (or shorter) gastric emptying study compared to a 4-hour gastric emptying study to evaluate for delayed gastric emptying [conditional recommendation, low certainty of evidence] Implementation considerations: - By performing a 4-hour study in all cases, patients with normal emptying at 2 hours may get reclassified as delayed at 4 hours. - GES is optimally done without confounders that could affect gastric emptying including hyperglycemia and medications that alter gastric emptying (e.g. opioids, glucagon-like peptide-1 receptor agonists, prokinetics). - The 4-hour gastric emptying should be measured directly at 4 hours and not mathematically derived from earlier time points to maximize accuracy. - Among the different meals used to measure gastric emptying over a 4-hour period, two have well-established normal values and are widely accepted: the low-fat EggBeaters meal (Tougas meal), consisting of liquid egg white with bread, jam and water (250 kcal, 2% fat); and the Mayo Clinic meal consisting of two whole eggs, bread and skim milk (320 kcal, 30% fat). The latter has been more thoroughly validated to date. gastroparesis and may continue to be symptomatic without appropriate treatment. They may also receive additional testing in an attempt to identify the cause of their symptoms. A 2-hour test had a pooled specificity of 0.94 (95%CI 0.93 to 0.95; Supplemental Figure 3) which correlates with 4% false positive rate based on prevalence of 30% using the 4-hour result as the standard. Those patients will be assumed to have gastroparesis and may receive inappropriate medical therapy with potential for adverse events and ultimately may be considered refractory to treatment and receive surgical intervention in an attempt to alleviate symptoms. A systematic review that evaluated the association between upper gastrointestinal symptoms and gastric emptying showed that utilizing optimal gastric emptying test methods, defined as a study that appraises emptying of a solid meal for at least 3 hours, showed association between gastric emptying and symptoms, compared to utilizing suboptimal testing methods (predominantly liquid meals, or assessment of gastric emptying over < 3 hours). 33 Another systematic review showed that when gastric emptying is measured optimally with a solid meal over > 3 hours, pharmacological treatments that improved gastric half emptying (T1/2) by 26.8 min were associated with a clinical improvement in symptoms.32 3.1.3 Certainty in Evidence of Effects The overall certainty in the evidence across the critical outcomes and considering both benefits and harms was low. This certainty applies mainly to the diagnostic test accuracy as we could not identify any study to compare the impact of different diagnostic approaches on patient important outcomes. The certainty of evidence was rated down due to concerns regarding serious risk of bias, as outlined above, and indirectness, as not all the included patients had suspected gastroparesis. The evidence profile and evidence to decision framework are presented in Supplemental Tables 9 and 10. 3.1.4 Discussion The gastric emptying test is usually performed with images taken at 0, 1, 2, 4 hours after meal ingestion to determine the amount remaining in the stomach at each time period. A 4-hour diagnostic test is time-consuming for patients and facilities and has increased costs related to the longer study requiring more personnel and facility time. Shortening the length of testing without sacrificing test accuracy would be beneficial to patients and health systems alike. A shortened test could improve access to testing for symptomatic patients. However, based on pooled results, the shortened test length would decrease the sensitivity of the study and may compromise specificity. Extending the time to 4 hours is important to detect cases with normal emptying at 2 hours but delay at 4 hours. Testing for four hours increased the number of cases classified as gastroparesis by 25%.27 In clinical practice, delayed gastric emptying at 2 hours with normal or borderline delay at 4 hours leads to a therapeutic trial for gastroparesis. Presently, there are two types of meals that have well-defined normal values. One is the “Tougas meal” which is a low-fat meal consisting of liquid egg white (EggBeaters), 2 slices of toasted bread, jelly, and a cup of water.34 A second meal is the “Mayo Clinic meal” which is a higher caloric, higher fat meal consisting of two eggs, toast, and a glass of skim milk.35 Regardless of the test meal chosen for evaluation, all panelists agreed on the importance of a 4-hour study to ensure diagnostic accuracy. The Tougas liquid egg white (EggBeaters) meal with imaging at 0, 1, 2, and 4 hours is currently the most commonly used assessment of gastric emptying nationwide. This low-fat, egg-white meal was chosen to measure gastric emptying without confounding by hormonal release from fatty meals and utilizes standardized, “store-bought” ingredients for reproducibility across centers. The initial, multi-institutional study that developed the EggBeaters meal protocol investigated 123 normal subjects from 11 medical institutions in the United States, Canada, and Europe. Based on that study, using this method, delayed gastric emptying (increased gastric retention) was determined to be > 90% at 1 hour, >60% at 2 hours, and >10% gastric retention at 4 hours.34 In an observational study of 75 patients seen at 5 tertiary centers, moderate gastric retention of greater than 20% at 4 hours using the Tougas meal predicted which patients respond to G-POEM in the treatment of medically refractory gastroparesis.36 One potential limitation of the Tougas protocol is reproducibility in symptomatic patients which has not been assessed. In a multicenter cohort assessing the disease course of patients with chronic upper GI symptoms receiving clinically-indicated treatment, 42% of patients with an initial diagnosis of gastroparesis (defined as gastric retention >10% at 4 hours) using the Tougas protocol were reclassified at 48 weeks as having functional dyspepsia with normal gastric emptying (defined as <10% retention at 4 hours) and 37% of patients with functional dyspepsia were reclassified as having gastroparesis in the same timeframe.37 The ”Mayo Clinic meal” with imaging at 0, 1, 2, and 4 hours has normative data based on 319 healthy controls from a single center.38 The normal values are <75% retained at 2h, <25% retained at 4h and gastric emptying T1/2 <175 minutes. Robust data supporting its use emanate from 214 symptomatic patients, including those with functional dyspepsia, post-fundoplication, rumination and diabetic dyspepsia as well as 1,287 patients with upper gastrointestinal symptoms evaluated in clinical practice.39, 40 In single-center cohorts of patients with dyspeptic symptoms and either baseline functional dyspepsia (normal gastric emptying <25% retention at 4h) or gastroparesis (>25% retention at 4h), approximately 85% of the patients randomized to placebo retained their diagnosis upon repeat testing performed between 3 days and 4 weeks later.41 Results using the Mayo Clinic protocol also appear to be responsive to prokinetic medications in placebo-controlled clinical trials. Using the 5-HT4 receptor agonists feclisetrag in patients with gastroparesis and velusetrag in healthy controls, the investigators found significant changes in gastric emptying using the Mayo Clinic meal.42, 43 The magnitude of these changes in gastric emptying in response to those prokinetics conforms with the threshold for improvement in symptoms of gastroparesis in a meta-analysis that included patients assessed using multiple methods of gastric emptying measurement.32 Therefore this shows that the Mayo Clinic meal has been used to document responsiveness to pharmacological agents. 3.1.5 Evidence Gaps and Future Research Additional studies are needed to compare the impact of the utilization of different meals and thresholds on relevant patient outcomes. Initial studies could compare the diagnostic accuracy of the two different meals (Tougas meal and Mayo Clinic meal) performed in the same patients with symptoms of gastroparesis in diagnosing gastroparesis by gastric emptying scintigraphy. Further studies might compare outcomes among patients treated for suspected gastroparesis based on the results of one test protocol versus the other. The duration of the test could be optimized: we advocate for a 4-hour test over a 2 hour test; would a 3 hour test suffice? The panel also acknowledges that current gastric emptying testing does not necessarily correlate well with symptom severity and additional strategies are needed to bridge this gap. Finally, more longitudinal data is needed to understand the stability of gastric emptying results over time, because current data using the Tougas meal suggest that some patients may switch between normal and delayed emptying over time with retesting.37 General Considerations for Implementing the Treatment Recommendations in Clinical Practice - It is important to optimize glycemic control in individuals with gastroparesis in the setting of diabetes mellitus. - It is important to review the patient’s current medications to eliminate medications that may alter gastrointestinal tract motility; medications that can delay gastric emptying include opiate pain medications and the GLP-1 receptor agonists for diabetes management and weight loss. One should be cognizant for potential drug-drug interactions when considering pharmacological interventions. - The guideline panel content experts use dietary interventions with small-particle and low-fat, low-residue diets prior to initiation of pharmacological interventions or as a concomitant intervention.44 - Patients with gastroparesis may need to use anti-emetics as needed including selective serotonin receptor (5-HT3) antagonists (e.g., ondansetron), histamine H1 antagonists (e.g., promethazine), and/or dopamine D2 antagonists (e.g., prochlorperazine). - The guidelines panel content experts assess the efficacy of pharmacological interventions at 4 to 8 weeks to decide whether the treatment should be continued or adjusted. 3.2 Should metoclopramide be used for the treatment of individuals with gastroparesis? 3.2.1 Summary of the Evidence Evidence informing the recommendation to use metoclopramide was derived from 7 clinical trials that compared the use of metoclopramide to placebo in individuals with gastroparesis. Five of the trials utilized oral metoclopramide and two trials utilized intranasal metoclopramide. The oral metoclopramide trials were published between 1979 and 1985, and two of them were cross-over trials.45-49 They included a total of 150 patients with gastroparesis. The diagnosis of gastroparesis was based on a barium burger study or scintigraphy. The oral metoclopramide trials informed the evidence on the effect of metoclopramide on individual gastroparesis symptoms. The two intranasal metoclopramide trials were published in 2015 and 2024, and only one of them required scintigraphy to confirm the diagnosis of gastroparesis.50, 51 They included 492 patients. They only reported the changes in composite symptom severity score. Overall, the metoclopramide trials were limited by lack of information regarding randomization and deviations from the intended interventions for the older trials, and by selection of the reported outcome for the newer trials which reported the composite symptoms score rather than individual symptom scores. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 11 and 12. 3.2.2 Benefits and Harms Metoclopramide led to a reduction in nausea by 0.24 points (95% CI -0.96 to 0.45; N/n = 2/41, I2 = 63%) and vomiting by 0.92 points (95% CI -1.46 to -0.38; N/n = 3/51, I2 = 86%), which was considered a clinically important reduction based on the MCID of -0.44, which was a threshold that we defined a priori. It also led to a reduction in early satiety by 0.12 points (95% CI -0.95 to 0.71; N/n = 2/41, I2 = 50%), postprandial fullness by 0.2 points (95% CI -1.01 to 0.61; N/n = Recommendation 2: In individuals with gastroparesis, the AGA suggests using metoclopramide over no metoclopramide [conditional recommendation; very low certainty of evidence]. Comment: As part of shared decision making, patients who place a higher value on the potential risk of adverse events and lower value on the symptomatic improvement may reasonably select not to use metoclopramide. Implementation Considerations: - Potential side effects of metoclopramide should be discussed with patients prior to starting treatment. - Patients on psychotropic agents and older individuals may be at higher risk for adverse events leading to discontinuation, but the absolute risk is low, including the risk of tardive dyskinesia. - Typical dosing uses the oral (tablet or liquid) or intranasal formulation, starting at 5 mg before meals, which can be increased up to 10 mg before meals. - Medication efficacy and side effects are usually assessed at 4 to 8 weeks to decide whether the treatment should be continued. Efficacy and side effects are then monitored for the duration that a patient is receiving metoclopramide. 1/28), and the symptom severity score by 0.54 points (95% CI -0.98 to -0.11; N/n = 6/593, I2 = 79%). None of the studies reported the change in quality of life in individuals receiving metoclopramide compared to placebo. The risk of serious adverse events was lower in individuals that received metoclopramide compared to placebo (RR 0.64; 95% CI 0.17 to 2.41; N/n = 3/439, I2 = 45%), while the risk of adverse events leading to discontinuation of treatment was higher in patients that received metoclopramide compared to placebo (RR 2.15; 95% CI 0.89 to 5.19; N/n = 5/501, I2 = 3%). The forest plots are presented in Supplemental Figures 4 to 8. 3.2.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for serious risk of bias (arising from randomization process, deviation from intended intervention, and in measurement of the outcome), serious inconsistency (studies varied in their conclusion based on MCID with I2 ≥ 50%), and serious to very serious imprecision (wide confidence intervals crossing one or more effect size thresholds). Publication bias was not suspected. The evidence profile and evidence to decision framework are presented in Tables 2 and 3. 3.2.4 Discussion The panel noted the significant reduction in the major symptoms with treatment with metoclopramide experienced by patients with gastroparesis including nausea, vomiting, early satiety, and postprandial fullness as well as the reduction in symptom severity score. The forest plots showed overall trends in favor of metoclopramide, and the 95% confidence interval in reduction in vomiting almost exceeds the threshold for minimally important difference. However the panel appreciated the heterogeneity in the results between studies, and the fact that the clinical trials were performed on relatively small numbers more than 4 decades ago using diagnostic tests that are not recommended in the current guideline. Nevertheless, including the larger, more recent studies using the intranasal formulation of metoclopramide showed statistically significant benefit over placebo, leading to the conclusion that treatment with metoclopramide is indicated over no metoclopramide. Because of the FDA black box warning regarding tardive dyskinesia and recommendation to prescribe for up to 12 weeks, however, the panel expressed concern regarding implementation as well as concern about the acceptability of the recommendation given perceived risks. Increased risk for adverse events leading to discontinuation may be seen in older individuals or those on psychotropic agents.52 However, the absolute risk of tardive dyskinesia with metoclopramide use is low: prescription database studies from Scandinavia and the United Kingdom estimated prevalences of 1 in 17,800 or 1 in about 35,000 prescriptions, respectively.53, 54 A 2019 article that reviewed the literature estimated the risk of tardive dyskinesia from metoclopramide to be only in the range of 0.1% per 1000 patient years.55 The guideline panel content experts usually assess the efficacy of the intervention at 4 to 8 weeks to decide whether the treatment should be continued. Efficacy and side effects are monitored for the duration of the time that a patient is receiving metoclopramide. 3.2.5 Evidence Gaps and Future Research Better definition of the types of patients that might respond to metoclopramide are needed – the type and severity of symptoms, the type of gastroparesis, and the severity of the delay in gastric emptying. There is an opportunity to further appraise efficacy of metoclopramide, particularly in relation to the effects on individual symptoms in the large clinical trials that were performed in the last decade with the intranasal formulation. In addition, the safety of prolonged use of metoclopramide in a chronic condition such as gastroparesis needs further evaluation, particularly in relation to the risk of developing irreversible, chronic tardive dyskinesia (in contrast to reversible involuntary movements that cease when metoclopramide is discontinued). Additional research is also needed to determine whether the risk of tardive dyskinesia is modified by intermittent use or “drug holidays”, such as temporarily stopping the medication after 3 months of treatment. Finally, research should examine the specific role of concomitant treatment with psychotropic agents on the reported occurrence of tardive dyskinesia in patients treated with metoclopramide. 3.3 Should erythromycin be used for the treatment of individuals with gastroparesis? 3.3.1 Summary of the Evidence There were no well-designed placebo-controlled RCTs studying the effect of erythromycin on symptoms of gastroparesis. Evidence informing the recommendation to use erythromycin was derived from a trial that compared the use of erythromycin to metoclopramide and 7 trials that compared the use of metoclopramide to placebo in individuals with gastroparesis.45-51, 56 The results of the trials were included in a network meta-analysis to provide an estimate for the effect of erythromycin compared to placebo. The trial comparing erythromycin to metoclopramide was a crossover trial which assessed the efficacy of the medications using a composite symptom severity score composed of the combined severity scores for nausea, vomiting, abdominal pain, abdominal bloating, early satiety, diarrhea, constipation, and anorexia, rated from 0 to 3 with a maximum score of 24 for the composite score. The trial was limited to patients with severe symptoms of gastroparesis in the settings of diabetes and evidence of delayed gastric emptying on a semisolid gastric emptying study. The trial was overall at high risk of bias due to limited reporting of the randomization process, missing Recommendation 3: In individuals with gastroparesis, the AGA suggests using erythromycin over no erythromycin [conditional recommendation; very low certainty of evidence]. Implementation Considerations: - Erythromycin is thought to improve symptoms of gastroparesis due to its gastric prokinetic effects, but it has no direct effect on nausea and vomiting. - The effect of erythromycin on symptoms and gastric motility is unrelated to its antibiotic properties. - Low doses (e.g. 100-150 mg by mouth, 30 min before meals) are used to reduce potential side effects seen with higher doses (250-500 mg). - Because erythromycin tablets are only available in high doses, using erythromycin ethylsuccinate oral suspension allows the administration of smaller, better tolerated doses. - Tachyphylaxis often develops with prolonged, continuous use of erythromycin; in practice, efficacy is maintained using drug holidays (e.g. 3 weeks on therapy, 1 week off therapy). - Safety concerns with erythromycin include development of antibiotic resistance (with long-term use), drug interactions (it is a CYP3A isoform inhibitor), and QT-prolongation. - Azithromycin has been used as a substitute for erythromycin in practice, and many of the same safety considerations such as antibiotic resistance are also applicable. outcomes data, and potential bias in the measurement of the outcome and selection of the reported outcomes. In this study, erythromycin was more efficacious than metoclopramide. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 13 and 14. 3.3.2 Benefits and Harms Based on the network meta-analysis, erythromycin reduced GCSI-DD by 0.81 points (95% CI -1.64 to 0.01) compared to placebo (Supplemental Figures 9 and 11), and this exceeds the MCID total score for clinical relevance GCSI-DD total score of 0.58. There were no adverse effects reported in either erythromycin or metoclopramide groups of the clinical trial. However, the FDA package has warnings regarding hepatotoxicity, QT segment prolongation, and drug interactions due to inhibition of Cytochrome P450 3A4 (involved in the metabolism of colchicine, some HMG-CoA reductase inhibitors, calcium channel blockers, and other medications). The FDA package insert also reports nausea, vomiting, abdominal pain, and diarrhea as potential adverse side effects.57 3.3.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for serious risk of bias (bias in measurement of the outcome), serious indirectness (indirect comparison from network meta-analysis), and very serious imprecision (very wide confidence interval crossing more than one effect size threshold). Publication bias was not suspected. The evidence profile and evidence to decision framework are presented in in Supplemental Tables 15 and 16. 3.3.4 Discussion The motilin agonists, classically erythromycin, are a class of drugs that are used to treat gastroparesis on an off-label basis. Motilin receptors mediate the gastric portion of the phase III migrating motor complex (MMC) of the GI tract, increasing antral contractility and improving gastric emptying. While erythromycin may improve symptoms of gastroparesis due to its gastric prokinetic effects, it has no direct effect on nausea and vomiting, unlike dopamine receptor antagonists (metoclopramide, domperidone). Erythromycin is a strong prokinetic agent; IV erythromycin (3mg/kg iv infusion over 45 minutes) is occasionally used to clear the stomach in patients hospitalized because of acute gastroparesis symptoms with evidence of food retention or in those undergoing upper endoscopy for upper gastrointestinal bleeding.58, 59 The gastric prokinetic effects of erythromycin result from stimulation of the motilin receptor, and are unrelated to its antibiotic properties. Low doses (e.g. 100-150 mg by mouth, 30 min before meals) are used to reduce potential side effects, such as nausea, vomiting, abdominal pain, and diarrhea that are seen with higher doses (250-500 mg). However, erythromycin tablets are only available in 250 and 500 mg tablets that are difficult to divide due to tablet coating. Erythromycin ethylsuccinate oral suspension (EryPed®) allows the administration of smaller, better tolerated doses. Motilin receptors are known to undergo internalization leading to tachyphylaxis with prolonged exposure, reducing efficacy over time.60 Because of this tachyphylaxis, these drugs are often administered for a limited period of time with drug holidays, such as using a 3 week on therapy, 1 week off therapy approach. Safety concerns with erythromycin include development of antibiotic resistance (with long-term use), drug interactions (it is a CYP3A isoform inhibitor), and QT-prolongation. Anecdotal evidence suggests these side effects are less with oral, compared to intravenous, administration of erythromycin. Azithromycin has been used as a substitute to erythromycin and is effective in accelerating GE with a similarly favorable clinical response.61 3.3.5 Evidence Gaps and Future Research The evidence for the efficacy of erythromycin in gastroparesis was rated as a conditional recommendation due to low evidence. The main study from which this recommendation was derived compared erythromycin to metoclopramide for three weeks in patients with diabetic gastroparesis. A significant improvement in gastrointestinal symptoms was observed with both drugs but was more pronounced with erythromycin. To better understand the efficacy of erythromycin in gastroparesis, placebo-controlled trials would be valuable. Additional data is also needed on which specific symptoms of gastroparesis improve with erythromycin treatment. The panel suggests using this primarily for symptoms attributed to gastric stasis, particularly early satiety and postprandial fullness. In addition to patients with diabetic gastroparesis, other subgroups should be studied as well, such as patients with idiopathic and postsurgical gastroparesis. Because of the risk of tachyphylaxis with continued use, studies that would identify the best duration of treatment interruption to restore efficacy would be helpful. Finally, the development of motilin receptor agonists without antibiotic properties would be helpful in mitigating the risk of antibiotic resistance; several have been studied in the past but have not been shown to be efficacious in RCTs, possibly due to tachyphylaxis of the motilin receptor. 3.4 Should domperidone be used for the treatment of individuals with gastroparesis? 3.4.1 Summary of the Evidence Evidence informing the recommendation to use domperidone was derived from 2 randomized controlled trials that compared the use of domperidone to placebo in individuals with gastroparesis.62-64 One of the trials was an enriched enrollment randomized withdrawal trial. The trials included 224 patients with gastroparesis diagnosed based on scintigraphy, and one of them was limited to patients with diabetes while the other was limited to patients with idiopathic gastroparesis. Only one of the trials reported the effect of domperidone on individual symptoms. Overall, the trials were limited by lack of information regarding the randomization process and any bias arising from the measurement of outcomes. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 17 and 18. 3.4.2 Benefits and Harms Domperidone led to a reduction in nausea score by 0.28 points (95% CI -0.61 to 0.05; N/n = 1/208), vomiting by 0.19 points (95% CI -0.41 to 0.04; N/n = 1/208), early satiety by 0.41 points (95% CI -0.73 to -0.10; N/n = 1/208), and the symptoms severity score by 0.37 points (95% CI -0.58 to -0.16; N/n = 2/228, I2 = 84%). Although the improvement in symptoms was statistically significant, the reduction in these scores was not clinically significant based on the predefined MCID thresholds. Clinically important changes in the abdominal pain score and the quality of life Recommendation 4: In individuals with gastroparesis, the AGA suggests against use of domperidone over no domperidone [conditional recommendation; very low certainty of evidence]. Comment: As part of shared decision making, patients who place a higher value on the potential improvement in nausea, vomiting and early satiety and lower value on the potential adverse events may reasonably select to use domperidone, when available. Implementation Considerations: - Domperidone may be especially helpful for patients with gastroparesis who had neurological side effects with metoclopramide and for patients with neurologic movement disorders such as Parkinson’s disease. - Domperidone is not approved by the FDA in the United States for human use. The prescription of domperidone requires an expanded access use investigational new drug (IND) application from the FDA. The supplier for domperidone in the United States at the time of writing is exiting the business and will no longer supply the medication, based on information available from the FDA. - Patients starting domperidone should have blood work for serum potassium and magnesium and undergo an EKG at baseline for measurement of the QTc interval. - The guideline panel content experts usually assess the efficacy of domperidone at 4 to 8 weeks to decide whether the treatment should be continued. Serum potassium, magnesium and EKG QTc interval are monitored while a patient is treated with domperidone. scores were not observed. No serious adverse events or adverse events were reported in the trials. Forest plots are presented in Supplemental Figures 11 to 16. Based on observational studies, the use of domperidone, compared to non-use, is associated with increased risk of sudden cardiac death and ventricular arrhythmia (adjusted odds ratio 1.69; 95% CI 1.46 to 1.95; I2 = 0%).65 3.4.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for serious risk of bias (arising from randomization process, and in measurement of the outcome), serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold), and suspected publication bias. The evidence profile and evidence to decision framework are included in Supplemental Tables 19 and 20. 3.4.4 Discussion The panel acknowledged significant reductions in early satiety and symptom severity scores with a trend toward improvement in nausea and vomiting within the data. Clinical experience of those on the panel suggests that domperidone confers meaningful benefits to a subset of patients with gastroparesis, but there are no data to help identify the subset of patients likely to improve prior to initiating treatment. Panelists noted that the benefit of domperidone may be related to its unique mechanism of action as both a prokinetic and antiemetic—a class shared only with metoclopramide—with the antiemetic effect in particular providing benefits to the responders. In contrast to metoclopramide, domperidone does not cross the blood brain barrier and has less neurologic side effects than metoclopramide. The panel acknowledged a small but serious risk of cardiac arrhythmias and sudden death with domperidone use, particularly among those older than 60 years old, those taking higher does, and those using QT-prolonging medications or CYP3A4 inhibitors concurrently.66 In clinical use, panelists check blood work for serum potassium and magnesium and an EKG at baseline with subsequent monitoring of the safety and efficacy at four and eight weeks after initiating treatment obtaining blood work and EKG on follow up visits. Panelists expressed concerns about implementation, as domperidone is not approved for sale in the United States and requires an investigational new drug (IND) application from the FDA. While several gastroenterology practices maintain this program for their patients, there were concerns that the current supplier for the FDA IND program is exiting the business, therefore jeopardizing the current pathway for domperidone access in the US. 3.4.5 Evidence Gaps and Future Research The panel suggested the need to identify new agents that are mechanistically similar to domperidone (providing both prokinetic and antiemetic effects with minimal blood-brain barrier penetration) with a better efficacy and cardiac risk profile. Initial studies with domperidone were primarily performed in diabetic gastroparesis; studies in different etiologies of gastroparesis would be of benefit to assess the type of patient in whom domperidone might be best suited. Further research should also continue to reassess the real-world risk of domperidone use with any eye toward potentially expanding access in the future. 3.5 Should prucalopride be used for the treatment of individuals with gastroparesis? 3.5.1 Summary of the Evidence Evidence informing the recommendation to use prucalopride was derived from two cross-over randomized controlled clinical trials that compared the use of prucalopride to placebo in individuals with gastroparesis.67, 68 The studies included 49 patients with gastroparesis diagnosed based on scintigraphy. One study was limited to individuals with idiopathic gastroparesis, and the other was limited to patients with gastroparesis attributed to diabetes mellitus or mixed connective tissue disease. Overall, the trials were at low risk for bias. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 21 and 22. 3.5.2 Benefits and Harms Prucalopride led to a reduction in the composite nausea and vomiting score by 0.18 points (95% CI -0.53 to 0.17; N/n = 2/42, I2 = 31%), composite early satiety and postprandial fullness score by 0.2 points (95% CI -0.58 to 0.18; N/n = 2/42, I2 = 75%), and the symptom severity score by Recommendation 5: In individuals with gastroparesis, the AGA suggests against the use of prucalopride compared to no prucalopride [conditional recommendation; very low certainty of evidence]. Comment: As part of shared decision making, patients who place a higher value on the potential improvement in symptoms and a lower value on the potential risk of adverse effects may reasonably elect to use prucalopride, particularly if they have idiopathic gastroparesis. Implementation Considerations: - Patients with gastroparesis and chronic idiopathic constipation (CIC) may reasonably elect to use prucalopride for efficiency of care given prucalopride’s established efficacy for CIC. - Patients with gastroparesis attributed to diabetes or connective tissue diseases may be less likely to respond to prucalopride treatment than patients with idiopathic gastroparesis. - The mechanism of action for prucalopride is purely prokinetic, without any direct anti-emetic effects. - Prescribers recommending prucalopride should note the warning to monitor patients for depression and suicidal thoughts and behavior when considering use of prucalopride, but the evidence for this risk is quite low. 0.22 points (95% CI -0.50 to 0.07; N/n = 2/42, I2 = 82%). However, the reduction in the scores did not meet the threshold for clinical importance. There was an increased risk of serious adverse effects observed (RR 2.8, 95% CI 0.12 to 0.66.05; N/n = 2/42) and adverse events leading to discontinuation of treatment (RR 1.87; 95% CI 0.18 to 19.47; N/n = 2/42) in patients that received prucalopride compared to placebo. Prucalopride led to a reduction in abdominal pain and improvement in quality of life, but these changes did not meet the threshold for clinical importance. Forest plots are presented in Supplemental Figures 17 to 22. 3.5.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for serious risk of bias (arising due to missing outcome data), and serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold). Publication bias was not suspected. The evidence profile and evidence to decision frameworks are included in Supplemental Tables 23 and 24. 3.5.4 Discussion There are a number of serotonergic agonists that have been evaluated for use in gastroparesis, including prucalopride, a selective 5HT-4 agonist, which has no appreciable cardiac side effects as was seen with cisapride and perhaps tegaserod. Prucalopride is currently approved for use in chronic idiopathic constipation (CIC) and works exclusively as a promotility agent. As such, the rationale for use in gastroparesis is related to improvement in gastric emptying, which may result in a reduction of symptoms due to gastric retention such as nausea, vomiting, and post-prandial fullness. In fact, the two randomized controlled trials of prucalopride versus placebo that were included in the current assessment revealed significant improvements in these areas. However, when judged on the basis of meaningful improvement, they did not reach a pre-defined threshold of clinical significance. Other 5HT-4 agonists, such as velusetrag and naronapride, have been shown to accelerate gastric emptying as well.43, 69 The current recommendation comes with an acknowledgement that many of the patients who demonstrated symptom improvement had idiopathic gastroparesis. Together with the existing approval for patients with CIC, it is possible that some patients–especially those with coexisting constipation that would be candidates for prucalopride therapy–would reasonably elect to treat both conditions with one medication. Whether specific counseling is needed to patients regarding the potential risk for mood disturbances and suicidal ideation and/or behavior is unknown, as there has been no evidence of a causal association between prucalopride treatment and these outcomes.70-72 3.5.5 Evidence Gaps and Future Research Given the small number of patients on which the current recommendations are based, future research should focus on generating larger randomized controlled trials with longer durations of treatment from which to better assess the potential benefits and harms of prucalopride. These studies should also focus on assessing the effects of prucalopride on different etiologies of gastroparesis, to better identify whether specific subgroups may more likely benefit from therapy. Finally, these studies should specifically include an analysis of the potential psychological implications of prucalopride to better inform patients and clinicians about any true risks. 3.6 Should human neurokinin-1 (NK-1) receptor agonists be used for the treatment of individuals with gastroparesis? 3.6.1 Summary of the Evidence Evidence informing the recommendation to use aprepitant was derived from 1 clinical trial that compared the use of aprepitant to placebo in individuals with nausea and symptoms suggestive of gastric origin.73 The trial included 126 patients, and only 57% of patients had delayed gastric emptying. The trial was at low risk for bias. Additionally, there were two clinical trials that evaluated the use of tradipitant in patients with gastroparesis.12, 74 However, they were not used in developing the recommendation as tradipitant has not yet been approved by the FDA and is not presently available. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 25 and 26. 3.6.2 Benefits and Harms Compared to placebo, aprepitant led to a reduction in nausea by 0.3 points (95% CI -0.61 to 0.01; N/n = 1/122), vomiting by 0.2 points (95% CI -0.44 to 0.04; N/n = 1/122), early satiety and postprandial fullness by 0.2 points (95% CI -0.49 to 0.09; N/n = 1/117), and the symptom severity score by 0.1 points (95% CI -0.36 to 0.16; N/n = 1/122) which did not represent clinically important reductions based on the predefined MCIDs thresholds. Abdominal pain had a clinically important reduction by 0.4 points (95% CI -0.69 to -0.11; N/n = 1/122). There was no clinically important change in quality of life. The risk of serious adverse events (RR 3.00; 95% CI 0.12 to 72.26; N/n = 1/126) and adverse events leading to discontinuation of treatment (RR 2.00; 95% CI 0.52 to 7.65; N/n = 1/126) were possibly higher in patients who received aprepitant compared to placebo. It is important to note that the lack of benefit was based on the outcome predefined by the panel as we relied on GCSI-DD when reported. The trial also reported results based on PAGI-SYM severity index that showed more improvement in nausea and vomiting. Forest plots are presented in Supplemental Figures 23 to 27. 3.6.3 Certainty in Evidence of Effects The overall CoE was low. The CoE was rated down due to concerns for serious indirectness (trial included patients with chronic unexplained nausea and vomiting without requiring evidence of delayed gastric emptying), and serious to very serious imprecision (wide confidence intervals Recommendation 6: In individuals with gastroparesis, the AGA suggests against the use of aprepitant compared to no aprepitant [conditional recommendation; low certainty of evidence]. Comment: As part of shared decision making, patients who place a higher value on the potential improvement in nausea, vomiting, early satiety, and abdominal pain and lower value on the potential adverse effects may reasonably elect to use aprepitant. Implementation Considerations: - Aprepitant is an anti-emetic approved for chemotherapy-induced vomiting but has no effect on gastric motor function. - Aprepitant, a NK1 receptor antagonist, may be helpful in patients with gastroparesis and nausea/vomiting in whom 5HT3 receptor antagonists, such as ondansetron, are not helpful. crossing one or more effect size threshold). Publication bias was not suspected. The evidence profile and evidence to decision framework are included in Supplemental Tables 27 and 28. 3.6.4 Discussion The AGA panel suggested against the use of aprepitant for management of gastroparesis due to its small benefit in producing clinically meaningful reductions in nausea and vomiting. Aprepitant is currently approved for management of chemotherapy induced nausea and vomiting as well as for post-operative nausea and vomiting or chemotherapy-induced emesis. In chemotherapy induced nausea and vomiting, aprepitant is often used as an additive agent to 5HT3 antagonists, such as ondansetron. In patients with gastroparesis, it may be helpful in patients in whom 5HT3 antagonists are not helpful. Access to its use for management of gastroparesis can be difficult due to limited insurance coverage and its high cost. 3.6.5 Evidence Gaps and Future Research Larger studies focusing on use of neurokinin-1 inhibitors in patients with gastroparesis are needed to justify regular clinical use. The evaluated trial of aprepitant included patients with chronic unexplained nausea and vomiting and the results are extrapolated from the 57% of patients with delayed gastric emptying. Additionally, it appeared that the clinical effect of aprepitant differed based on the choice of outcome measure. Aprepitant did not produce a meaningful reduction in nausea based on visual analog scale, however, it did seem to decrease symptom severity for nausea, vomiting, and overall symptoms using the GCSI based on a two week recall. Given limited effective treatments for gastroparesis, future trials utilizing NK-1 receptor antagonists will need the correct choice of outcome to ensure success and clinical relevance. 3.7 Should neuromodulators be used for the treatment of individuals with gastroparesis? 3.7.1. Nortriptyline 3.7.1.1 Summary of the Evidence Recommendation 7.1: In patients with gastroparesis, the AGA suggests against the use of nortriptyline compared to no nortriptyline [conditional recommendation, low certainty of evidence] Comment: As part of shared decision making, patients who place a higher value on the potential improvement in early satiety, post-prandial fullness, abdominal pain, and quality of life and lower value on the potential adverse effects may reasonably elect to use nortriptyline. Implementation Considerations: - Patients with irritable bowel syndrome or significant abdominal pain associated with gastroparesis may benefit from the use of tricyclic antidepressants such as nortriptyline. - When using tricyclic antidepressants in patients with gastroparesis, the panel suggests starting at low doses and increasing slowly to ensure tolerance. - The guideline panel content experts inform patients being started on nortriptyline that this agent is in the class of antidepressants, however, in gastroparesis, it is being used as a peripheral neuromodulator to reduce afferent sensory transmission to the central nervous system. Evidence informing the recommendation to use nortriptyline was derived from one clinical trial that compared the use of nortriptyline to placebo in individuals with idiopathic gastroparesis and moderate symptoms severity.75 The trial included 130 patients with moderate to severe idiopathic gastroparesis and evidence of delayed gastric emptying. The trial was at low risk for bias overall. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 29 and 30. 3.7.1.2 Benefits and Harms Nortriptyline led to worsening of the nausea score by 0.05 points (95% CI -0.33 to 0.43; N/n = 1/118) compared to placebo, which was not a clinically important change. It led to a reduction in early satiety and postprandial fullness by 0.34 points (95% CI -0.72 to 0.04; N/n = 1/118) and symptom severity score by 0.14 points (95% CI -0.47 to 0.18; N/n = 1/118) which were not considered as clinically important changes compared to placebo. There was no change in abdominal pain (mean difference 0; 95% CI -0.47 to 0.47; N/n = 1/118). Nortriptyline led to a possible clinically important improvement in quality of life based on Short Form 36 Physical Component Score (mean difference 2.1; 95% CI -1.01 to 5.21; N/n = 1/118) but not the Mental Component Score. The risk of serious adverse events (RR 5.00; 95% CI 0.6 to 41.63; N/n = 1/130) and adverse events leading to discontinuation of treatment (RR 3.17; 95% CI 1.35 to 7.42; N/n = 1/130) were higher in patients that received nortriptyline compared to placebo. Forest plots are presented in Supplemental Figures 28 to 33. 3.7.1.3 Certainty in Evidence of Effects The overall CoE was low. The CoE was rated down serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold). Publication bias was not suspected. The evidence profile and evidence to decision framework are included in Supplemental Tables 31 and 32. 3.7.1.4 Discussion Tricyclic antidepressants such as nortriptyline have central and peripheral neuromodulatory mechanisms. Peripherally, nortriptyline functions as a neuromodulator, reducing sensory afferent transmission to the central nervous system (CNS). Although the cardinal symptoms of gastroparesis including nausea and or vomiting were not significantly improved with nortriptyline, there was a numeric reduction in early satiety and postprandial fullness as well as improvement in symptom severity and quality of life. Nevertheless, the observed trends toward symptom improvement together with the clinical experience of the experts on the panel led the group to comment that some patients may derive a benefit from nortriptyline treatment. Concurrently, the panel acknowledged a trend toward increased risk of serious adverse effects with a significantly increased risk of adverse events causing discontinuation. The panel felt that nortriptyline and other tricyclic antidepressants could be useful in patients with gastroparesis who also suffered from irritable bowel syndrome (primarily the diarrhea-predominant subtype) given stronger evidence supporting its use in IBS.76 Clinical experience of the content experts on the panel suggests that tricyclic antidepressants should be started at low doses (10-25 mg), given at night to help avoid side effects, and increased slowly to maximize effect while minimizing risk of side effects. Use of secondary imines (nortriptyline, desipramine) over tertiary amines (amitriptyline, imipramine) is also thought to minimize anticholinergic side effects. Other types of antidepressants are also used in gastroparesis; mirtazapine is often used for nausea/vomiting. Panelists highlighted the common misconception that tricyclic antidepressants would further prolong gastric emptying in patients with gastroparesis; slowing of gastric emptying was not observed in healthy controls treated with up to 50 mg daily of nortriptyline or in patients with functional dyspepsia treated with up to 50 mg of amitriptyline.77, 78 3.7.1.5 Evidence Gaps and Future Research Patients with overlapping disorders of gut-brain interaction—including gastroparesis and irritable bowel syndrome—are commonly encountered in clinical practice. Further randomized control trials of nortriptyline should focus on patients with pain-predominant manifestations of gastroparesis or overlapping irritable bowel syndrome to determine whether there is a differential response in these patient populations with known high symptom burden and healthcare resource utilization. 3.7.2. Buspirone 3.7.2.1 Summary of the Evidence Evidence informing the recommendation to use buspirone was derived from one clinical trial that compared the use of buspirone to placebo in individuals with early satiety and postprandial fullness with other symptoms suggestive of gastric origin.79 The trial included 96 patients with moderate to severe symptoms. Delayed gastric emptying was not required for inclusion in the trial, and only around 50% of the included patients had delayed gastric emptying. The trial was at low risk for bias overall. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 33 and 34. 3.7.2.2 Benefits and Harms Compared to placebo, buspirone led to a reduction in the nausea and vomiting score by 0.04 points (95% CI -0.51 to 0.43; N/n = 1/78), early satiety and postprandial fullness score by 0.32 points (95% CI -0.68 to 0.04; N/n = 1/96), and total symptom severity score by 0.45 points (95% CI -0.79 to 0.11; N/n = 1/78) which were not clinically important changes based on the predefined MCIDs. There was possible worsening in the abdominal pain score by 0.14 points (95% CI -0.43 to 0.70; N/n = 1/78) as well as quality of life as measured by the Short Form 36 Physical (mean difference -2.07; 95% CI -5.0 to 0.86; N/n = 1/78) and Mental components. There was no clinically important difference in the risk of serious adverse events between patients that received buspirone and placebo (RR 1.04; 95% CI 0.07 to 16.19; N/n = 1/96). Forest plots are presented in Supplemental Figures 34 to 39. Recommendation 7.2: In patients with gastroparesis, the AGA suggests against the use of buspirone compared to no buspirone [conditional recommendation, very low certainty of evidence] Comment: As part of shared decision making, patients who place a higher value on the potential improvement in postprandial fullness and early satiety may reasonably elect to use buspirone Implementation Considerations: - Use of buspirone may be helpful in patients with gastroparesis and predominant early satiety and bloating. 3.7.2.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for serious indirectness (delayed gastric emptying was not required for inclusion), and serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold). Publication bias was not suspected. The evidence profile is included in Table X. 3.7.2.4 Discussion Buspirone is a 5HT1 receptor agonist approved for treatment of anxiety due to its effects on the central nervous system. Several studies have been performed with buspirone in functional dyspepsia where it has peripheral actions enhancing gastric accommodation. The single clinical trial of buspirone in gastroparesis comparing buspirone to placebo did not show significant improvements in nausea, vomiting, early satiety, postprandial fullness, or overall symptoms severity score. Despite these findings, panel content experts felt that buspirone could be useful in patients with gastroparesis whose predominant symptoms are early satiety and bloating, especially in the absence of any signal suggesting increased risk of serious adverse events among those treated with buspirone. 3.7.2.5 Evidence Gaps and Future Research Evidence underpinning the current recommendation comes from a single clinical trial that assessed patients with gastroparesis and moderate-to-severe symptoms of post-prandial fullness and early satiety. Of note, the subgroup of patients with severe bloating at baseline seemed to derive a benefit in the early satiety/postprandial fullness symptoms from buspirone compared to those with lesser degrees of bloating. As bloating is a particularly difficult symptom to treat across GI conditions, a larger trial adequately powered to examine this subgroup could be of great clinical utility. Studies could be performed in gastroparesis patients with impaired fundic relaxation to help augment patients with defined pathophysiologic abnormalities that might respond to buspirone. 3.8 Should cannabidiol be used for the treatment of individuals with gastroparesis? 3.8.1 Summary of the Evidence Evidence informing the recommendation to use cannabidiol was derived from one clinical trial that compared the use of cannabidiol to placebo in individuals with idiopathic or diabetic gastroparesis.80 The trial included 44 patients. The trial was at low risk for bias overall. The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 37 and 38. 3.8.2 Benefits and Harms Cannabidiol led to a clinically important reduction in the vomiting score by 0.7 points (95% CI -1.13 to -0.27; N/n = 1/44), early satiety by 0.77 points (95% CI -1.45 to -0.09; N/n = 1/44), symptom severity score by 0.61 points (95% CI -0.97 to 0.25; N/n = 1/44), and abdominal pain score by 0.64 points (95% CI -1.4 to 0.12; N/n = 1/44) compared to placebo. There was also a trivial reduction in the nausea and postprandial fullness scores. There was no clinically important difference in the risk of serious adverse events (RR 1.10; 95% CI 0.07 to 16.43; N/n = 1/44) and no adverse events leading to discontinuation of treatment were reported. Forest plots are presented in Supplemental Figures 40 to 44. 3.8.3 Certainty in Evidence of Effects The overall CoE was low. The CoE was rated down due to serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold; and the optimal information size was not met). Publication bias was not suspected. The evidence profile and evidence to decision frameworks are included in Supplemental Tables 39 and 40. 3.8.4 Discussion Some patients with refractory nausea and vomiting use marijuana to treat their symptoms, particularly nausea, vomiting, and lack of appetite. The main pharmacologically active chemicals in cannabis are tetrahydrocannabinol (THC) and cannabidiol (CBD), which may explain the orexigenic, antiemetic, and pain-relieving properties that underpin marijuana’s frequent use for Recommendation 8: In individuals with gastroparesis, the AGA suggests against the use of cannabidiol except in the context of a clinical trial [conditional recommendation, low certainty of evidence] Implementation Considerations: - The single study demonstrating efficacy for cannabidiol in gastroparesis used a pharmaceutical grade cannabidiol, Epidiolex, that is approved to treat seizures due to Lennox-Gastaut syndrome and Dravet syndrome, in patients two years of age and older, and the medication is not available for clinical use in gastroparesis. - Available cannabidiol formulations are not regulated with varying potencies that make their impact on patients with gastroparesis unknown. - Among cannabidiol formulations that include tetrahydrocannabinol (THC), there is concern about increased risk of cannabinoid hyperemesis syndrome. the treatment of GI symptoms. Our current understanding is that the actions of cannabis are mediated through the endocannabinoid system (ECS), which includes the cannabinoid receptors CB1 and CB2—receptors activated by cannabis. THC works by binding to and activating the cannabinoid (CB) receptors: CB1 and CB2. Cannabinoids are thought to mediate their antiemetic effects via CB1 receptors in the dorsal vagal complex of the brainstem. The single study demonstrating efficacy for cannabidiol (CND) in gastroparesis was performed in patients with idiopathic and diabetic gastroparesis, and used the pharmaceutical grade CBD, Epidiolex, that is approved to treat seizures associated with Lennox-Gastaut syndrome, Dravet syndrome, or tuberous sclerosis. The clinically important reduction in vomiting, early satiety, and abdominal pain seen with cannabidiol in this study was tempered by the fact that it is unlikely that the form of CBD used in the study would be available for gastroparesis. The panel felt additional studies, particularly larger multicenter studies, were needed to document the efficacy of CBD for symptoms of gastroparesis. Finally, there are numerous formulations of CBD with varying potency with unknown effects in gastroparesis. A stronger endorsement of cannabidiol for use in gastroparesis (despite evidence in its favor) was deferred due to concerns that patients and their clinicians may use such a recommendation to justify use of cannabidiol outside of the studied formulation. 3.8.5 Evidence Gaps and Future Research Although the study evaluated of CBD for gastroparesis was well-designed with positive results, the panel felt that larger studies are needed, specifically looking at different formulations more likely to be available to patients. The study showed that CBD improves some symptoms of gastroparesis, however, marijuana and CBD can slow gastric emptying which might aggravate symptoms in some patients, particularly those with severely delayed gastric emptying. Since gastroparesis is a chronic condition, prolonged treatment with CBD might be needed. Chronic daily administration of cannabis, or marijuana is associated with cannabinoid hyperemesis syndrome, a variant of cyclic vomiting syndrome, which would be deleterious for patients with gastroparesis with nausea and vomiting. Chronic marijuana use has also been associated with depression, anxiety, and respiratory issues. Although there are fewer side effects with CBD, long-term use of CBD may be linked to liver damage and thyroid issues. Further research is needed to understand any potential adverse effects from chronic CBD use specifically. 3.9 Should gastric peroral endoscopic myotomy be used for the treatment of individuals with gastroparesis refractory to medical therapy? 3.9.1 Summary of the Evidence Evidence informing the recommendation to use gastric peroral endoscopic myotomy (G-POEM) was derived from one clinical trial that compared the use of G-POEM to a sham intervention in individuals with gastroparesis refractory to medical therapy.81 The trial included 41 patients with either idiopathic, diabetes-associated, or postoperative gastroparesis. The patients had to have severe and refractory symptoms for more than 6 months despite treatment with antiemetics and prokinetic agents. The evidence was partially informed by a second clinical trial that compared the use of G-POEM to pyloric botulinum toxin injection in patients with gastroparesis refractory to medical therapy for more than a year.82 The trial included 40 patients with idiopathic, diabetes-associated, and/or postoperative gastroparesis. Both trials were at low risk for bias overall. We also identified multiple observational studies that compared the use of G-POEM to surgical pyloric interventions, gastric electric stimulators, and in combination with other interventions.83-87 The observational studies were at high risk for bias overall and were not used to inform the recommendation. The details of the studies characteristics and risk of bias assessment are reported in Supplemental Tables 41 and 42. 3.9.2 Benefits and Harms After 3 months, compared to sham intervention, G-POEM led to a clinically important reduction in the nausea and vomiting score by 0.80 points (95% CI -1.35 to -0.25; N/n = 1/41), early satiety and postprandial fullness score by 1.28 points (95% CI -1.78 to -0.78; N/n = 1/41), and symptom severity score by 1.2 points (95% CI -1.84 to -0.56; N/n = 1/41). Improvements were similar at 6-month follow-up. There was no clinically important change in quality of life as measured by Patient Assessment of Upper Gastrointestinal Disorders-Quality of Life (PAGI-QOL) at 3 months, but there was possible improvement at 6 months (mean change -0.7; 95% CI -1.53 to 0.13; N/n = 1/41). The risk of serious adverse events was possibly increased (RR 1.51; 95% CI 0.42 to 5.48; N/n = 1/41) as well as an increased risk of procedural and peri-procedural Recommendation 9: In patients with gastroparesis, the AGA suggests against the use of gastric peroral endoscopic myotomy (G-POEM) compared to no G-POEM, except for selected patients with medically refractory disease [conditional recommendation, low certainty of evidence]. Implementation Considerations: - Candidates for G-POEM should have had a diagnosis of gastroparesis with an appropriately-done, 4-hour gastric emptying study, generally with at least a moderate delay in gastric emptying (20% retention at 4 hours with the egg beaters meal). - Candidates for G-POEM should have at least six months of moderate symptoms with the cardinal symptoms of nausea, vomiting, and/or postprandial fullness. - Prior to consideration of G-POEM, patients with gastroparesis should undergo a trial of other treatments for gastroparesis, such as a prokinetic agent like metoclopramide and one antiemetic agent. adverse events in patients that had G-POEM compared to the sham intervention. Forest plots are presented in Supplemental Figures 45 to 49. Compared to pyloric botulinum toxin injection, at 3 months there was no clinically important difference between the two interventions in the nausea and vomiting score (mean change 0.0; 95% CI –0.75 to 0.75; N/n = 1/37), early satiety and postprandial fullness score (mean change -0.08; 95% CI -0.78 to 0.51; N/n = 1/37), or symptom severity score (mean change -0.24; 95% CI -0.86 to 0.38; N/n = 1/37) at 3 months. The findings were similar when assessed again at 12 months.82 3.9.3 Certainty in Evidence of Effects The overall CoE was low. The CoE was rated down due to concerns for serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold; and small sample size). Publication bias was not suspected. The evidence profile and evidence to decision framework are included in Supplemental Tables 43 and 44. 3.9.4 Discussion The panel acknowledges that numerous recent studies have demonstrated efficacy of G-POEM, but these studies rely on retrospective cohort data reporting a center’s experience comparing G-POEM to surgical techniques such as robotic pyloroplasty or gastric electric stimulator placement. Furthermore, there may be important and unmeasured differences in expected outcomes between surgical techniques of pyloromyotomy and pyloroplasty that are not addressed when compared with G-POEM. Importantly, the panel’s recommendation seemingly contrasts with the results of numerous observational studies, which suggest benefit of G-POEM. However, the guidelines panel excluded these studies due to the high risk of bias. The panel did acknowledge that select patient populations may benefit from G-POEM, specifically those with refractory disease. Compared to sham and pyloric botulinum toxin injection, G-POEM did provide at least modest duration (12 months) benefit among patients with refractory disease. The panel’s clinical experts defined refractory disease—and therefore those patients appropriate for consideration of G-POEM—as chronic symptoms of nausea, vomiting, and/or postprandial fullness of at least 6 months duration having failed at least three months of a prokinetic agent like metoclopramide and at least one antiemetic treatment. The panel’s more cautious approach to G-POEM recognizes the increased risk of serious adverse events with an invasive, irreversible procedure for which the evidence of benefit was derived from a single clinical trial. Importantly, the panel felt that a blanket positive recommendation could lead to hasty, inappropriate referrals for G-POEM without a more concerted trial of medical therapy. 3.9.5 Evidence Gaps and Future Research Because the apparent benefit of G-POEM was derived from a single clinical trial, the most important research priority is replicating these positive findings in other randomized controlled trials. Furthermore, longevity of the efficacy to G-POEM needs to be delineated. Acknowledging the heterogenous patient population diagnosed with gastroparesis, additional research is also needed to determine the patients most likely to benefit from the procedure, especially subgroups defined by gastroparesis etiology (e.g. diabetic, idiopathic, post-surgical) and predominant symptom type. Similarly, additional studies providing best practices and normative data for pyloric functional lumen imaging probe (FLIP) would be helpful to determine if any measurements from this technology or others could clarify those most likely to benefit from G-POEM. Similarly, the response to botulinum toxin or endoscopic balloon dilation of the pylorus prior to G-POEM as a potential predictor of treatment response needs further study and validation. 3.10 Should surgical pyloric interventions (pyloromyotomy or pyloroplasty) be used for the treatment of individuals with gastroparesis refractory to medical therapy? 3.10.1 Summary of the Evidence Evidence informing the recommendation to use surgical pyloric interventions was derived from observational clinical studies. The studies compared the use of surgical pyloromyotomy to G-POEM, surgical pyloroplasty to G-POEM, any surgical pyloric interventions to gastric electric stimulators, and surgical pyloroplasty combined with gastric electrical stimulation compared to surgical pyloroplasty alone.85, 86, 88, 89 The studies were at high risk for bias overall mainly due to confounding, selection of participants, and measurement of outcomes. Due to the lack of evidence comparing the surgical interventions to sham interventions and the lack of clinical trials comparing them to other interventions, we were unable to estimate their efficacy. Thus, this area was considered to be a knowledge gap. The details of the studies characteristics and risk of bias assessment are reported in Supplemental Tables 44 and 45. 3.10.2 Discussion Current evidence on surgical pyloric interventions for gastroparesis consisted of observational clinical studies with high risk of bias rather than RCTs. Therefore, the panel determined that there is insufficient high-quality evidence to support a recommendation for or against the routine use of surgical pyloric interventions in clinical practice. As a result, the panel emphasizes the need for further clinical trials on the use of pyloromyotomy and pyloroplasty. The guidelines for G-POEM suggesting its use for medically refractory patients that have tried treatment with metoclopramide and antiemetic agents would apply to this therapy as well. Pyloromyotomy and pyloroplasty involve incision of both the circular and longitudinal muscles of the pylorus, whereas endoscopic pyloromyotomy cuts only the circular muscle. Thus, there may be benefit of performing surgical pyloromyotomy or pyloroplasty in patients that do not respond to G-POEM. Surgical pyloromyotomy and pyloroplasty can also be used for pyloric therapies when intraabdominal surgery is being performed, such as placement of a gastric electric stimulator or performing fundoplication. Preliminary studies suggest an added clinical benefit of performing pyloroplasty at the time of placement of a gastric electric stimulator.90, 91 Recommendation 10: In patients with gastroparesis refractory to medical therapy, the AGA recommends the use of surgical pyloric interventions (pyloromyotomy or pyloroplasty) only in the context of clinical trials [no recommendation, knowledge gap]. 3.10.3 Evidence Gaps and Future Research There is a need for sham-controlled RCTs evaluating surgical pyloric interventions in patients with gastroparesis. Head-to-head clinical trials comparing surgical pyloric interventions to endoscopic pyloric interventions in patients with refractory gastroparesis are also needed. The role of gastric electrical stimulation in combination with surgical pyloric interventions also needs to be addressed in prospective studies. Further research is also needed to identify subgroups of patients who will benefit from surgical pyloric interventions, for example those with post-operative gastroparesis or gastric retention. 3.11 Should botulinum toxin injection be used for the treatment of individuals with gastroparesis refractory to medical therapy? 3.11.1 Summary of the Evidence Evidence informing the recommendation to use botulinum toxin injection (BTI) was derived from two clinical trials that compared the use of BTI to sham intervention (saline injection).92, 93 Art et al conducted a crossover randomized clinical trial which included 23 patients with gastroparesis. The patients were followed for one month only after treatment and crossover occurred at one month which raised concerns regarding carryover effect. Friedenberg et al conducted a parallel randomized controlled trial which included 32 patients with moderate to severe symptoms who were also followed for one month after treatment only. In both trials, it was not clear if the patients were required to be refractory to treatment to be included. The evidence was also informed by a trial conducted by Gonzalez et al, summarized above, in Recommendation 9.82 The details of the trial characteristics and risk of bias assessment are reported in Supplemental Tables 46 and 47. 3.11.2 Benefits and Harms After one month, compared to saline injection, BTI was associated with a worsening in the symptoms severity score which was not clinically important (pooled mean difference 0.17; 95% CI -0.34 to 0.67; N/n = 2/44; I2 = 0%). There was a possible decrease in the risk of serious Recommendation 11: In patients with gastroparesis refractory to medical management, the AGA suggests against the use of botulinum toxin injection (BTI) compared to no BTI [conditional recommendation, very low certainty of evidence] Comments: - As part of decision making, patients and clinicians who place a higher value on endoscopic intervention and lower value on chronic medical therapy may reasonably attempt treatment with botulinum toxin injection. Implementation Considerations: - The guideline panel only evaluated the use of BTI in patients with refractory GP. However, the studies were not specific to refractory GP. - The need for retreatment as often as every 3 months could limit implementation from both a cost-effectiveness perspective and concern for diminishing clinical effect. - Repeat treatment may predispose to pyloric scarring, which could complicate attempts at other interventions such as G-POEM. adverse events in patients that received saline injection compared to BTI (RR 0.67; 95% CI 0.13 to 3.47; N/n = 2/44) while the risk of periprocedural adverse events was higher in patients that received BTI compared to saline injection (RR 2.0; 95% CI 0.6 to 6.64; N/n = 2/44). The trials did not report the impact of BTI on individual symptoms. As outlined above in the section of GPOEM, BTI injection had comparable efficacy to GPOEM based on the trial by Gonzalez et al. Forest plots are presented in Supplemental Figures 50 to 51. 3.11.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for very serious risk of bias (arising from randomization process, period and carry over effect, and selection of reported results), and serious to very serious imprecision (wide confidence intervals crossing one or more effect size threshold, and small sample size). Publication bias was not suspected. The evidence profile and evidence to decision framework are included in Supplemental Tables 49 and 50. 3.11.4 Discussion Although BTI is performed in patients with gastroparesis, the panel noted that the 2 included studies had significant limitations, including small sample sizes and limited follow-up periods. Notably, while the current recommendation suggests the use of BTI for refractory gastroparesis, the included clinical trials did not impose this restriction on participants. However, the suggestion against the use of BTI in refractory gastroparesis acknowledges that current clinical practice sees BTI largely utilized for refractory disease. Panelists acknowledged the results of the Gonzalez et al clinical trial showing a similar benefit to G-POEM and BTI at 3 and 12 months in medically-refractory gastroparesis, which seemingly conflicts with the guideline’s more favorable consideration of G-POEM relative to BTI in refractory disease.82 The panel’s recommendations were based off the placebo-controlled trial showing the relative benefit of G-POEM, which was not seen in the two RCTs for BTI. The panel noted the common use of BTI in clinical practice as a diagnostic strategy for the prediction of G-POEM responders, but this strategy lacks robust supporting evidence. In fact, prior studies suggesting response to BTI as a predictor for G-POEM response were small and predominantly retrospective, and so the panel is unable to rigorously assess existing evidence for use of BTI in this context.94 While feasibility is high and risk of adverse events relatively low, reliance on BTI may necessitate repeat treatments as often as every 3 months with risk for diminishing efficacy. Moreover, repeat treatments have the potential to cause pyloric scarring, which could complicate attempts at other interventions such as G-POEM. A retrospective analysis of different botulinum toxin doses, suggested the 200 unit dose produced a more favorable response.95 Improved patient selection for use of BTI may produce a more favorable response. Reduced pyloric compliance as measured by FLIP has been suggested to yield a more favorable response.96 In patients with post-surgical gastroparesis, there may be a beneficial response to BTI.97 3.11.5 Evidence Gaps and Future Research Future high-quality, sham-controlled trials with adequate follow-up time performed specifically in medically refractory gastroparesis patients would be beneficial to provide further evidence for or against the use of BTI in refractory gastroparesis. In addition, prospective comparative data on the use of BTI in predicting response to other interventions like G-POEM and surgical pyloromyotomy would help to further inform its current use in this setting. Sham-controlled trials of BTI in specific gastroparesis etiologies that may mechanistically be hypothesized to benefit from BTI, including post-vagal injury or post-fundoplication, may also aid in identifying subpopulations in whom BTI may be more effective. Other knowledge gaps in the utilization of BTI include the role of FLIP in predicting response to BTI, which has been reported previously but needs further validation. Lastly, the optimal effective dose of botulinum toxin has not been clearly established and warrants further investigation. Prior studies have utilized dosages of both 100 units and 200 units, both of which are supported by the experts on the panel. 3.12 Should gastric electrical stimulation be used for the treatment of individuals with gastroparesis refractory to medical therapy? 3.12.1 Summary of the Evidence Evidence informing the recommendation to use gastric electrical stimulation (GES) was derived from four crossover randomized clinical trials that compared having GES turned on to GES turned off after placement of the stimulator.98-101 The trial included 277 with gastroparesis refractory to medical therapy including antiemetics and prokinetics. The patients had idiopathic, diabetic, or postoperative gastroparesis. None of the trials had a washout period between having the stimulator turned on and off which raised concerns for carryover effect. The trials were at high risk for bias overall mainly due to the concerns regarding carryover effect and selection of the reported outcomes. We also identified multiple observational studies that compared the use of GES to G-POEM, surgical pyloric interventions, and combined with other interventions.83, 84, 87-89 The observational studies were at high risk for bias overall and were not used to inform the recommendation. The details of the studies characteristics and risk of bias assessment are reported in Supplemental Tables 51 and 52. 3.12.2 Benefits and Harms Compared to having GES off, having GES on led to only minimal improvements in the vomiting score by 0.18 points (95% CI -0.41 to 0.06; N/n = 3/229; I2 = 74.4%), nausea score by 0.28 points (95% CI -0.47 to 0.08; N/n = 3/229; I2 = 76.1%), early satiety score by 0.11 points (95% CI -0.28 to 0.07; N/n = 3/229; I2 = 48.9%), postprandial fullness score by 0.15 points (95% CI -0.54 to 0.24; N/n = 2/57; I2 = 34.1%), symptom severity score by 0.09 points (95% CI -0.35 to 0.17; N/n = 3/90; I2 = 20.5%), and abdominal pain score by 0.02 points (95% CI -0.20 to 0.16; N/n = 3/229; I2 = 0%). GES had no clinically important impact on quality of life. Around 7% of Recommendation 12: In patients with gastroparesis refractory to medical therapy, the AGA suggests against the use of gastric electrical stimulation (GES) compared to no GES [conditional recommendation, very low certainty of evidence] Comments: As part of shared decision making, patients and clinicians who place a higher value on the potential improvement in nausea and vomiting and lower value on the increase in serious adverse events may reasonably select to have GES placement. Implementation Considerations: - When used, patients should be informed that GES aims to primarily improve nausea and vomiting, not abdominal pain. patients who had the gastric electrical stimulator placement developed serious adverse events with a relative risk of 11 (95%CI 2.62 to 46.12; N/n = 4/292; I2 = 0%) compared to having no stimulator placed (assuming that none of the patients developed serious adverse events with GES). Forest plots are presented in Supplemental Figures 52 to 57. 3.12.3 Certainty in Evidence of Effects The overall CoE was very low. The CoE was rated down due to concerns for risk of bias (arising from carry over effects, missing outcome data, and selection of reported outcomes), inconsistency (studies varied in their conclusion based on MCID), and imprecision (wide confidence intervals crossing one or more effect size threshold). Publication bias was suspected for some of the outcomes. The evidence profile and evidence to decision frameworks are included in Supplemental Tables 52 and 53. 3.12.4 Discussion GES presently involves high frequency stimulation of the stomach using parameters that were chosen to improve gastric emptying in preclinical studies. Patients may see improvement in symptoms of nausea and vomiting with GES without an improvement in gastric emptying; some studies suggest it has an effect on vagal sensory afferent nerves.102 Earlier studies suggested diabetic patients had a better response to GES than idiopathic gastroparesis. Despite not meeting the minimal clinically important difference for outcomes, the panel felt that the numeric reduction in nausea and weekly vomiting frequency suggests that GES may provide a benefit to certain patients. After placement of a gastric electrical stimulator, patients responding well can have the gastric stimulation can be increased by changing the current, frequency, and/or duration of stimulation. When to increase or how to increase, has not been well studied, though there are empiric suggestions available. The panel felt that use of GES in gastroparesis should be balanced by an acknowledgement of the significant risk of serious adverse events (7%) posed by an invasive surgical treatment, including events necessitating a second procedure for battery change or even device removal. A recent study suggests added efficacy when a pyloroplasty is performed with placement of the gastric electric stimulator—this combination of two complementary treatments increases gastric emptying and may further reduce symptoms.88 Note should be made of the data being derived from two RCTs, several comparative observational studies, and one retrospective cohort study. In the latter two study designs, GES was compared to G-POEM, pyloric surgical interventions, or was used as salvage therapy after failure of pyloric surgical intervention or G-POEM was performed after failure of GES. Thus, there were critical risks of bias in the included observational or retrospective studies, reducing the enthusiasm for the use of GES among some of the panel, especially compared to procedural interventions (i.e. G-POEM) with better quality data and apparent lower risk of SAEs. Implementation considerations were also a concern, given the limited number of centers currently placing and/or maintaining GES devices. 3.12.5 Evidence Gaps and Future Research The panel suggested the need for future studies utilizing a parallel group design to generate better quality data. Different subgroups of patients (diabetic, idiopathic, postsurgical) should be studied. Use of temporary gastric stimulation with endoscopically placed stimulating wires might help determine which patients might benefit from surgically placed gastric stimulators. Stimulator settings and location of the stimulating wires in the stomach were derived empirically and have not been well studied. The presently available device uses high frequency stimulation that does not entrain gastric slow waves. Clinically, the panel felt that additional studies combining GES with other treatment modalities, such as pyloroplasty, might offer more complementary mechanisms of action. Finally, the panel hoped to see studies elucidating the physiologic effects of GES on gastric sensorimotor function beyond just the ideal device parameters to better understand which patient subgroups may respond to therapy and build a plausible physiologic rationale for use. Figure 1. Clinical Decision Support Tool Table 1. Summary of Recommendations and Implementation Considerations Number Recommendation Strength CoE 1 In individuals with suspected gastroparesis, the AGA suggests against the use of a 2-hour (or shorter) gastric emptying study compared to a 4-hour gastric emptying study to evaluate for delayed gastric emptying Implementation considerations: - By performing a 4-hour study in all cases, patients with normal emptying at 2 hours may get reclassified as delayed at 4 hours. - GES is optimally done without confounders that could affect gastric emptying including hyperglycemia and medications that alter gastric emptying (e.g. opioids, glucagon-like peptide-1 receptor agonists, prokinetics). - The 4-hour gastric emptying should be measured directly at 4 hours and not mathematically derived from earlier time points to maximize accuracy. - Among the different meals used to measure gastric emptying over a 4-hour period, two have well-established normal values and are widely accepted: the low-fat EggBeaters meal (Tougas meal), consisting of liquid egg white with bread, jam and water (250 kcal, 2% fat); and the Mayo Clinic meal consisting of two whole eggs, bread and skim milk (320 kcal, 30% fat). The latter has been more thoroughly validated to date. Conditional Low General Considerations for Implementing the Treatment Recommendations in Clinical Practice - It is important to optimize glycemic control in individuals with gastroparesis in the setting of diabetes mellitus. - It is important to review the patient’s current medications to eliminate medications that may alter gastrointestinal tract motility; medications that can delay gastric emptying include opiate pain medications and the GLP-1 receptor agonists for diabetes management and weight loss. One should be cognizant for potential drug-drug interactions when considering pharmacological interventions. - The guideline panel content experts use dietary interventions with small-particle and low-fat, low-residue diets prior to initiation of pharmacological interventions or as a concomitant intervention. - Patients with gastroparesis may need to use anti-emetics as needed including selective serotonin receptor (5-HT3) antagonists (e.g., ondansetron), histamine H1 antagonists (e.g., promethazine), and/or dopamine D2 antagonists (e.g., prochlorperazine). - The guidelines panel content experts assess the efficacy of pharmacological interventions at 4 to 8 weeks to decide whether the treatment should be continued or adjusted. 2 In individuals with gastroparesis, the AGA suggests using metoclopramide over no metoclopramide. Comment: As part of shared decision making, patients who place a higher value on the potential risk of adverse events and lower value on the symptomatic improvement may reasonably select not to use metoclopramide. Implementation Considerations: - Potential side effects of metoclopramide should be discussed with patients prior to starting treatment. - Patients on psychotropic agents and older individuals may be at higher risk for adverse events leading to discontinuation, but the absolute risk is low, including the risk of tardive dyskinesia. - Typical dosing uses the oral (tablet or liquid) or intranasal formulation, starting at 5 mg before meals, which can be increased up to 10 mg before meals. - Medication efficacy and side effects are usually assessed at 4 to 8 weeks to decide whether the treatment should be continued. Efficacy and side effects are then monitored for the duration that a patient is receiving metoclopramide. Conditional Very Low 3 In individuals with gastroparesis, the AGA suggests using erythromycin over no erythromycin. Implementation Considerations: Conditional Very Low - Erythromycin is thought to improve symptoms of gastroparesis due to its gastric prokinetic effects, but it has no direct effect on nausea and vomiting. - The effect of erythromycin on symptoms and gastric motility is unrelated to its antibiotic properties. - Low doses (e.g. 100-150 mg by mouth, 30 min before meals) are used to reduce potential side effects seen with higher doses (250-500 mg). - Because erythromycin tablets are only available in high doses, using erythromycin ethylsuccinate oral suspension allows the administration of smaller, better tolerated doses. - Tachyphylaxis often develops with prolonged, continuous use of erythromycin; in practice, efficacy is maintained using drug holidays (e.g. 3 weeks on therapy, 1 week off therapy). - Safety concerns with erythromycin include development of antibiotic resistance (with long-term use), drug interactions (it is a CYP3A isoform inhibitor), and QT-prolongation. - Azithromycin has been used as a substitute for erythromycin in practice, and many of the same safety considerations such as antibiotic resistance are also applicable. 4 In individuals with gastroparesis, the AGA suggests against use of domperidone over no domperidone. Comment: As part of shared decision making, patients who place a higher value on the potential improvement in nausea, vomiting and early satiety and lower value on the potential adverse events may reasonably select to use domperidone, when available. Implementation Considerations: - Domperidone may be especially helpful for patients with gastroparesis who had neurological side effects with metoclopramide and for patients with neurologic movement disorders such as Parkinson’s disease. - Domperidone is not approved by the FDA in the United States for human use. The prescription of domperidone requires an expanded access use investigational new drug (IND) application from the FDA. The supplier for domperidone in the United States at the time of writing is exiting the business and will no longer supply the medication, based on information available from the FDA. - Patients starting domperidone should have blood work for serum potassium and magnesium and undergo an EKG at baseline for measurement of the QTc interval. - The guideline panel content experts usually assess the efficacy of domperidone at 4 to 8 weeks to decide whether the treatment should be continued. Serum potassium, magnesium and EKG QTc interval are monitored while a patient is treated with domperidone. Conditional Very Low 5 In individuals with gastroparesis, the AGA suggests against the use of prucalopride compared to no prucalopride. Comment: As part of shared decision making, patients who place a higher value on the potential improvement in symptoms and a lower value on the potential risk of adverse effects may reasonably elect to use prucalopride, particularly if they have idiopathic gastroparesis. Implementation Considerations: - Patients with gastroparesis and chronic idiopathic constipation (CIC) may reasonably elect to use prucalopride for efficiency of care given prucalopride’s established efficacy for CIC. - Patients with gastroparesis attributed to diabetes or connective tissue diseases may be less likely to respond to prucalopride treatment than patients with idiopathic gastroparesis. - The mechanism of action for prucalopride is purely prokinetic, without any direct anti-emetic effects. - Prescribers recommending prucalopride should note the warning to monitor patients for depression and suicidal thoughts and behavior when considering use of prucalopride, but the evidence for this risk is quite low. Conditional Very Low 6 In individuals with gastroparesis, the AGA suggests against the use of aprepitant compared to no aprepitant. Conditional Low Comment: As part of shared decision making, patients who place a higher value on the potential improvement in nausea, vomiting, early satiety, and abdominal pain and lower value on the potential adverse effects may reasonably elect to use aprepitant. Implementation Considerations: - Aprepitant is an anti-emetic approved for chemotherapy-induced vomiting but has no effect on gastric motor function. - Aprepitant, a NK1 receptor antagonist, may be helpful in patients with gastroparesis and nausea/vomiting in whom 5HT3 receptor antagonists, such as ondansetron, are not helpful. 7.1 In patients with gastroparesis, the AGA suggests against the use of nortriptyline compared to no nortriptyline Comment: As part of shared decision making, patients who place a higher value on the potential improvement in early satiety, post-prandial fullness, abdominal pain, and quality of life and lower value on the potential adverse effects may reasonably elect to use nortriptyline. Implementation Considerations: - Patients with irritable bowel syndrome or significant abdominal pain associated with gastroparesis may benefit from the use of tricyclic antidepressants such as nortriptyline. - When using tricyclic antidepressants in patients with gastroparesis, the panel suggests starting at low doses and increasing slowly to ensure tolerance. - The guideline panel content experts inform patients being started on nortriptyline that this agent is in the class of antidepressants, however, in gastroparesis, it is being used as a peripheral neuromodulator to reduce afferent sensory transmission to the central nervous system. Conditional Low 7.2 In patients with gastroparesis, the AGA suggests against the use of buspirone compared to no buspirone Comment: As part of shared decision making, patients who place a higher value on the potential improvement in postprandial fullness and early satiety may reasonably elect to use buspirone Implementation Considerations: - Use of buspirone may be helpful in patients with gastroparesis and predominant early satiety and bloating. Conditional Very Low 8 In individuals with gastroparesis, the AGA suggests against the use of cannabidiol except in the context of a clinical trial Implementation Considerations: - The single study demonstrating efficacy for cannabidiol in gastroparesis used a pharmaceutical grade cannabidiol, Epidiolex, that is approved to treat seizures due to Lennox-Gastaut syndrome and Dravet syndrome, in patients two years of age and older, and the medication is not available for clinical use in gastroparesis. - Available cannabidiol formulations are not regulated with varying potencies that make their impact on patients with gastroparesis unknown. - Among cannabidiol formulations that include tetrahydrocannabinol (THC), there is concern about increased risk of cannabinoid hyperemesis syndrome. Conditional Low 9 In patients with gastroparesis, the AGA suggests against the use of gastric peroral endoscopic myotomy (G-POEM) compared to no G-POEM, except for selected patients with medically refractory disease. Implementation Considerations: - Candidates for G-POEM should have had a diagnosis of gastroparesis with an appropriately-done, 4-hour gastric emptying study, generally with at least a moderate delay in gastric emptying (20% retention at 4 hours with the egg beaters meal). - Candidates for G-POEM should have at least six months of moderate symptoms with the cardinal symptoms of nausea, vomiting, and/or postprandial fullness. - Prior to consideration of G-POEM, patients with gastroparesis should undergo a trial of other treatments for gastroparesis, such as a prokinetic agent like metoclopramide and one antiemetic agent. Conditional Low 10 In patients with gastroparesis refractory to medical therapy, the AGA recommends the use of surgical pyloric interventions (pyloromyotomy or pyloroplasty) only in the context of clinical trials. None Knowledge Gap 11 In patients with gastroparesis refractory to medical management, the AGA suggests against the use of botulinum toxin injection (BTI) compared to no BTI. Comments: - As part of decision making, patients and clinicians who place a higher value on endoscopic intervention and lower value on chronic medical therapy may reasonably attempt treatment with botulinum toxin injection. Implementation Considerations: - The guideline panel only evaluated the use of BTI in patients with refractory GP. However, the studies were not specific to refractory GP. - The need for retreatment as often as every 3 months could limit implementation from both a cost-effectiveness perspective and concern for diminishing clinical effect. - Repeat treatment may predispose to pyloric scarring, which could complicate attempts at other interventions such as G-POEM. Conditional Very Low 12 In patients with gastroparesis refractory to medical therapy, the AGA suggests against the use of gastric electrical stimulation (GES) compared to no GES Comments: As part of shared decision making, patients and clinicians who place a higher value on the potential improvement in nausea and vomiting and lower value on the increase in serious adverse events may reasonably select to have GES placement. Implementation Considerations: - When used, patients should be informed that GES aims to primarily improve nausea and vomiting, not abdominal pain. Conditional Very Low Table 2. GRADE evidence profile: should metoclopramide compared to no metoclopramide be used for patients with gastroparesis? Certainty assessment № of patients Effect Certainty № of studies (Design) Risk of bias Inconsistency Indirectness Imprecision Other considerations metoclopramide no metoclopramide Relative (95% CI) Absolute (95% CI) Nausea (follow-up: 3 weeks; assessed with: scaled to GCSI-DD; MID 0.44; Scale from: 0 to 4) 2 (RCTs) seriousa seriousb not serious seriousc none 27 27 - MD 0.24 lower (0.93 lower to 0.45 higher) ⨁◯◯◯ Very lowa,b,c Vomiting (follow-up: 3 weeks; assessed with: scaled to GCSI-DD; MID 0.44; Scale from: 0 to 4) 3 (RCTs) seriousa seriousd not serious not seriouse none 37 37 - MD 0.92 lower (1.46 lower to 0.38 higher) ⨁⨁◯◯ Lowa,d,e Early satiety (follow-up: 2 weeks; assessed with: scaled to GCSI-DD; MID 0.62; Scale from: 0 to 4) 2 (RCTs) seriousa seriousb not serious seriousc none 27 27 - MD 0.12 lower (0.95 lower to 0.71 higher) ⨁◯◯◯ Very lowa,b,c Postprandial fullness (follow-up: 3 weeks; assessed with: scaled to GCSI-DD; MID 0.62; Scale from: 0 to 4) 1 (RCT) seriousa not serious not serious very seriousf none 14 14 - MD 0.2 lower (1.01 lower to 0.61 higher) ⨁◯◯◯ Very lowa,f Symptoms severity score (follow-up: range 3 to 4 weeks; assessed with: scaled to GCSI-DD; MID 0.58; Scale from: 0 to 4) 6 (RCTs) seriousg not serioush not serious seriousi none 357 262 - MD 0.54 lower (0.98 lower to 0.11 lower) ⨁⨁◯◯ Lowg,h,i Abdominal pain (follow-up: 3 weeks; assessed with: scaled to GCSI-DD; MID 0.24) 1 (RCT) seriousa not serious not serious seriousj none 14 14 - MD 1.2 lower (2.01 lower to 0.39 lower) ⨁⨁◯◯ Lowa,j Quality of life - not reported - - - - - - - - - - - Serious adverse events (follow-up: range 3 to 4 weeks) 3 (RCTs) not serious seriousk not serious seriousl none 3/217 (1.4%) 5/222 (2.3%) RR 0.64 (0.17 to 2.41) 8 fewer per 1,000 (from 19 fewer to 32 more) ⨁⨁◯◯ Lowk,l Adverse events leading to discontinuation of treatment (follow-up: range 3 to 4 weeks) 5 (RCTs) not seriousm not serious not serious seriousl none 14/250 (5.6%) 6/251 (2.4%) RR 2.15 (0.89 to 5.19) 27 more per 1,000 (from 3 fewer to 100 more) ⨁⨁⨁◯ Moderatel,m Abbreviation: CI: confidence interval; MD: mean difference; RR: risk ratio Explanations: a. Possible bias arising from randomization process, deviation from intended intervention, and measurement of outcome. b. One of the studies small improvement and the other study showed no-to-trivial harm. Also, I2 was 50 to 63%. c. Serious because CI included small benefit, no-to-trivial benefits and harms, and small harm. Although the confidence interval included numerous thresholds, we rated down once only because the wide CI probably resulted from the inconsistency. d. Two studies showed small improvement and one study showed small harm. Also, I2 was 86%. e. Although the CI included small benefit and no-to-trivial benefit, we did not rate down for serious imprecision because the wide confidence interval was probably because of the inconsistency. f. Very serious because CI included small benefit, no-to-trivial benefit, and no-to-trivial harm. g. Possible bias arising from randomization process, deviation from intended intervention, and measurement of outcome in four studies (Perkel 1979, Perkel 1980, Snape 1982, and Ricci 1985) which contributed around 60% of pooled estimate. h. Only study, Snape 1982 which contributed 13% of the pooled estimate, showed extreme results. The remaining studies showed consistent findings thus we did not rate down for inconsistency. i. CI included small benefit and no-to-trivial benefit. j. Very small sample size, 28. k. One study showed increased risk of adverse events and the other showed decreased risk. l. CI included both increased and decreased risk of adverse events. m. Possible bias arising from randomization process, deviation from intended intervention, and measurement of outcome in three studies (Perkel 1979, Perkel 1980, and Ricci 1985) which contributed less than 15% of pooled estimate. Thus, we did not rate down. Tabe 3. Evidence to decision framework: should metoclopramide compared to no metoclopramide be used for patients with gastroparesis? JUDGEMENT DESIRABLE EFFECTS Trivial Small Moderate Large Varies Don't know UNDESIRABLE EFFECTS Large Moderate Small Trivial Varies Don't know CERTAINTY OF EVIDENCE Very low Low Moderate High No included studies VALUES Important uncertainty or variability Possibly important uncertainty or variability Probably no important uncertainty or variability No important uncertainty or variability BALANCE OF EFFECTS Favors the comparison Probably favors the comparison Does not favor either the intervenƟon or the comparison Probably favors the intervenƟon Favors the intervenƟon Varies Don't know EQUITY Reduced Probably reduced Probably no impact Probably increased Increased Varies Don't know ACCEPTABILITY No Probably no Probably yes Yes Varies Don't know FEASIBILITY No Probably no Probably yes Yes Varies Don't know 4. References 1. Ye Y, Yin Y, Huh SY, et al. 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Aliment Pharmacol Ther 2007;26:1251-8. 93. Friedenberg FK, Palit A, Parkman HP, et al. Botulinum toxin A for the treatment of delayed gastric emptying. Am J Gastroenterol 2008;103:416-23. 94. Wadhwa V, Gonzalez A, Azar F, et al. Response to botulinum toxin may predict response to peroral pyloromyotomy in patients with gastroparesis. Endoscopy 2023;55:508-514. 95. Coleski R, Anderson MA, Hasler WL. Factors associated with symptom response to pyloric injection of botulinum toxin in a large series of gastroparesis patients. Dig Dis Sci 2009;54:2634-42. 96. Jacques J, Pagnon L, Hure F, et al. Peroral endoscopic pyloromyotomy is efficacious and safe for refractory gastroparesis: prospective trial with assessment of pyloric function. Endoscopy 2019;51:40-49. 97. Reddymasu SC, Singh S, Sankula R, et al. Endoscopic pyloric injection of botulinum toxin-A for the treatment of postvagotomy gastroparesis. Am J Med Sci 2009;337:161-4. 98. Abell T, McCallum R, Hocking M, et al. Gastric electrical stimulation for medically refractory gastroparesis. Gastroenterology 2003;125:421-8. 99. McCallum RW, Snape W, Brody F, et al. Gastric electrical stimulation with Enterra therapy improves symptoms from diabetic gastroparesis in a prospective study. Clin Gastroenterol Hepatol 2010;8:947-54; quiz e116. 100. McCallum RW, Sarosiek I, Parkman HP, et al. Gastric electrical stimulation with Enterra therapy improves symptoms of idiopathic gastroparesis. Neurogastroenterol Motil 2013;25:815-e636. 101. Ducrotte P, Coffin B, Bonaz B, et al. Gastric Electrical Stimulation Reduces Refractory Vomiting in a Randomized Crossover Trial. Gastroenterology 2020;158:506-514 e2. 102. McCallum RW, Dusing RW, Sarosiek I, et al. Mechanisms of symptomatic improvement after gastric electrical stimulation in gastroparetic patients. Neurogastroenterol Motil 2010;22:161-7, e50-1.
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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up Back To Orden de Números RealesBack 2.13 Real Number Line Graphs Written by:Brenda Meery | Kaitlyn Spong Fact-checked by:The CK-12 Editorial Team Last Modified: Aug 01, 2025 Can you describe the number 13? Can you say what number sets the number 13 belongs to? Real Number Line Graphs All of the numbers you have learned about so far in math belong to the real number system. Positives, negatives, fractions, and decimals are all part of the real number system. The diagram below shows how all of the numbers in the real number system are grouped. In practice, there may be more than one symbol or combination of symbols used to describe a particular set of numbers. For example, the set of all irrational numbers may be defined as ¯Q, or I, or even R−Q (Real Numbers minus Rational Numbers). The most important thing is to be consistent, and to state the definition of the symbol(s) you use. Any number in the real number system can be plotted on a real number line. You can also graph inequalities on a real number line. In order to graph inequalities, make sure you know the following symbols: The symbol > means “is greater than.” The symbol < means “is less than.” The symbol ≥ means “is greater than or equal to.” The symbol ≤ means “is less than or equal to.” The inequality symbol indicates the type of dot that is placed on the beginning point and the number set indicates whether an arrow is drawn on the number line or if points are used. Let's practice using a number line: Represent x>4 where x is an integer, on a number line. The open dot on the four means that 4 is not included in the graph of all integers greater than 4. The closed dots on 5, 6, 7, 8 means that these numbers are included in the set of integers greater than 4. The arrow pointing to the right means that all integers to the right of 8 are also included in the graph of all integers greater than 4. Represent this inequality statement on a number line {x≥−2|x ∈ R}. The statement can be read as “x is greater than or equal to –2, such that x belongs to or is a member of the real numbers.” In other words, represent all real numbers greater than or equal to –2. The inequality symbol says that x is greater than or equal to –2. This means that –2 is included in the graph. A solid dot is placed on –2 and on all numbers to the right of –2. The line is on the number line to indicate that all real numbers greater than –2 are also included in the graph. Represent this inequality statement, also known as set notation, on a number line {x|2<x≤7,x ∈ N}. This inequality statement can be read as x such that x is greater than 2 and less than or equal to 7 and x belongs to the natural numbers. In other words, all natural numbers greater than 2 and less than or equal to 7. The inequality statement that was to be represented on the number line had to include the natural numbers greater than 2 and less than or equal to 7. These are the only numbers to be graphed. There is no arrow on the number line. Examples Example 1 Earlier, you were asked to describe the number 13 and to identify what number set(s) 13 belongs to. The number 13 is a natural number because it is in the set N={1,2,3,4…}. The number 13 is a whole number because it is in the set W={0,1,2,3…}. The number 13 is an integer because it is in the set Z={…,−3,−2,−1,0,1,2,3,…}. The number 13 is a rational number because it is in the set Q={ab,b≠0}. The number 13 belongs to the real number system. Example 2 Check the set(s) to which each number belongs. The number may belong to more than one set. | Number | N | W | Z | Q | ¯Q | | --- --- --- | 5 | | | | | | | | −473 | | | | | | | | 1.48 | | | | | | | | √7 | | | | | | | | 0 | | | | | | | | | π | | | | | | | | Review the definitions for each set of numbers. | Number | N | W | Z | Q | ¯Q | --- --- --- | | 5 | X | X | X | X | | | −473 | | | | X | | | 1.48 | | | | X | | | √7 | | | | | X | | 0 | | X | X | X | | | π | | | | | X | Example 3 Graph {x|−3≤x≤8,x ∈ R} on a number line. {x|−3≤x≤8,x ∈ R} The set notation means to graph all real numbers between –3 and +8. The line joining the solid dots represents the fact that the set belongs to the real number system. Example 4 Use set notation to describe the set shown on the number line. The closed dot means that –3 is included in the answer. The remaining dots are to the right of –3. The open dot means that 2 is not included in the answer. This means that the numbers are all less than 2. Graphing on a number line is done from smallest to greatest or from left to right. There is no line joining the dots so the variable does not belong to the set of real numbers. However, negative whole numbers, zero and positive whole numbers make up the integers. The set notation that is represented on the number line is {x|−3≤x<2,x ∈ Z}. Review Describe each set notation in words. {x|x>8,x ∈ R} {x|x≤−3,x ∈ Z} {x|−4≤x≤6,x ∈ R} {x|5≤x≤11,x ∈ W} {x|x≥6,x ∈ N} Represent each graph using set notation . . . . . For each of the following situations, use set notations to represent the limits. To ride the new tilt-a whirl at the fairgrounds, a child can be no taller than 4.5 feet. A dance is being held at the community hall to raise money for breast cancer. The dance is only for those people 19 years of age or older. A sled driver in the Alaska Speed Quest must start the race with no less than 10 dogs and no more than 16 dogs. The residents of a small community are planning a skating party at the local lake. In order for the event to take place, the outdoor temperature needs to be above −6∘C and not above −1∘C. Juanita and Hans are planning their wedding supper at a local venue. To book the facility, they must guarantee that at least 100 people will have supper but no more than 225 people will eat. Represent the following set notations on a number line. {x|x>6,x ∈ N} {x|x≤8,x ∈ R} {x|−3≤x<6,x ∈ Z} Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. | Image | Reference | Attributions | --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. 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https://www.droracle.ai/articles/333742/summarize-the-latest-hyperlipidemia-guidelines-workup-for-me
What is the initial workup and management for hyperlipidemia? Select Language​▼ What is the initial workup and management for hyperlipidemia? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: September 16, 2025 • View editorial policy Latest Hyperlipidemia Guidelines Workup The initial workup for hyperlipidemia should include a fasting lipid panel (total cholesterol, LDL-C, HDL-C, triglycerides), assessment of secondary causes, and evaluation of cardiovascular risk factors to guide treatment decisions.1 Initial Laboratory Assessment Fasting Lipid Panel: Total cholesterol LDL cholesterol HDL cholesterol Triglycerides Non-HDL cholesterol (calculated) Additional Laboratory Tests: Fasting glucose or HbA1c Liver function tests (AST, ALT) Renal function (creatinine, eGFR) Thyroid function (TSH) Evaluation for Secondary Causes Secondary causes of hyperlipidemia must be identified and treated before initiating specific lipid-lowering therapy 2. Common secondary causes include: Endocrine disorders: Diabetes mellitus (uncontrolled) Hypothyroidism Cushing's syndrome Renal disorders: Chronic kidney disease Nephrotic syndrome Liver disease: Obstructive liver disease Non-alcoholic fatty liver disease (NAFLD) Medications: Thiazide diuretics Beta-blockers Estrogens Corticosteroids Isotretinoin Antiretroviral protease inhibitors Antipsychotics Immunosuppressants Lifestyle factors: Excessive alcohol intake Poor diet high in saturated fats Physical inactivity Cardiovascular Risk Assessment Risk stratification is essential for determining treatment goals and intensity 1: Very High Risk: Established ASCVD Diabetes with target organ damage Severe CKD (eGFR <30 mL/min) SCORE ≥10% LDL-C Target:<70 mg/dL or ≥50% reduction from baseline High Risk: Markedly elevated single risk factors Diabetes without target organ damage Moderate CKD (eGFR 30-59 mL/min) SCORE ≥5% to <10% LDL-C Target:<100 mg/dL or ≥50% reduction from baseline Moderate Risk: Young patients with diabetes SCORE ≥1% to <5% LDL-C Target:<115 mg/dL Low Risk: SCORE <1% LDL-C Target:<115 mg/dL Family History Assessment Screen for premature CVD in first-degree relatives (men <55 years, women <65 years) Evaluate for familial hypercholesterolemia patterns: Heterozygous FH: LDL-C typically >190 mg/dL Homozygous FH: LDL-C typically >500 mg/dL Family history of premature ASCVD Special Populations Considerations Children and Adolescents Initial screening at age ≥2 years if family history of early CVD exists 1 Without family history, screen at puberty (≥10 years) 1 Optimal goals: LDL-C <100 mg/dL, HDL-C >35 mg/dL, triglycerides <150 mg/dL 3 Diabetic Patients Screen when initial glycemic control has been achieved and annually thereafter 3, 1 Consider less frequent screening (every 2 years) for patients with consistently low-risk lipid values 1 Patients with Chronic Kidney Disease Assess lipid profile at diagnosis and annually Consider more aggressive treatment targets due to higher cardiovascular risk 4 Management Algorithm Lifestyle Modifications (First-line for all patients): Heart-healthy diet (reduced saturated fat <7% of calories) Reduced dietary cholesterol (<200 mg/day) Elimination of trans fats Weight management if overweight/obese Regular physical activity Smoking cessation Limited alcohol intake Pharmacological Therapy: Moderate Hyperlipidemia: Statins as first-line therapy (high-intensity for high-risk patients) Consider ezetimibe as add-on therapy if LDL-C goals not achieved Severe Hypertriglyceridemia (>500 mg/dL): Fibrates are first-line treatment to reduce pancreatitis risk 2 Consider omega-3 fatty acids as adjunctive therapy Reduce dietary fat and simple carbohydrates Familial Hypercholesterolemia: High-intensity statins Consider combination therapy with ezetimibe PCSK9 inhibitors for those not achieving targets 5 Monitoring Check lipid levels 4-12 weeks after initiating or changing therapy 1 Monitor liver enzymes 8-12 weeks after starting statin therapy Assess for muscle symptoms at follow-up visits Annual lipid profile monitoring once target levels are achieved 1 Common Pitfalls to Avoid Failing to identify secondary causes before initiating lipid-lowering therapy Not assessing overall cardiovascular risk when determining treatment intensity Inadequate monitoring of response to therapy and potential side effects Overlooking familial hypercholesterolemia in patients with very high LDL-C levels Using statins alone in patients with severe hypertriglyceridemia (>500 mg/dL) Discontinuing therapy due to minor side effects without attempting dose adjustments or alternative medications By following this structured approach to hyperlipidemia workup, clinicians can effectively identify patients at risk, determine appropriate treatment goals, and implement evidence-based management strategies to reduce cardiovascular morbidity and mortality. References 1 Guideline Lipid Management in Diabetic Patients Praxis Medical Insights: Practical Summaries of Clinical Guidelines, 2025 2 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 3 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 4 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 5 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 Related Questions What is the management for a patient with hyperlipidemia?What are the targets for managing hyperlipidaemia?What is the initial treatment approach for hyperlipidemia?What are the recommended treatment options for hyperlipidemia?What are the treatment goals for hyperlipidemia according to ACC (American College of Cardiology) guidelines?What laboratory tests should be ordered for a patient presenting with low energy?What is considered a controlled heart rate in patients with atrial fibrillation?What is the role of clopidogrel (Plavix) in managing essential thrombocytosis in a patient with Chronic Kidney Disease (CKD) stage 2?What is the appropriate workup for a patient presenting with neck nerve pain?What is the recommended diagnostic approach for a patient presenting with intermittent pulsatile tinnitus?Can a workup for Cushing's syndrome be done while a patient is on insulin? Professional Medical Disclaimer This information is intended for healthcare professionals. Any medical decision-making should rely on clinical judgment and independently verified information. The content provided herein does not replace professional discretion and should be considered supplementary to established clinical guidelines. Healthcare providers should verify all information against primary literature and current practice standards before application in patient care. Dr.Oracle assumes no liability for clinical decisions based on this content. Have a follow-up question? Our Medical A.I. is used by practicing medical doctors at top research institutions around the world. Ask any follow up question and get world-class guideline-backed answers instantly. Ask Question Original text Rate this translation Your feedback will be used to help improve Google Translate
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http://www.survivorlibrary.com/library/a_treatise_on_plane_and_spherical_trigonometry_and_its_applications_to_astronomy_and_geodesy_1892.pdf
B O W S E R ' S M A T H E M A T I C S . A C A D E M I C A L G E B R A . W i t h n u m e r o u s E x a m p l e s . C O L L E G E A L G E B B A . W i t h n u m e r o u s E x a m p l e s . P L A N E A N D S O L I D G E O M E T B Y . W i t h n u m e r o u s E x e r c i s e s . E L E M E N T S O F P L A N E A N D S P H E R I C A L T B I G O N O M E . T B T . W i t h n u m e r o u s E x a m p l e s . A T B E A T I S E O N P L A N E A N D S P H E B I C A L T B I G O N O M E -T R Y , a n d i t s a p p l i c a t i o n s t o A s t r o n o m y a n d G e o d e s y . W i t h n u m e r o u s E x a m p l e s . A N E L E M E N T A R Y T R E A T I S E O N A N A L Y T I C G E O M E T R Y , e m b r a c i n g P l a n e G e o m e t r y , a n d a n I n t r o d u c t i o n t o G e o m e t r y o f T h r e e D i m e n s i o n s . A N E L E M E N T A R Y T R E A T I S E O N T H E D I F F E R E N T I A L A N D I N T E G R A L C A L C U L U S . W i t h n u m e r o u s E x a m p l e s . A N E L E M E N T A B Y T B E A T I S E O N A N A L Y T I C M E C H A N I C S . W i t h n u m e r o u s E x a m p l e s . A N E L E M E N T A B Y T R E A T I S E O N H Y D R O M E C H A N I C S . W i t h n u m e r o u s E x a m p l e s . A T R E A T I S E O N P L A N E A N D S P H E R I C A L -O F T E I G O N O M E T E T , A N D I T S A P P L I C A T I O N S T O A S T B O J T O M Y A N D G E O D E S Y , W I T H N U M E R O U S E X A M P L E S . E D W A R D A . B O W S E R , L L . D . , P r o f e s s o r o f M a t h e m a t i c s a n d E n g i n e e r i n g i n R u t g e r s C o l l e g e . B O S T O N , U . S . A . : P U B L I S H E D B Y D . C . H E A T H & C O . 1 8 9 2 . C O P Y R I G H T , 1 8 9 2 , B t E . A . B O W S E R . C A J O R I T v p o g r a p h v b y J , S . C u s h i n g & C o . , B o s t o n , U . S . A . P r e s s w o r k b y B e r w i c k & S m i t h , B o s t o n , U . S . A . P E E F A C E . T h e p r e s e n t t r e a t i s e o n P l a n e a n d S p h e r i c a l T r i g o n o m e t r y i s d e s i g n e d a s a t e x t - b o o k f o r C o l l e g e s , S c i e n t i f i c S c h o o l s , a n d I n s t i t u t e s o f T e c h n o l o g y . T h e a i m h a s b e e n t o p r e s e n t t h e s u b j e c t i n a s c o n c i s e a f o r m a s i s c o n s i s t e n t w i t h c l e a r n e s s , t o m a k e i t a t t r a c t i v e a n d e a s i l y i n t e l l i g i b l e t o t h e s t u d e n t , a n d a t t h e s a m e t i m e t o p r e s e n t t h e f u l l e s t c o u r s e o f T r i g o n o m e t r y w h i c h i s u s u a l l y g i v e n i n t h e b e s t T e c h n o l o g i c a l S c h o o l s . C o n s i d e r a b l e c a r e h a s b e e n t a k e n t o i n s t r u c t t h e s t u d e n t i n t h e t h e o r y a n d u s e o f L o g a r i t h m s , a n d t h e i r p r a c t i c a l a p p l i c a t i o n t o t h e s o l u t i o n o f t r i a n g l e s . I t i s h o p e d t h a t t h e w o r k m a y c o m m e n d i t s e l f , n o t o n l y t o t h o s e w h o w i s h t o c o n f i n e t h e m s e l v e s t o t h e n u m e r i c a l c a l c u l a t i o n s w h i c h o c c u r i n T r i g o n o m e t r y , b u t a l s o t o t h o s e w h o i n t e n d t o p u r s u e t h e s t u d y o f t h e h i g h e r m a t h e m a t i c s . T h e e x a m p l e s a r e v e r y n u m e r o u s a n d a r e c a r e f u l l y s e l e c t e d . M a n y a r e p l a c e d i n i m m e d i a t e c o n n e c t i o n w i t h t h e s u b j e c t - m a t t e r w h i c h t h e y i l l u s t r a t e . T h e n u m e r i c a l s o l u t i o n o f t r i a n g l e s h a s r e c e i v e d m u c h a t t e n t i o n , e a c h c a s e b e i n g t r e a t e d i n d e t a i l . T h e i l i 9 1 4 _ 3 8 4 i v P R E F A C E . e x a m p l e s a t t h e e n d s o f t h e c h a p t e r s h a v e b e e n c a r e f u l l y g r a d e d , b e g i n n i n g w i t h t h o s e w h i c h a r e e a s y , a n d e x t e n d i n g t o t h o s e w h i c h a r e m o r e a n d m o r e d i f f i c u l t . T h e s e e x a m p l e s i l l u s t r a t e e v e r y p a r t o f t h e s u b j e c t , a n d a r e i n t e n d e d t o t e s t , n o t o n l y t h e s t u d e n t , s k n o w l e d g e o f t h e u s u a l m e t h o d s o f c o m p u t a t i o n , b u t h i s a b i l i t y t o g r a s p t h e m i n t h e m a n } r f o r m s t h e y m a y a s s u m e i n p r a c t i c a l a p p l i c a t i o n s . A m o n g t h e s e e x a m p l e s a r e s o m e o f t h e m o s t e l e g a n t t h e o r e m s i n P l a n e a n d S p h e r i c a l T r i g o n o m e t r y . T h e C h a p t e r s o n D e M o i v r e , s T h e o r e m , a n d A s t r o n o m y , G e o d e s y , a n d P o l y e d r o n s , w i l l s e r v e t o i n t r o d u c e t h e s t u d e n t t o s o m e o f t h e h i g h e r a p p l i c a t i o n s o f T r i g o n o m e t r y , r a r e l y f o u n d i n A m e r i c a n t e x t - b o o k s . I n w r i t i n g t h i s b o o k , t h e b e s t E n g l i s h a n d F r e n c h a u t h o r s h a v e b e e n c o n s u l t e d . I a m i n d e b t e d e s p e c i a l l y t o t h e w o r k s o f T o d h u n t e r , C a s e y , L o c k , H o b s o n , C l a r k e , E u s t i s , S n o w b a l l , M , C l e l l a n d a n d P r e s t o n , S m i t h , a n d S e r r e t . I t r e m a i n s f o r m e t o e x p r e s s m y t h a n k s t o m y c o l l e a g u e s , P r o f . R . W . P r e n t i s s f o r r e a d i n g t h e M S . , a n d M r . I . S . U p s o n f o r r e a d i n g t h e p r o o f - s h e e t s . A n y c o r r e c t i o n s o r s u g g e s t i o n s , e i t h e r i n t h e t e x t o r t h e e x a m p l e s , w i l l b e t h a n k f u l l y r e c e i v e d . E . A . B . R u t g e r s C o l l e g e , N e w B r u n s w i c k , N . J . , A p r i l , 1 8 0 2 . T A B L E O F C O N T E N T S . P A R T I . P L A N E T R I G O N O M E T R Y . C H A P T E R I . M e a s u r e m e n t o f A n g l e s . A R T . 1 . T r i g o n o m e t r y 1 2 . T h e M e a s u r e o f a Q u a n t i t y 1 3 . A n g l e s 2 4 . P o s i t i v e a n d N e g a t i v e A n g l e s 3 5 . T h e M e a s u r e o f A n g l e s 3 6 . T h e S e x a g e s i m a l M e t h o d 5 7 . T h e C e n t e s i m a l o r D e c i m a l M e t h o d 0 8 . T h e C i r c u l a r M e a s u r e S ) . C o m p a r i s o n o f t h e S e x a g e s i m a l a n d C e n t e s i m a l M e a s u r e s . . 8 1 0 . C o m p a r i s o n o f t h e S e x a g e s i m a l a n d C i r c u l a r M e a s u r e s 9 1 1 . G e n e r a l M e a s u r e o f a n A n g l e 1 1 1 2 . C o m p l e m e n t a n d S u p p l e m e n t o f a n A n g l e 1 2 E x a m p l e s 1 3 C H A P T E R I I . T h e T r i g o n o m e t r i c F u n c t i o n s . 1 3 . D e f i n i t i o n s o f t h e T r i g o n o m e t r i c F u n c t i o n s 1 6 1 4 . T h e F u n c t i o n s a r e a l w a y s t h e S a m e f o r t h e S a m e A n g l e 1 8 1 5 . F u n c t i o n s o f C o m p l e m e n t a l A n g l e s 2 0 1 6 . R e p r e s e n t a t i o n o f t h e F u n c t i o n s b y S t r a i g h t L i n e s 2 0 1 7 . P o s i t i v e a n d N e g a t i v e L i n e s 2 3 v v i C O N T E N T S . A R T . P A O E 1 8 . F u n c t i o n s o f A n g l e s o f A n y M a g n i t u d e 2 3 1 9 . C h a n g e s i n S i n e a s t h e A n g l e i n c r e a s e s f r o m 0 ° t o 3 6 0 ° 2 5 2 0 . C h a n g e s i n C o s i n e a s t h e A n g l e i n c r e a s e s f r o m 0 ° t o 3 6 0 ° . . . 2 6 2 1 . C h a n g e s i n T a n g e n t a s t h e A n g l e i n c r e a s e s f r o m 0 ° t o 3 6 0 ° . 2 7 2 2 . T a b l e g i v i n g C h a n g e s o f F u n c t i o n s i n F o u r Q u a d r a n t s 2 8 2 3 . R e l a t i o n s b e t w e e n t h e F u n c t i o n s o f t h e S a m e A n g l e 2 9 2 4 . U s e o f t h e P r e c e d i n g F o r m u l a e 3 0 2 5 . G r a p h i c M e t h o d o f f i n d i n g t h e F u n c t i o n s i n T e r m s o f O n e . . 3 0 2 6 . T o f i n d t h e T r i g o n o m e t r i c F u n c t i o n s o f 4 5 ° 3 1 2 7 . T o f i n d t h e T r i g o n o m e t r i c F u n c t i o n s o f 6 0 ° a n d 3 0 ° 3 1 2 8 . R e d u c t i o n o f F u n c t i o n s t o 1 s t Q u a d r a n t 3 3 2 9 . F u n c t i o n s o f C o m p l e m e n t a l A n g l e s 3 4 3 0 . F u n c t i o n s o f S u p p l e m e n t a l A n g l e s 3 4 3 1 . T o p r o v e s i n ( 9 0 ° + A ) = c o s A , e t c 3 5 3 2 . T o p r o v e s i n ( 1 8 0 ° + A ) = -s i n A , e t c 3 5 3 3 . T o p r o v e s i n ( — A ) = — s i n A , e t c 3 6 3 4 . T o p r o v e s i n ( 2 7 0 ° + A ) = s i n ( 2 7 0 ° -A ) = -c o s A , e t c . . . . 3 6 3 5 . T a b l e g i v i n g t h e R e d u c e d F u n c t i o n s o f A n y A n g l e 3 7 3 6 . P e r i o d i c i t y o f t h e T r i g o n o m e t r i c F u n c t i o n s 3 8 3 7 . A n g l e s c o r r e s p o n d i n g t o G i v e n F u n c t i o n s 3 9 3 8 . G e n e r a l E x p r e s s i o n f o r A l l A n g l e s w i t h a G i v e n S i n e 4 0 3 9 . A n E x p r e s s i o n f o r A l l A n g l e s w i t h a G i v e n C o s i n e 4 1 4 0 . A n E x p r e s s i o n f o r A l l A n g l e s w i t h a G i v e n T a n g e n t 4 1 4 1 . T r i g o n o m e t r i c I d e n t i t i e s 4 3 E x a m p l e s 4 4 C H A P T E R I I I . T r i g o n o m e t r i c F u n c t i o n s o f T w o A n g l e s . 4 2 . F u n d a m e n t a l F o r m u l a e 5 0 4 3 . T o f i n d t h e V a l u e s o f s i n ( x + y ) a n d c o s ( x + y ) 5 0 4 4 . T o f i n d t h e V a l u e s o f s i n ( x — y ' ) a n d c o s ( x — y ) 5 2 4 5 . F o r m u l a e f o r t r a n s f o r m i n g S u m s i n t o P r o d u c t s 5 5 4 6 . U s e f u l F o r m u l a e 5 6 4 7 . T a n g e n t o f S u m a n d D i f f e r e n c e o f T w o A n g l e s 5 7 4 8 . F o r m u l a e f o r t h e S u m o f T h r e e o r M o r e A n g l e s 5 8 4 9 . F u n c t i o n s o f D o u b l e A n g l e s 6 0 5 0 . F u n c t i o n s o f 3 x i n T e r m s o f t h e F u n c t i o n s o f x 6 1 5 1 . F u n c t i o n s o f H a l f a n A n g l e 6 3 5 2 . D o u b l e V a l u e s o f S i n e a n d C o s i n e o f H a l f a n A n g l e 6 3 C O N T E N T S . v i i A B T . P A O e 5 3 . Q u a d r u p l e V a l u e s o f S i n e a n d C o s i n e o f H a l f a n A n g l e 6 5 5 4 . D o u b l e V a l u e o f T a n g e n t o f H a l f a n A n g l e 6 6 5 5 . T r i p l e V a l u e o f S i n e o f O n e - t h i r d a n A n g l e 6 7 5 6 . F i n d t h e V a l u e s o f t h e F u n c t i o n s o f 2 2 £ ° 6 9 5 7 . F i n d t h e S i n e a n d C o s i n e o f 1 8 ° 6 9 5 8 . F i n d t h e S i n e a n d C o s i n e o f 3 6 ° 7 0 5 9 . I f A + B + C = 1 8 0 ° , t o A n d s i n A + s i n B + s i n C , e t c 7 0 6 0 . I n v e r s e T r i g o n o m e t r i c F u n c t i o n s 7 2 6 1 . T a b l e o f U s e f u l F o r m u l a s 7 5 E x a m p l e s 7 7 C H A P T E R I V . L o g a r i t h m s a n d L o g a r i t h m i c T a b l e s . — T r i g o n o m e t r i c T a b l e s . 6 2 . N a t u r e a n d U s e o f L o g a r i t h m s 8 7 6 3 . P r o p e r t i e s o f L o g a r i t h m s 8 7 6 4 . C o m m o n S y s t e m o f L o g a r i t h m s 9 1 6 5 . C o m p a r i s o n o f T w o S y s t e m s o f L o g a r i t h m s 9 3 6 6 . T a b l e s o f L o g a r i t h m s 9 5 6 7 . U s e o f T a b l e s o f L o g a r i t h m s o f N u m b e r s 9 8 6 8 . T o f i n d t h e L o g a r i t h m o f a G i v e n N u m b e r 9 9 6 9 . T o f i n d t h e N u m b e r c o r r e s p o n d i n g t o a G i v e n L o g a r i t h m . . . 1 0 2 6 9 a . A r i t h m e t i c C o m p l e m e n t 1 0 3 7 0 . U s e o f T r i g o n o m e t r i c T a b l e s 1 0 5 7 1 . U s e o f T a b l e s o f N a t u r a l T r i g o n o m e t r i c F u n c t i o n s 1 0 6 7 2 . T o f i n d t h e S i n e o f a G i v e n A n g l e 1 0 6 7 3 . T o f i n d t h e C o s i n e o f a G i v e n A n g l e ' . 1 0 6 7 4 . T o f i n d t h e A n g l e w h o s e S i n e i s G i v e n 1 0 8 7 5 . T o f i n d t h e A n g l e w h o s e C o s i n e i s G i v e n 1 0 8 7 6 . U s e o f T a b l e s o f L o g a r i t h m i c T r i g o n o m e t r i c F u n c t i o n s 1 1 0 7 7 . T o f i n d t h e L o g a r i t h m i c S i n e o f a G i v e n A n g l e 1 1 2 7 8 . T o f i n d t h e L o g a r i t h m i c C o s i n e o f a G i v e n A n g l e 1 1 2 7 9 . T o f i n d t h e A n g l e w h o s e L o g a r i t h m i c S i n e i s G i v e n 1 1 4 8 0 . T o f i n d t h e A n g l e w h o s e L o g a r i t h m i c C o s i n e i s G i v e n 1 1 5 8 1 . A n g l e s n e a r t h e L i m i t s o f t h e Q u a d r a n t 1 1 6 E x a m p l e s 1 1 7 v i i i C O N T E N T S . C H A P T E R V . S o l u t i o n o f T r i g o n o m e t r i c E q u a t i o n s , a r t . p a c k 8 2 . T r i g o n o m e t r i c E q u a t i o n s 1 2 6 8 3 . T o s o l v e t o s i n < £ — a , m c o s # = 6 1 2 8 8 4 . T o s o l v e a s i n < p + b c o s 0 = c 1 2 9 8 5 . T o s o l v e s i n ( a + x ) = m s i n x 1 3 1 8 0 . T o s o l v e t a n ( « + x ) = m t a n x 1 3 2 8 7 . T o s o l v e t a n ( a + x ) t a n x = m 1 3 3 8 8 . T o s o l v e m s i n ( 0 + x ) = a , t o s i n ( < j > + x ) — . b 1 3 4 8 9 . T o s o l v e x c o s a + y s i n a = t o , x s i n a — y c o s a = n 1 3 5 9 0 . A d a p t a t i o n t o L o g a r i t h m i c C o m p u t a t i o n 1 3 5 9 1 . T o s o l v e r c o s $ c o s 9 — a , r c o s < p s i n 9 = 6 , r s i n 0 = c 1 3 7 9 2 . T r i g o n o m e t r i c E l i m i n a t i o n 1 3 8 E x a m p l e s 1 4 0 C H A P T E R V I . R e l a t i o n s b e t w e e n t h e S i d e s o f a T r i a n g l e a n d t h e F u n c t i o n s o f i t s A n g l e s . 9 3 . F o r m u l a e 1 4 0 9 4 . R i g h t T r i a n g l e s 1 4 6 9 5 . O b l i q u e T r i a n g l e s — L a w o f S i n e s 1 4 7 9 6 . L a w o f C o s i n e s 1 4 8 9 7 . L a w o f T a n g e n t s 1 4 9 9 8 . T o p r o v e c = a c o s B + b c o s A 1 4 9 9 9 . F u n c t i o n s o f H a l f a n A n g l e i n T e r m s o f t h e S i d e s 1 5 0 1 0 0 . T o e x p r e s s t h e S i n e o f a n A n g l e i n T e r m s o f t h e S i d e s 1 5 2 1 0 1 . E x p r e s s i o n s f o r t h e A r e a o f a T r i a n g l e 1 5 3 1 0 2 . I n s c r i b e d C i r c l e 1 5 4 1 0 3 . C i r c u m s c r i b e d C i r c l e 1 5 4 1 0 4 . E s c r i b e d C i r c l e I 5 5 1 0 5 . D i s t a n c e b e t w e e n t h e I n - c e n t r e a n d t h e C i r c u m c e n t r e 1 5 5 1 0 6 . T o f i n d t h e A r e a o f a C y c l i c Q u a d r i l a t e r a l 1 5 7 E x a m p l e s 1 5 9 C H A P T E R V I I . S o l u t i o n o f T r i a n g l e s . 1 0 7 . D e f i n i t i o n s I 6 5 1 0 8 . F o u r C a s e s o f R i g h t T r i a n g l e s 1 6 5 C O N T E N T S . i x A B T . I " A O K 1 0 9 . C a s e L — G i v e n a S i d e a n d t h e H y p o t e n u s e 1 6 6 1 1 0 . C a s e I I . — G i v e n a n A c u t e A n g l e a n d t h e H y p o t e n u s e 1 6 7 1 1 1 . C a s e I I I . — G i v e n a S i d e a n d a n A c u t e A n g l e 1 6 8 1 1 2 . C a s e I V . — G i v e n t h e T w o S i d e s 1 6 9 1 1 3 . W h e n a S i d e a n d t h e H y p o t e n u s e a r e n e a r l y E q u a l 1 6 9 1 1 4 . F o u r C a s e s o f O b l i q u e T r i a n g l e s 1 7 2 1 1 5 . C a s e I . — G i v e n a S i d e a n d T w o A n g l e s 1 7 2 1 1 6 . C a s e I I . — G i v e n T w o S i d e s a n d t h e A n g l e o p p o s i t e O n e o f t h e m , 1 7 3 1 1 7 . C a s e I I I . — G i v e n T w o S i d e s a n d t h e I n c l u d e d A n g l e 1 7 6 1 1 8 . C a s e I V . — G i v e n t h e T h r e e S i d e s 1 7 7 1 1 9 . A r e a o f a T r i a n g l e 1 8 0 1 2 0 . H e i g h t s a n d D i s t a n c e s — D e f i n i t i o n s 1 8 1 1 2 1 . H e i g h t s o f a n A c c e s s i b l e O b j e c t 1 8 2 1 2 2 . H e i g h t a n d D i s t a n c e o f a n I n a c c e s s i b l e O b j e c t 1 8 2 1 2 3 . A n I n a c c e s s i b l e O b j e c t a b o v e a H o r i z o n t a l P l a n e 1 8 4 1 2 4 . O b j e c t o b s e r v e d f r o m T w o P o i n t s i n S a m e V e r t i c a l L i n e . . . . 1 8 5 1 2 5 . D i s t a n c e b e t w e e n T w o I n a c c e s s i b l e O b j e c t s 1 8 6 1 2 6 . T h e D i p o f t h e H o r i z o n 1 8 6 1 2 7 . P r o b l e m o f P o t h e n o t o r o f S n e l l i u s 1 8 8 E x a m p l e s 1 8 9 C H A P T E R V I I I . C o n s t r u c t i o n o f L o g a r i t h m i c a n d T r i g o n o m e t r i c T a b l e s . 1 2 8 . L o g a r i t h m i c a n d T r i g o n o m e t r i c T a b l e s 2 0 4 1 2 9 . E x p o n e n t i a l S e r i e s 2 0 4 1 3 0 . L o g a r i t h m i c S e r i e s 2 0 5 1 3 1 . C o m p u t a t i o n o f L o g a r i t h m s 2 0 7 1 3 2 . S i n 9 a n d t a n f l a r e i n A s c e n d i n g O r d e r o f M a g n i t u d e 2 0 8 1 3 3 . T h e L i m i t o f ^ i s U n i t y 2 0 9 e 1 3 4 . L i m i t i n g V a l u e s o f s i n B a n d c o s 0 2 0 9 1 3 5 . T o c a l c u l a t e t h e S i n e a n d C o s i n e o f 1 0 " a n d o f 1 ' 2 1 1 1 3 6 . T o c o n s t r u c t a T a b l e o f N a t u r a l S i n e s a n d C o s i n e s 2 1 3 1 3 7 . A n o t h e r M e t h o d 2 1 3 1 3 8 . T h e S i n e s a n d C o s i n e s f r o m 3 0 ° t o 0 0 ° 2 1 4 1 3 9 . S i n e s o f A n g l e s G r e a t e r t h a n 4 5 ° 2 1 5 1 4 0 . T a b l e s o f T a n g e n t s a n d S e c a n t s 2 1 5 1 4 1 . F o r m u l a ? o f V e r i f i c a t i o n 2 1 6 1 4 2 . T a b l e s o f L o g a r i t h m i c T r i g o n o m e t r i c F u n c t i o n s 2 1 7 1 4 3 . T h e P r i n c i p l e o f P r o p o r t i o n a l P a r t s 2 1 8 X C O N T E N T S . A R T . P A O I 1 4 4 . T o p r o v e t h e R u l e f o r t h e T a b l e o f C o m m o n L o g a r i t h m s . . . 2 1 8 1 4 5 . T o p r o v e t h e R u l e f o r t h e T a b l e o f N a t u r a l S i n e s 2 1 9 1 4 6 . T o p r o v e t h e R u l e f o r a T a b l e o f N a t u r a l C o s i n e s 2 1 9 1 4 7 . T o p r o v e t h e R u l e f o r a T a b l e o f N a t u r a l T a n g e n t s 2 2 0 1 4 8 . T o p r o v e t h e R u l e f o r a T a b l e o f L o g a r i t h m i c S i n e s 2 2 1 1 4 9 . T o p r o v e t h e R u l e f o r a T a b l e o f L o g a r i t h m i c C o s i n e s 2 2 2 1 5 0 . T o p r o v e t h e R u l e f o r a T a b l e o f L o g a r i t h m i c T a n g e n t s 2 2 2 1 5 1 . C a s e s o f I n a p p l i c a b i l i t y o f R u l e o f P r o p o r t i o n a l P a r t s 2 2 3 1 5 2 . T h r e e M e t h o d s t o r e p l a c e t h e R u l e o f P r o p o r t i o n a l P a r t s . . . 2 2 4 E x a m p l e s 2 2 6 C H A P T E R I X . D e M o i v r e ' s T h e o r e m . — A p p l i c a t i o n s . 1 6 3 . D e M o i v r e ' s T h e o r e m 2 2 9 p 1 5 4 . T o f i n d a l l t h e V a l u e s o f ( c o s O + V -1 s i n 0 ) « 2 3 1 1 5 5 . T o d e v e l o p c o s n 0 a n d s i n n 0 i n P o w e r s o f s i n 8 a n d c o s O 2 3 3 1 5 6 . T o d e v e l o p s i n O a n d c o s 8 i n S e r i e s o f P o w e r s o f O 2 3 4 1 5 7 . C o n v e r g e n c e o f t h e S e r i e s 2 3 5 1 5 8 . E x p a n s i o n o f c o s " 8 i n T e r m s o f C o s i n e s o f M u l t i p l e s o f O . . . 2 3 5 1 5 9 . E x p a n s i o n o f s i n " 0 i n T e r m s o f C o s i n e s o f M u l t i p l e s o f O . . . 2 3 6 1 6 0 . E x p a n s i o n o f s i n " 8 i n T e r m s o f S i n e s o f M u l t i p l e s o f O 2 3 7 1 6 1 . E x p o n e n t i a l V a l u e s o f S i n e a n d C o s i n e 2 3 8 1 6 2 . G r e g o r y ' s S e r i e s 2 3 9 1 6 3 . E u l e r ' s S e r i e s 2 4 0 1 6 4 . M a c h i n ' s S e r i e s 2 4 1 1 6 5 . G i v e n s i n 0 = x s i n ( O + a ) ; e x p a n d O i n P o w e r s o f x 2 4 2 1 6 6 . G i v e n t a n x = n t a n 6 ; e x p a n d x i n P o w e r s o f n 2 4 2 1 6 7 . R e s o l v e x " — 1 i n t o F a c t o r s 2 4 3 1 0 8 . R e s o l v e x " + 1 i n t o F a c t o r s 2 4 4 1 6 9 . R e s o l v e x 2 " — 2 « " c o s O + 1 i n t o F a c t o r s 2 4 5 1 7 0 . D e M o i v r e ' s P r o p e r t y o f t h e C i r c l e 2 4 7 1 7 1 . C o t e ' s P r o p e r t i e s o f t h e C i r c l e 2 4 8 1 7 2 . R e s o l v e s i n O i n t o F a c t o r s 2 4 8 1 7 3 . R e s o l v e c o s f l i n t o F a c t o r s 2 5 0 1 7 4 . S u r a t h e S e r i e s s i n a + s i n ( a + 0 ) + e t c 2 5 1 1 7 5 . S u m t h e S e r i e s c o s a + c o s ( a + 0 ) + e t c 2 5 2 1 7 6 . S u m t h e S e r i e s s i n ™a + s i n r a ( « + 0 ) + e t c 2 5 2 1 7 7 . S u m t h e S e r i e s s i n « — s i n ( « + / 3 ) + e t c 2 5 4 1 7 8 . S u m t h e S e r i e s c o s e c 8 + c o s e c 2 8 + c o s e c 4 8 + e t c 2 5 4 C O N T E N T S . x i A R T . W O " 1 7 9 . S u m t h e S e r i e s t a n 9 + \ t a n - + \ t a n - + e t c 2 5 5 1 8 0 . S u m t h e S e r i e s s i n « + x s i n ( a + 0 ) + e t c 2 5 5 1 8 1 . S u m m a t i o n o f I n f i n i t e S e r i e s 2 5 6 E x a m p l e s 2 5 7 P A R T I I . S P H E R I C A L T R I G O N O M E T R Y . C H A P T E R X . F O R M U L / E R E L A T I V E T O S P H E R I C A L T R I A N G L E S . 1 8 2 . S p h e r i c a l T n i g o n o m e t r y 2 6 7 1 8 3 . G e o m e t r i c P r i n c i p l e s 2 6 7 1 8 4 . F u n d a m e n t a l D e f i n i t i o n s a n d P r o p e r t i e s 2 6 8 1 8 5 . F o r m u l a e f o r R i g h t S p h e r i c a l T r i a n g l e s 2 7 0 1 8 6 . N a p i e r ' s R u l e s 2 7 2 1 8 7 . T h e S p e c i e s o f t h e P a r t s 2 7 3 1 8 8 . A m b i g u o u s S o l u t i o n 2 7 4 1 8 9 . Q u a d r a n t a l T r i a n g l e s 2 7 4 1 9 0 . L a w o f S i n e s 2 7 6 1 9 1 . L a w o f C o s i n e s 2 7 7 1 9 2 . R e l a t i o n b e t w e e n a S i d e a n d t h e T h r e e A n g l e s 2 7 8 1 9 3 . T o f i n d t h e V a l u e o f c o t a s i n 6 , e t c 2 7 9 1 9 4 . U s e f u l F o r m u l a e 2 8 0 1 9 5 . F o r m u l a e f o r t h e H a l f A n g l e s 2 8 1 1 9 6 . F o r m u l a e f o r t h e H a l f S i d e s 2 8 4 1 9 7 . N a p i e r ' s A n a l o g i e s 2 8 6 1 9 8 . D e l a m b r e ' s ( o r G a u s s ' s ) A n a l o g i e s 2 8 7 E x a m p l e s 2 8 8 C H A P T E R X I . S o l u t i o n o f S p h e r i c a l T r i a n g l e s . 1 9 9 . P r e l i m i n a r y O b s e r v a t i o n s 2 9 7 2 0 0 . S o l u t i o n o f R i g h t S p h e r i c a l T r i a n g l e s 2 9 7 x i i C O N T E N T S . A R T . F A O E 2 0 1 . C a s e I . — G i v e n t h e H y p o t e n u s e a n d a n A n g l e 2 9 8 2 0 2 . C a s e I I . — G i v e n t h e H y p o t e n u s e a n d a S i d e 2 9 9 2 0 3 . C a s e I I I . — G i v e n a S i d e a n d t h e A d j a c e n t A n g l e ; i 0 0 2 0 4 . C a s e I V . — G i v e n a S i d e a n d t h e O p p o s i t e A n g l e 3 0 1 2 0 5 . C a s e V . — G i v e n t h e T w o S i d e s 3 0 2 2 0 6 . C a s e V I . — G i v e n t h e T w o A n g l e s 3 0 2 2 0 " . Q u a d r a n t a l a n d I s o s c e l e s T r i a n g l e s 3 0 3 2 0 8 . S o l u t i o n o f O b l i q u e S p h e r i c a l T r i a n g l e s 3 0 4 2 0 9 . C a s e I . — G i v e n T w o S i d e s a n d t h e I n c l u d e d A n g l e 3 0 5 2 1 0 . C a s e I I . — G i v e n T w o A n g l e s a n d t h e I n c l u d e d S i d e 3 0 7 2 1 1 . C a s e I I I . — G i v e n T w o S i d e s a n d O n e O p p o s i t e A n g l e 3 0 9 2 1 2 . C a s e I V . — G i v e n T w o A n g l e s a n d O n e O p p o s i t e S i d e 3 1 2 2 1 3 . C a s e V . — G i v e n t h e T h r e e S i d e s 3 1 3 2 1 4 . C a s e V I . — G i v e n t h e T h r e e A n g l e s 3 1 4 E x a m p l e s 3 1 6 C H A P T E R X I I . T h e I n - C i r c x e s a n d E x - C i r c l e s . — A r e a s . 2 1 5 . T h e I n s c r i b e d C i r c l e 3 2 4 2 1 6 . T h e E s c r i b e d C i r c l e s 3 2 5 2 1 7 . T h e C i r c u m s c r i b e d C i r c l e 3 2 6 2 1 8 . C i r c u m c i r c l e s o f C o l u n a r T r i a n g l e s 3 2 8 2 1 9 . A r e a s o f T r i a n g l e s . — G i v e n t h e T h r e e A n g l e s 3 2 9 2 2 0 . A r e a s o f T r i a n g l e s . — G i v e n t h e T h r e e S i d e s 3 3 0 2 2 1 . A r e a s o f T r i a n g l e s . — G i v e n T w o S i d e s a n d t h e I n c l u d e d A n g l e , 3 3 1 E x a m p l e s 3 3 2 C H A P T E R X I I I . A p p l i c a t i o n s o f S p h e r i c a l T r i g o n o m e t r y . 2 2 2 . A s t r o n o m i c a l D e f i n i t i o n s 3 3 8 2 2 3 . S p h e r i c a l C o o r d i n a t e s 3 3 9 2 2 4 . G r a p h i c R e p r e s e n t a t i o n o f t h e S p h e r i c a l C o o r d i n a t e s 3 4 1 2 2 5 . P r o b l e m s 3 4 2 2 2 6 . T h e C h o r d a l T r i a n g l e . 3 4 6 2 2 7 . L e g e n d r e ' s T h e o r e m ' . 3 4 8 2 2 8 . R o y ' s R u l e 3 5 0 2 2 9 . R e d u c t i o n o f a n A n g l e t o t h e H o r i z o n 3 5 2 C O N T E N T S . x i i i A R T . P A G E 2 3 0 . S m a l l V a r i a t i o n s i n P a r t s o f a S p h e r i c a l T r i a n g l e 3 5 3 2 3 1 . I n c l i n a t i o n o f A d j a c e n t F a c e s o f P o l y e d r o n s 3 5 6 2 3 2 . V o l u m e o f P a r a l l e l o p i p e d : 5 5 7 2 3 3 . D i a g o n a l o f a P a r a l l e l o p i p e d 3 5 8 2 3 4 . T a b l e o f F o r m u l a e i n S p h e r i c a l T r i g o n o m e t r y 3 5 9 E x a m p l e s 3 6 2 T R E A T I S E O N T R I G O M M E T ^ . : P A R T I . P L A N E T R I G O N O M E T R Y . C H A P T E R I . M E A S U E E M E N T 0 1 A N G L E S . 1 . T r i g o n o m e t r y i s t h a t b r a n c h o f m a t h e m a t i c s w h i c h t r e a t s ( 1 ) o f t h e s o l u t i o n o f p l a n e a n d s p h e r i c a l t r i a n g l e s , a n d ( 2 ) o f t h e g e n e r a l r e l a t i o n s o f a n g l e s a n d c e r t a i n f u n c t i o n s o f t h e m c a l l e d t h e t r i g o n o m e t r i c f u n c t i o n s . P l a n e T r i g o n o m e t r y c o m p r i s e s t h e s o l u t i o n o f p l a n e t r i a n g l e s a n d i n v e s t i g a t i o n s o f p l a n e a n g l e s a n d t h e i r f u n c t i o n s . T r i g o n o m e t r y w a s o r i g i n a l l y t h e s c i e n c e w h i c h t r e a t e d o n l y o f t h e s i d e s a n d a n g l e s o f p l a n e a n d s p h e r i c a l t r i a n g l e s ; h u t i t h a s b e e n r e c e n t l y e x t e n d e d s o a s t o i n c l u d e t h e a n a l y t i c t r e a t m e n t o f a l l t h e o r e m s i n v o l v i n g t h e c o n s i d e r a t i o n o f a n g u l a r m a g n i t u d e s . 2 . T h e M e a s u r e o f a Q u a n t i t y . — A l l m e a s u r e m e n t s o f l i n e s , a n g l e s , e t c . , a r e m a d e i n t e r m s o f s o m e f i x e d s t a n d a r d o r u n i t , a n d t h e m e a s u r e o f a q u a n t i t y i s t h e n u m b e r o f t i m e s t h e q u a n t i t y c o n t a i n s t h e u n i t . T t i s e v i d e n t t h a t t h e s a m e q u a n t i t y w i l l b e r e p r e s e n t e d b y d i f f e r e n t n u m b e r s w h e n d i f f e r e n t u n i t s a r e a d o p t e d . F o r e x a m p l e , t h e d i s t a n c e o f a m i l e w i l l b e r e p r e s e n t e d b y t h e n u m b e r 1 w h e n a m i l e i s t h e u n i t o f l e n g t h , b y t h e n u m b e r 1 7 6 0 w h e n a y a r d i s t h e u n i t o f l e n g t h , b y t h e n u m b e r 5 2 8 0 w h e n a f o o t i s t h e u n i t o f l e n g t h , a n d s o o n . I n l i k e m a n -1 2 P L A N E T R I G O N O M E T R Y . " j i p j G t h e n U n l l l e i : E x p r e s s i n g t h e m a g n i t u d e o f a n a n g l e w i l l . . d e p e n d . o n . t h e u n i . t o f a n g l e . , . v -; , --. / --» --t , E X A M P L E S . 1 . W h a t i s t h e m e a s u r e o f 2 £ m i l e s w h e n a y a r d i s t h e u n i t ? 2 $ m i l e s = f X 1 7 6 0 y a r d s = 4 4 0 0 y a r d s = 4 4 0 0 x 1 y a r d . . - . t h e m e a s u r e i s 4 4 0 0 w h e n a y a r d i s t h e u n i t . 2 . W h a t i s t h e m e a s u r e o f a m i l e w h e n a c h a i n o f 6 6 f e e t i s t h e u n i t ? A n s . 8 0 . 3 . W h a t i s t h e m e a s u r e o f 2 a c r e s w h e n a s q u a r e w h o s e s i d e i s 2 2 y a r d s i s t h e u n i t ? A n s . 2 0 . 4 . T h e m e a s u r e o f a c e r t a i n f i e l d i s 4 4 a n d t h e u n i t i s 1 1 0 0 s q u a r e y a r d s ; e x p r e s s t h e a r e a o f t h e f i e l d i n a c r e s . A n s . 1 0 a c r e s . 5 . I f 7 i n c h e s b e t a k e n a s t h e u n i t o f l e n g t h , b y w h a t n u m b e r w i l l 1 5 f e e t 2 i n c h e s b e r e p r e s e n t e d ? A n s . 2 6 . 6 . I f 1 9 2 s q u a r e i n c h e s b e r e p r e s e n t e d b y t h e n u m b e r 1 2 , w h a t i s t h e u n i t o f l i n e a r m e a s u r e m e n t ? A n s . 4 i n c h e s . 3 . A n g l e s . — A n a n g l e i s t h e o p e n i n g b e t w e e n t w o s t r a i g h t l i n e s d r a w n f r o m t h e s a m e p o i n t . T h e p o i n t i s c a l l e d t h e v e r t e x o f t h e a n g l e , a n d t h e s t r a i g h t l i n e s a r e c a l l e d t h e s i d e s o f t h e a n g l e . A n a n g l e m a y b e g e n e r a t e d b y r e v o l v i n g a l i n e f r o m c o i n c i d e n c e w i t h a n o t h e r l i n e a b o u t a f i x e d p o i n t . T h e i n i t i a l a n d f i n a l p o s i t i o n s o f t h e l i n e a r e t h e s i d e s o f t h e a n g l e ; t h e a m o u n t o f r e v o l u t i o n m e a s u r e s t h e m a g n i t u d e o f t h e a n g l e ; a n d t h e a n g l e m a y b e t r a c e d o u t b y a n y n u m b e r o f r e v o l u t i o n s o f t h e l i n e . O A T h u s , t o f o r m t h e a n g l e A O B , O B m a y b e s u p p o s e d t o h a v e r e v o l v e d f r o m O A t o O B ; a n d i t i s o b v i o u s t h a t O B T H E C I R C U L A R M E A S U R E . 3 m a y g o o n r e v o l v i n g u n t i l i t c o m e s i n t o t h e s a m e p o s i t i o n O B a s m a n y t i m e s a s w e p l e a s e ; t h e a n g l e A O B , h a v i n g t h e s a m e b o u n d i n g l i n e s O A a n d O B , m a y t h e r e f o r e b e g r e a t e r t h a n 2 , 4 , 8 , o r a n y n i l m b e r o f r i g h t a n g l e s . T h e l i n e O A f r o m w h i c h O B m o v e s i s c a l l e d t h e i n i t i a l l i n e , a n d O B i n i t s f i n a l p o s i t i o n , t h e t e r m i n a l l i n e . T h e r e v o l v i n g l i n e O B i s c a l l e d t h e g e n e r a t r i x . T h e p o i n t O i s c a l l e d t h e o r i g i n , v e r t e x , o r p o l e . 4 . P o s i t i v e a n d N e g a t i v e A n g l e s . — W e s u p p o s e d i n A r t . 3 t h a t O B r e v o l v e d i n t h e d i r e c t i o n o p p o s i t e t o t h a t o f t h e h a n d s o f a w a t c h . B u t a n g l e s m a y , o f c o u r s e , b e d e s c r i b e d b y a l i n e r e v o l v i n g i n t h e s a m e d i r e c t i o n a s t h e h a n d s o f a w a t c h , a n d i t i s o f t e n n e c e s s a r y t o d i s t i n g u i s h b e t w e e n t h e t w o d i r e c t i o n s i n w h i c h a n g l e s m a y b e m e a s u r e d f r o m t h e s a m e f i x e d l i n e . T h i s i s c o n v e n i e n t l y e f f e c t e d b y a d o p t i n g t h e c o n v e n t i o n t h a t a n g l e s m e a s u r e d i n o n e d i r e c t i o n s h a l l b e c o n s i d e r e d p o s i t i v e , a n d a n g l e s m e a s u r e d i n t h e o p p o s i t e d i r e c t i o n , n e g a t i v e . I n a l l b r a n c h e s o f m a t h e m a t i c s a n g l e s d e s c r i b e d b y t h e r e v o l u t i o n o f a s t r a i g h t l i n e i n t h e d i r e c t i o n o p p o s i t e t o t h a t i n w h i c h t h e h a n d s o f a w a t c h m o v e a r e u s u a l l y c o n s i d e r e d p o s i t i v e , a n d a l l a n g l e s d e s c r i b e d b y t h e r e v o l u t i o n o f a s t r a i g h t l i n e i n t h e s a m e d i r e c t i o n a s t h e h a n d s o f a w a t c h m o v e a r e c o n s i d e r e d n e g a t i v e . T h u s , t h e r e v o l v i n g l i n e O B s t a r t s f r o m t h e i n i t i a l l i n e O A . W h e n i t r e v o l v e s i n t h e d i r e c t i o n c o n t r a r y t o t h a t o f t h e h a n d s o f a w a t c h , a n d c o m e s i n t o t h e p o s i t i o n O B , i t t r a c e s o u t t h e p o s i t i v e a n g l e A O B ( m a r k e d + a ) ; a n d w h e n i t r e v o l v e s i n t h e s a m e d i r e c t i o n a s t h e h a n d s o f a w a t c h , i t t r a c e s t h e n e g a t i v e a n g l e A O B ( m a r k e d — b ) . T h e r e v o l v i n g l i n e i s a l w a y s c o n s i d e r e d n e g a t i v e . 5 . T h e M e a s u r e o f A n g l e s . — A n a n g l e i s m e a s u r e d b y t h e a r c o f a c i r c l e w h o s e c e n t r e i s a t t h e v e r t e x o f t h e 4 P L A N E T R I G O N O M E T R Y . a n g l e a n d w h o s e e n d s a r e o n t h e s i d e s o f t h e a n g l e ( G e o m . , A r t . 2 3 6 ) . L e t t h e l i n e O P o f f i x e d l e n g t h g e n e r a t e a n a n g l e b y r e v o l v i n g i n t h e p o s i t i v e d i r e c t i o n r o u n d a f i x e d p o i n t O f r o m a n i n i t i a l p o s i t i o n O A . S i n c e O P i s o f c o n s t a n t l e n g t h , t h e p o i n t P w i l l t r a c e o u t t h e c i r c u m f e r e n c e A B A ' B ' w h o s e c e n t r e A ' | -i s O . T h e t w o p e r p e n d i c u l a r d i a m e t e r s A A ' a n d B B ' o f t h i s c i r c l e w i l l i n c l o s e t h e f o u r r i g h t a n g l e s A O B , B O A ' , A ' O B ' , a n d B ' O A . T h e c i r c u m f e r e n c e i s d i v i d e d a t t h e p o i n t s A , B , A ' , B ' i n t o f o u r q u a d r a n t s , o f w h i c h A B i s c a l l e d t h e f i r s t q u a d r a n t . B A ' " " " s e c o n d q u a d r a n t . A ' B ' " " " t h i r d q u a d r a n t . " " " f o u r t h q u a d r a n t . I n t h e f i g u r e , t h e a n g l e A O P u b e t w e e n t h e i n i t i a l l i n e O A a n d t h e r e v o l v i n g l i n e O P ^ i s l e s s t h a n a r i g h t a n g l e , a n d i s s a i d t o b e a n a n g l e i n t h e f i r s t q u a d r a n t . A O P 2 i s g r e a t e r t h a n o n e a n d l e s s t h a n t w o r i g h t a n g l e s , a n d i s s a i d t o b e a n a n g l e i n t h e s e c o n d q u a d r a n t . A O P 3 i s g r e a t e r t h a n t w o a n d l e s s t h a n t h r e e r i g h t a n g l e s , a n d i s s a i d t o b e a n a n g l e i n t h e t h i r d q u a d r a n t . A O P 4 i s g r e a t e r t h a n t h r e e a n d l e s s t h a n f o u r r i g h t a n g l e s , a n d i s s a i d t o b e a n a n g l e i n t h e f o u r t h q u a d r a n t . W h e n t h e r e v o l v i n g l i n e r e t u r n s t o t h e i n i t i a l p o s i t i o n O A , t h e a n g l e A O A i s a n a n g l e o f f o u r r i g h t a n g l e s . B y s u p p o s i n g O P t o c o n t i n u e r e v o l v i n g , t h e a n g l e d e s c r i b e d w i l l b e c o m e g r e a t e r t h a n a n a n g l e o f f o u r r i g h t a n g l e s . T h u s , w h e n O P c o i n c i d e s w i t h t h e l i n e s O B , O A ' , O B ' , O A , i n t h e s e c o n d r e v o l u t i o n , t h e a n g l e s d e s c r i b e d , m e a s u r e d f r o m t h e b e g i n n i n g o f t h e f i r s t r e v o l u t i o n , a r e a n g l e s o f f i v e r i g h t a n g l e s , s i x r i g h t a n g l e s , s e v e n r i g h t a n g l e s , e i g h t r i g h t T I I E S E X A G E S I M A L M E T I I O T ) . 5 a n g l e s , r e s p e c t i v e l y , a n d s o o n . B y t h e c o n t i n u e d r e v o l u t i o n o f O P t h e a n g l e b e t w e e n t h e i n i t i a l l i n e O A a n d t h e r e v o l v i n g l i n e O P m a y b e c o m e o f a n y m a g n i t u d e w h a t e v e r . I n t h e s a m e w a y O P m a y r e v o l v e i n t h e n e g a t i v e d i r e c t i o n a b o u t O a n y n u m b e r o f t i m e s , g e n e r a t i n g a n e g a t i v e a n g l e ; a n d t h i s n e g a t i v e a n g l e m a y o b v i o u s l y h a v e a n y m a g n i t u d e w h a t e v e r . T h e a n g l e A O P m a y b e t h e g e o m e t r i c r e p r e s e n t a t i v e o f a n y o f t h e T r i g o n o m e t r i c a n g l e s f o r m e d b y a n y n u m b e r o f c o m p l e t e r e v o l u t i o n s , e i t h e r i n t h e p o s i t i v e d i r e c t i o n a d d e d t o t h e p o s i t i v e a n g l e A O P , o r i n t h e n e g a t i v e d i r e c t i o n a d d e d t o t h e n e g a t i v e a n g l e A O P . I n a l l c a s e s t h e a n g l e i s s a i d t o b e i n t h e q u a d r a n t i n d i c a t e d b y i t s t e r m i n a l l i n e . T h e r e a r e t h r e e m e t h o d s o f m e a s u r i n g a n g l e s , c a l l e d r e s p e c t i v e l y t h e S e x a g e s i m a l , t h e C e n t e s i m a l , a n d t h e C i r c u l a r m e t h o d s . 6 . T h e S e x a g e s i m a l M e t h o d . — T h i s i s t h e m e t h o d i n g e n e r a l u s e . I n t h i s m e t h o d t h e r i g h t a n g l e i s d i v i d e d i n t o 9 0 e q u a l p a r t s , e a c h o f w h i c h i s c a l l e d a d e g r e e . E a c h d e g r e e i s s u b d i v i d e d i n t o 6 0 e q u a l p a r t s , e a c h o f w h i c h i s c a l l e d a m i n u t e . E a c h m i n u t e i s s u b d i v i d e d i n t o 6 0 e q u a l p a r t s , e a c h o f w h i c h i s c a l l e d a s e c o n d . T h e n t h e m a g n i t u d e o f a n a n g l e i s e x p r e s s e d b y t h e n u m b e r o f d e g r e e s , m i n u t e s , a n d s e c o n d s w h i c h i t c o n t a i n s . D e g r e e s , m i n u t e s , a n d s e c o n d s a r e d e n o t e d r e s p e c t i v e l y b y t h e s y m b o l s ° , ' , " : t h u s , t o r e p r e s e n t 1 8 d e g r e e s , 6 m i n u t e s , 3 4 . 5 8 s e c o n d s , w e w r i t e 1 8 ° 6 ' 3 4 " . 5 8 . A d e g r e e o f a r c i s o f t h e c i r c u m f e r e n c e t o w h i c h t h e a r c b e l o n g s . T h e d e g r e e o f a r c i s s u b d i v i d e d i n t h e s a m e m a n n e r a s t h e d e g r e e o f a n g l e . T h e n 1 c i r c u m f e r e n c e = 3 6 0 ° = 2 1 6 0 0 ' = 1 2 9 6 0 0 0 " . 1 q u a d r a n t o r r i g h t a n g l e = 9 0 ° . I n s t r u m e n t s u s e d f o r m e a s u r i n g a n g l e s a r e s u b d i v i d e d a c c o r d i n g l y . 6 P L A N E T R I G O N O M E T R Y . 7 . T h e C e n t e s i m a l o r D e c i m a l M e t h o d . — I n t h i s m e t h o d t h e r i g h t a n g l e i s d i v i d e d i n t o 1 0 0 e q u a l p a r t s , e a c h o f w h i c h i s c a l l e d a g r a d e . E a c h g r a d e i s s u b d i v i d e d i n t o 1 0 0 e q u a l p a r t s , e a c h o f w h i c h i s c a l l e d a m i n u t e . E a c h m i n u t e i s s u b d i v i d e d i n t o 1 0 0 e q u a l p a r t s , e a c h o f w h i c h i s c a l l e d a s e c o n d . T h e m a g n i t u d e o f a n a n g l e i s t h e n e x p r e s s e d b y t h e n u m b e r o f g r a d e s , m i n u t e s , a n d s e c o n d s w h i c h i t c o n t a i n s . G r a d e s , m i n u t e s , a n d s e c o n d s a r e d e n o t e d r e s p e c t i v e l y b y t h e s y m b o l s g , x , v v : t h u s , t o r e p r e s e n t 3 4 g r a d e s , 4 8 m i n u t e s , 8 6 . 4 7 s e c o n d s , w e w r i t e 3 4 4 8 v 8 6 v \ 4 7 . T h e c e n t e s i m a l o r d e c i m a l m e t h o d w a s p r o p o s e d b y t h e F r e n c h m a t h e m a t i c i a n s i n t h e b e g i n n i n g o f t h e p r e s e n t c e n t u r y . B u t a l t h o u g h i t p o s s e s s e s m a n y a d v a n t a g e s o v e r t h e e s t a b l i s h e d m e t h o d , t h e y w e r e n o t c o n s i d e r e d s u f f i c i e n t t o c o u n t e r b a l a n c e t h e e n o r m o u s l a b o r w h i c h w o u l d h a v e b e e n n e c e s s a r y t o r e a r r a n g e a l l t h e m a t h e m a t i c a l t a b l e s , b o o k s o f r e f e r e n c e , a n d r e c o r d s o f o b s e r v a t i o n s , w h i c h w o u l d h a v e t o b e t r a n s f e r r e d i n t o t h e d e c i m a l s y s t e m b e f o r e i t s a d v a n t a g e s c o u l d b e f e l t . T h u s , t h e c e n t e s i m a l m e t h o d h a s n e v e r b e e n u s e d e v e n i n F r a n c e , a n d i n a l l p r o b a b i l i t y n e v e r w i l l b e u s e d i n p r a c t i c a l w o r k . 8 . T h e C i r c u l a r M e a s u r e . — T h e u n i t o f B c i r c u l a r m e a s u r e i s t h e a n g l e s u b t e n d e d a t t h e c e n t r e o f a c i r c l e b y a n a r c e q u a l i n l e n g t h t o t h e r a d i u s . T h i s u n i t o f c i r c u l a r m e a s u r e i s c a l l e d a r a d i a n . L e t O b e t h e c e n t r e o f a c i r c l e w h o s e r a d i u s i s r . L e t t h e a r c A B b e e q u a l t o t h e r a d i u s O A = r . T h e n , s i n c e a n g l e s a t t h e c e n t r e o f a c i r c l e a r e i n t h e s a m e r a t i o a s t h e i r i n t e r c e p t e d a r c s ( G e o m . , A r t . 2 3 4 ) , a n d s i n c e t h e r a t i o o f t h e c i r c u m f e r e n c e o f a c i r c l e t o i t s d i a m e t e r i s i r = 3 . 1 4 1 5 9 2 6 5 ( G e o m . , A r t . 4 3 6 ) , T H E C I R C U L A R M E A S U R E . 7 . - . a n g l e A O B : 4 r t . a n g l e s : : a r c A B : c i r c u m f e r e n c e , : : r : 2 - r r r : : l : 2 i r . , . a n g l e A O B = 4 r t a n g l e s = 2 r t a " g l e s -. : a r a d i a n = a n g l e A O B = — 8 ° ° - = 5 7 ° . 2 9 5 7 7 9 5 O . 1 4 1 5 9 z o 5 = 3 4 3 7 ' . 7 4 6 7 7 = 2 0 6 2 6 4 " . 8 0 6 . T h e r e f o r e , t h e r a d i a n i s t h e s a m e f o r a l l c i r c l e s , a n d = 5 7 ° . 2 9 5 7 7 9 o . L e t A B P b e a n y c i r c l e ; l e t t h e a n g l e A O B b e t h e r a d i a n ; a n d l e t A O P b e a n y o t h e r a n g l e . T h e n a r c A B = r a d i u s O A . . - . a n g l e A O P : a n g l e A O B : : a r c A P : a r c A B ; o r a n g l e A O P : r a d i a n : : a r c A P : r a d i u s . . - . a n g l e A O P = a r c ^ x r a d i a n , r a d i u s T h e m e a s u r e o f a n y q u a n t i t y i s t h e n u m b e r o f t i m e s i t c o n t a i n s t h e u n i t o f m e a s u r e ( A r t . 2 ) . . - . t h e c i r c u l a r m e a s u r e o f a n g l e A O P = a i c — _ . r a d i u s N o t e 1 . — T h e s t u d e n t w i l l n o t i c e t h a t a r a d i a n i s a l i t t l e l e s s t h a n a n a n g l e o f a n e q u i l a t e r a l t r i a n g l e , i . e . , o f 6 0 ° . A n g l e s e x p r e s s e d i n c i r c u l a r m e a s u r e a r e u s u a l l y d e n o t e d h y G r e e k l e t t e r s , a , / 3 , y , < f , , 9 , T h e c i r c u l a r m e a s u r e i s e m p l o y e d i n t h e v a r i o u s b r a n c h e s o f A n a l y t i c a l M a t h e m a t i c s , i n w h i c h t h e a n g l e u n d e r c o n s i d e r a t i o n i s a l m o s t a l w a y s e x p r e s s e d h y a l e t t e r . N o t e 2 . — T h e s t u d e n t c a n n o t t o o c a r e f u l l y n o t i c e t h a t u n l e s s a n a n g l e i s o b v i o u s l y r e f e r r e d t o , t h e l e t t e r s a , 3 9 , 4 > , . . . s t a n d f o r m e r e n u m b e r s . T h u s , n - s t a n d s f o r a n u m b e r , a n d a n u m b e r o n l y , v i z . , 3 . 1 4 1 5 9 b u t i n t h e e x p r e s s i o n 1 t h e a n g l e i r , ' t h a t i s , ' t h e a n g l e 3 . 1 4 1 5 9 t h e r e m u s t b e s o m e u n i t u n d e r s t o o d . T h e u n i t u n d e r s t o o d h e r e l s a r a d i a n , a n d t h e r e f o r e ' t h e a n g l e t t ' s t a n d s f o r ' i t r a d i a n s ' o r 3 . 1 4 1 5 9 . . . r a d i a n s , t h a t i s , t w o r i g h t a n g l e s . H e n c e , w h e n a n a n g l e i s r e f e r r e d t o , i r i s a v e r y c o n v e n i e n t a b b r e v i a t i o n f o r t w o r i g h t a n g l e s . 8 0 a l s o ' t h e a n g l e a o r & ' m e a n s 1 a r a d i a n s o r 9 r a d i a n s . ' T h e u n i t s i n t h e t h r e e s y s t e m s , w h e n e x p r e s s e d i n t e r m s o f o n e c o m m o n s t a n d a r d , t w o r i g h t a n g l e s , s t a n d t h u s : 8 P L A N E T R I G O N O M E T R Y . T h e u n i t i n t h e S e x a g e s i m a l M e t h o d = o f 2 r i g h t a n g l e s . " " " " C e n t e s i m a l " = — " " " " 2 0 0 « « « « C i r c u l a r " = -" " " " T T I f D , G , a n d 6 d e n o t e t h e n u m b e r o f d e g r e e s , g r a d e s , a a d r a d i a n s r e s p e c t i v e l y i n a n y a n g l e , t h e n D _ = G - = ( 1 ) 1 8 0 2 0 0 t t , v ' b e c a u s e e a c h f r a c t i o n i s t h e r a t i o o f t h e a n g l e t o t w o r i g h t a n g l e s . 9 . C o m p a r i s o n o f t h e S e x a g e s i m a l a n d C e n t e s i m a l M e a s u r e s o f a n A n g l e . — A l t h o u g h t h e c e n t e s i m a l m e t h o d w a s n e v e r i n g e n e r a l u s e a m o n g m a t h e m a t i c i a n s , a n d i s n o w t o t a l l y a b a n d o n e d e v e r y w h e r e , y e t i t s t i l l p o s s e s s e s s o m e i n t e r e s t , a s i t s h o w s t h e a p p l i c a t i o n o f t h e d e c i m a l s y s t e m t o t h e m e a s u r e m e n t o f a n g l e s . F r o m ( 1 ) o f A r t . 8 w e h a v e D = 1 8 0 2 0 0 ' . - . D = — G , a n d G = — D . 1 0 9 E X A M P L E S . 1 . E x p r e s s 4 9 ° 1 5 ' 3 5 " i n c e n t e s i m a l m e a s u r e . F i r s t e x p r e s s t h e a n g l e i n d e g r e e s a n d d e c i m a l s o f a d e g r e e t h u s : 6 0 ) 3 5 " 6 0 ) 1 5 ' . 5 8 3 4 9 ° . 2 5 9 7 2 " 1 0 9 ) 4 9 2 . 5 9 7 2 5 4 g . 7 3 3 0 2 4 - . . . . - . 4 9 ° 1 5 ' 3 5 " = 5 4 7 3 v 3 0 v \ 2 4 . . . . C O M P A R I S O N O F M E A S U R E S . 9 2 . E x p r e s s 8 7 2 v 2 5 u i n d e g r e e s , e t c . F i r s t e x p r e s s t h e a n g l e i n g r a d e s a n d d e c i m a l s o f a g r a d e t h u s : 8 7 2 v 2 5 v v = 8 7 0 2 2 5 . 9 7 8 . 3 2 0 2 5 6 0 1 9 . 2 1 5 6 0 1 2 . 9 . - . 8 7 2 v 2 5 " = 7 8 ° 1 9 ' 1 2 " . 9 . F i n d t h e n u m b e r o f g r a d e s , m i n u t e s , a n d s e c o n d s i n t h e f o l l o w i n g a n g l e s : F i n d t h e n u m b e r o f d e g r e e s , m i n u t e s , a n d s e c o n d s i n t h e f o l l o w i n g a n g l e s : 1 0 . C o m p a r i s o n o f t h e S e x a g e s i m a l a n d C i r c u l a r M e a s u r e s o f a n A n g l e . F r o m ( 1 ) o f A r t . 8 w e h a v e J D = 6 1 8 0 x ' 3 . ' 5 1 ° 4 ' 3 0 " . 4 . 4 5 ° 3 3 ' 3 " . 5 . 2 7 ° 1 5 ' 4 6 " . 6 . 1 5 7 ° 4 ' 9 " . A n s . 5 6 7 5 v 0 v \ 5 0 6 1 v 2 0 v \ 3 7 . 3 0 2 9 v 1 9 v \ 7 5 - - - . 1 7 4 5 2 v 1 2 v \ 9 6 2 - . . 7 . 1 9 4 5 x 9 5 v \ 8 . 1 2 4 5 v 8 v \ 9 . 5 5 1 8 v 3 5 v \ A n s . 1 7 ° 3 0 ' 4 8 " . 7 8 . 1 1 1 ° 3 8 ' 4 4 " . 5 9 2 . 4 9 ° 3 9 ' 5 4 " . 5 4 . . - . D = 1 8 0 0 i T 1 0 P L A N E T R I G O N O M E T R Y . E X A M P L E S . 1 . F i n d t h e n u m b e r o f d e g r e e s i n t h e a n g l e w h o s e c i r c u l a r m e a s u r e i s H e r e 6 = \ -. . . D = 1 8 ° X l = 9 - 5 I T 2 i T = ? ^ - I = 2 8 0 3 8 ' 1 0 " } i , w h e r e ^ i s u s e d f o r i r . 2 . F i n d t h e c i r c u l a r m e a s u r e o f t h e a n g l e 5 9 ° 5 2 ' 3 0 " . E x p r e s s t h e a n g l e i n d e g r e e s a n d d e c i m a l s o f a d e g r e e t h u s : 6 0 ) 5 2 . 5 5 9 . 8 7 5 . - . 6 = s ^ | x s w = ( . 3 3 3 - ) i t = 1 . 0 4 5 3 - . 3 . E x p r e s s , i n d e g r e e s , t h e a n g l e s w h o s e c i r c u l a r m e a s u r e s i T i T i T T T o a r e - , - , - , - , i r . 2 , 3 4 , 6 , " N o t e 1 . — T h e s t u d e n t s h o u l d e s p e c i a l l y a c c u s t o m h i m s e l f t o e x p r e s s r e a d i l y i n c i r c u l a r m e a s u r e a n a n g l e w h i c h i s g i v e n i n d e g r e e s . 4 . E x p r e s s i n c i r c u l a r m e a s u r e t h e f o l l o w i n g a n g l e s : 6 0 ° , 2 2 ° 3 0 ' , 1 1 ° 1 5 ' , 2 7 0 ° . A n s . - , — , ' ' ' 3 8 1 6 2 5 . E x p r e s s i n c i r c u l a r m e a s u r e 3 ° 1 2 ' , a n d f i n d t o s e c o n d s t h e a n g l e w h o s e c i r c u l a r m e a s u r e i s . 8 . ( T a k e , r = — \ A n s . — , 4 5 ° 4 9 ' 5 " A -V 7 J 2 2 5 1 1 6 . O n e a n g l e o f a t r i a n g l e i s 4 5 ° , a n d t h e c i r c u l a r m e a s u r e o f a n o t h e r i s 1 . 5 . F i n d t h e t h i r d a n g l e i n d e g r e e s . A n s . 4 9 ° 5 ' 2 7 " T V N o t e 2 . — Q u e s t i o n s i n w h i c h a n g l e s a r e e x p r e s s e d i n d i f f e r e n t s y s t e i u s o f m e a s u r e m e n t a r e e a s i l y s o l v e d b y e x p r e s s i n g e a c h a n g l e i n r i g h t a n g l e s . G E N E R A L M E A S U R E O F A N A N G L E . 1 1 7 . T h e s u m o f t h e m e a s u r e o f a n a n g l e i n d e g r e e s a n d t w i c e i t s m e a s u r e i n r a d i a n s i s 2 3 f ; f i n d i t s m e a s u r e i n d e g r e e s ( t t = t y ) . L e t t h e a n g l e c o n t a i n x r i g h t a n g l e s . T h e n t h e m e a s u r e o f t h e a n g l e i n d e g r e e s = 9 0 x . " " " " " " " r a d i a n s = x . 2 . - . 9 0 a ; + i r x = 2 3 £ ; . - . 9 0 + 2 ^ x = 5 y s . - . 6 5 2 a ; = 1 6 3 , . - . x = - -4 . - . t h e a n g l e i s \ o f 9 0 ° = 2 2 £ ° . 8 . T h e d i f f e r e n c e b e t w e e n t w o a n g l e s i s - , a n d t h e i r s u m i s 5 6 ° ; f i n d t h e a n g l e s i n d e g r e e s . A n s . 3 8 ° , 1 8 ° . 1 1 . G e n e r a l M e a s u r e o f a n A n g l e . — I n E u c l i d i a n g e o m e t r y a n d i n p r a c t i c a l a p p l i c a t i o n s o f t r i g o n o m e t r y , a n g l e s a r e g e n e r a l l y c o n s i d e r e d t o b e l e s s t h a n t w o r i g h t a n g l e s ; b u t i n t h e t h e o r e t i c a l p a r t s o f m a t h e m a t i c s , a n g l e s a r e t r e a t e d a s q u a n t i t i e s w h i c h m a y b e o f a n y m a g n i t u d e w h a t e v e r . T h u s , w h e n w e a r e t o l d t h a t a n a n g l e i s i n s o m e p a r t i c u l a r q u a d r a n t , s a y t h e s e c o n d ( A r t . 5 ) , w e k n o w t h a t t h e p o s i t i o n i n w h i c h t h e r e v o l v i n g l i n e s t o p s i s i n t h e s e c o n d q u a d r a n t . B u t t h e r e i s a n u n l i m i t e d n u m b e r o f a n g l e s h a v i n g t h e s a m e f i n a l p o s i t i o n , O P . T h e r e v o l v i n g l i n e O F m a y p a s s f r o m B O A t o O P , n o t o n l y b y d e s c r i b i n g t h e a r c A B P , b u t b y m o v i n g t h r o u g h a w h o l e r e v o l u t i o n p l u s t h e a r c A B P , o r t h r o u g h a n y n u m b e r o f r e v o l u t i o n s p l u s t h e a r c A B P . F o r e x a m p l e , t h e f i n a l p o s i t i o n o f O P m a y r e p r e s e n t g e o m e t r i c a l l y a l l t h e f o l l o w i n g a n g l e s : 1 2 P L A N E T R I G O N O M E T R Y . A n g l e A O P = 1 3 0 ° , o r 3 6 0 ° + 1 3 0 ° , o r 7 2 0 ° + 1 3 0 ° , o r -3 6 0 ° + 1 3 0 ° , o r -7 2 0 ° 4 - 1 3 0 ° , e t c . L e t A b e a n a n g l e b e t w e e n 0 a n d 9 0 ° , a n d l e t n b e a n y w h o l e n u m b e r , p o s i t i v e o r n e g a t i v e . T h e n ( 1 ) 2 n x 1 8 0 ° + A r e p r e s e n t s a l g e b r a i c a l l y a n a n g l e i n t h e f i r s t q u a d r a n t . ( 2 ) 2 n x 1 8 0 ° — A r e p r e s e n t s a l g e b r a i c a l l y a n a n g l e i n t h e f o u r t h q u a d r a n t . ( 3 ) ( 2 n + 1 ) 1 8 0 ° — A r e p r e s e n t s a l g e b r a i c a l l y a n a n g l e i n t h e s e c o n d q u a d r a n t . ( 4 ) ( 2 n + 1 ) 1 8 0 ° + A r e p r e s e n t s a l g e b r a i c a l l y a n a n g l e i n t h e t h i r d q u a d r a n t . I n c i r c u l a r m e a s u r e t h e c o r r e s p o n d i n g e x p r e s s i o n s a r e ( 1 ) 2 n i r + 6 , ( 2 ) 2 n i i — 0 , ( 3 ) ( 2 n + l ) - 6 , ( 4 ) ( 2 n + l ) + 6 . E X A M P L E S . S t a t e i n w h i c h q u a d r a n t t h e r e v o l v i n g l i n e w i l l b e a f t e r d e s c r i b i n g t h e f o l l o w i n g a n g l e s : ( 1 ) 1 2 0 ° , ( 2 ) 3 4 0 ° , ( 3 ) 4 9 0 ° , ( 4 ) -1 0 0 ° , ( 5 ) - 3 8 0 ° , ( 6 ) f x , ( 7 ) 1 0 , r + J . 1 2 . C o m p l e m e n t a n d S u p p l e m e n t o f a n A n g l e o r A r c . — T h e c o m p l e m e n t o f a n a n g l e o r a r c i s t h e r e m a i n d e r o b t a i n e d b y s u b t r a c t i n g i t f r o m a r i g h t a n g l e o r 9 0 ° . T h e s u p p l e m e n t o f a n a n g l e o r a r c i s t h e r e m a i n d e r o b t a i n e d b y s u b t r a c t i n g i t f r o m t w o r i g h t a n g l e s o r 1 8 0 ° . T h u s , t h e c o m p l e m e n t o f A i s ( 9 0 ° — A ) . T h e c o m p l e m e n t o f 1 9 0 ° i s ( 9 0 ° -1 9 0 ° ) = - 1 0 0 ° . T h e s u p p l e m e n t o f A i s ( 1 8 0 ° -A ) . T h e s u p p l e m e n t o f 2 0 0 ° i s ( 1 8 0 ° -2 0 0 ° ) = -2 0 ° . T h e c o m p l e m e n t o f ^ i r i s V 2 1 T h e s u p p l e m e n t o f ^ i r i s ( t t — | t t ) = \ n . E X A M P L E S . 1 3 E X A M P L E S . 1 . I f 1 9 2 s q u a r e i n c h e s b e r e p r e s e n t e d b y t h e n u m b e r 1 2 , w h a t i s t h e u n i t o f l i n e a r m e a s u r e m e n t ? A n s . 4 i n c h e s . 2 . I f 1 0 0 0 s q u a r e i n c h e s b e r e p r e s e n t e d b y t h e n u m b e r 4 0 , w h a t i s t h e u n i t o f l i n e a r m e a s u r e m e n t ? A n s . 5 i n c h e s . 3 . I f 2 0 0 0 c u b i c i n c h e s b e r e p r e s e n t e d b y t h e n u m b e r 1 6 , w h a t i s t h e u n i t o f l i n e a r m e a s u r e m e n t ? A n s . 5 i n c h e s . 4 . T h e l e n g t h o f a n A t l a n t i c c a b l e i s 2 3 0 0 m i l e s a n d t h e l e n g t h o f t h e c a b l e f r o m E n g l a n d t o F r a n c e i s 2 1 m i l e s . E x p r e s s t h e l e n g t h o f t h e f i r s t i n t e r m s o f t h e s e c o n d a s u n i t . A n s . 1 0 9 £ f 5 . F i n d t h e m e a s u r e o f a m i l e s w h e n b y a r d s i s t h e u n i t . A 1 7 6 0 a A n s . b 6 . T h e r a t i o o f t h e a r e a o f o n e f i e l d t o t h a t o f a n o t h e r i s 2 0 : 1 , a n d t h e a r e a o f t h e f i r s t i s h a l f a s q u a r e m i l e . F i n d t h e n u m b e r o f s q u a r e y a r d s i n t h e s e c o n d . A n s . 7 7 4 4 0 . 7 . A c e r t a i n w e i g h t i s 3 . 1 2 5 t o n s . W h a t i s i t s m e a s u r e i n t e r m s o f 4 c w t . ? A n s . 1 5 . 6 2 5 . E x p r e s s t h e f o l l o w i n g 1 2 a n g l e s i n c e n t e s i m a l m e a s u r e : 8 . 4 2 ° 1 5 ' 1 8 " . A n s . 4 6 9 5 \ 9 . 6 3 ° 1 9 ' 1 7 " . 7 0 . 3 5 v 7 0 v \ 9 8 1 0 . 1 0 3 ° 1 5 ' 4 5 " . 1 1 4 7 3 v 6 1 1 . 1 1 . 1 9 ° 0 ' 1 8 " . 2 1 1 F 6 6 v \ 6 . 1 2 . 1 4 3 ° 9 ' 0 " . 1 5 9 5 v 5 5 v \ 5 . 1 3 . 3 0 0 ° 1 5 ' 5 8 " . 3 3 3 6 2 v 9 0 v \ i 2 3 4 5 6 7 8 9 0 . 1 4 . 2 7 ° 4 1 ' 5 1 " , 3 0 7 7 5 . 1 5 . 6 7 ° . 4 3 2 5 . 7 4 9 2 5 . 1 6 . 8 ° 1 5 ' 2 7 " . 9 1 7 v 5 0 v \ 1 7 . 9 7 ° 5 ' 1 5 " . 1 0 7 8 7 v 5 0 v \ 1 8 . 1 6 ° 1 4 ' 1 9 " . 1 8 4 v 2 9 v v - . . 1 9 . 1 3 2 ° 6 ' . 1 4 6 7 7 v 7 7 v \ 7 . 1 4 P L A N E T R I G O N O M E T R Y . E x p r e s s t h e f o l l o w i n g 1 1 a n g l e s i n d e g r e e s , m i n u t e s , a n d s e c o n d s : 2 0 . 1 0 5 5 2 v 7 5 v \ A n s . 9 4 ° 5 8 ' 2 9 " . l . 2 1 . 8 2 9 v 5 4 X \ 7 3 ° 5 3 ' 9 " . 0 9 6 . 2 2 . 7 0 1 5 v 9 2 v \ 6 3 ° 8 ' 3 5 " . 8 0 8 . 2 3 . 1 5 0 v 1 5 v v . 1 3 ° 3 0 ' 4 " . 8 6 . 2 4 1 5 4 7 ^ 2 4 " . 1 3 8 ° 3 9 ' 5 4 " . 5 7 6 . 2 5 . 3 2 4 1 3 v 8 8 v \ 7 . 2 9 1 ° 4 3 ' 2 9 " . 9 3 8 8 . 2 6 . 1 0 4 2 v 5 0 . 9 ° 2 2 ' 5 7 " . 2 7 . 2 0 7 7 v 5 0 " . 1 8 ° 4 1 ' 5 1 " . 2 8 . 8 s 7 5 \ 7 ° 5 2 ' 3 0 " . 2 9 . 1 7 0 4 5 v 3 5 " . 1 5 3 ° 2 4 ' 2 9 " . 3 4 . 3 0 . 2 4 0 v 2 5 v \ 2 1 ° 3 6 ' 8 " . l . E x p r e s s i n c i r c u l a r m e a s u r e t h e f o l l o w i n g a n g l e s : 3 1 . 3 1 5 ° , 2 4 ° 1 3 ' . A n s . U , U 5 S " . T ' 1 0 8 0 0 3 2 . 9 5 ° 2 0 ' , 1 2 ° 5 ' 4 " . 1 4 3 t t 2 7 1 9 t t 2 7 0 , 4 0 5 0 0 3 3 . 2 2 1 ° , 1 ° , 5 7 ° . 2 9 5 . I , i i o , 1 r a d i a n -3 4 . 1 2 0 ° , 4 5 ° , 2 7 0 ° . 2 . 0 9 4 3 9 , J f „ - . 3 5 . 3 6 0 ° , 3 ^ r t . a n g l e s . 2 t t , f a - . E x p r e s s i n d e g r e e s , e t c . , t h e a n g l e s w h o s e c i r c u l a r m e a s u r e s a r e : 3 6 -i r A n s . 1 1 2 ° . 5 , 1 2 0 ° , f H d e g r e e s . ^ I T q t 1 1 2 4 5 , 3 0 , 1 2 0 , S 1 -a > f t , q ' — d e g r e e s , — d e g r e e s , _ d e g r e e s . 1 0 O T T I T . „ 3 8 . I . 7 8 5 4 . 4 7 ° 4 3 ' 3 8 " ^ , 4 5 ° . E X A M P L E S . 1 5 4 | , j V t t , 2 . 5 0 4 . A . i s . 2 5 7 ° 4 9 ' 4 3 " . 3 9 , 1 5 ° , 1 4 3 ° . 4 6 8 . 4 0 . . 0 2 3 4 , 1 . 2 3 4 , ? -1 ° 2 0 ' 2 7 " , 7 0 ° 4 2 ' 1 1 " , 3 8 ° 1 1 ' 5 0 " . 3 4 1 . F i n d t h e n u m b e r o f r a d i a n s i n a n a n g l e a t t h e c e n t r e o f a c i r c l e o f r a d i u s 2 5 f e e t , w h i c h i n t e r c e p t s a n a r c o f 3 7 | f e e t . A n s . 1 £ . 4 2 . F i n d t h e n u m b e r o f d e g r e e s i n a n a n g l e a t t h e c e n t r e o f a c i r c l e o f r a d i u s 1 0 f e e t , w h i c h i n t e r c e p t s a n a r c o f 5 i r f e e t . A n s . 9 0 ° . 4 3 . F i n d t h e n u m b e r o f r i g h t a n g l e s i n a n a n g l e a t t h e c e n t r e o f a c i r c l e o f r a d i u s 3 - ^ i n c h e s , w h i c h i n t e r c e p t s a n a r c o f 2 f e e t . A n s . 4 $ . 4 4 . F i n d t h e l e n g t h o f t h e a r c s u b t e n d i n g a n a n g l e o f 4 £ r a d i a n s a t t h e c e n t r e o f a c i r c l e w h o s e r a d i u s i s 2 5 f e e t . A n s . 1 1 2 £ f t . 4 5 . F i n d t h e l e n g t h o f a n a r c o f § 0 ° o n a c i r c l e o f 4 f e e t r a d i u s . A n s . 5 f £ f t . 4 6 . T h e a n g l e s u b t e n d e d b y t h e d i a m e t e r o f t h e S u n a t t h e e y e o f a n o b s e r v e r i s 3 2 ' : f i n d a p p r o x i m a t e l y t h e d i a m e t e r o f t h e S u n i f i t s d i s t a n c e f r o m t h e o b s e r v e r b e 9 0 0 0 0 0 0 0 m i l e s . A n s . 8 3 8 0 0 0 m i l e s . 4 7 . A r a i l w a y t r a i n i s t r a v e l l i n g o n a c u r v e o f h a l f a m i l e r a d i u s a t t h e r a t e o f 2 0 m i l e s a n h o u r : t h r o u g h w h a t a n g l e h a s i t t u r n e d i n 1 0 s e c o n d s ? A n s . 6 T 4 T d e g r e e s . 4 8 . I f t h e r a d i u s o f a c i r c l e b e 4 0 0 0 m i l e s , f i n d t h e l e n g t h o f a n a r c w h i c h s u b t e n d s a n a n g l e o f 1 " a t t h e c e n t r e o f t h e c i r c l e . A n s . A b o u t 3 4 y a r d s . 4 9 . O n a c i r c l e o f 8 0 f e e t r a d i u s i t w a s f o u n d t h a t a n a n g l e o f 2 2 ° 3 0 ' a t t h e c e n t r e w a s s u b t e n d e d b y a n a r c 3 1 f t . 5 i n . i n l e n g t h : h e n c e c a l c u l a t e t o f o u r d e c i m a l p l a c e s t h e n u m e r i c a l v a l u e o f t h e r a t i o o f t h e c i r c u m f e r e n c e o f a c i r c l e t o i t s d i a m e t e r . A n s . 3 . 1 4 1 6 . 5 0 . F i n d t h e n u m b e r o f r a d i a n s i n 1 0 " c o r r e c t t o f o u r s i g n i f i c a n t f i g u r e s ( u s e f f f f o r » . > A n s . . 0 0 0 0 4 8 4 8 . 1 6 P L A N E T R I G O N O M E T R Y . C H A P T E R I I . T H E T R I G O N O M E T R I C E U N O T I O N S . 1 3 . D e f i n i t i o n s o f t h e T r i g o n o m e t r i c F u n c t i o n s . — L e t R A D b e a n a n g l e ; i n A D , o n e o f t h e l i n e s c o n t a i n i n g t h e a n g l e , t a k e a n y p o i n t B , a n d f r o m B d r a w B C p e r p e n d i c u l a r t o t h e o t h e r l i n e A R , t h u s f o r m i n g a r i g h t t r i a n g l e A B C , r i g h t - a n g l e d a t C . T h e n d e n o t i n g t h e a n g l e s b y t h e c a p i t a l l e t t e r s A , B , C , r e s p e c t i v e l y , a n d t h e t h r e e s i d e s o p p o s i t e t h e s e a n g l e s b y t h e c o r r e s p o n d i n g s m a l l i t a l i c s , a , b , c , w e h a v e t h e f o l l o w i n g d e f i n i t i o n s : a = o p p o s i t e s i d e ^ c a U e d ^ ^ o f ^ a n g l e A c h y p o t e n u s e - = a a J a c e n t s i d e j s c a l l e d t h e c o s i n e o f t h e a n g l e A . c h y p o t e n u s e a = o p p o s i t e s i d e j s c a U e d ^ t a n g m t o f t h e a n g l e A b a d j a c e n t s i d e - = a ( t i a c e i r k s ^ e j s c a l l e d t h e c o t a n g e n t o f t h e a n g l e A . a o p p o s i t e s i d e - = ^ Y P o t e n u s e i s c a l l e d t h e s e c a n t o f t h e a n g l e A . b a d j a c e n t s i d e - = k v f > 0 j ' e i m s e i s c a l l e d t h e c o s e c a n t o f t h e a n g l e A . a o p p o s i t e s i d e I f t h e c o s i n e o f A b e s u b t r a c t e d f r o m u n i t y , t h e r e m a i n d e r i s c a l l e d t h e v e r s e d s i n e o f A . I f t h e s i n e o f A b e s u b - T h e l e t t e r s a , b , c a r e n u m b e r s , b e i n g t h e n u m b e r o f t i m e s t h e l e n g t h s o f t h e s i d e s c o n t a i n s o m e c h o s o n u n i t o f l e n g t h . T R I G O N O M E T R I C F U N C T I O N S . 1 7 t r a c t e d f r o m u n i t y , t h e r e m a i n d e r i s c a l l e d t h e c o v e r s e d s i n e o f A ; t h e l a t t e r t e r m i s h a r d l y e v e r u s e d i n p r a c t i c e . T h e w o r d s s i n e , c o s i n e , e t c . , a r e a b b r e v i a t e d , a n d t h e f u n c t i o n s o f a n a n g l e A a r e w r i t t e n t h u s : s i n A , c o s A , t a n A , c o t A , s e c A , c o s e c A , v e r s A , c o v e r s A . T h e f o l l o w i n g i s t h e v e r b a l e n u n c i a t i o n o f t h e s e d e f i n i t i o n s : T h e s i n e o f a n a n g l e i s t h e r a t i o o f t h e o p p o s i t e s i d e t o t h e h y p o t e n u s e ; o r s i n A = - -c T h e c o s i n e o f a n a n g l e i s t h e r a t i o o f t h e a d j a c e n t s i d e t o t h e h y p o t e n u s e ; o r c o s A = — c T h e t a n g e n t o f a n a n g l e i s t h e r a t i o o f t h e o p p o s i t e s i d e t o t h e a d j a c e n t s i d e ; o r t a n A = - -T h e c o t a n g e n t o f a n a n g l e i s t h e r a t i o o f t h e a d j a c e n t s i d e t o t h e o p p o s i t e s i d e ; o r c o t A = - -a T h e s e c a n t o f a n a n g l e i s t h e r a t i o o f t h e h y p o t e n u s e t o t h e a d j a c e n t s i d e ; o r s e c A = - . T h e c o s e c a n t o f a n a n g l e i s t h e r a t i o o f t h e h y p o t e n u s e t o t h e o p p o s i t e s i d e ; o r c o s e c A = —a T h e v e r s e d s i n e o f a n a n g l e i s u n i t y m i n u s t h e c o s i n e o f t h e a n g l e ; o r v e r s A = 1 — c o s A = 1 — - -c T h e c o v e r s e d s i n e o f a n a n g l e i s u n i t y m i n u s t h e s i n e o f t h e a n g l e ; o r c o v e r s A = 1 — s i n A = 1 — - -c T h e s e r a t i o s a r e c a l l e d T r i g o n o m e t r i c F u n c t i o n s . T h e s t u d e n t s h o u l d c a r e f u l l y c o m m i t t h e m t o m e m o r y , a s u p o n t h e m i s f o u n d e d t h e w h o l e t h e o r y o f T r i g o n o m e t r y . T h e s e f u n c t i o n s a r e , i t w i l l b e o b s e r v e d , n o t l e n g t h s , b u t 1 8 P L A N E T R I G O N O M E T R Y . r a t i o s o f o n e l e n g t h t o a n o t h e r ; t h a t i s , t h e y a r e a b s t r a c t n u m b e r s , s i m p l y n u m e r i c a l q u a n t i t i e s ; a n d t h e y r e m a i n u n c h a n g e d s o l o n g a s t h e a n g l e r e m a i n s u n c h a n g e d , a s w i l l b e p r o v e d i n A r t . 1 4 . I t i s c l e a r f r o m t h e a b o v e d e f i n i t i o n s t h a t c o s e c A = — — - , o r s i n A : s i n A c o s e c A s e c A = — ^ — - , o r c o s A : ^ c o s A ' s e c A , t a n A = — - — , o r c o t A = -^ c o t A t a n A T h e p o w e r s o f t h e T r i g o n o m e t r i c f u n c t i o n s a r e e x p r e s s e d a s f o l l o w s : ( s i n A ) 2 i s w r i t t e n s i n 2 A , ( c o s A ) 3 i s w r i t t e n c o s 3 A , a n d s o o n . N o t e . — T h e s t u d e n t m u s t n o t i c e t h a t ' s i n A ' i s a s i n g l e s y m b o l , t h e n a m e o f a n u m b e r , o r f r a c t i o n b e l o n g i n g t o t h e a n g l e A . A l s o s i n 2 A i s a n a b b r e v i a t i o n f o r ( s i n A ) 2 , i . e . , f o r ( s i n A ) x ( s i n A ) . S u c h a b b r e v i a t i o n s a r e u s e d f o r c o n v e n i e n c e . 1 4 . T h e T r i g o n o m e t r i c F u n c t i o n s a r e a l w a y s t h e S a m e f o r t h e S a m e A n g l e . — L e t B A D b e a n y a n g l e ; i n A D t a k e P , P ' , a n y P ^ < ^ t w o p o i n t s , a n d d r a w P C , P ' C p e r -p e n d i c u l a r t o A B . T a k e P " , a n y p o i n t i n A B , a n d d r a w P " C " p e r p e n d i c u l a r t o A D . A T h e n t h e t h r e e t r i a n g l e s P A C , P ' A C , P " A C " a r e e q u i a n g u l a r , s i n c e t h e y a r e r i g h t - a n g l e d , a n d h a v e a c o m m o n a n g l e a t A : t h e r e f o r e t h e y a r e s i m i l a r . P C = P ' C = P " C " " A P A P ' A P " ' B u t e a c h o f t h e s e r a t i o s i s t h e s i n e o f t h e a n g l e A . T h u s , s i n A i s t h e s a m e w h a t e v e r b e t h e p o s i t i o n o f t h e p o i n t P o n e i t h e r o f t h e l i n e s c o n t a i n i n g t h e a n g l e A . F U N C T I O N S O F C O M P L E M E N T A L A N G L E S . 1 9 T h e r e f o r e s i n A i s a l w a y s t h e s a m e . A s i m i l a r p r o o f m a y b e g i v e n f o r e a c h o f t h e o t h e r f u n c t i o n s . I n t h e r i g h t t r i a n g l e o f A r t . 1 3 , s h o w t h a t a = c s i n A = c c o s B = b t a n A = b c o t B , b = a c o t A = a t a n B = c c o s A = c s i n B , c = a c o s e c A = a s e c B = b s e c A = b c o s e c B . N o t e . — T h e s e r e s u l t s s h o u l d b e c a r e f u l l y n o t i c e d , a s t h e y a r e o f f r e q u e n t u s e i n t h e s o l u t i o n o f r i g h t t r i a n g l e s a n d e l s e w h e r e . 1 . C a l c u l a t e t h e v a l u e o f t h e f u n c t i o n s , s i n e , c o s i n e , e t c . , o f t h e a n g l e A i n t h e r i g h t t r i a n g l e s w h o s e s i d e s a , b , c a r e r e s p e c t i v e l y ( 1 ) 8 , 1 5 , 1 7 ; ( 2 ) 4 0 , 9 , 4 1 ; ( 3 ) 1 9 6 , 3 1 5 , 3 7 1 ; ( 4 ) 4 8 0 , 3 1 , 4 8 1 ; ( 5 ) 1 7 0 0 , 9 4 5 , 1 9 4 5 . A m . ( 1 ) s i n A = T 8 T , c o s A = | f , t a n A = ^ j , e t c . ; E X A M P L E S . I n a r i g h t t r i a n g l e , g i v e n : 2 . a = V m 2 + n 2 , b = V 2 m n ; c a l c u l a t e s i n A . A n s . V « i 2 + n 2 m + n 3 . a ^ m 2 — m n , b = n ; c a l c u l a t e s e c A . m — n n 5 . a 4 . a V m 2 + m n , c = m + n ; c a l c u l a t e t a n A . 2 w i n . , b = m 2 — ? i 2 ; c a l c u l a t e c o s A . f m 2 + m n m n + n 2 m 2 — n 2 6 . s i n A = f , c = 2 0 0 . 5 ; c a l c u l a t e a . 7 . c o s A = . 4 4 , c = 3 0 . 5 ; c a l c u l a t e b . 8 . t a n A = J g L , b = ; c a l c u l a t e c . m 2 + n 2 1 2 0 . 3 . 1 3 . 4 2 . 3 f t - V l 3 0 . 2 0 P L A N E T R I G O N O M E T R Y . 1 5 . F u n c t i o n s o f C o m p l e m e n t a l A n g l e s . — I n t h e r t . A A B C w e h a v e s i n A : , a n d c o s B = ® . ( A r t . 1 3 . ) . s i n A = c o s B . B u t B i s t h e c o m p l e m e n t o f A , s i n c e t h e i r s u m i s a r i g h t a n g l e , o r 9 0 ° ; i . e . , B = 9 0 ° -A . A l s o , s i n A = c o s B = c o s ( 9 0 ° - A ) — a . c c o s A = s i n B = s i n ( 9 0 ° - A ) b — s c t a n A = c o t B = c o t ( 9 0 ° - A ) a ~ b ' c o t A = t a u B = t a n ( 9 0 ° -A ) b — > a s e c A = c o s e c B = c o s e c ( 9 0 ° -A ) c c o s e c A = s e c B = s e c ( 9 0 ° -A ) _ c a v e r s A = c o v e r s B = c o v e r s ( 9 0 ° - A ) c c o v e r s A -v e r s B = v e r s ( 9 0 ° — A ) = 1 — - -c T h e r e f o r e t h e s i n e , t a n g e n t , s e c a n t , a n d v e r s e d s i n e o f a n a n g l e a r e e q u a l r e s p e c t i v e l y t o t h e c o s i n e , c o t a n g e n t , c o s e c a n t , a n d c o v e r s e d s i n e o f t h e c o m p l e m e n t o f t h e a n g l e . 1 6 . R e p r e s e n t a t i o n o f t h e T r i g o n o m e t r i c F u n c t i o n s b y S t r a i g h t L i n e s . — T h e T r i g o n o m e t r i c f u n c t i o n s w e r e f o r m e r l y d e f i n e d a s b e i n g c e r t a i n s t r a i g h t l i n e s g e o m e t r i c a l l y c o n n e c t e d w i t h t h e a r c s u b t e n d i n g t h e a n g l e a t t h e c e n t r e o f a c i r c l e o f g i v e n r a d i u s . T h u s , l e t A P b e t h e a r c o f a c i r c l e s u b t e n d i n g t h e a n g l e A O P a t t h e c e n t r e . R E P R E S E N T A T I O N O F F U N C T I O N ; 3 £ F L I N E S . 2 1 D r a w t h e t a n g e n t s A T , B T ' m e e t i n g O P p r o d u c e d t o T ' , a n d d r a w P C , P D _ L t o O A , O B . c o t a n g e n t . — J i s \ f T ' t a m i n e H T h e n P C w a s c a l l e d t h e s i n e o f - t h e a r c A P . o c « c o s i n e t t A T a t a n g e n t i t B T ' a c o t a n g e n t a O T a s e c a n t t t O T ' a c o s e c a n t i t A C t t v e r s e d s i n e t t B D a c o v e r s e d s i n e i t S i n c e a n y a r c i s t h e m e a s u r e o f t h e a n g l e a t t h e c e n t r e w h i c h t h e a r c s u b t e n d s ( A r t . 5 ) , t h e a b o v e f u n c t i o n s o f t h e a r c A P a r e a l s o f u n c t i o n s o f t h e a n g l e A O P . I t s h o u l d b e n o t i c e d t h a t t h e o l d f u n c t i o n s o f t h e a r c a b o v e g i v e n , w h e n d i v i d e d b y t h e r a d i u s o f t h e c i r c l e , b e c o m e t h e m o d e r n f u n c t i o n s o f t h e a n g l e w h i c h t h e a r c s u b t e n d s a t t h e c e n t r e . I f , t h e r e f o r e , t h e r a d i u s b e t a k e n a s u n i t y , t h e o l d f u n c t i o n s o f t h e a r c A P b e c o m e t h e m o d e r n f u n c t i o n s o f t h e a n g l e A O P . T h u s , r e p r e s e n t i n g t h e a r c A P , o r t h e a n g l e A O P b y 6 , w e h a v e , w h e n O A . = O P = 1 , 2 2 P L A N E T R I G O N O M E T R Y . s m O = ^ = ^ = P C , o r 1 ' t a n . = A T = f - A T , a n d s i m i l a r l y f o r t h e o t h e r f u n c t i o n s . T h e r e f o r e , i n a c i r c l e w h o s e r a d i u s i s u n i t y , t h e T r i g o n o m e t r i c f u n c t i o n s o f a n a r c , o r o f t h e a n g l e a t t h e c e n t r e m e a s u r e d b y t h a t a r c , m a y b e d e f i n e d a s f o l l o w s : T h e s i n e i s t h e p e r p e n d i c u l a r l e t f a l l f r o m o n e e x t r e m i t y o f t h e a r c u p o n t h e d i a m e t e r p a s s i n g t h r o u g h t h e o t h e r e x t r e m i t y . T h e c o s i n e i s t h e d i s t a n c e f r o m t h e c e n t r e o f t h e c i r c l e t o t h e f o o t o f t h e s i n e . T h e t a n g e n t i s t h e l i n e w h i c h t o u c h e s o n e e x t r e m i t y o f t h e a r c a n d i s t e r m i n a t e d b y t h e d i a m e t e r p r o d u c e d p a s s i n g t h r o u g h t h e o t h e r e x t r e m i t y . T h e s e c a n t i s t h e p o r t i o n o f t h e d i a m e t e r p r o d u c e d t h r o u g h o n e e x t r e m i t y o f t h e a r c w h i c h i s i n t e r c e p t e d b e t w e e n t h e c e n t r e a n d t h e t a n g e n t a t t h e o t h e r e x t r e m i t y . T h e v e r s e d s i n e i s t h e p a r t o f t h e d i a m e t e r i n t e r c e p t e d b e t w e e n t h e b e g i n n i n g o f t h e a r c a n d t h e f o o t o f t h e s i n e . S i n c e t h e l i n e s P D o r O C , B T ' , O T ' , a n d B D a r e r e s p e c t i v e l y t h e s i n e , t a n g e n t , s e c a n t , a n d v e r s e d s i n e o f t h e a r e B P , w h i c h ( A r t . 1 2 ) i s t h e c o m p l e m e n t o f A P , w e s e e t h a t t h e c o s i n e , t h e c o t a n g e n t , t h e c o s e c a n t , a n d t h e c o v e r s e d s i n e o f a n a r c a r e r e s p e c t i v e l y t h e s i n e , t h e t a n g e n t , t h e s e c a n t , a n d t h e v e r s e d s i n e o f i t s c o m p l e m e n t . E X A M P L E S . 1 . P r o v e t a n A s i n A + c o s A = s e c A . 2 . " c o t A c o s A + s i n A = c o s e c A . 3 . " ( t a n A - s i n A ) 2 + ( l -c o s A ) 2 = ( s e c A -l ) 2 . 4 . " t a n A + c o t A = s e c A c o s e c A . 5 . " ( s i n A + c o s A ) - r - ( s e c A + c o s e c A ) = s i n A c o s A . 6 . " ( l + t a n A ) 2 + ( l + c o t A ) 2 = ( s e c A + c o s e c A ) 2 . P O S I T I V E A N D N E G A T I V E L I N E S . 2 3 7 . G i v e n t a n A = c o t 2 A ; f i n d A . 8 . " s i n A = e o s 3 A ; f i n d A . 9 . " s i n A = c o s ( 4 5 ° -\ A ) ; f i n d A . 1 0 . " t a n A = c o t 0 A ; f i n d A . 1 1 . " c o t A = t a n ( 4 5 ° + A ) ; f i n d A . B M O M 1 7 . P o s i t i v e a n d N e g a t i v e L i n e s . — L e t A A ' a n d B B ' b e t w o p e r p e n d i c u l a r r i g h t l i n e s i n t e r s e c t i n g a t t h e p o i n t 0 . T h e n t h e p o s i t i o n o f a n y p o i n t i n t h e l i n e A A ' o r B B ' w i l l b e d e t e r m i n e d i f w e k n o w t h e d i s t a n c e o f t h e p o i n t f r o m O , a n d i f w e k n o w a l s o u p o n w h i c h s i d e o f O t h e A -p o i n t l i e s . I t i s t h e r e f o r e c o n v e n i e n t t o e m p l o y t h e a l g e b r a i c s i g n s + a n d — , s o t h a t i f d i s t a n c e s m e a s u r e d a l o n g t h e f i x e d l i n e O A o r O B f r o m O i n o n e d i r e c t i o n ' b e c o n s i d e r e d p o s i t i v e , d i s t a n c e s m e a s u r e d a l o n g O A ' o r O B ' i n t h e o p p o s i t e d i r e c t i o n f r o m O w i l l b e c o n s i d e r e d n e g a t i v e . T h i s c o n v e n t i o n , a s i t i s c a l l e d , i s e x t e n d e d t o l i n e s p a r a l l e l t o A A ' a n d B B ' ; a n d i t i s c u s t o m a r y t o c o n s i d e r d i s t a n c e s m e a s u r e d f r o m B B ' t o w a r d s t h e r i g h t a n d f r o m A A ' u p w a r d s a s p o s i t i v e , a n d c o n s e q u e n t l y d i s t a n c e s m e a s u r e d f r o m B B ' t o w a r d s t h e l e f t a n d f r o m A A ' d o w n w a r d s a s n e g a t i v e . B 1 8 . T r i g o n o m e t r i c F u n c t i o n s o f A n g l e s o f A n y M a g n i t u d e . — I n t h e d e f i n i t i o n s o f t h e t r i g o n o m e t r i c f u n c t i o n s g i v e n i n A r t . 1 3 w e c o n s i d e r e d o n l y a c u t e a n g l e s , i . e . , a n g l e s i n t h e f i r s t q u a d r a n t ( A r t . 5 ) , s i n c e t h e a n g l e w a s a s s u m e d t o b e o n e o f t h e a c u t e a n g l e s o f a r i g h t t r i a n g l e . W e s h a l l n o w s h o w t h a t t h e s e d e f i n i t i o n s a p p l y t o a n g l e s o f a n y m a g n i t u d e , a n d t h a t t h e f u n c t i o n s v a r y i n s i g n a c c o r d i n g t o t h e q u a d r a n t i n w h i c h t h e a n g l e h a p p e n s t o b e . 2 4 P L A N E T R I G O N O M E T R Y . L e t A O P b e a n a n g l e o f a n y m a g n i t u d e f o r m e d b y O P r e v o l v i n g f r o m a n i n i t i a l p o s i t i o n O A . D r a w P M ± t o A A ' . C o n s i d e r O P a s a l w a y s p o s i t i v e . L e t t h e a n g l e A O P b e d e n o t e d b y A ; t h e n w h a t e v e r b e t h e m a g n i t u d e o f t h e a n g l e A , t h e d e f i n i t i o n s o f t h e t r i g o n o m e t r i c f u n c t i o n s a r e / S n d \ O u a d . p 1 s t \ Q u a d r a n t \ \ m \ \ M 0 / \ 3 r d \ Q u a d . i t h \ Q u a d S P P s i n A s e c A : M P ' O P , O P O M , c o s A = c o t A : O M O P , O M M P , t a n A = c o s e c A -B M P O M , O P M P ' I . W h e n A l i e s i n t h e 1 s t q u a d r a n t , M P i s p o s i t i v e b e c a u s e m e a s u r e d f r o m M u p w a r d s , O M i s p o s i t i v e b e c a u s e m e a s u r e d f r o m O t o w a r d s t h e r i g h t ( A r t . 1 7 ) , a n d O P i s p o s i t i v e . H e n c e i n t h e f i r s t q u a d r a n t a l l t h e f u n c t i o n s a r e p o s i t i v e . I I . W h e n A l i e s i n t h e 2 d q u a d r a n t , a s t h e o b t u s e a n g l e A O P , M P i s p o s i t i v e b e c a u s e m e a s u r e d f r o m M u p w a r d s , O M i s n e g a t i v e b e c a u s e m e a s u r e d f r o m O t o w a r d s t h e l e f t ( A r t . 1 7 ) , a n d O P i s p o s i t i v e . H e n c e i n t h e s e c o n d q u a d r a n t M P s i n A = i s p o s i t i v e : O P 1 a O M . c o s A = i s n e g a t i v e ; 4 . a M P -t a n A = i s n e q a t i v e : O M y a n d t h e r e f o r e s e c A a n d c o t A a r e n e g a t i v e , a n d c o s e c A i s p o s i t i v e ( A r t . 1 3 ) . F U N C T I O N S O F A N G L E S . 2 5 I I I . W h e n A l i e s i n t h e 3 d q u a d r a n t , a s t h e r e f l e x a n g l e A O P , M P i s n e g a t i v e b e c a u s e m e a s u r e d f r o m M d o w n w a r d s , O M i s n e g a t i v e , a n d O P i s p o s i t i v e . H e n c e i n t h e t h i r d q u a d r a n t t h e s i n e , c o s i n e , s e c a n t , a n d c o s e c a n t , a r e n e g a t i v e , b u t t h e t a n g e n t a n d c o t a n g e n t a r e p o s i t i v e . I V . W h e n A l i e s i n t h e 4 t h q u a d r a n t , a s t h e r e f l e x a n g l e A O P , M P i s n e g a t i v e , O M i s p o s i t i v e , a n d O P i s p o s i t i v e . H e n c e i n t h e f o u r t h q u a d r a n t t h e s i n e , t a n g e n t , c o t a n g e n t , a n d c o s e c a n t a r e n e g a t i v e , b u t t h e c o s i n e a n d s e c a n t a r e p o s i t i v e . T h e s i g n s o f t h e d i f f e r e n t f u n c t i o n s a r e s h o w n i n t h e a n n e x e d t a b l e . Q u a d r a n t . I . I I . i n . I V . S i n a n d c o s e c + + --C o s a n d s e c + --+ T a n a n d c o t + -+ -N o r a . — I t i s a p p a r e n t f r o m t h i s t a b l e t h a t t h e s i g n s o f a l l t h e f u n c t i o n s i n a n y q u a d r a n t a r e k n o w n w h e n t h o s e o f t h e l i n e a n d c o s i n e a r e k n o w n . T h e t a n g e n t a n d c o t a n g e n t a r e + o r — , a c c o r d i n g a s t h e s i n e a n d c o s i n e h a v e l i k e o r d i f f e r e n t s i g n s . 1 9 . C h a n g e s i n t h e V a l u e o f t h e S i n e a s t h e A n g l e i n c r e a s e s f r o m 0 ° t o 3 6 0 ° . — L e t A d e n o t e t h e a n g l e A O P d e s c r i b e d b y t h e r e v o l u t i o n o f O P f r o m i t s i n i t i a l p o s i t i o n O A t h r o u g h 3 6 0 ° . T h e n , P M b e i n g d r a w n p e r p e n d i c u l a r t o A A ' , s i n A = M P O P , w h a t e v e r b e t h e m a g n i t u d e o f t h e a n g l e A . 2 6 P L A N E T R I G O N O M E T R Y . W h e n t h e a n g l e A i s 0 ° , P c o i n c i d e s w i t h A , a n d M P i s z e r o ; t h e r e f o r e s i n 0 ° = 0 . A s A i n c r e a s e s f r o m 0 ° t o 9 0 ° , M P i n c r e a s e s f r o m z e r o t o O B o r O P , a n d i s p o s i t i v e ; t h e r e f o r e s i n 9 0 ° = 1 . H e n c e i n t h e 1 s t q u a d r a n t s i n A i s p o s i t i v e , a n d i n c r e a s e s f r o m 0 t o 1 . A s A i n c r e a s e s f r o m 9 0 ° t o 1 8 0 ° , M P d e c r e a s e s f r o m O P t o z e r o , a n d i s p o s i t i v e ; t h e r e f o r e s i n 1 8 0 ° = 0 . H e n c e i n t h e 2 d q u a d r a n t s i n A i s p o s i t i v e , a n d d e c r e a s e s f r o m 1 t o 0 . A s A i n c r e a s e s f r o m 1 8 0 ° t o 2 7 0 ° , M P i n c r e a s e s f r o m z e r o t o O P , a n d i s n e g a t i v e ; t h e r e f o r e s i n 2 7 0 ° = — 1 . H e n c e i n t h e 3 d q u a d r a n t s i n A i s n e g a t i v e , a n d d e c r e a s e s a l g e b r a i c a l l y f r o m 0 t o — 1 . A s A i n c r e a s e s f r o m 2 7 0 ° t o 3 6 0 ° , M P d e c r e a s e s f r o m O P t o z e r o , a n d i s n e g a t i v e ; t h e r e f o r e s i n 3 0 0 ° = 0 . H e n c e i n t h e 4 t h q u a d r a n t s i n A i s n e g a t i v e , a n d i n c r e a s e s a l g e b r a i c a l l y f r o m — 1 t o 0 . 2 0 . C h a n g e s i n t h e C o s i n e a s t h e A n g l e i n c r e a s e s f r o m 0 ° t o 3 6 0 ° . — I n t h e f i g u r e o f A r t . 1 9 W h e n t h e a n g l e A i s 0 ° , P c o i n c i d e s w i t h A , a n d O M = O P ; t h e r e f o r e c o s 0 ° = 1 . A s A i n c r e a s e s f r o m 0 ° t o 9 0 ° , O M d e c r e a s e s f r o m O P t o . z e r o a n d i s p o s i t i v e ; t h e r e f o r e c o s 9 0 ° = 0 . H e n c e i n t h e 1 s t q u a d r a n t c o s A i s p o s i t i v e , a n d d e c r e a s e s f r o m 1 t o 0 . A s A i n c r e a s e s f r o m 9 0 ° t o 1 8 0 ° , O M i n c r e a s e s f r o m z e r o t o O P , a n d i s n e g a t i v e ; t h e r e f o r e c o s 1 8 0 ° = — 1 . H e n c e i n t h e 2 d q u a d r a n t c o s A i s n e g a t i v e , a n d d e c r e a s e s a l g e b r a i c a l l y f r o m 0 t o — 1 . A s A i n c r e a s e s f r o m 1 8 0 ° t o 2 7 0 ° , O M d e c r e a s e s f r o m O P t o z e r o , a n d i s n e g a t i v e ; t h e r e f o r e c o s 2 7 0 ° = 0 . C H A N G E S I N T H E T A N G E N T . 2 7 H e n c e i n t h e 3 d q u a d r a n t c o s A i s n e g a t i v e , a n d i n c r e a s e s a l g e b r a i c a l l y f r o m — 1 t o 0 . A s A i n c r e a s e s f r o m 2 7 0 ° t o 3 6 0 ° , O M i n c r e a s e s f r o m z e r o t o O P , a n d i s p o s i t i v e ; t h e r e f o r e c o s 3 6 0 ° = 1 . H e n c e i n t h e 4 t h q u a d r a n t c o s A i s p o s i t i v e , a n d i n c r e a s e s f r o m 0 t o 1 . 2 1 . C h a n g e s i n t h e T a n g e n t a s t h e A n g l e i n c r e a s e s f r o m 0 ° t o 3 6 0 ° . — I n t h e f i g u r e o f A r t . 1 9 ^ a M P t a n A = O M W h e n A i s 0 ° , M P i s z e r o , a n d O M = O P ; t h e r e f o r e t a n 0 ° = 0 . A s A i n c r e a s e s f r o m 0 ° t o 9 0 ° , M P i n c r e a s e s f r o m z e r o t o O P , a n d O M d e c r e a s e s f r o m O P t o z e r o , s o t h a t o n b o t h a c c o u n t s t a n A i n c r e a s e s n u m e r i c a l l y ; t h e r e f o r e t a n 9 0 ° = o o . H e n c e i n t h e 1 s t q u a d r a n t t a n A i s p o s i t i v e , a n d i n c r e a s e s f r o m 0 t o o o . A s A i n c r e a s e s f r o m 9 0 ° t o 1 8 0 ° , M P d e c r e a s e s f r o m O P t o z e r o , a n d i s p o s i t i v e , O M b e c o m e s n e g a t i v e a n d d e c r e a s e s a l g e b r a i c a l l y f r o m z e r o t o — 1 ; t h e r e f o r e t a n 1 8 0 ° = 0 . H e n c e i n t h e 2 d q u a d r a n t t a n A i s n e g a t i v e , a n d i n c r e a s e s a l g e b r a i c a l l y f r o m — o o t o 0 . W h e n A p a s s e s i n t o t h e 2 d q u a d r a n t , a n d i s o n l y j u s t g r e a t e r t h a n 9 0 ° , t a n A c h a n g e s f r o m + < x > t o — o o . A s A i n c r e a s e s f r o m 1 8 0 ° t o 2 7 0 ° , M P i n c r e a s e s f r o m z e r o t o O P , a n d i s n e g a t i v e , O M d e c r e a s e s f r o m O P t o z e r o , a n d i s n e g a t i v e ; t h e r e f o r e t a n 2 7 0 ° = o o . H e n c e i n t h e 3 d q u a d r a n t t a n A i s p o s i t i v e , a n d i n c r e a s e s f r o m 0 t o o o . A s A i n c r e a s e s f r o m 2 7 0 ° t o 3 6 0 ° , M P d e c r e a s e s f r o m O P t o z e r o , a n d i s n e g a t i v e , O M i n c r e a s e s f r o m z e r o t o O P , a n d i s p o s i t i v e ; t h e r e f o r e t a n 3 6 0 ° = 0 . H e n c e i n t h e 4 t h q u a d r a n t t a n A i s n e g a t i v e , a n d i n c r e a s e s a l g e b r a i c a l l y f r o m — o o t o 0 . 2 8 P L A N E T R I G O N O M E T R Y . T h e s t u d e n t i s r e c o m m e n d e d t o t r a c e i n a m a n n e r s i m i l a r t o t h e a b o v e t h e c h a n g e s i n t h e o t h e r f u n c t i o n s , i . e . , t h e c o t a n g e n t , s e c a n t , a n d c o s e c a n t , a n d t o s e e t h a t h i s r e s u l t s a g r e e w i t h t h o s e g i v e n i n t h e f o l l o w i n g t a b l e . 2 2 . T a b l e g i v i n g t h e C h a n g e s o f t h e T r i g o n o m e t r i c F n n e t i o n s i n t h e F o u r Q u a d r a n t s . Q u a d r a n t . I . I I . I I I . I V . + 0 t o 1 + s i n v a r i e s f r o m 1 t o 0 0 t o -1 -1 t o 0 c o s " " + 1 t o 0 0 t o -1 -1 t o O + . 0 t o 1 + + 0 t o o o t a n " -0 t o c o — o o t o 0 — o o t o 0 + c o t o 0 + c o t " 0 t o c o o o t o 0 0 t o — o o + + s e c " " 1 t o C O — 0 0 t o — 1 — 1 t o — 0 0 o o t o 1 + + c o s e e " " C O t o 1 1 t o C O — o o t o — 1 — 1 t o — 0 0 + + + + v e r s " " 0 t o 1 1 t o 2 2 t o 1 1 t o 0 N o t e 1 . — T h e c o s e c a n t , s e c a n t , a n d c o t a n g e n t o f a n a n g l e A h a v e t h e s a m e s i g n a s t h e s i n e , c o s i n e , a n d t a n g e n t o f A r e s p e c t i v e l y . T h e s i n e a n d c o s i n e v a r y f r o m 1 t o — 1 , p a s s i n g t h r o u g h t h e v a l u e 0 . T h e y a r e n e v e r g r e a t e r t h a n u n i t y . T h e s e c a n t a n d c o s e c a n t v a r y f r o m 1 t « — 1 , p a s s i n g t h r o u g h t h e v a l u e o o . T h e y a r e n e v e r n u m e r i c a l l y l e s s t h a n u n i t y . T h e t a n g e n t a n d c o t a n g e n t a r e u n l i m i t e d i n v a l u e . T h e y h a v e a l l v a l u e s f r o m — c o t O + o o . T h e v e r s e d s i n e a n d c o v e r s e d s i n e v a r y f r o m 0 t o 2 , a n d a r e a l w a y s p o s i t i v e . T h e t r i g o n o m e t r i c f u n c t i o n s c h a n g e s i g n i n p a s s i n g t h r o u g h t h e v a l u e s 0 a n d o o , a n d t h r o u g h n o o t h e r v a l u e s . I n t h e l s t q u a d r a n t t h e f u n c t i o n s i n c r e a s e , a n d t h e c o f u n c t i o n s d e c r e a s e . N o t e 2 . — F r o m t h e r e s u l t s g i v e n i n t h e a b o v e t a b l e , i t w i l l b e s e e n t h a t , i f t h e v a l u e o f a t r i g o n o m e t r i c f u n c t i o n b e g i v e n , w e c a n n o t f i x o n o n e a n g l e t o w h i c h i t b e l o n g s e x c l u s i v e l y . T h u s , i f t h e g i v e n v a l u e o f s i n A b e w e k n o w s i n c e s i n A p a s s e s t h r o u g h a l l v a l u e s f r o m 0 t o 1 a s A i n c r e a s e s f r o m 0 ° t o 9 0 % t h a t o n e v a l u e o f A l i e s b e t w e e n 0 ° R E L A T I O N S B E T W E E N F U N C T I O N S . 2 9 a n d 9 0 ° . B u t s i n c e w e a l s o k n o w t h a t t h e v a l u e o f s i n A p a s s e s t h r o u g h a l l v a l u e s b e t w e e n 1 a n d 0 a s A i n c r e a s e s f r o m 9 0 ° t o 1 8 0 ° , i t i s e v i d e n t t h a t t h e r e i s a n o t h e r v a l u e o f A b e t w e e n 9 0 ° a n d 1 8 0 ° f o r w h i c h s i n A = J . 2 3 . R e l a t i o n s b e t w e e n t h e T r i g o n o m e t r i c F u n c t i o n s o f t h e S a m e A n g l e . — L e t t h e r a d i u s s t a r t f r o m t h e i n i t i a l p o s i t i o n O A , a n d r e v o l v e i n e i t h e r d i r e c t i o n , t o t h e p o s i t i o n O P . L e t 6 d e n o t e t h e a n g l e t r a c e d o u t , a n d l e t t h e l e n g t h s o f t h e s i d e s P M , M O , O P b e d e n o t e d b y t h e l e t t e r s a , b , c . T h e f o l l o w i n g r e l a t i o n s a r e e v i d e n t f r o m t h e d e f i n i t i o n s ( A r t . 1 3 ) : 1 : J -c o s ( f a M b O c o s e c 6 -I . t a n 6 1 = F o r t a n 0 = . s e c 0 = — ^ . , c o t 6 = - -s i n 6 c o s 6 t a n 0 s i n 6 c o s 6 a a c b c s i n 6 c o s 6 I I . s i n 2 0 + c o s 2 0 = 1 . F o r s i n 2 0 + c o s 2 0 = - „ + = ^ 1 ± 6 ! = 1 . c 2 c 2 c 2 I I I . s e c 2 0 = l + t a n 2 t f . F o r s e c 2 0 = - = ^ - ± ^ = l + - = l + t a n 2 0 . b 2 V b 2 I V . c o s e c 2 0 = l + c o t 2 0 . . l + c o t 2 & F o r c o s e c 2 0 = ^ = ^ = l + ^ S i l l ( ( i l l F o r m u l a e I . , I I . , I I I . , I V . a r e v e r y i m p o r t a n t , a n d m u s t b e r e m e m b e r e d . a , b , c a r e n u m b e r s , b e i n g t h e n u m b e r o f t i m e s t h e l e n g t h s o f t h e s i d e s c o n t a i n s o m e c h o s e n u n i t o f l e n g t h . 3 0 P L A N E T R I G O N O M E T R Y . 2 4 . U s e o f t h e P r e c e d i n g F o r m u l a e . I . T o e x p r e s s a l l t h e o t h e r f u n c t i o n s i n t e r m s o f t h e s i n e . S i n c e s i n 2 0 + c o s 2 0 = 1 , . - . c o s 0 = ± V l — s i n 2 0 . t a n f l = s - l ^ = ± s i n 6 c o s 6 V I -s i n 2 6 t a n 0 s i n 6 s e c O — = ± c o s 0 " " " V l - s i n 2 ? c o s e c 0 = — i — . s i n 0 I I . 7 b e x p r e s s a l l t h e o t h e r f u n c t i o n s i n t e r m s o f t h e t a n g e n t . S i n c e t a n 0 = S l n ^ , c o s g c . ; „ d + o « t a n g t a n < > s i n 0 = t a n 0 c o s f > -= ± — — s e c 0 V l + t a n 2 0 c o s 0 = ^ L = ± . 1 s e c 6 J " V l + t a n ! f l c o t e = - i — s e c 0 = ± V 1 + t a n 2 0 . t a n 0 . 1 , V l + t a n 2 0 c o s e c f l = — — - = ± s i n 0 t a n 9 S i m i l a r l y , a n y o n e o f t h e f u n c t i o n s o f a n a n g l e m a y b e e x p r e s s e d i n t e r m s o f a n y o t h e r f u n c t i o n o f t h a t a n g l e . T h e s i g n o f t h e r a d i c a l w i l l i n a l l c a s e s d e p e n d u p o n t h e q u a d r a n t i n w h i c h t h e a n g l e 6 l i e s . 2 5 . G r a p h i c M e t h o d o f f i n d i n g A l l t h e F u n c t i o n s i n T e r m s o f O n e o f t h e m . T o e x p r e s s a l l t h e o t h e r f u n c t i o n s i n t e r m s o f t h e c o s e c a n t . C o n s t r u c t a r i g h t t r i a n g l e A B C , h a v i n g t h e s i d e B C = 1 . T h e n V o w e c - i R E L A T I O N S B E T W E E N F U N C T I O N S . 3 1 c o s e c A : : A B = A B = A R B C 1 . - . A C = ± V c o s e c 2 A — 1 . N o w s i n A = - — = r , A B c o s e c A A A C , V c o s e c 2 A — 1 c o s A = T 5 = ± . A B c o s e c A t a n A = | § = ± -V c o s e c 2 A — 1 a n d s i m i l a r l y t h e o t h e r f u n c t i o n s m a y b e e x p r e s s e d i n t e r m s o f c o s e c A . 2 6 . T o f i n d t h e T r i g o n o m e t r i c F u n c t i o n s o f 4 5 ° . — L e t A B C b e a n i s o s c e l e s r i g h t t r i a n g l e i n w h i c h C A = C B . T h e n C A B = C B A = 4 5 ° . L e t A C = m = C B . T h e n A B 2 = A C 2 + C B 2 = m J + m 2 = 2 m 2 . A . - . A B = m V 2 . s i n 4 5 ° = B C A B A K O A C c o s 4 5 = — = A B m m V 2 m m V 2 t a n 4 5 ° = 5 C = ™ = l . A C m s e c 4 5 ° = V 2 . c o s e c 4 5 ° = V 2 . 2 7 . T o f i n d t h e T r i g o n o m e t r i c F u n c t i o n s o f 6 0 ° a n d 3 0 ° . — L e t A B b e a n e q u i l a t e r a l t r i a n g l e . D r a w A D p e r p e n d i c u l a r t o B C . T h e n A D b i s e c t s t h e a n g l e B A C a n d t h e s i d e B C . T h e r e f o r e B A D = 3 0 ° , a n d A B D = 6 0 ° . 3 2 P L A N E T R I G O N O M E T R Y . L e t T h e n B A = 2 m . . - . B D = m . A D = V i m 2 — m 2 = m V 3 . s i n 6 0 ° = A D . V 3 c o s 6 0 ° A B 2 m B D = 1 B A 2 ' = £ V 3 . . - . c o s e c 6 0 ° = ^ . V 3 s e c 6 0 ° = 2 . t a n 6 0 ° = A D > V 3 B D m = V 3 . c o t 6 0 ° = — -V 3 v e r s 6 0 ° = 1 -c o s 6 0 ° = 1 -» i . o n o B D m 1 A l s o s i n 3 0 ° = - — = - — = - -A B 2 m 2 1 1 2 = " Y - . c o s e c 3 0 ° = 2 . c o s 3 0 ° = A D m V 3 t a n 3 0 ° = A B D B 2 m m = V 3 . D A m V 3 V 3 . s e c 3 0 ° = — -V 3 c o t 3 0 ° = . V 3 . E X A M P L E S . 1 . G i v e n s i n 0 = - ; f i n d t h e o t h e r o t r i g o n o m e t r i c f u n c t i o n s . L e t B A C b e t h e a n g l e , a n d B C b e p e r p e n d i c u l a r t o A C . R e p r e s e n t B C b y 3 , A B b y 5 , a n d c o n s e q u e n t l y A C £ b y V 2 6 - 9 = 4 . A C 4 T h e n c o s 6 -1 A A B 5 ' B C 3 R E D U C T I O N O F F U N C T I O N S . 3 3 3 3 5 2 . G i v e n s i n 6 = -; f i n d t a n 6 a n d c o s e c f t A n s . - > - -5 4 o 3 . G i v e n c o s 0 = - ; f i n d s i n 6 a n d c o t f t & V 2 , — i --3 2 V 2 4 . G i v e n s e c 0 = 4 ; f i n d c o t 0 a n d s i 1 1 f t 1 V l T > V l 5 4 5 . G i v e n t a n 6 = V 3 ; f i n d s i n 6 a n d c o s f t £ V 3 , - -6 . G i v e n s i n 6 — — ; f i n d c o s 6 1 2 £ - i a 5 — : f i n d c o s f t — -1 3 , 1 3 7 . G i v e n c o s e c 0 = 5 ; f i n d s e c 6 a n d t a n 0 . 1 2 V 6 2 V 6 4 1 4 0 9 8 . G i v e n s e c 6 = — ; f i n d s i n 6 a n d c o t f t — > — . 9 4 1 4 1 2 " \ / 5 3 9 . G i v e n c o t 6 = — ; f i n d s i n 6 a n d s e c f t — — > - -V 5 3 2 1 0 . G i v e n s i n 0 = f ; f i n d c o s f t t a n f t a n d c o t f t 4 V 7 3 V 7 V 7 x , T ~ , - 3 -1 1 . G i v e n s i n 6 = -; f i n d t a n f t c V c ^ T 2 1 2 . G i v e n s i n 0 = — - ; f i n d t a n f t 2 8 . K e d u c t i o n o f T r i g o n o m e t r i c F u n c t i o n s t o t h e 1 s t Q u a d r a n t . — A l l m a t h e m a t i c a l t a b l e s g i v e t h e t r i g o n o m e t r i c f u n c t i o n s o f a n g l e s b e t w e e n 0 ° a n d 9 0 ° o n l y , b u t i n p r a c t i c e w e c o n s t a n t l y h a v e t o d e a l w i t h a n g l e s g r e a t e r t h a n 9 0 ° . T h e o b j e c t o f t h e f o l l o w i n g s i x A r t i c l e s i s t o s h o w t h a t t h e t r i g o n o m e t r i c f u n c t i o n s o f a n y a n g l e , p o s i t i v e o r n e g a t i v e , c a n b e e x p r e s s e d i n t e r m s o f t h e t r i g o n o m e t r i c f u n c t i o n s o f a n a n g l e l e s s t h a n 9 0 ° , s o t h a t , i f a g i v e n a n g l e i s g r e a t e r t h a n 9 0 ° , w e c a n f i n d a n a n g l e i n t h e 1 s t q u a d r a n t w h o s e t r i g o n o m e t r i c f u n c t i o n h a s t h e s a m e a b s o l u t e v a l u e . 3 4 P L A N E T R I G O N O M E T R Y . 2 9 . F u n c t i o n s o f C o m p l e m e n t a l A n g l e s . — L e t A A ' , B B ' b e t w o d i a m e t e r s o f a c i r c l e a t r i g h t a n g l e s , a n d l e t O P a n d O P ' b e t h e p o s i t i o n s o f t h e r a d i u s f o r a n y a n g l e A O P = A , a n d i t s c o m p l e m e n t A O P ' = 9 0 ° - A ( A r t . 1 2 ) . D r a w P M a n d P ' M ' a t r i g h t a n g l e s t o O A . A n g l e O P ' M ' = B O P ' = A O P = A . A l s o O P = O P ' . H e n c e t h e t r i a n g l e s O P M a n d O P ' M ' a r e e q u a l i n a l l r e s p e c t s . . - . P ' M ' = O M . . . . ™. ' = O M O P ' O P . - . s i n ( 9 0 ° -A ) = c o s A O P = c o s A . O M ' _ P M " O P ' ~ ~ O P ' . - . c o s ( 9 0 ° -A ) = s i n A O P = s i n A . S i m i l a r l y , t a n ( 9 0 ° - A ) = t a n A O P ' = 5 ! M . ' = < M = e o t A . J > \ J . Q p , Q p T h e o t h e r r e l a t i o n s a r e o b t a i n e d b y i n v e r t i n g t h e a b o v e . 3 0 . F u n c t i o n s o f S u p p l e m e n t a l A n g l e s . — L e t O P a n d O P ' b e t h e p o s i t i o n s o f t h e r a d i u s f o r a n y a n g l e A O P = A , P / a n d i t s s u p p l e m e n t A O P ' = 1 8 0 ° - A , ( A r t . 1 2 ) . S i n c e O P = O P ' , a n d P O A = P ' O A ' , t h e t r i a n g l e s P O M a n d P ' O M ' a r e g e o m e t r i c a l l y e q u a l . A l s o , O M ' = P M . . - . s i n ( 1 8 0 ° - A ) = s i n A O P ' = c o s ( 1 8 0 ° - A ) = c o s A O P ' = = ^ M = - c o s A , P ' M ' O P ' O M ' O P ' P M . . = = s i n A , O P , R E D U C T I O N O F F U N C T I O N S . 3 5 t a n ( 1 8 0 ° - A ) = t a n A O F ' = = — = -t a n A . v ' O M ' O M S i m i l a r l y t h e o t h e r r e l a t i o n s m a y b e o b t a i n e d . 3 1 . T o p r o v e s i n ( 9 0 ° + A ) = c o s A , c o s ( 9 0 ° + A ) = -s i n A , a n d t a n ( 9 0 ° + A ) = -c o t A . „ , B L e t O P a n d O P ' b e t h e p o s i t i o n s o f t h e r a d i u s f o r a n y a n g l e A O P = A , a n d A O P ' = 9 0 ° + A . S i n c e O P = O P ' , a n d A O P = P ' O B = O P ' M ' , t h e t r i a n g l e s P O M a n d P ' O M ' a r e e q u a l i n a l l r e s p e c t s . . - . s i n ( 9 0 ° + A ) = s i n A O P ' = . 2 ? J = c o s A , D M ' _ V l f c o s ( 9 0 ° + A ) = c o s A O P ' = ^ 7 = = = -s i n A , p ! O M l t a n ( 9 0 ° + A ) = t a n A O P ' = P ' M ' O M O M ' - P M = — c o t A . 3 2 . T o p r o v e s i n ( 1 8 0 ° + A ) = - s i n A , c o s ( 1 8 0 ° + A ) = -c o s A , a n d t a n ( 1 8 0 ° + A ) = t a n A . L e t t h e a n g l e A O P = A ; t h e n t h e a n g l e A O P ' , m e a s u r e d i n t h e p o s i t i v e A d i r e c t i o n , = ( 1 8 0 ° + A ) . p T h e t r i a n g l e s P O M a n d P ' O M ' a r e e q u a l . . - . s i n ( 1 8 0 ° + A ) = s i n A O P ' = c o s ( 1 8 0 ° + A ) = c o s A O P ' = — = ^ ° M v o p ' O P O P ' O M ' O P ' P ' M ' - P M t a n ( 1 8 0 ° + A ) = t a n A O P ' = ~ L ' = ~ ™ = t a n A . 3 6 P L A N E T R I G O N O M E T R Y . 3 3 . T o p r o v e s i n ( — A ) = — s i n A , c o s ( — A ) = c o s A , t a n ( — A ) = — t a n A . L e t O P a n d O P ' b e t h e p o s i t i o n s o f t h e r a d i u s f o r a n y e q u a l a n g l e s A O P a n d A O P ' m e a s u r e d f r o m t h e i n i t i a l l i n e A O i n o p p o s i t e d i r e c t i o n s . T h e n i f t h e a n g l e A O P b e d e n o t e d b y A , t h e n u m e r i c a l l y e q u a l a n g l e A O P ' w i l l b e d e n o t e d b y — A ( A r t . 4 ) . T h e t r i a n g l e s P O M a n d P ' O M a r e g e o m e t r i c a l l y e q u a l . P ' M -P M . - . s i n ( - A ) = s i n A O P ' = O P ' O P : — s i n A , / a \ a r v o i O M O M . c o s ( -A ) = c o s A O P ' = ( — = ^ p - = c o s A , t a n ( -A ) = t a n A O P ' = ^ = — ~ = -t a n A . O M O M 3 4 . T o p r o v e s i n ( 2 7 0 ° + A ) = s i n ( 2 7 0 ° -A ) = -c o s A , a n d c o s ( 2 7 0 ° + A ) = -c o s ( 2 7 0 ° - A ) B = s i n A . L e t t h e a n g l e A O P = A ; t h e n t h e a n g l e s A O Q a n d A O R , m e a s u r e d i n t h e p o s i t i v e d i r e c t i o n , = ( 2 7 0 ° - A ) A l a n d ( 2 7 0 ° + A ) r e s p e c t i v e l y . T h e t r i a n g l e s P O M , Q O N , a n d R O L a r e g e o m e t r i c a l l y e q u a l . N O b 0 R L = Q N = O M . R L _ _ Q 1 S T O R O Q : O M O P A l s o , - . s i n ( 2 7 0 ° + A ) = s i n ( 2 7 0 ° - A ) = -c o s A . O L = - O N = P M O R O Q O P ' - . c o s ( 2 7 0 ° + A ) = -c o s ( 2 7 0 ° - A ) = s i n A . V A L U E S O F T H E F U N C T I O N S . 3 7 3 5 . T a b l e g i v i n g t h e V a l u e s o f t h e F u n c t i o n s o f A n y A n g l e i n T e r m s o f t h e F u n c t i o n s o f a n A n g l e l e s s t h a n 9 0 ° . — T h e f o r e g o i n g r e s u l t s , a n d o t h e r s i m i l a r o n e s , w h i c h m a y b e p r o v e d i n t h e s a m e m a n n e r , a r e h e r e c o l l e c t e d f o r r e f e r e n c e . Q u a d r a n t I I . s i n ( 1 8 0 ° -A ) = s i n A . c o s ( 1 8 0 ° - A ) = -c o s A . t a n ( 1 8 0 ° - A ) = - t a n A . c o t ( 1 8 0 ° - A ) = — c o t A . s e c ( 1 8 0 ° - A ) = -s e c A . c o s e c ( 1 8 0 ° — A ) = c o s e c A . s i n ( 9 0 ° + A ) = c o s A . c o s ( 9 0 ° + A ) = -s i n A . t a n ( 9 0 ° + A ) = - c o t A . c o t ( 9 0 ° + A ) = -t a n A . s e c ( 9 0 ° + A ) = — c o s e c A . c o s e c ( 9 0 ° + A ) = s e c A . s i n ( 1 8 0 ° + A ) : c o s ( 1 8 0 ° + A ) : t a n ( 1 8 0 ° + A ) : c o t ( 1 8 0 ° + A ) : s e c ( 1 8 0 ° + A ) = c o s e c ( 1 8 0 ° + A ) : Q u a d r a n t I I I . . s i n A . s i n ( 2 7 0 ° — A ) = — c o s A . . c o s A . c o s ( 2 7 0 ° — A ) = — s i n A . t a n A . t a n ( 2 7 0 ° -A ) = c o t A . c o t A . c o t ( 2 7 0 ° -A ) = t a n A . s e c A . s e c ( 2 7 0 ° — A ) = — c o s e c A . . c o s e c A . c o s e c ( 2 7 0 ° — A ) = — s e c A . Q u a d r a n t I V . s i n ( 2 7 0 ° + A ) = -c o s A . c o s ( 2 7 0 ° + A ) = s i n A . t a n ( 2 7 0 ° + A ) = - c o t A . c o t ( 2 7 0 ° + A ) = -t a n A . s e c ( 2 7 0 ° + A ) = c o s e c A . s e c A . - s i n A . c o s A . . t a n A . - c o t A . s e c A . s i n ( 3 6 0 ° - A ) c o s ( 3 6 0 ° - A ) t a n ( 3 6 0 ° - A ) c o t ( 3 6 0 ° - A ) s e c ( 3 6 0 ° - A ) c o s e c ( 3 6 0 ° - A ) = -c o s e c A . c o s e c ( 2 7 0 ° + A ) = N o t e . — T h e s e r e l a t i o n s m a y b e r e m e m b e r e d b y n o t i n g t h e f o l l o w i n g r u l e s : W h e n A i s a s s o c i a t e d w i t h a n e v e n m u l t i p l e o f 9 0 ° , a n y f u n c t i o n o f t h e a n g l e i s n u m e r i c a l l y e q u a l t o t h e s a m e f u n c t i o n o f A . W h e n A i s a s s o c i a t e d w i t h a n o d d m u l t i p l e o f 9 0 ° , a n y f u n c t i o n o f t h e a n g l e i s n u m e r i c a l l y e q u a l t o t h e c o r r e s p o n d i n g c o f u n c t i o n o f t h e a n g l e A . T h e s i g n t o b e p r e f i x e d w i l l d e p e n d u p o n t h e q u a d r a n t t o w h i c h t h e a n g l e b e l o n g s ( A r t . 5 ) , r e g a r d i n g A a s a n a c u t e a n g l e . A l t h o u g h t h e s e r e l a t i o n s h a v e b e e n p r o v e d o n l y i n c a s e o f A , a n a c u t e a n g l e , t h e y a r e t r u e w h a t e v e r A m a y b e . 3 8 P L A N E T R I G O N O M E T R Y . T h u s , c o s ( 2 7 0 ° - A ) = — s i n A ; t h e a n g l e 2 7 0 ° — A b e i n g i n t h e 3 d q u a d r a n t , a n d i t s c o s i n e n e g a t i v e i l l c o n s e q u e n c e . F o r a n a n g l e i n t h e F i r s t q u a d r a n t a l l t h e f u n c t i o n s a r e p o s i t i v e . S e c o n d q u a d r a n t a l l a r e n e g a t i v e e x c e p t t h e s i n e a n d c o s e c a n t . T h i r d q u a d r a n t a l l a r e n e g a t i v e e x c e p t t h e t a n g e n t a n d c o t a n g e n t . F o u r t h q u a d r a n t a l l a r e n e g a t i v e e x c e p t t h e c o s i n e a u d s e c a n t . 3 6 . P e r i o d i c i t y o f t h e T r i g o n o m e t r i c F u n c t i o n s . — L e t A O P b e a n a n g l e o f a n y m a g n i t u d e , a s i n t h e f i g u r e o f A r t . 1 8 ; t h e n i f O P r e v o l v e i n t h e p o s i t i v e o r t h e n e g a t i v e d i r e c t i o n t h r o u g h a n a n g l e o f 3 6 0 ° , i t w i l l r e t u r n t o t h e p o s i t i o n f r o m w h i c h i t s t a r t e d . H e n c e i t i s c l e a r f r o m t h e d e f i n i t i o n s t h a t t h e t r i g o n o m e t r i c f u n c t i o n s r e m a i n u n c h a n g e d w h e n t h e a n g l e i s i n c r e a s e d o r d i m i n i s h e d b y 3 6 0 " , o r a n y m u l t i p l e o f 3 6 0 ° . T h u s t h e f u n c t i o n s o f t h e a n g l e 4 0 0 ° a r e t h e s a m e b o t h i n n u m e r i c a l v a l u e a n d i n a l g e b r a i c s i g n a s t h e f u n c t i o n s o f t h e a n g l e o f 4 0 0 ° — 3 6 0 ° , i . e . , o f t h e a n g l e o f 4 0 ° . A l s o t h e f u n c t i o n s o f 3 6 0 ° + A a r e t h e s a m e i n n u m e r i c a l v a l u e a n d i n s i g n a s t h o s e o f A . I n g e n e r a l , i f n d e n o t e a n y i n t e g e r , e i t h e r p o s i t i v e o r n e g a t i v e , t h e f u n c t i o n s o f n x 3 6 0 ° + A a r e t h e s a m e a s t h o s e o f A . T h u s t h e f u n c t i o n s o f 1 4 7 0 ° = t h e f u n c t i o n s o f 3 0 ° . I f 6 d e n o t e s a n y a n g l e i n c i r c u l a r m e a s u r e , t h e f u n c t i o n s o f ( 2 n i r + 6 ) a r e t h e s a m e a s t h o s e o f 6 . T h u s s i n ( 2 n i r + 0 ) = s i n 0 , c o s ( 2 n t t + 6 ) = c o s 6 , e t c . B y t h i s p r o p o s i t i o n w e c a n r e d u c e a n a n g l e o f a n y m a g n i t u d e t o a n a n g l e l e s s t h a n 3 6 0 ° w i t h o u t c h a n g i n g t h e v a l u e s o f t h e f u n c t i o n s . I t i s t h e r e f o r e u n n e c e s s a r y t o c o n s i d e r t h e f u n c t i o n s o f a n g l e s g r e a t e r t h a n 3 6 0 ° ; t h e f o r m u l a e a l r e a d y e s t a b l i s h e d a r e t r u e f o r a n g l e s o f a n y m a g n i t u d e w h a t e v e r . E X A M P L E S . E x p r e s s s i n 7 0 0 ° i n t e r m s o f t h e f u n c t i o n s o f a n a c u t e a n g l e . s i n 7 0 0 ° = s i n ( 3 6 0 ° + 3 4 0 ° ) = s i n 3 4 0 ° = s i n ( 1 8 0 ° + 1 6 0 ° ) = s i n l 6 0 ° = - s i n 2 0 ° . E X A M P L E S . 3 9 E x p r e s s t h e f o l l o w i n g f u n c t i o n s i n t e r m s o f t h e f u n c t i o n s o f a c u t e a n g l e s : 1 . s i n 2 0 4 ° , s i n 5 1 0 ° . A n s . - s i n 2 4 ° , s i n 3 0 ° . 2 . c o s ( - 8 0 0 ° ) , c o s 3 5 9 ° . c o s 8 0 ° , c o s l ° . 3 . t a n 5 0 0 ° , t a n 3 0 0 ° . - t a n 4 0 ° , - c o t 3 0 ° . F i n d t h e v a l u e o f t h e s i n e , c o s i n e , a n d t a n g e n t o f t h e f o l l o w i n g a n g l e s : 4 . 1 5 0 ° . A n s . - , - I V S , 2 " V 3 5 . - 2 4 0 ° . V 3 , - \ , - V 3 . L i 6 . 3 3 0 ° . - \ $ V 3 , L . 2 V 3 7 . 2 2 5 ° . — , - — , 1 . V 2 V 2 F i n d t h e v a l u e s o f t h e f o l l o w i n g f u n c t i o n s : 8 . s i n 8 1 0 ° , s i n ( - 2 4 0 ° ) , c o s 2 1 0 ° . A n s . 1 , £ V 3 , - £ V 3 . 9 . t a n ( - 1 2 0 ° ) , c o t 4 2 0 ° , c o t 5 1 0 ° . V 3 , — , 1 . V 3 1 0 . s i n 9 3 0 ° t a n 0 4 2 0 ° . - - , J — 2 ' v a i l , c o t 1 0 3 5 ° , c o s e c 5 7 0 ° . - 1 , - 2 . 3 7 . A n g l e s c o r r e s p o n d i n g t o G i v e n F u n c t i o n s . — W h e n a n a n g l e i s g i v e n , w e c a n f i n d i t s t r i g o n o m e t r i c f u n c t i o n s , a s i n A r t s . 2 6 a n d 2 7 ; a n d t o e a c h v a l u e o f t h e a n g l e t h e r e i s b u t o n e v a l u e o f e a c h o f t h e f u n c t i o n s . B u t i n t h e c o n v e r s e p r o p o s i t i o n — b e i n g g i v e n t h e v a l u e o f t h e t r i g o n o m e t r i c f u n c t i o n s , t o f i n d t h e c o r r e s p o n d i n g a n g l e s — w e h a v e s e e n ( A r t . 3 6 ) t h a t t h e r e a r e m a n y a n g l e s o f d i f f e r e n t m a g n i t u d e w h i c h h a v e t h e s a m e f u n c t i o n s . I f t w o s u c h a n g l e s a r e i n t h e s a m e q u a d r a n t , t h e y a r e r e p r e s e n t e d g e o m e t r i c a l l y b y t h e s a m e p o s i t i o n o f O P , s o t h a t t h e y d i f f e r b y s o m e m u l t i p l e o f f o u r r i g h t a n g l e s . 4 0 P L A N E T R I G O N O M E T R Y . I f w e a r e g i v e n t h e v a l u e o f t h e s i n e o f a n a n g l e , i t i s i m p o r t a n t t o b e a b l e t o f i n d a l l t h e a n g l e s w h i c h h a v e t h a t v a l u e f o r t h e i r s i n e . 3 8 . G e n e r a l E x p r e s s i o n f o r A l l A n g l e s w h i c h h a v e a G i v e n S i n e a . — L e t O b e t h e c e n t r e o f t h e u n i t c i r c l e . D r a w t h e d i a m e t e r s A A ' , B B ' , a t r i g h t a n g l e s . F r o m O d r a w o n O B a l i n e O N , s o t h a t i t s m e a s u r e i s a . B l w o m I A A ' . J o i n O P , O P ' , a n d d r a w P M , P ' M ' , p e r p e n d i c u l a r t o A A ' . T h e n s i n c e M P = M ' P ' = O N = a , t h e s i n e o f A O P i s e q u a l t o t h e s i n e o f A O P ' . H e n c e t h e a n g l e s A O P a n d A O P ' a r e s u p p l e m e n t a l ( A r t . 3 0 ) , a n d i f A O P b e d e n o t e d b y a , A O P ' w i l l = j r -a . N o w i t i s c l e a r f r o m t h e f i g u r e t h a t t h e o n l y p o s i t i v e a n g l e s w h i c h h a v e t h e s i n e e q u a l t o a a r e a a n d - n — a , a n d t h e a n g l e s f o r m e d b y a d d i n g a n y m u l t i p l e o f f o u r r i g h t a n g l e s t o a a n d j r — a . H e n c e , i f 6 b e t h e g e n e r a l v a l u e o f t h e r e q u i r e d a n g l e , w e h a v e 6 = 2 n i r + a , o r 6 = 2 m r + i r — a , . . . . ( 1 ) w h e r e n i s z e r o o r a n y p o s i t i v e i n t e g e r . A l s o t h e o n l y n e g a t i v e a n g l e s w h i c h h a v e t h e s i n e e q u a l t o a a r e — ( i r + « ) , a n d — ( 2 i r — a ) , a n d t h e a n g l e s f o r m e d b y a d d i n g t o t h e s e a n y m u l t i p l e o f f o u r r i g h t a n g l e s t a k e n n e g a t i v e l y ; t h a t i s , w e h a v e 0 = 2 n i r - ( j r + « ) , 0 = 2 m i r - ( 2 i r - « ) , . . ( 2 ) w h e r e n i s z e r o o r a n y n e g a t i v e i n t e g e r . N o w t h e a n g l e s i n ( 1 ) a n d ( 2 ) m a y b e a r r a n g e d t h u s : 2 m r + a , ( 2 ? l + l ) i r — a , ( 2 j l — 1 ) i r — a , ( 2 n — 2 ) i r + « , a l l o f w h i c h , a n d n o o t h e r s , a r e i n c l u d e d i n t h e f o r m u l a 6 = m r + ( - ! ) " « , ( 3 ) G E N E R A L E X P R E S S I O N F O R A N G L E S . 4 1 w h e r e n i s z e r o , o r a n y p o s i t i v e o r n e g a t i v e i n t e g e r . T h e r e f o r e ( 3 ) i s t h e g e n e r a l e x p r e s s i o n f o r a l l a n g l e s w h i c h h a v e a g i v e n s i n e . N o t e . — T h e s a m e f o r m u l a d e t e r m i n e s a l l t h e a n g l e s w h i c h h a v e t h e s a m e c o s e c a n t a s a . 3 9 . A n E x p r e s s i o n f o r A l l A n g l e s w i t h a G i v e n C o s i n e a . — L e t O b e t h e c e n t r e o f t h e u n i t c i r c l e . B D r a w A A ' , B B ' , a t r i g h t a n g l e s . F r o m O d r a w O M , s o t h a t i t s m e a s u r e i s a . T h r o u g h M d r a w P P ' p a r a l l e l t o B B ' . J o i n O P , O P ' . T h e n s i n c e O M = a , t h e c o s i n e o f A O P i s e q u a l t o t h e c o s i n e o f A O P ' . H e n c e , i f A O P = a , A O P ' = -« . N o w i t i s c l e a r t h a t t h e o n l y a n g l e s w h i c h h a v e t h e c o s i n e e q u a l t o a a r e a a n d — a , a n d t h e a n g l e s w h i c h d i f f e r f r o m e i t h e r b y a m u l t i p l e o f f o u r r i g h t a n g l e s . H e n c e i f 6 b e t h e g e n e r a l v a l u e o f a l l a n g l e s w h o s e c o s i n e i s a , w e h a v e 6 = 2 n i r ± a , n V V w h e r e n i s z e r o , o r a n y p o s i t i v e o r n e g a t i v e i n t e g e r . N o t e . — T h e s a m e f o r m u l a d e t e r m i n e s a l l t h e a n g l e s w h i c h h a v e t h e s a m e s e c a n t o r t h e s a m e v e r s e d s i n e a s a . 4 0 . A n E x p r e s s i o n f o r A l l A n g l e s w i t h a G i v e n T a n g e n t a . — L e t O b e t h e c e n t r e o f t h e u n i t c i r c l e . D r a w A T , t o u c h i n g t h e c i r c l e a t A , a n d t a k e A T s o t h a t i t s m e a s u r e i s a . J o i n O T , c u t t i n g t h e c i r c l e a t P A ' a n d P ' . T h e n i t i s c l e a r f r o m t h e f i g u r e t h a t t h e o n l y a n g l e s w h i c h h a v e t h e t a n g e n t e q u a l t o a a r e a a n d n + a , a n d t h e a n g l e s w h i c h d i f f e r 4 2 P L A N E T R I G O N O M E T R Y . f r o m e i t h e r b y a m u l t i p l e o f f o u r r i g h t a n g l e s . H e n c e i f 6 b e t h e g e n e r a l v a l u e o f t h e r e q u i r e d a n g l e , w e h a v e 6 = 2 n v + a , a n d 2 n i r + i r + « ( 1 ) A l s o , t h e o n l y n e g a t i v e a n g l e s w h i c h h a v e t h e t a n g e n t e q u a l t o a a r e — ( i r — a ) , a n d — ( 2 - — « ) , a n d t h e a n g l e s w h i c h d i f f e r f r o m e i t h e r b y a m u l t i p l e o f f o u r r i g h t a n g l e s t a k e n n e g a t i v e l y ; t h a t i s , w e h a v e 0 = 2 M i r - ( i r - a ) , a n d 2 ? u r -( 2 - -« ) , . . ( 2 ) w h e r e n i s z e r o o r a n y n e g a t i v e i n t e g e r . N o w t h e a n g l e s i n ( 1 ) a n d ( 2 ) m a y b e a r r a n g e d t h u s : 2 n i r + a , ( 2 n + l ) i r + « , ( 2 n — l ) i r + « , ( 2 n — 2 ) v + a , a l l o f w h i c h , a n d n o o t h e r s , a r e i n c l u d e d i n t h e f o r m u l a 6 = n - i r + a , ( 3 ) w h e r e n i s z e r o , o r a n y p o s i t i v e o r n e g a t i v e i n t e g e r . T h e r e f o r e ( 3 ) i s t h e g e n e r a l e x p r e s s i o n f o r a l l a n g l e s w h i c h h a v e a g i v e n t a n g e n t . N o t e . — T h e b u m f o r m u l a d e t e r m i n e s a l l t h e a n g l e s w h i c h h a v e t h e s a m e c o t a n g e n t a s a . E X A M P L E S . 1 . F i n d s i x a n g l e s b e t w e e n — 4 r i g h t a n g l e s a n d + 8 r i g h t a n g l e s w h i c h s a t i s f y t h e e q u a t i o n s i n A = s i n 1 8 ° . W e h a v e f r o m ( 3 ) o f A r t . 3 8 , O = m r + ( - l ) " j ^ , o r A = n x 1 8 0 ° + ( -1 ) " 1 8 ° . P u t f o r n t h e v a l u e s — 2 , — 1 , 0 , 1 , 2 , 3 , s u c c e s s i v e l y , a n d w e g e t A = -3 6 0 ° + 1 8 ° , -1 8 0 ° -1 8 ° , 1 8 ° , 1 8 0 ° -1 8 ° , 3 6 0 ° + 1 8 ° , 5 4 0 ° -1 8 ° ; t h a t i s , -3 4 2 ° , -1 9 8 ° , 1 8 ° , 1 6 2 ° , 3 7 8 ° , 5 2 2 ° . N o t e . — T b e s t u d e n t s h o u l d d r a w a f i g u r e I n t h e a b o v e e x a m p l e , a n d i n e a c h e x a m p l e o f t h i s k i n d w h i c h h e w o r k s . T R I G O N O M E T R I C I D E N T I T I E S . 4 8 2 . F i n d t h e f o u r s m a l l e s t a n g l e s w h i c h s a t i s f y t h e e q u a t i o n s ( 1 ) s i n A = i , ( 2 ) s i n A = - — , ( 3 ) s i n A = ^ - j i 2 V 2 2 ( 4 ) s i n A = - i . v ' 2 A n a . ( 1 ) 3 0 ° , 1 5 0 ° , -2 1 0 ° , -3 3 0 ° ; ( 2 ) 4 5 ° , 1 3 5 ° , -2 2 5 ° , -3 1 5 ° ; ( 3 ) 6 0 ° , 1 2 0 ° , -2 4 0 ° , -3 0 0 ° ; ( 4 ) _ 3 0 ° , -1 5 0 ° , 2 1 0 ° , 3 3 0 ° . 4 1 . T r i g o n o m e t r i c I d e n t i t i e s . — A t r i g o n o m e t r i c i d e n t i t y i s a n e x p r e s s i o n w h i c h s t a t e s i n t h e f o r m o f a n e q u a t i o n a r e l a t i o n w h i c h i s t r u e f o r a l l v a l u e s o f t h e a n g l e i n v o l v e d . T h u s , t h e r e l a t i o n s o f A r t s . 1 3 a n d 2 3 , a n d a l l o t h e r s t h a t m a y b e d e d u c e d f r o m t h e m b y t h e a i d o f t h e o r d i n a r y f o r m u l a e o f A l g e b r a , a r e u n i v e r s a l l y t r u e , a n d a r e t h e r e f o r e c a l l e d i d e n t i t i e s ; b u t s u c h r e l a t i o n s a s s i n 6 = % , c o s 0 = i , a r e n o t i d e n t i t i e s . E X A M P L E S . 1 . P r o v e t h a t s e c 6 — t a n 6 - s i n 6 = c o s 6 . H e r e s e c 6 — t a n 0 s i n 6 = — s _ 1 — s i n 0 ( A r t . 2 4 ) c o s 6 c o s 6 l - s i n 2 0 c O s 0 c O S 2 0 ( A r t . 2 4 ) c O s 0 = c o s 6 . 2 . P r o v e t h a t c o t 6 — s e c 6 c o s e c 6 ( 1 — 2 s i n 2 6 ) = t a n 6 . c o t O — s e c 6 c o s e c 6 ( 1 — 2 s i n 2 0 ) = c o s f l ? _ - j L M _ 2 s i n 2 0 ) ( A r t . 2 4 ) s i n ^ c o s 6 s i n 6 ' _ c o 8 ! f l — l + 2 s i n 2 0 s i n 6 c o s 6 4 4 P L A N E T R I G O N O M E T R Y . c o s 2 0 -( s i n 2 0 + c o s 2 f l ) + 2 s i n 2 6 s i n 0 c o s 0 ( A r t . 2 4 ) - s i n 2 6 s i n 0 S i l l 1 7 , „ = — t a n 6 . s i n 0 c o s 6 c o s 6 N o t e . — I t w i l l b e o b s e r v e d t h a t i n s o l v i n g t h e s e e x a m p l e s w e f i r s t e x p r e s s t h e o t h e r f u n c t i o n s i n t e r m s o f t h e s i n e a n d c o s i n e , a n d i n m o s t c a s e s t h e b e g i n n e r w i l l f i n d t h i s t h e s i m p l e s t c o u r s e . I t i s g e n e r a l l y a d v i s a b l e t o b e g i n w i t h t h e m o s t c o m p l i c a t e d s i d e a n d w o r k t o w a r d s t h e o t h e r . P r o v e t h e f o l l o w i n g i d e n t i t i e s : 3 . c o s 6 t a n 6 = s i n 6 . 4 . c o s 6 = s i n 6 c o t 6 . 5 . ( t a n 0 + c o t 0 ) s i n 0 c o s 6 = 1 . 6 . ( t a n 6 — c o t 6 ) s i n 6 c o s 6 = s i n 2 6 — c o s 2 6 . 7 . s i n 2 6 s - c o s e c 2 6 = s i n 4 6 . 8 . s e c 4 6 — t a n 4 6 = s e c 2 6 + t a n 2 6 . 9 . ( s i n 6 - c o s 6 ) - = 1 - 2 s i n 6 c o s 6 . 1 0 . 1 -t a n 4 6 = 2 s e c 2 6 -s e c 4 6 . 1 1 . 1 + c o s 6 = ( c o s e c 6 + c o t 6 ) 1 . l - c o s f l v ; 1 2 . ( s i n e + c o s 0 ) 2 + ( s i n 0 -c o s 0 ) 2 = 2 . 1 3 . s i n 4 0 — c o s 4 6 = s i n 2 6 — c o s 2 6 . 1 4 . s i n 2 0 + v e r s 2 6 = 2 ( 1 - c o s 0 ) . 1 5 . c o t 2 6 -c o s 2 0 = c o t 2 6 c o s 2 0 . I n a r i g h t t r i a n g l e A B C ( s e e f i g u r e o f A r t . 1 5 ) g i v e n : 1 . a = p 2 - \ - p q , c = q 2 + p q ; c a l c u l a t e c o t A . — — 2 . b = I m - t - n , c = l n - r - m ; c a l c u l a t e c o s e c A . E X A M P L E S . A n s . P V n — m E X A M P L E S . 4 5 3 s i n A = - , c = 2 0 . 5 ; c a l c u l a t e a . A n s . 1 2 . 3 . 5 4 . G i v e n c o t £ A = t a n A ; f i n d A . 5 . a s i n A = c o s 2 A ; f i n d A . 6 . u c o t A = t a n 6 A ; f i n d A . 7 . t t t a n A = c o t 8 A ; f i n d A . 8 . a s i n 2 A = c o s 3 A ; f i n d A . V S 2 3 , V . > ' 9 . t t s i n A = -; f i n d c o s A a n d t a n A . 3 , 3 3 5 , 4 1 0 . a c o s A = f i n d s i n A a n d t a n A 5 , V 7 3 4 , W 1 1 . u c o s e c A = -: f i n d c o s A a n d t a n A . 3 , 1 2 . t t s i n A = - ! - : f i n d c o s A a n d t a n A . V 3 S l 1 3 . t t c o s A = 6 ; f i n d t a n A a n d c o s e c A . V l - 6 2 b , 1 1 4 u s i n A = . 6 ; f i n d c o s A a n d c o t A . V i -v 4 4 5 , 3 ' 4 1 5 . a t a n A = - : f i n d s i n A . 5 , V i l 1 6 . t t c o t A = — ; f i n d s e c A a n d s i n A . 1 5 , c 8 1 5 1 7 , 1 7 " 1 7 . t t s i n A = — ; f i n d c o s A . 1 3 1 2 5 1 3 " 1 8 . t t c o s A = . 2 8 ; f i n d s i n A . . 9 6 . 1 9 . t t t a n A = f i n d s i n A . 3 , 4 5 ' 2 0 . t t s i n A = i ; f i n d c o s A . 3 , f V 2 . 4 6 P L A N E T R I G O N O M E T R Y . 2 1 . G i v e n t a n A = f i n d s i n A a n d s e c A . A n s . - , ^ . 3 5 3 2 2 . 2 3 . t a ñ 0 = - ; f i n d s i n 0 a n d c o s 0 . b „ " c o s 6 = -; f i n d s i n 6 a n d c o t a V e i 2 + V V a 2 + 6 2 f t V a 2 ^ ! 1 _ 2 4 . I f s i n 0 = a , a n d t a n 0 = b , p r o v e t h a t ( 1 - a 2 ) ( 1 + 6 2 ) = 1 . E x p r e s s t h e f o l l o w i n g f u n c t i o n s i n t e r m s o f t h e f u n c t i o n s o f a c u t e a n g l e s l e s s t h a n 4 5 ° : 2 5 . s i n 1 6 8 ° , s i n 2 1 0 ° . 2 6 . t a n 1 2 5 ° , t a n 3 1 0 ° . 2 7 . s e c 2 4 4 ° , c o s e c 2 8 1 ° . 2 8 . s e c 9 3 0 ° , c o s e c ( - 6 0 0 ° ) . 2 9 . c o t 4 6 0 ° , s e c 2 9 9 ° . 3 0 . t a n 1 4 0 0 ° , c o t ( - 1 4 0 0 ° ) . A n s . s i n 1 2 ° , - s i n 3 0 ° . - c o t 3 5 ° , - c o t 4 0 ° . - c o s e c 2 6 ° , - s e c 1 1 ° . - s e c 3 0 ° , s e c 3 0 ° . — t a n 1 0 ° , c o s e c 2 9 ° . - t a n 4 0 ° , c o t 4 0 ° . F i n d t h e v a l u e s o f t h e f o l l o w i n g f u n c t i o n s 3 1 . s i n 1 2 0 ° , s i n 1 3 5 ° , s i n 2 4 0 ° . 3 2 . c o s 1 3 5 ° , t a n 3 0 0 ° , c o s e c 3 0 0 ° . 3 3 . s e c 3 1 5 ° , c o t 3 3 0 ° , t a n 7 8 0 ° . 3 4 . s i n 4 8 0 ° , s i n 4 9 5 ° , s i n 8 7 0 ° . 3 5 . t a n 1 0 2 0 ° , s e c 1 3 9 5 ° , s i n 1 4 8 5 ' A n s . A 2 V 2 2 _ J _ _ / o _ 2 V 2 , V 3 , V 3 V 2 , - V 3 , V 3 . V 3 J _ 1 2 , V 2 2 " - V 3 , V 2 , V 2 E X A M P L E S . 4 7 3 6 . s i n ( - 2 4 0 ° ) , c o t ( - 6 7 5 ° ) , c o s e c ( -6 9 0 ° ) . A n s . 2 @ 1 2 . 2 , , 3 7 . c o s ( - 3 0 0 ° ) , c o t ( - 3 1 5 ° ) , c o s e c ( -1 7 4 0 ° ) . I 1 - 2 -2 , , V 3 3 8 . t a n 3 6 6 0 ° , c o s 3 1 0 2 0 ° . -3 V 3 , -F i n d t h e v a l u e o f t h e s i n e , c o s i n e , a n d t a n g e n t o f t h e f o l l o w i n g a n g l e s : A n s . £ I V 3 . 3 9 . - 3 0 0 ° . 4 0 . -1 3 5 ° . - — , - — , I -V 2 V 2 4 1 . 7 5 0 ° . 1 V 3 1 2 , 2 , V 3 ' 4 2 . - 8 4 0 ° . ~ I V 3 . 4 3 . 1 0 2 0 ° . _ . V 3 1 _ 2 , 2 , V J -4 4 . ( 2 n + l ) - | . V 3 _ 1 _ 2 , 2 ' V 3 ' 1 V 3 1 4 5 . ( 2 n - l ) w + : -2 ' 2 , V 5 " P r o v e , d r a w i n g a s e p a r a t e f i g u r e i n e a c h c a s e , t h a t 4 6 . s i n 3 4 0 ° = s i n ( - 1 6 0 ° ) . 4 7 . s i n ( - 4 0 ° ) = s i n 2 2 0 ° . 4 8 . c o s 3 2 0 ° = - c o s ( 1 4 0 ° ) . 4 9 . c o s ( - 3 8 0 ° ) = - c o s 5 6 0 ° . 5 0 . c o s l 9 5 ° = - c o s ( - 1 5 ° ) . 5 1 . c o s 3 8 0 ° = - c o s 5 6 0 ° . 4 8 P L A N E T R I G O N O M E T R Y . 5 2 . c o s ( -2 2 5 " ) = -c o s ( -4 5 ° ) . 5 3 . c o s 1 0 0 5 ° = - c o s 1 1 8 5 ° . 5 4 . D r a w a n a n g l e w h o s e s i n e i s - -L i 5 5 . " " " " e o s e c a n t i s 2 . 5 6 . " " " " t a n g e n t i s 2 . 5 7 . C a n a n a n g l e b e d r a w n w h o s e t a n g e n t i s 4 2 7 ? 5 8 . " " " " " " c o s i n e i s -? 4 5 9 . " " " " " " s e c a n t i s 7 ? 6 0 . F i n d f o u r a n g l e s b e t w e e n z e r o a n d + 8 r i g h t a n g l e s w h i c h s a t i s f y t h e e q u a t i o n s ( 1 ) s i n A = s i n 2 0 ° , ( 2 ) s i n A = - ^ , ( 3 ) s i n A = - - . V 2 7 A r t s . ( 1 ) 2 0 ° , 1 6 0 ° , 3 8 0 ° , 5 2 0 ° ; , q \ 5 w 7 t T 1 3 i T 1 5 t T ( 2 ) T , T , ~ T , ~ r > 8 t t I S t t 2 2 i r 2 7 i t V ) y , — . - y -6 1 . S t a t e t h e s i g n o f t h e s i n e , c o s i n e , a n d t a n g e n t o f e a c h o f t h e f o l l o w i n g a n g l e s : ( 1 ) 2 7 5 ° ; ( 2 ) -9 1 ° ; ( 3 ) -1 9 3 ° ; ( 4 ) -3 5 0 ° ; ( 5 ) - 1 0 0 0 ° ; ( 6 ) 2 n t t + — . A n s . ( 1 ) - , + , - ; ( 2 ) - , - , + ; ( 3 ) + , - , - ; ( 4 ) + , + , + ; ( 5 ) + , + , + ; ( 6 ) + , - , P r o v e t h e f o l l o w i n g i d e n t i t i e s : 6 2 . ( s i n 2 0 + c o s 2 0 ) 2 c = l . 6 3 . ( s i n 2 6 -c o s 2 6 ) 2 = 1 -4 c o s 2 0 + 4 c o s 4 6 . 6 4 . ( s i n 0 + c o s 0 ) L ' = l + 2 s i n 0 c o s f t 6 5 . ( s e c f l — t a n 0 ) ( s e c 0 + t a n 0 ) = 1 . E X A M P L E S . 4 9 6 6 . ( c o s e c 0 — c o t 0 ) ( c o s e c 0 + c o t 0 ) = 1 . 6 7 . s i n 3 6 + c o s 3 0 = ( s i n 0 + c o s 0 ) ( 1 -s i n 0 c o s 0 ) . 6 8 . s i n 6 0 + c o s 6 0 = s i n 4 0 + c o s 4 0 -s i n 2 0 c o s 2 0 . 6 9 . s i u 2 0 t a n 2 0 + c o s 2 0 c o t 2 0 = t a n 2 0 + c o t 2 0 -1 . 7 0 . s i n 0 t a n 2 0 + c o s e c 0 s e c 2 0 = 2 t a n 0 s e c 0 — c o s e c 0 + s i n 0 . 7 1 . c o s 3 0 -s i n 3 0 = ( c o s 0 — s i n 0 ) ( 1 + s i n 0 c o s 0 ) . 7 2 . s i n 6 0 + c o s 6 0 = 1 — 3 s i n 2 0 c o s 2 0 . 7 3 . t a n a + t a n / 3 = t a n a t a n / ? ( c o t a + c o t / 8 ) . 7 4 . c o t a + t a n / ? = c o t a t a n / ? ( t a n a + c o t / J ) . 7 5 . 1 — s i n a = ( 1 + s i n a ) ( s e c a — t a n a ) . 7 6 . 1 + c o s a = ( 1 — c o s a ) ( c o s e c a — c o t a ) . 7 7 . ( 1 + s i n a + c o s a ) 2 = 2 ( 1 + s i n a ) ( 1 + c o s a ) . 7 8 . ( 1 — s i n a — c o s « ) 2 ( l + s i n « + c o s « ) 2 = 4 s i n 2 a c o s 2 a . 7 9 . 2 v e r s a — v e r s 2 a = s i n 2 a . 8 0 . v e r s a ( 1 4 - c o s a ) = s i n 2 a . 5 0 P L A N E T R I G O N O M E T R Y . C H A P T E R m . T R I G O N O M E T R I O F U N C T I O N S O P T W O A N G L E S . 4 2 . F u n d a m e n t a l F o r m u l a e . — W e n o w p r o c e e d t o e x p r e s s t h e t r i g o n o m e t r i c f u n c t i o n s o f t h e s u m a n d d i f f e r e n c e o f t w o a n g l e s i n t e r m s o f t h e t r i g o n o m e t r i c f u n c t i o n s o f t h e a n g l e s t h e m s e l v e s . T h e f u n d a m e n t a l f o r m u l a e f i r s t t o b e e s t a b l i s h e d a r e t h e f o l l o w i n g : s i n ( x + y ) = s i n x c o s y + c o s x s i n y . . . . ( 1 ) c o s ( x + y ) = c o s x c o s y — s i n x s i n y . . . . ( 2 ) s i n ( x — = s i n a ; c o s ? / — c o s z s i n y . . . . ( 3 ) c o s ( x — y ) = c o s x c o s y + s i n x s i n y . . . . ( 4 ) N o t s . — H e r e x a n d y a r e a n g l e s ; s o t h a t ( x + y ) a n d ( x — y ) a r e a l s o a n g l e s . H e n c e , s i n ( x + y ) i s t h e s i n e o f a n a n g l e , a n d i s n o t t h e s a m e a s s i n x + s i n y . S i n ( x + y ) i s a s i n g l e f r a e t i o n . S i n x + s i n y i s t h e s u m o f t w o f r a c t i o n s . 4 3 . T o p r o v e t h a t s i n ( a ; + y ) = s i n x c o s y + c o s x s i n y , a n d c o s ( a ; + y ) = c o s a ; c o s y — s i n a ; s i n y . I n O C , t h e b o u n d i n g l i n e o f t h e a n g l e ( x + y ) , t a k e a n y p o i n t P , a n d d r a w P D , P E , p e r p e n d i c u l a r t o O A a n d O B , r e s p e c t i v e l y ; d r a w L e t t h e a n g l e A O B = x , a n d t h e a n g l e B O C = y ; t h e n t h e a n g l e A O C = x + y . B A D K E H , E K , p e r p e n d i c u l a r t o P D a n d O A . F U N D A M E N T A L F O R M U L A E . 5 1 T h e n a n g l e E P H = 9 0 ° - H E P = H E O = A O E = x . 8 i n ( a ; + y ) = D g = = E K + P H = E K P H = E K O E P H P E O E ' O P P E ' O P = = s i n x c o s y + c o s x s i n y . , . , O D _ O K - H E O K H E v + w O P 0 P O P O P = o k o e h e p e o e " o p p e ' o p = c o s x c o s y — s i n x s i n y . N o t b . — T h e s e t w o f o r m u l a ' h a v e b e e n o b t a i n e d b y a c o n s t r u c t i o n i n w h i c h x + y i s a n a c u t e a n g l e ; b u t t h e p r o o f i s p e r f e c t l y g e n e r a l , a n d a p p l i e s t o a n g l e s o f a n y m a g n i t u d e w h a t e v e r , b y p a y i n g d u e r e g a r d t o t h e a l g e b r a i c s i g n s . F o r e x a m p l e , L e t A O B = x , a s b e f o r e , a n d B 0 C = y ; t h e n A O C , m e a s u r e d i n t h e p o s i t i v e d i r e c t i o n , i s t h e a n g l e x + y . I n O C , t a k e a n y p o i n t P , a n d d r a w P D , P E , p e r p e n d i c u l a r t o O A a n d O B p r o d u c e d ; d r a w E H a n d E K p e r p e n d i c u l a r t o P D a n d O A . T h e n , a n g l e E O K = 1 8 0 ° - x ; E P H = E O K = 1 8 0 ° -a n d C O E = » - 1 8 0 ° . i n < • + ) — 5 ? E K n Z S = _ E K P H O P O P O P O P = _ E K O E ( P H P E O E O P P E ' O P = -s i n ( 1 8 0 ° - x ) c o s ( y -1 8 0 ° ) + c o s ( 1 8 0 ° - x ) s i n ( y - 1 8 0 ° ) = s i n x c o s y + c o s x s i n y . ( A r t . 3 5 ) T O W r + ^ - O D - O K + E H O K E H O P O P O P O P _ O K O E + E H P E O E O P P E ' O P T h e i n t r o d u c e d l i n e O E i s t h e o n l y l i n e i n t h e f i g u r e w h i c h i s a t o n c e a s i d e o f t w o r i g h t t r i a n g l e s ( O E K a n d O E P ) i n t o w h i c h E K a n d O P e n t e r . A s i m i l a r r e m a r k a p p l i e s t o P E . 5 2 P L A N E T R I G O N O M E T R Y . = c o s ( 1 8 0 ° - x ) c o s ( y -1 8 0 ° ) + s i n ( 1 8 0 ° -x ) s i n ( y -1 8 0 ° ) = c o s x c o s y — s i n x s i n y . ( A r t . 3 5 ) T h e s t u d e n t s h o u l d n o t i c e t h a t t h e w o r d s o f t h e t w o p r o o f s a r e v e r y n e a r l y t h e 4 4 . T o p r o v e t h a t s i n ( a ; — y ) = s i n x c o s y — c o s x s i n y , a n d c o s ( x — y ) = c o s c o s y + s i n a ; s i n y . L e t t h e a n g l e A O B b e d e n o t e d b y x , a n d C O B b y y ; t h e n t h e a n g l e A O C = x - y . I n O C t a k e a n y p o i n t P , a n d d r a w P D , P E , p e r p e n d i c u l a r t o O A a n d O B r e s p e c t i v e l y ; d r a w E H , E K , p e r p e n d i c u l a r t o P D a n d O O A r e s p e c t i v e l y . T h e n t h e a n g l e E P H = 9 0 ° - H E P = B E H = A O B = x . . , . D P E K - H P E K H P s i n ( x — y ) = — — = — = v O P O P O P O P = E K O E H P P E O E ' O P P E ' O P = s i n x c o s y — c o s x s i n y . , , O D O K + E H O K , E H c o s ( X — V ) = = . . v O P O P O P O P = O K O E E H P E O E ' O P P E ' O P = c o s x c o s y + s i n x s i n y . N o t e 1 . — T h e s i g n i n t h e e x p r e s s i o n o f t h e s i n e i s t h e s a m e a s i t i s i n t h e a n g l e e x p a n d e d ; i n t h e c o s i n e i t i s t h e o p p o s i t e . V i s t a k e n i n t h e l i n e h o u n d i n g t h e a n g l e u n d e r c o n s i d e r a t i o n ; i . e . , A O C . S I N E A N D C O S I N E . 6 8 N o t e 2 . — I n t b i s p r o o f t h e a n g l e x — y i s a c u t e ; b u t t b e p r o o f , l i k e t h e o n e g i v e n i n A r t . 4 3 , a p p l i e s t o a n g l e s o f a n y m a g n i t u d e w h a t e v e r . F o r e x a m p l e , L e t A O B , m e a s u r e d i n t h e p o s i t i v e d i r e c t i o n , = X , a n d B O C = y . T h e n A O C = x - y I n O C t a k e a n y p o i n t P , a n d d r a w P D , P E , p e r p e n d i c u l a r t o O A a n d O B p r o d u c e d : d r a w E H , E K , p e r p e n d i c u l a r t o D P a n d A O p r o d u c e d . T h e n , a n g l e E P H - E O K = A O B = 3 6 0 ° -x t a n d P O E = 1 8 0 ° -y . " O P O P = E K O E H P P R O E O P P E ' O P = s i n ( 3 8 0 ° - x ) c o s ( 1 8 0 ° -y ) -c o s ( 3 6 0 ° - x ) s i n ( 1 8 0 ° - y ) — ( — s i n x ) ( - c o s y ) — c o s x s i n y = s i n x c o s y — c o s x s i n y . c o b ( x — y ) = O D = O P O K + H E O P _ _ O K O E _ H E P E O E O P P E ' O P = -c o s ( 3 6 0 ° - x ) c o s ( 1 8 0 ° -y ) -s i n ( 3 6 0 ° - x ) s i n ( 1 8 0 ° - y ) - ( — c o s x ) ( — c o s y ) — ( — s i n x ) s i n y = c o s x c o s y + s i n x s i n y . N o t e 3 . — T h e f o u r f u n d a m e n t a l f o r m u l a s j u s t p r o v e d a r e v e r y i m p o r t a n t , u n d m u s t b e c o m m i t t e d t o m e m o r y . I t w i l l b e c o n v e n i e n t t o r e f e r t o t h e m a s t b e ' x , y 1 f o r m u l a ? . F r o m a n y o n e o f t h e m , a l l t h e o t h e r s c a n b e d e d u c e d i n t b e f o l l o w i n g T h u s , f r o m c o s ( x — y ) t o d e d u c e s i n ( a ; + y ) . W e h a v e c o s ( x — y ) = c o s x c o s y + s i n x s i n y . . . . ( 1 ) S u b s t i t u t e 9 0 ° — x f o r x i n ( 1 ) , a n d i t b e c o m e s c o s { 9 0 ° — ( x + y ) \ = c o s ( 9 0 ° -a ; ) c o s y + s i n ( 9 0 ° — x ) s i n y . . : s i n ( a ; + 2 / ) = s i n a ; c o s ? / + c o s a ; s i n ? / . ( A r t . 2 9 ) T h e s t u d e n t s h o u l d m a k e t h e s u b s t i t u t i o n s i n d i c a t e d b e l o w , a n d s a t i s f y h i m s e l f t h a t t h e c o r r e s p o n d i n g r e s u l t s f o l l o w : 5 4 P L A N E T R I G O N O M E T R Y . F r o m s i n ( x + y ) t o d e d u c e c o s ( x + y ) s u b s t i t u t e ( 9 0 ° + x ) f o r x . " " " c o s ( x - y ) " ( 9 0 ° - x ) f o r c o s ( a ; + y ) " " s i n ( x + y ) " ( 9 0 ° + x ) f o r x . " « " s i n ( x - i / ) " ( 9 0 ° -x ) f o r . " " " c o s ( x — y ) " — y f o r y . e t c . e t c . e t c . E X A M P L E S . 1 . T o f i n d t h e v a l u e o f s i n 1 5 ° . s i n l 5 ° = s i n ( 4 5 o - 3 0 ° ) = s i n 4 5 ° c o s 3 0 ° -c o s 4 5 ° s i n 3 0 ° _ J L V 5 1 l V 2 ' ' 2 V 2 2 V 3 - 1 4 . S h o w t h a t c o s 1 5 ° = 2 V 2 V 3 + 1 o 2 V 2 V 3 - 1 ; ° _ 2 V 2 V 3 + 1 ; ° — 2 V 2 3 5 5 . I f s i n x = - , a n d c o s y = — f i n d s i n ( x - + - y ) a n d « » ( ■- ) . " " 6 3 , a n d 6 . 6 5 ' 6 5 6 . I f s i n a ; = i , a n d c o s y = \ , f i n d s i n ( a ; + ? / ) a n d c o s ( a ' - 2 / ) -' 1 A n e . l , a n d # F O R M U L A E F O R T R A N S F O R M A T I O N . 5 5 4 5 . F o r m u l a e f o r t h e T r a n s f o r m a t i o n o f S u m s i n t o P r o d u c t s . — F r o m t h e f o u r f u n d a m e n t a l f o r m u l a } o f A r t s . 4 3 a n d 4 4 w e h a v e , b y a d d i t i o n a n d s u b t r a c t i o n , t h e f o l l o w i n g : s i n ( x + y ) + s i n ( x — y ) = 2 s i n x c o s y . . . ( 1 ) s i n ( x + y ) — s i n ( x — y ) = 2 c o s a ; s i n y . . . ( 2 ) c o s ( x + y ) + c o s ( x — y ) = 2 c o s x c o s y . . . ( 3 ) c o s ( a ; — y ) — c o s ( x + y ) = 2 s i n a ; s i n y . . . ( 4 ) T h e s e f o r m u l a e a r e u s e f u l i n p r o v i n g i d e n t i t i e s b y t r a n s f o r m i n g p r o d u c t s i n t o t e r m s o f f i r s t d e g r e e . T h e y e n a b l e u s , w h e n r e a d f r o m r i g h t t o l e f t , t o r e p l a c e t h e p r o d u c t o f a s i n e o r a c o s i n e i n t o a s i n e o r a c o s i n e b y h a l f t h e s u m o r h a l f t h e d i f f e r e n c e o f t w o s u c h r a t i o s . L e t x + y = A , a n d x — y = B . . : x = \ ( A + B ) , a n d y = | ( A - B ) . S u b s t i t u t i n g t h e s e v a l u e s i n t h e a b o v e f o r m u l a e , a n d p u t t i n g , f o r t h e s a k e o f u n i f o r m i t y o f n o t a t i o n , x , y i n s t e a d o f A , B , w e g e t s i n x + s i n y = 2 s i n £ ( x + y ) c o s £ ( a ; — ? / ) . . ( 5 ) s i n a ; — s i n y = 2 c o s £ ( a ; + y ) s i n \ ( x — y ) . . ( 6 ) c o s x + c o s y = 2 c o s J ( a ; + y ) c o s £ ( a ; — y ) . . ( 7 ) c o s y — c o s a ; = 2 s i n £ ( a ; + y ) s i n £ ( a ; — y ) . . ( 8 ) T h e f o r m u l a e a r e o f g r e a t i m p o r t a n c e i n m a t h e m a t i c a l i n v e s t i g a t i o n s ( e s p e c i a l l y i n c o m p u t a t i o n s b y l o g a r i t h m s ) ; t h e y e n a b l e u s t o e x p r e s s t h e s u m o r t h e d i f f e r e n c e o f t w o s i n e s o r t w o c o s i n e s i n t h e f o r m o f a p r o d u c t . T h e s t u d e n t i s r e c o m m e n d e d t o b e c o m e f a m i l i a r w i t h t h e m , a n d t o c o m m i t t h e f o l l o w i n g e n u n c i a t i o n s t o m e m o r y : O f a n y t w o a n g l e s , t h e S u m o f t h e s i n e s = 2 s i n £ s u m - c o s ^ d i f f . D i f f . " " " = 2 c o s £ s u m - s i n | d i f f . 5 6 P L A N E T R I G O N O M E T R Y . S u m o f t h e c o s i n e s = 2 c o s £ s u m - c o s i d i f f . D i f f . " " " = 2 s i n £ s u m . s i n £ d i f f . E X A M P L E S . 1 . s i n 5 x c o s 3 a ; = £ ( s i n 8 a ; + s i n 2 x ) . F o r , s i n 5 x c o s 3 a ; = £ { s i n ( 5 x + 3 x ) + s i n ( 5 a ; — 3 a ; ) \ = £ ( s i n 8 x + s i n 2 x ) . 2 . P r o v e s i n 0 s i n 3 0 = : | ( c o s 2 0 -c o s 4 0 ) . 3 . » 2 s i n 0 c o s < £ = s i n ( 0 + $ ) + s i n ( 0 - < £ ) . 4 . u 2 s i n 2 0 c o s 3 < £ = : s i n ( 2 0 + 3 < f > ) + s i n ( 2 0 - 3 < £ ) 5 . a s i n 6 0 ° + s i n 3 0 ° = 2 s i n 4 5 ° c o s 1 5 ° . 6 . u s i n 4 0 ° - s i n 1 0 ° = 2 c o s 2 5 ° s i n 1 5 ° . 7 . i t s i n l 0 0 + s i n 6 0 = 2 s i n 8 0 c o s 2 0 . 8 . a s i n 8 « — s i n 4 « = 2 c o s 6 a s i n 2 a . 9 . a s i n 3 a ; + s i n c e = 2 s i n 2 a ; c o s a ; . 1 0 . t t s i n 3 a ; — s i n a ; = 2 c o s 2 a ; s i n . 1 1 . u s i n 4 a ; + s i n 2 x = 2 s i n 3 a ; c o s a ; . 4 6 . U s e f u l F o r m u l a e . — T h e f o l l o w i n g f o r m u l a e , w h i c h a r e o f f r e q u e n t u s e , m a y b e d e d u c e d b y t a k i n g t h e q u o t i e n t o f e a c h p a i r o f t h e f o r m u l a e ( 5 ) t o ( 8 ) o f A r t . 4 5 a s f o l l o w s : j s i n a ; + s i n y _ 2 s i n ^ ( x + y ) c o s \ ( x — y ) s i n a ; — s i n ? / 2 c o s | - ( a ; + y ) s i n £ ( x — y ) = t a n £ ( a ; + ? / ) c o t % ( x — y ) = t a n | ( a ; + y ) . ( A r t 2 4 ) t a n £ ( a ; - ? / ) T h e f o l l o w i n g m a y b e p r o v e d b y t h e s t u d e n t i n a s i m i l a r m a n n e r : „ s i n x + s i n y , , , , . 2 . -2 = t a n £ ( a ; + y ) , c o s x + c o s y T H E T A N G E N T O F T W O A N G L E S . 5 7 „ s i n x + s i n y , , , , 3 . 2 = c o t £ ( a ; - y ) , c o s y — c o s a ; . s i n a ; — s i n y . , , N 4 . = t a n i ( x — y ) , c o s a ; + c o s y e s i n x — s i n y . . . . 5 . ? = c o t $ ( x + y ) , c o s y — c o s 6 . c o s + c o s y = « | - ( x + y ) c o t - ) . c o s y — c o s a ; 4 7 . T h e T a n g e n t o f t h e S u m a n d D i f f e r e n c e o f T w o A n g l e s . — E x p r e s s i o n s f o r t h e v a l u e o f t a n ( a ; + y ) , t a n ( x — y ) , e t c . , m a y b e e s t a b l i s h e d g e o m e t r i c a l l y . I t i s s i m p l e r , h o w e v e r , t o d e d u c e t h e m f r o m t h e f o r m u l a e a l r e a d y e s t a b l i s h e d , a s f o l l o w s : D i v i d i n g t h e f i r s t o f t h e ' x , y , f o r m u l a e b y t h e s e c o n d , w e h a v e , b y A r t . 2 3 , t a n ( x + y ) = s i n ( x + V ) _ s i n a ; c o s y + c o s a > s i n y c o s ( x + y ) c o s x c o s y — s i n x s i n y D i v i d i n g b o t h t e r m s o f t h e f r a c t i o n b y c o s x c o s y , s i n x c o s y , c o s a ; s i n y c o s x c o s ? / c o s x c o s w t a n ( x + y ) = -: r - ^ c o s x c o s y s i n a ; s i n y c o s a ; c o s y c o s a ; c o s x = J a n + t a n y _ 2 3 ) ( 1 ) 1 — t a n a ; t a n y I n t h e s a m e m a n n e r m a y b e d e r i v e d t a n ( a ; - y ) = - t ^ ^ s l M . ( 2 ) 1 + t a n x t a n y A l s o , o o t i x + y ) ^ ^ -1 ( 3 ) c o t a ; + c o t y , . , , c o t a ; c o t y + 1 / A \ a n d c o t ( x — y ) = ( 4 ) c o t y — c o t a ; 5 8 P L A N E T R I G O N O M E T R Y . E X E R C I S E S . P r o v e t h e f o l l o w i n g : t a n a ; + _ l _ 1 . t a n ( a ; + 4 5 ° ) = ^ ^ v ' 1 — t a n 2 . . / A t . 0 \ t a n a ; — 1 t a n ( x — 4 5 ) = v 1 + t a n 2 s i n ( x + y ) _ t a n x + t a n y s i n ( x — y ) t a n x — t a n y ^ c o s ( x — y ) _ t a n x t a n y + 1 c o s ( a ; + y ) 1 — t a n a ; t a n 3 / 5 . s i n ( x + y ) s i n ( x — y ) = s ' m 2 x — s i n 2 y = c o s 2 y — c o s 2 a ; . 6 . c o s ( x - f y ) c o s ( a ; — y ) = c o s 2 a ; — s i n 2 y = c o s 2 ? / — s i n 2 a ; . 7 . t a n » ± t a n y = s i n c o s a ; c o s y 8 . c o t a ; ± c o t y = s i n ( 3 / ± » ) . s i n a ; s i n y r , s i n 2 a ; c o s 2 x 9 . — = s e c x . s i n x c o s x 1 0 . I f t a n x = I a n d t a n y = ^ , p r o v e t h a t t a n ( a ; - f y ) = f , a n d t a n ( a ; — y ) = 1 1 . P r o v e t h a t t a n 1 5 ° = 2 - V 3 . 1 2 . I f t a n a ; = $ , a n d t a n y = T l T > p r o v e t h a t t a n ( a ; + y ) = 1 . W h a t i s ( a ; + y ) i n t h i s c a s e ? 4 8 . F o r m u l a e f o r t h e S u m o f T h r e e o r M o r e A n g l e s . — L e t x , y , z b e a n y t h r e e a n g l e s ; w e h a v e b y A r t . 4 3 , E X A M P L E S . 5 9 s i n ( a ; + y + z ) = s i n ( x + y ) c o s z + c o s ( x + y ) s i n z = s i n x c o s y c o s z + c o s s i n y c o s z + c o s a ; c o s y s i u z — s i n x s i n y s i n z . . ( 1 ) I n l i k e m a n n e r , c o s ( x + y + z ) = c o s a ; c o s y c o s z — s i n x s i n ? / c o s z — s i n x c o s y s i n z — c o s x s i n y s i n z . . ( 2 ) D i v i d i n g ( 1 ) b y ( 2 ) , a n d r e d u c i n g b y d i v i d i n g b o t h t e r m s o f t h e f r a c t i o n b y c o s a ; c o s y c o s z , w e g e t t a n ( x + y + z ) — t a n + t a n y + t a n z — t a n x t a n y t a n z 1 — t a n a ; t a n y — t a n y t a n z — t a n z t a n a ; E X A M P L E S . 1 . P r o v e t h a t s i n x + s i n y + s i n z — s i n ( a ; + y + z ) = 4 s i n £ ( x + y ) s i n £ ( y + z ) s i n £ ( z + x ) . B y ( 6 ) o f A r t . 4 4 w e h a v e s i n a ; — s i n ( a ; + y + z ) = — 2 c o s $ ( 2 x + y + z ) s i n | ( y + z ) , a n d s i n y + s i n z = 2 s i n £ ( y + z ) c o s £ ( y — z ) . . - . s i n x + s i n y 4 - s i n z — s i n ( x + y + z ) = 2 s i n | ( y + z ) c o s £ ( y — z ) — 2 c o s | ( 2 x + y + z ) s i n £ ( y + z ) = 2 s i n £ ( y + z ) { e o s £ ( y — z ) — c o s £ ( 2 x + y + z ) \ = 2 s i n £ ( y + z ) 2 s i n £ ( a ; + y ) s i n | ( x + z ) = 4 s i n I ( a ; + y ) s i n £ ( y + z ) s i n | ( z + a ; ) . P r o v e t h e f o l l o w i n g : 2 . c o s x + c o s y + c o s z + c o s ( x + y + z ) = 4 e o s £ ( y + z ) c o s £ ( z + x ) c o s £ ( x + y ) . 6 0 P L A N E T R I G O N O M E T R Y . 3 . s i n ( x + y — 2 ) = s i n x c o s y c o s 2 4 - c o s a ; s i n y c o s 2 — c o s a ; c o s y s i 2 + s i n a ; s i n 3 / s i n 2 . 4 . s i n x + s i n y — s i n 2 — s i n ( a ; + y — 2 ) = 4 s i n £ ( a ; — z ) s i n | ( y — 2 ) s i n £ ( a ; - f y ) . 5 . s i n ( y — 2 ) + s i n ( 2 — a ; ) + s i n ( a ; — ? / ) + 4 s i n £ ( ? / — 2 ) s i n £ ( 2 — a ; ) s i n J ( a ; — y ) = 0 . 4 9 . F u n c t i o n s o f D o u b l e A n g l e s . — T o e x p r e s s t h e t r i g o n o m e t r i c f u n c t i o n s o f t h e a n g l e 2 x i n t e r m s o f t h o s e o f t h e a n g l e x . P u t y = x i n ( 1 ) o f A r t . 4 2 , a n d i t b e c o m e s s i n 2 x = s i n a ; c o s a ; + c o s a ; s i n a ; , o r s i n 2 x = 2 s i n a ; c o s x ( 1 ) P u t y = x i n ( 2 ) o f A r t . 4 2 , a n d i t b e c o m e s c o s 2 x = c o s 2 x — s i n 2 x ( 2 ) = l - 2 s i n 2 a ; ( 3 ) o r = 2 c o s 2 a ; — 1 ( 4 ) P u t y = x i n ( 1 ) a n d ( 3 ) o f A r t . 4 7 , a n d t h e y b e c o m e . 0 2 t a n a ; / C N t a n 2 a ; = -— ( 5 ) l - t a n 2 a ; v ' c o t 2 a , - = c o t 2 a ; ~ 1 ( 6 ) 2 c o t a ; v ' T r a n s p o s i n g 1 i n ( 4 ) , a n d d i v i d i n g i t i n t o ( 1 ) , w e h a v e E X A M P L E S . 6 1 N o t e . — T h e s e s e v e n f o r m u l a e a r e v e r y i m p o r t a n t . T h e s t u d e n t m u s t n o t i c e t h a t x i s a n y a n g l e , a n d t h e r e f o r e t h e s e f o r m u l a ? w i l l b e t r u e w h a t e v e r w e p u t f o r x . T h u s , i f w e w r i t e ^ f o r x , w e g e t s i n z = 2 s i n ? e o s -( 8 ) 2 2 c o s x = c o s 2 - - s i n -( 9 1 2 2 o r = l - 2 s i n 2 5 = 2 c o s 2 ? - l ( 1 0 ) 2 2 a n d s o o n . E X A M P L E S . P r o v e t h e f o l l o w i n g : 1 . 2 c o s e c 2 a ; = s e c a ; c o s e c a ; . „ c o s e c 2 0 2 . = s e c 2 x . c o s e c x — 2 3 2 t & n . ? / -o . — -s i n 2 x . 1 + t a n 2 a ; . 1 — t a n 2 a ; „ 4 . — = c o s 2 x . 1 + t a n 2 x 5 . t a n x + c o t x = 2 c o s e c 2 a ; . 6 . c o t — t a n a ; = 2 c o t 2 w . „ -s i n , x 7 . = t a n — 1 + c o s x 2 Q s i n x , x 8 . = c o t — 1 — c o s x 2 9 . G i v e n s i n 4 5 ° = f i n d t a n 2 2 A ° . ^ I n s . V 2 -1 . V 2 3 2 4 2 4 1 0 . G i v e n t a n x = -: f i n d t a n 2 a ; , a n d s i n 2 a ; . — , — 4 , , 7 , 2 5 5 0 . T o E x p r e s s t h e F u n c t i o n s o f 3 a ; i n T e r m s o f t h e F u n c t i o n s o f x . P u t y = 2 a ; i n ( 1 ) o f A r t . 4 2 , a n d i t b e c o m e s 6 2 P L A N E T R I G O N O M E T R Y . s i n 3 a ; = s i n ( 2 a ; + x ) = s i n 2 x c o s x + c o s 2 x s i n x = 2 s i n a ; c o s 2 a ; + ( 1 — 2 s i n 2 a ; ) s i n a ; ( A r t . 4 9 ) = 2 s i n a ; ( 1 — s i n 2 a ; ) 4 - s i n a ; — 2 s i n 3 x = 3 s i n a ; — 4 s i n 3 x . c o s 3 a ; = c o s ( 2 a ; + x ) = c o s 2 x c o s x — s i n 2 x s i n x . = ( 2 c o s 2 a ; — 1 ) c o s a ; — 2 s i n 2 a ; c o s a ; ( A r t . 4 9 ) = 4 c o s 3 x — 3 c o s x . t a n 3 x = t a n 2 x + x t a n 2 a ; + t a n x 1 — t a n 2 a ; t a n x 2 t a n a ; 1 — t a n 2 a ; + t a n a ; j 2 t a n 2 a ; 1 — t a n 2 a ; 3 t a n x — t a n 3 a ; 1 — 3 t a n 2 a ; E X A M P L E S . P r o v e t h e f o l l o w i n g : 1 s i n 3 a ; s i n a ; = 2 c o s 2 a ; + 1 . 2 . s i n 3 x - s i n = t a n ¡ B c o s 3 a ; + c o s 3 . s i n 3 a ; + c o s 3 a ; = 2 s i n 2 a , _ L c o s a ; — s i n x F U N C T I O N S O F T H E H A L F A N G L E . 6 3 4 . 1 = c o t 2 x . t a n 3 a ; — t a n x c o t x — c o t 3 x 5 . 1 ~ c o s 3 x = ( l + 2 c o s x Y . 1 — c O S X 5 1 . F u n c t i o n s o f H a l f a n A n g l e . — T o e x p r e s s t h e f u n c -x 2 t i o n s o f -i n t e r m s o f t h e f u n c t i o n s o f x . S i n c e c o s x = 1 — 2 s i n 2 - , 2 o r = 2 c o s 2 - - l . . . [ A r t . 4 9 , ( 1 0 ) ] 2 . , x 1 — c o s . r / 1 N . - . s i n 2 - = ( 1 ) 2 2 w a n d c o s 2 | = 1 + ° o s a ; ( 2 ) -X / l — c o s , Q , O r s m - = ± ^ ( 3 ) j X / 1 + c o s , i s a n d c o s - = ± y j ^ _ _ ( 4 ) t , j -. -. X , / l — c o s X , 1 — c o s X , B y d i v i s i o n , t a n - = ± \ -= ± . . ( 5 ) 2 \ l + c o s a ; s i n a ; B y f o r m u l a e ( 3 ) , ( 4 ) , a n d ( 5 ) t h e f u n c t i o n s o f h a l f a n a n g l e m a y b e f o u n d w h e n t h e c o s i n e o f t h e w h o l e a n g l e i s g i v e n . 5 2 . I f t h e C o s i n e o f a n A n g l e b e g i v e n , t h e S i n e a n d t h e C o s i n e o f i t s H a l f a r e e a c h T w o - V a l u e d . B y A r t . 5 1 , e a c h v a l u e o f c o s a ; ( n o t h i n g e l s e b e i n g k n o w n a b o u t t h e a n g l e x ) g i v e s t w o v a l u e s e a c h f o r s i n - a n d c o s ^ , . 2 2 o n e p o s i t i v e a n d o n e n e g a t i v e . B u t i f t h e v a l u e o f x b e 6 4 P L A N E T R I G O N O M E T R Y . g i v e n , w e k n o w t h e q u a d r a n t i n w h i c h -l i e s , a n d h e n c e L i w e k n o w w h i c h s i g n i s t o b e t a k e n . T h u s , i f x l i e s b e t w e e n 0 ° a n d 3 6 0 ° , -l i e s b e t w e e n 0 ° a n d 1 8 0 ° , a n d t h e r e f o r e s i n -i s p o s i t i v e ; b u t i f x l i e s b e t w e e n 2 3 6 0 ° a n d 7 2 0 ° , -l i e s b e t w e e n 1 8 0 ° a n d 3 6 0 ° , a n d h e n c e . 2 A l s o , i f x l i e b e t w e e n 0 ° a n d 1 8 0 ° , c o s -a n d 3 6 0 ° , c o s -i s 2 s i n -i s n e g a t i v e . i s p o s i t i v e ; b u t i f x l i e b e t w e e n 1 8 0 ° n e g a t i v e . T h e c a s e m a y b e i n v e s t i g a t e d g e o m e t r i c a l l y t h u s : L e t O M = t h e g i v e n c o s i n e ( r a d i u s b e i n g u n i t y , A r t . 1 6 ) , = c o s x . T h r o u g h M d r a w P Q p e r p e n d i c u l a r t o O A ; a n d d r a w O P , O Q . T h e n a l l a n g l e s w h o s e c o s i n e s a r e e q u a l t o c o s a ; a r e t e r m i n a t e d e i t h e r b y O P o r O Q , a n d t h e h a l v e s o f t h e s e a n g l e s a r e t e r m i n a t e d b y t h e d o t t e d l i n e s O p , O q , O r , o r O s . T h e s i n e s o f a n g l e s e n d i n g a t O p a n d O q a r e t h e s a m e , a n d e q u a l n u m e r i c a l l y t o t h o s e o f a n g l e s e n d i n g a t O r a n d O s ; b u t i n t h e f o r m e r c a s e t h e y a r e p o s i t i v e , a n d i n t h e l a t t e r , n e g a t i v e ; h e n c e w e o b t a i n t w o , a n d o n l y t w o , v a l u e s o f s i n ^ f r o m a g i v e n v a l u e o f c o s x . A l s o , h e c o s i n e s o f a n g l e s e n d i n g a t O p a n d O s a r e t h e s a m e , a n d h a v e t h e p o s i t i v e s i g n . T h e y a r e e q u a l n u m e r i c a l l y t o t h e c o s i n e s o f t h e a n g l e s e n d i n g a t O q a n d O r , b u t t h e l a t t e r a r e n e g a t i v e ; h e n c e w e o b t a i n t w o , a n d o n l y t w o , v a l u e s o f c o s ^ f r o m a g i v e n v a l u e o f c o s a ; . A l s o , t h e t a n g e n t o f h a l f t h e a n g l e w h o s e c o s i n e i s g i v e n i s t w o - v a l u e d . T h i s f o l l o w s i m m e d i a t e l y f r o m ( 5 ) o f A r t . 5 1 . — ^ 3 r M J 2 A T T ' F U N C T I O N S O F T H E H A L F A N G L E . 6 5 5 3 . I f t h e S i n e o f a n A n g l e b e g i v e n , t h e S i n e a n d t h e C o s i n e o f i t s H a l f a r e e a c h F o u r - V a l u e d . W e h a v e a n d 2 s i n -c o s - = s i n a ; . 2 2 s i n 2 - + c o s 2 - = 1 . . 2 2 ( A r t . 4 9 ) ( A r t . 2 3 ) B y a d d i t i o n , ^ s i n ? + c o s | ^ = 1 B y s u b t r a c t i o n , ^ s i n ? — - s i n x . c o s -) = 1 — s i n a ; . 9 , . - . s i n r + c o s - « = ± V l + s i n a ; . . ( 1 ) 2 2 a n d a n d s i n c o s - = ± V l — s i n a ; . . ( 2 ) 2 2 - . 2 s i n r = ± V l + s i n ± V l — s i n x . . ( 3 ) 2 c o s ^ = ± V I + s i n x T V l — s i n a ; . . ( 4 ) T h u s , i f w e a r e g i v e n t h e v a l u e o f s i n a ; ( n o t h i n g e l s e b e i n g k n o w n a b o u t t h e a n g l e a ; ) , i t f o l l o w s f r o m ( 3 ) a n d 1 X O S ( 4 ) t h a t s i n -a n d c o s -h a v e e a c h f o u r v a l u e s e q u a l , t w o 2 2 b y t w o , i n a b s o l u t e v a l u e , b u t o f c o n t r a r y s i g n s . T h e c a s e m a y b e i n v e s t i g a t e d g e o m e t r i c a l l y t h u s : L e t O N = t h e g i v e n s i n e ( r a d i u s b e i n g u n i t y ) = s i n x . T h r o u g h N d r a w P Q p a r a l l e l t o O A ; a n d d r a w O P , O Q . T h e n a l l a n g l e s w h o s e s i n e s a r e e q u a l t o s i n a ; a r e t e r m i n a t e d e i t h e r b y O P o r O Q , a n d t h e h a l v e s o f t h e s e a n g l e s a r e t e r m i n a t e d b y t h e d o t t e d l i n e s O p , O q , O r , o r O s . T h e s i n e s o f a n g l e s e n d i n g a t O p , O q , O r , a n d O s a r e a l l d i f f e r e n t 0 . / / \ p 7 \ N A . " " \ 6 6 P L A N E T R I G O N O M E T R Y . i n v a l u e ; a n d s o a r e t h e i r c o s i n e s . H e n c e w e o b t a i n f o u r v a l u e s f o r s i n ^ , a n d f o u r a l s o f o r c o s ^ , i n t e r m s o f x . W h e n t h e a n g l e x i s g i v e n , t h e r e i s n o a m b i g u i t y i n t h e c a l c u l a t i o n s ; f o r -i s t h e n k n o w n , a n d t h e r e f o r e t h e s i g n s a n d r e l a t i v e m a g n i t u d e s o f s i n -a n d c o s -a r e k n o w n . T h e n 2 2 e q u a t i o n s ( 1 ) a n d ( 2 ) , w h i c h s h o u l d a l w a y s b e u s e d , i m m e d i a t e l y d e t e r m i n e t h e s i g n s t o b e t a k e n i n e q u a t i o n s ( 3 ) a n d ( 4 ) . T h u s , w h e n -l i e s b e t w e e n — 4 5 ° a n d + 4 5 ° , c o s - > s i n - , 2 2 2 a n d i s p o s i t i v e . T h e r e f o r e ( 1 ) i s p o s i t i v e , a n d ( 2 ) i s n e g a t i v e ' a n d h e n c e ( 3 ) a n d ( 4 ) b e c o m e 2 s i n - = V l + s i n x — V T ^ s i n x , 2 2 c o s - = V T + s i n - f V l — s i n a ; . " W h e n -l i e s b e t w e e n 4 5 ° a n d 1 3 5 ° , s i n 5 > c o s - , a n d i s 2 2 2 p o s i t i v e . T h e r e f o r e ( 1 ) a n d ( 2 ) a r e b o t h p o s i t i v e ; a n d h e n c e ( 3 ) a n d ( 4 ) b e c o m e 2 s i n - = V l 4 - s i n a ; + V l — s i n a ; , 2 2 c o s - = V l + s i n x — V l — s i n x . 2 A n d s o o n . 5 4 . I f t h e T a n g e n t o f a n A n g l e b e g i v e n , t h e T a n g e n t o f i t s H a l f i s T w o - V a l u e d . 2 t a n -W e h a v e t a n 0 = 1 ( A r t . 4 9 ) l - t a n 2 £2 F U N C T I O N S O F T H E H A L F A N G L E . 6 7 Q P u t t a n - = x ; t h u s ( 1 — x 2 ) t a n 0 = 2 x , x + — x = l . t a n 0 . , t a n g = a ; = - i ± v r + ™. 2 t a n 0 T h u s , g i v e n t a n 0 , w e f i n d f w o u n e q u a l v a l u e s f o r t a n - , o n e p o s i t i v e a n d o n e n e g a t i v e . T h i s r e s u l t m a y b e p r o v e d g e o m e t r i c a l l y , a n e x e r c i s e w h i c h w e l e a v e f o r t h e s t u d e n t . 5 5 . I f t h e S i n e o f a n A n g l e h e g i v e n , t h e S i n e o f O n e -T h i r d o f t h e A n g l e i s T h r e e - V a l u e d . W e h a v e s i n 3 x = 3 s i n x — 4 s i n 3 x . . ( A r t . 5 0 ) a P u t x = a n d w e g e t 3 s i n 6 = 3 s i n - — 4 s i n 3 - , 3 3 a c u b i c e q u a t i o n , w h i c h t h e r e f o r e h a s t h r e e r o o t s . E X A M P L E S . 1 . D e t e r m i n e t h e l i m i t s b e t w e e n w h i c h A m u s t l i e t o s a t i s f y t h e e q u a t i o n 2 s i n A = — V l + s i n 2 A — V l — s i n 2 A . B y ( 1 ) a n d ( 2 ) o f A r t . 5 3 , 2 s i n A c a n h a v e t h i s v a l u e o n l y w h e n s i n A + c o s A = — V l + s i n 2 A , a n d s i n A — c o s A = — V l — s i n 2 A ; i . e . , w h e n s i n A > c o s A a n d n e g a t i v e . 6 8 P L A N E T R I G O N O M E T R Y . T h e r e f o r e A l i e s b e t w e e n 2 2 5 ° a n d 3 1 5 ° , o r b e t w e e n t h e a n g l e s f o r m e d b y a d d i n g o r s u b t r a c t i n g a n y m u l t i p l e o f f o u r r i g h t a n g l e s t o e a c h o f t h e s e ; i . e . , A l i e s b e t w e e n 2 n i T + — a n d 2 n t t + — , 4 4 w h e r e n i s z e r o o r a n y p o s i t i v e o r n e g a t i v e i n t e g e r . 2 . D e t e r m i n e t h e l i m i t s b e t w e e n w h i c h A m u s t l i e t o s a t i s f y t h e e q u a t i o n 2 c o s A = V l + s i n 2 A — V l — s i n 2 A . B y ( 1 ) a n d ( 2 ) o f A r t . 5 3 , 2 c o s A c a n h a v e t h i s v a l u e o n l y w h e n c o s A + s i n A = V l + s i n 2 A , a n d c o s A — s i n A = — V l — s i n 2 A ; i . e . , w h e n s i n A > c o s A a n d p o s i t i v e . T h e r e f o r e A l i e s b e t w e e n 2 n i t + -a n d 2 n i r + — , 4 4 w h e r e n i s a n y p o s i t i v e o r n e g a t i v e i n t e g e r . 3 . S t a t e t h e s i g n s o f ( s i n 6 + c o s 6 ) a n d ( s i n 6 — c o s 6 ) w h e n 6 h a s t h e f o l l o w i n g v a l u e s : ( 1 ) 2 2 ° ; ( 2 ) 1 9 1 ° ; ( 3 ) 2 9 0 ° ; ( 4 ) 3 4 5 ° ; ( 5 ) - 2 2 ° ; ( 6 ) - 2 7 5 ° ; ( 7 ) - 4 7 0 ° ; ( 8 ) 1 0 0 0 ° . A n s . ( 1 ) + , - ; ( 2 ) - , + ; ( 3 ) - ; ( 4 ) + , - ; ( 5 ) + , - ; ( 6 ) + , + ; ( 7 ) - , ( 8 ) - . 4 . P r o v e t h a t t h e f o r m u l a e w h i c h g i v e t h e v a l u e s o f s i n -a n d o f c o s -i n t e r m s o f s i n a ; a r e u n a l t e r e d w h e n x 2 2 h a s t h e v a l u e s ( 1 ) 9 2 ° , 2 6 8 ° , 9 0 0 ° , 4 n i r + f i r , o r ( 4 n + 2 ) i r -f i r ; ( 2 ) 8 8 ° , -8 8 ° , 7 7 0 ° , -7 7 0 ° , o r I n ± ! . V A L U E S O F S P E C I A L A N G L E S . 6 9 5 . F i n d t h e l i m i t s b e t w e e n w h i c h A m u s t l i e w h e n 2 s i n A = V l + s i n 2 A — V 1 — s i n 2 A . 5 6 . F i n d t h e V a l u e s o f t h e F u n c t i o n s o f 2 2 £ ° . — I n ( 3 ) , ( 4 ) , a n d ( 5 ) o f A r t . 5 1 , p u t x = 4 5 ° . T h e n i n 2 2 £ ° = ^ i o s 2 2 r = ^ I \ 2 -V 2 2 V 2 + V 2 ' 2 ' t a n 2 2 £ ° = 1 - c o s 4 5 ° = V 2 - 1 . 7 s i n 4 5 ° S i n c e 2 2 £ ° i s a n a c u t e a n g l e , i t s f u n c t i o n s a r e a l l p o s i t i v e . T h e a b o v e r e s u l t s a r e a l s o t h e c o s i n e , s i n e , a n d c o t a n g e n t r e s p e c t i v e l y o f 6 7 £ ° , s i n c e t h e l a t t e r i s t h e c o m p l e m e n t o f 2 2 £ ° ( A r t . 1 5 ) . 5 7 . F i n d t h e S i n e a n d C o s i n e o f 1 8 ° . L e t x = 1 8 ° ; t h e n 2 x = 3 6 ° , a n d 3 x = 5 4 ° . . - . 2 x + 3 x = 9 0 ° . . - . s i n 2 x = c o s 3 x ( A r t . 1 5 ) . - . 2 s i n x c o s a ; = 4 c o s 3 a ; — 3 c o s a ; . . ( A r t . 5 0 ) o r 2 s i n x = 4 c o s 2 x — 3 = 1 — 4 s i n 2 x . S o l v i n g t h e q u a d r a t i c , a n d t a k i n g t h e u p p e r s i g n , s i n c e s i n 1 8 ° m u s t b e p o s i t i v e , w e g e t s i n l 8 ° = ^ = i . A l s o , c o s 1 8 ° = V I -s i n 2 1 8 ° = ^ 1 0 - ± ! ^ . ' 4 H e n c e w e h a v e a l s o t h e s i n e a n d c o s i n e o f 7 2 ° ( A r t . 1 5 ) . 7 0 P L A N E T R I G O N O M E T R Y . 5 8 . F i n d t h e S i n e a n d C o s i n e o f 3 6 ° . c o s 3 6 ° = l - 2 s i n 2 1 8 ° . . . [ ( 3 ) o f A r t . 4 9 ] = 1 6 - 2 V 5 = V 5 + l 8 ~ 4 . - . s i n 3 6 ° = V I -c o s 2 3 6 ° = ^ 1 0 ~ 2 V § -4 T h e a b o v e r e s u l t s a r e a l s o t h e s i n e a n d c o s i n e , r e s p e c t i v e l y , o f 5 4 ° ( A r t . 1 5 ) . O t h e r w i s e t h u s : L e t x = 3 6 ° ; t h e n 2 a ; = 7 2 ° , a n d 3 x = 1 0 8 ° . . - . 2 x + 3 x = 1 8 0 ° . . - . s i n 2 a ; = s i n 3 a ; ( A r t . 2 9 ) 2 s i n x c o s a ; = 3 s i n x — 4 s i n 3 , 2 c o s a ; = 3 — 4 s i n 2 a ; = 4 c o s 2 a ; — 1 . a i -± V 5 + 1 S o l v i n g , c o s x — — — - — '— B u t 3 6 ° i s a n a c u t e a n g l e , a n d t h e r e f o r e i t s c o s i n e i s p o s i t i v e . V 5 + 1 o o s d o = 5 9 . I f A + B + C = 1 8 0 ° , o r i f A , B , C a r e t h e A n g l e s o f a T r i a n g l e , p r o v e t h e F o l l o w i n g I d e n t i t i e s : ( 1 ) s i n A + s i n B + s i n C = 4 c o s ^ c o s 5 c o s A A Z A B C ( 2 ) c o s A + c o s B + c o s C = 1 + 4 s i n — s i n — s i n — -Z £ u ( 3 ) t a n A + t a n B + t a n C = t a n A t a n B t a n C . A N G L E S O F A T R I A N G L E . 7 1 W e h a v e A + B + C = 1 8 0 ° . . - . s i n ( A + B ) = s i n C , a n d s i n ^ - ± J = c o s — ( A r t s . l 5 a n d 3 0 ) 2 2 N o w s i n A + s i n B = 2 s i n — ^ c o s ^ — - . ( A r t . 4 5 ) Z 2 = 2 c o s ^ c o s ^ - = - ? . . ( A r t . 1 5 ) C C a n d s i n C = 2 s i n c o s -. . . . ( A r t . 4 9 ) 2 2 = 2 c o s ^ - ± - c o s ^ . . ( A r t . 1 5 ) 2 Z . - . s i n A + s i n B + s i n C = 2 c o s - c o s | - 2 c o s - c o s — - E — 2 2 2 2 o C / A -B , A + B \ - - 2 c o s -( c o s \ - c O S ' ] 2 V 2 2 y = 2 c o s ^ A j c o s — c o s — ] . ( A r t . 4 5 ) 2 V 2 2 J . A B C / 1 N = 4 c o s — c o s — c o s — . . . . ( 1 ) 2 2 2 K ' A g a i n , c o s A + c o s B = 2 c o s ' j " — c o s — ~ — . ( A r t . 4 5 ) 2 2 o -C A - B = 2 s i n -c o s : 2 2 , a n d c o s C = l - 2 s i n 2 -. . . . ( A r t . 4 9 ) 2 C / A — B C \ . - . c o s A + c o s B + c o s C = 1 + 2 s i n -( c o s — s i n " J = l + 2 s i n ° ( c o s ^ _ c o s ^ ) 2 V 2 2 J = l + 4 s i n | s i n | s i n | . . ( 2 ) 7 2 P L A N E T R I G O N O M E T R Y . A g a i n , t a n ( A + B ) = -t a n C ( A r t . 3 0 ) _ t a n A + t a n B , , , ~ l - t a n A t a n B ' ^ 1 ' ' . - . t a n A + t a n B = — t a n C ( 1 — t a n A t a n B ) . . . . t a n A + t a n B + t a n C = t a n A t a n B t a n C . . . . ( 3 ) N o t e . — T h e s t u d e n t w i l l o b s e r v e t h a t ( 1 ) , ( 2 ) , a n d ( 3 ) f o l l o w d i r e c t l y f r o m E x a m p l e s 1 a n d 2 , a n d f o r m u l a ( 3 ) , r e s p e c t i v e l y , o f A r t . 4 8 , b y p u t t i n g A + B + C = 1 8 0 ° . E X A M P L E S . P r o v e t h e f o l l o w i n g s t a t e m e n t s i f A + B + C = 1 8 0 ° : 1 . c o s ( A + B - C ) = - c o s 2 C . A B C 2 . s i n A + s i n B — s i n C = 4 s i n — s i n — c o s - -2 2 2 3 . s i n 2 A + s i n 2 B + s i n 2 C = 4 s i n A s i n B s i n C . 4 . s i n 2 A + s i n 2 B — s i n 2 C = 4 s i n C c o s A c o s B . 5 . t a n 7 A — t a n 4 A — t a n 3 A = t a n 7 A t a n 4 A t a n 3 A . A B C 6 . s i n A — s i n B + s i n C = 4 s i n — c o s -s i n — 2 2 2 7 . c o t — + c o t 5 + c o t - = c o t — c o t — c o t — . 2 2 2 2 2 2 8 . t a n A — c o t B = s e c A c o s e c B c o s C . 6 0 . I n v e r s e T r i g o n o m e t r i c F u n c t i o n s . — T h e e q u a t i o n s i n 6 = x m e a n s t h a t 6 i s t h e a n g l e w h o s e s i n e i s x ; t h i s m a y b e w r i t t e n 6 = s i n - 1 , w h e r e s i n - 1 x i s a n a b b r e v i a t i o n f o r t h e a n g l e ( o r a r c ) w h o s e s i n e i s x . S o t h e s y m b o l s c o s - 1 , t a n - 1 a ; , a n d s e c - 1 ? / , a r e r e a d " t h e a n g l e ( o r a r c ) w h o s e c o s i n e i s x , " " t h e a n g l e ( o r a r c ) w h o s e t a n g e n t i s x , " a n d " t h e a n g l e ( o r a r c ) w h o s e s e c a n t i s y . " T h e s e a n g l e s a r e s p o k e n o f a s b e i n g t h e i n v e r s e s i n e o f x , t h e I N V E R S E T R I G O N O M E T R I C F U N C T I O N S . 7 3 i n v e r s e c o s i n e o f x , t h e i n v e r s e t a n g e n t o f x , a n d t h e i n v e r s e s e c a n t o f y , r e s p e c t i v e l y . S u c h e x p r e s s i o n s a r e c a l l e d i n v e r s e t r i g o n o m e t r i c f u n c t i o n s . N o t e . — T h e s t u d e n t m u s t b e c a r e f u l t o n o t i c e t h a t — 1 i s n o t a n e x p o n e n t , s i n " 1 x i s n o t ( s i n x ) " 1 , w h i c h — — — s i n a ; N o t i c e a l s o t h a t s i n " 1 ~ = c o s ' 1 ~ i s n o t a n i d e n t i t y , b u t i s t r u e o n l y f o r t h e p a r t i c u l a r a n g l e 6 0 ° . T h i s n o t a t i o n i s o n l y a n a l o g o u s t o t h e u s e o f e x p o n e n t s i n m u l t i p l i c a t i o n , w h e r e w e h a v e a " 1 a = a 0 = 1 . T h u s , c o s - 1 ( c o s x ) = x , a n d s i n ( s i n " 1 x ) = x ; t h a t i s , c o s ' 1 i s i n v e r s e t o c o s , a n d a p p l i e d t o i t a n n u l s i t ; a n d s o f o r o t h e r f u n c t i o n s . T h e F r e n c h m e t h o d o f w r i t i n g i n v e r s e f u n c t i o n s i s a r c s i n x , a r c c o s x , a r c t a n x , a n d s o o n . E X A M P L E S . 1 . S h o w t h a t 3 0 ° i s o n e v a l u e o f s i n - 1 ^ . W e k n o w t h a t s i n 3 0 ° = . - . 3 0 ° i s a n a n g l e w h o s e s i n e i s I ; o r 3 0 ° = s i n - 1 £ . 2 . P r o v e t h a t t a n - 1 £ + t a n - 1 £ = 4 5 ° . t a n - 1 £ i s o n e o f t h e a n g l e s w h o s e t a n g e n t i s a n d t a n - 1 i s o n e o f t h e a n g l e s w h o s e t a n g e n t i s L e t a = t a n - I £ , a n d / J = t a n _ 1 £ ; t h e n t a n a = | a n d t a n / 3 = ^ . N o w t a n ( « + / j ) = t a n < + t a n / ? . . . ( A r t . 4 7 ) 1 — t a n « t a n / 3 1 = $ + i = i B u t t a n 4 5 ° = 1 , . - . « + / ? = 4 5 ° ; t h a t i s , t a n - 1 \ + t a n - 1 1 = 4 5 ° . T h e r e f o r e 4 5 ° i s o n e v a l u e o f t a n - 1 ^ + t a n - 1 £ . 7 4 P L A N E T R I G O N O M E T R Y . 3 . P r o v e t h a t t a n - 1 a ; + t a n ~ 1 y = t a n - 1 ^ - x i L 1 — x y L e t t a n - 1 a ; = A . . - . t a n A = x . t a n - 1 2 / = B . . - . t a n B = y . x r 4 . / , - o \ t a n A + t a n B N o w t a n ( A + B ) = - , ~ -v ' 1 - t a n A t a u B _ x + y 1 — x y . - . A + B = t a n - ] ^ - ± ^ -1 — x y . : t a n - 1 a ; + t a n - 1 y = t a n - 1 - " t ^ . 1 — x y A n y r e l a t i o n s w h i c h h a v e b e e n e s t a b l i s h e d a m o n g t h e t r i g o n o m e t r i c f u n c t i o n s m a y b e e x p r e s s e d b y m e a n s o f t h e i n v e r s e n o t a t i o n . T h u s , w e k n o w t h a t 4 . c o s a ; = V l — s i n 2 a ; . T h i s m a y b e w r i t t e n x = c o s - 1 V l — s i n 2 . . ( 1 ) P u t s i n x = 6 ; t h e n x = s i n - 1 6 . T h u s ( 1 ) b e c o m e s s i n - 1 6 = c o s - 1 V l — 6 . 5 . B y A r t . 4 9 , c o s 2 0 = 2 c o s 2 0 - l , w h i c h m a y b e w r i t t e n 2 6 = c o s - 1 ( 2 c o s 2 0 — 1 ) . P u t c o s 0 = x . . - . 2 c o s - 1 a ; = c o s - 1 ( 2 a ^ — 1 ) . 6 . B y A r t . 4 9 , s i n 2 6 = 2 s i n 6 c o s 6 , w h i c h m a y b e w r i t t e n 2 6 = s i n - 1 ( 2 s i n 6 c o s 6 ) . P u t s i n 0 = a : . . - . 2 s i n - 1 a ; = s i n - 1 ( 2 a ; V l -x 2 ) . T A B L E O F U S E F U L F O R M U L A . 7 5 7 . P r o v e s i n - 1 a ; = c o s - 1 V l — a r = t a n -. 1 V l - a r 1 8 . " t a n - 1 x = s i n - 1 x " " " - 1 — — 9 . " 2 t a n ~ 1 x = t a n 1 — a r V l + x V l + x 2 x 1 0 . " s i n ( 2 s i n - 1 ) = 2 x V l — x 1 . 1 1 . " t a n - ^ + t a n - 1 ^ ^ 7 6 4 1 2 . " c o s - 1 - + 2 S U 1 - 1 - = 1 2 0 ° . 2 2 1 3 . " c o t - 1 3 + c o s e c - 1 V B " = - -4 1 4 . « 3 s i n - 1 = s i n - 1 ( 3 x - 4 x 3 ) 6 1 . T a b l e o f U s e f u l F o r m u l a . — T h e f o l l o w i n g i s a l i s t o f i m p o r t a n t f o r m u l a e p r o v e d i n t h i s c h a p t e r , a n d s u m m e d u p f o r t h e c o n v e n i e n c e o f t h e s t u d e n t : 1 . s i n ( a ; + ? / ) = s i n x c o s y + c o s x s i n y . . ( A r t . 4 3 ) 2 . c o s ( x + y ) = c o s x c o s y — s i n x s i n y . 3 . s i n ( x — y ) = s i n x c o s y — c o s x s i n y . . ( A r t . 4 4 ) v 4 . c o s ( x — y ) = c o s a ; c o s y + s i n s i n y . 5 . 2 s i n a ; c o s y = s i n ( a ; + y ) + s i n ( a ; — y ) . ( A r t . 4 5 ) 6 . 2 c o s a ; s i n ? / = s i n ( a ; + y ) — s i n ( x — y ) . 7 . 2 c o s x c o s j / = c o s ( x + y ) + c o s ( x — y ) . 8 . 2 s i n a ; s i n y = c o s ( x — y ) — c o s ( x + y ) . 9 . s i n x + s i n « / = 2 s i n £ ( x + y ) c o s . £ ( a ; — y ) . 1 0 . s i n x — s i n y = 2 c o s £ ( a ; + y ) s i n i ( a ; — y ) . 7 6 P L A N E T R I G O N O M E T B Y . 1 1 . c o s x + c o s y = 2 c o s | ( a ; + y ) c o s £ ( a ; — y ) . 1 2 . c o s y — c o s a ; = 2 s i n £ ( a ; + y ) s i n £ ( x — y ) . 1 3 s i n x + s i n y _ t a n £ ( a ; + y ) ( A r t 4 6 ) s i n x — s i n y t a n £ ( a ¡ — y ) 1 4 . t a n ( a ; + y ) = t a n s + t a n y ( A r t . 4 7 ) 1 — t a n x t a n y i K . , -t a n x — t a n y 1 5 . t a n ( a ; — y ) = 1 + t a n x t a n y - u » . , , , c o t a ; c o t v — 1 1 6 . c o t ( a ; + y ) = c o t x + c o t y 1 7 . c o t ( a ; - y ) _ g g t g c g t j r + 1 . c o t a ; — c o t y 1 8 . t a n ( a ; ± 4 5 0 ) = t a n a ; : F l ( A r t . 4 7 ) t a n x ± 1 v 1 9 . s i n ( a ; + y ) s i n ( a ; — y ) = s i n 2 a ; — s i n 2 y = c o s 2 y — c o s 2 a ; . 2 0 . c o s ( a ; + y ) c o s ( a ; — y ) = c o s 2 a ; — s i n 2 y = c o s 2 y — s i n 2 a ; . 2 1 . t a n a ; ± t a n y = s i n c o s x c o s y 2 2 . c o t a ; ± c o t y = s i n ^ ± a ; ) . s i n a ; s i n y 2 3 . s i n 2 a ; = 2 s i n a ; c o s a ; = ^ 5 í ^ -. . . ( A r t . 4 9 ) 1 + t a n 2 a ; v ' 2 4 . c o s 2 a ; = c o s 2 x — s i n 2 a ; = 1 — 2 s i n 2 x = 2 c o s 2 a ; — 1 _ 1 — t a n 2 a ; 1 + t a n 2 a ; o r 1 — c o s 2 a ; 2 s t n 2 x . 2 ¿ 5 . = t a n , ' a ; . 1 + c o s 2 a ; 2 c o s 2 x E X A M P L E S . 7 7 2 6 . t a n 2 = 2 t a n 2 7 . c o t 2 a ; = 1 — t a n 2 c o t 2 — 1 2 c o t a ; 2 8 . s i n 3 = 3 s i n x — 4 s i n 3 x ( A r t . 5 0 ) 2 9 . c o s 3 a ; = 4 c o s 3 — 3 c o s . 3 t a n x — t a n 3 3 0 . t a n 3 = 1 — 3 t a n 2 3 1 . s i n 2 ; ? = 1 ~ , ° o s ( A r t . 5 1 ) — ■ — 3 2 . c o s 2 = 1 + c o s . 2 2 3 3 . t a n - 1 + t a n - 1 ? / = t a n - 1 ^ - ± ^ . . . . ( A r t . 6 0 ) 1 - x y E X A M P L E S . 1 2 1 . I f s i n a = - , a n d s i n B = - , f i n d a v a l u e f o r s i n ( r e + B ) , a n d s i n ( a -0 ) . ^ V 5 + 4 V 2 . V 5 -4 V 2 " 9 , ' 9 2 . I f c o s a = - , a n d c o s B = — , f i n d a v a l u e f o r s i n ( a + B ) , o 4 1 a n d c o s ( a + / 3 ) . . 1 5 6 1 3 3 2 0 5 , 2 0 5 ' 3 2 3 . I f c o s r e = ' - , a n d c o s B = - , f i n d a v a l u e f o r s i n ( r e + B ) , 4 5 a n d s i n ( a - 0 ) . . 2 V 7 + 3 V 2 1 2 V 7 - 3 V 2 1 A U S -2 0 , 2 0 4 . I f s i n a = - , a n d s i n 8 = % f i n d a v a l u e f o r s i n ( r e + B ) , 5 5 a n d c o s ( r e + B ) . ^ A n a 1 > 2 4 » 2 5 7 8 P L A N E T R I G O N O M E T R Y . g 5 . I f s i n c e = . 6 , a n d s i n / ? = — , f i n d a v a l u e f o r s i n ( « — / ? ) , a n d c o s ( a + / ? ) . j g A n s . — , — 6 5 , 6 5 6 . I f s i n a = = ^ ^ , a n d s i n = s h o w t h a t o n e v a l u e V 5 V l O o f a + $ i s 4 5 ° . 7 . P r o v e c o s 0 + c o s 3 0 = 2 c o s 2 0 c o s 0 . 8 . " 2 c o s a c o s / 3 = c o s ( « — / J ) + c o s ( a - f / ? ) . 9 . " 2 s i n 3 0 c o s 5 0 = s i n 8 0 -s i n 2 0 . 1 0 . " 2 c o s 0 c o s -= c o s 0 + c o s 2 0 . _ 2 1 1 . " s i n 4 0 s i n 0 = £ ( c o s 3 0 -c o s 5 0 ) . 1 2 . " 2 c o s 1 0 ° s i n 5 0 ° = s i n 6 0 ° + s i n 4 0 ° . 1 3 . S i m p l i f y 2 c o s 2 0 c o s 0 -2 s i n 4 0 s i n e 1 . A n s . 2 c o s 3 0 c o s 2 0 . 1 4 . S i m p l i f y s i n — c o s - — s i n — c o s — -— c o s 4 0 s i n 2 0 . v J 2 2 2 2 P r o v e t h e f o l l o w i n g s t a t e m e n t s : 1 5 . c o s 3 « — c o s 7 a = 2 s i n 5 « s i n 2 a . 1 6 . s i n 6 0 ° + s i n 2 0 ° = 2 s i n 4 0 ° c o s 2 0 ° . 1 7 . s i n 3 0 + s i n 5 0 = 2 s i n 4 0 c o s 0 . 1 8 . s i n 7 0 - s i n 5 0 = 2 c o s 6 0 s i n 0 . 1 9 . c o s 5 0 + c o s 9 0 = 2 c o s 7 0 c o s 2 0 . 2 0 . s i n 2 g + s i n = t a n ^ -c o s 0 + c o s 2 0 2 2 1 . c o s ( 6 0 ° + A ) + c o s ( 6 0 ° -A ) = c o s A . E X A M P L E S . 7 9 2 2 . c o s ( 4 5 ° + A ) + c o s ( 4 5 ° -A ) = V 2 c o s A . 2 3 . s i n ( 4 5 ° + A ) -s i n ( 4 5 ° - A ) = V 2 s i n A . 2 4 . c o s 2 0 + c o s 4 0 = 2 c o s 3 0 c o s 0 . 2 5 . c o s 4 0 — c o s 6 0 = 2 s i n 5 0 s i n 0 . 2 6 . c o s 6 + c o s 3 6 + c o s 5 6 + c o s 7 0 = 4 c o s 6 c o s 2 6 c o s 4 0 . 2 7 . c o t « + t a n / 3 = 0 0 8 ( " - f t ) . s i n a c o s ¡ 3 2 8 . c o t a -t a n 0 = c _ ? l í í L ± £ ) . s i n a c o s / J 2 9 . s i n ( A - 4 5 ° ) = ^ l A - c o s A . V 2 3 0 . V 2 s i n ( A + 4 5 ° ) = s i n A + c o s A . 3 1 . c o s ( A + 4 5 ° ) + s i n ( A - 4 5 ° ) = 0 . 3 2 . t a n ( g - ^ ) + t a n 0 = t a n f t 1 — t a n ( 6 — < £ ) t a n < f > 3 3 . t o n < g + ^ ) + t a n ^ = 1 + t a n ( 6 + c ¿ > ) t a n 0 3 4 . c o s ( 0 + < j > ) -s i n ( 0 -< ¿ > ) = 2 s i n ^ -O ^ j c o s -4 > j . 3 5 . s i n n O c o s 0 + c o s n O s i n 0 = s i n ( n + 1 ) 6 . 3 6 . c o t ^ - ^ = c o t g + 1 . V a ) i - c o t e 3 7 . t a n ^ - ^ + c o t ^ 0 + ^ = 0 . 3 8 . c o t ^ 0 - | ^ + t a n ^ 0 + ^ = 0 . 3 9 . t a n ( n + 1 ) 0 -t a n n 0 = t a n . 1 + t a n ( n + l ) < t > t a n 8 0 P L A N E T R I G O N O M E T R Y . 4 0 . I f t a , n x = 1 , a n d t a n y = - ^ - z , p r o v e t h a t V 3 t a n ( a ; + y ) = 2 + V 3 . 4 1 . I f t a n a = , a n d t a n / 3 = , p r o v e t h a t m + 1 2 m + 1 t a n ( « + / ? ) = l . 4 2 . I f t a n a = m , a n d t a n j 3 = n , p r o v e t h a t / . n \ 1 — i n n c o s ( a + / 3 ) = -V ( l + m 2 ) ( l + ? i 2 ) 4 3 . I f t a n 0 = ( a + 1 ) , a n d t a n < j > = ( a — 1 ) , p r o v e t h a t 2 c o t ( 0 -< £ ) = a 2 . P r o v e t h e f o l l o w i n g s t a t e m e n t s : 4 4 . c o s ( a ; — y + 2 ) = c o s x c o s y c o s 2 + c o s x s i n 2 / s i n 2 — s i n a ; c o s y s i n 2 + s i n x s i n t / c o s 2 . 4 5 . s i n ( x — y — 2 ) = s i n x + s i n y - f s i n 2 + 4 s i n £ ( a ; -y ) s i n % ( x — z ) s m i ( y + z ) . 4 6 . s i n ( a ; + y — 2 ) + s i n ( a ; + 2 — y ) + s i n ( y + 2 — a ; ) = s i n ( a ; + ? / + 2 ) + 4 s i n a ; s i n t / s i n 2 . 4 7 . s i n 2 a ; + s i n 2 t / + s i n 2 z — s i n 2 ( x + y + 2 ) = 4 s i n ( a ; + y ) s i n ( y + 2 ) s i n ( 2 + x ) . 4 8 . c o s 2 a ; + c o s 2 y + c o s 2 2 + c o s 2 ( x + y + z ) = 4 c o s ( x + y ) c o s ( y + 2 ) c o s ( 2 + x ) . 4 9 . c o s ( x + y — z ) + c o s ( y + z — a ; ) 4 - c o s ( 2 + x — y ) + c o s ( x + y + 2 ) = 4 c o s a ; c o s y c o s 2 . E X A M P L E S . 8 1 5 0 . s i n 2 a ; - f s i n 2 ? / + s i n 2 z + s i n 2 ( x - f y + z ) = 2 \ 1 — c o s ( x + y ) c o s ( y + z ) c o s ( z + x ) 5 1 . c o s 2 a ; + c o s 2 y + c o s 2 z + c o s 2 ( x + y — z ) = 2 { 1 + c o s ( a ; + y ) c o s ( x — z ) c o s ( y — z ) \ . 5 2 . c o s a ; s i n ( y — z ) + c o s y s i n ( z — x ) + c o s z s i n ( a ; — y ) = 0 . 5 3 . s i n a ; s i n ( y — z ) + s i n y s i n ( z — a ; ) + s i n z s i n ( a ; — y ) = 0 . 5 4 . c o s ( x + y ) c o s ( x — y ) + s i n ( y + z ) s i n ( y — z ) — c o s ( a ; + z ) c o s ( x — z ) = 0 . 2 — s e c 2 0 „ a 5 5 . — = c o s 2 0 . s e c 2 0 5 6 . c o s 2 0 ( 1 -t a n 2 0 ) = c o s 2 0 . 5 7 . c o t 2 0 = c o t ^ = i 2 c o t 0 c o o z ) C O t 2 0 + l 5 8 . s e c 2 0 = — — — — c o t 2 0 - l 5 9 . ^ s i n | + c o s | j = 1 + s i n 0 . 6 0 . f s i n - — c o s - ^ = 1 — s i n 0 . H -6 1 . = 2 c o s 2 ? . s e c 0 2 6 2 . c o s 2 0 1 — t a n 0 l + s i n 2 0 l + t a n 0 ~ 1 + t a n ^ g g c O s 0 _ 2 ' i - 8 i n < i ~ r ~ ^ ' 2 8 2 P L A N E T R I G O N O M E T R Y . c . 1 + s i n a ; + c o s a ; , x 6 4 . — ' -— : ! = c o t — 1 + s i n x — c o s x 2 g g c o s 3 a ; + s i n 3 a ; _ 2 — s i n 2 a ; c o s a ; + s i n a ; 2 g g c o s 3 a ; — s i n 3 a ; _ 2 + s i n 2 a ; c o s a ; — s i n a ; 2 6 7 . c o s 4 0 - s i n 4 0 = c o s 2 6 . g g s i n 3 0 _ c o s 3 6 _ 2 s i n 6 c o s 6 6 9 . ? 2 s 3 t f + 8 i n 3 f l = 2 c o t 2 0 . s i n 0 c o s 6 7 0 . s Í n i ^ = 2 c o s 2 g . s i n 2 0 s i n — . c o s — 7 1 . — = 2 V 3 . s i n — c o s — 1 2 1 2 7 2 . t a n ( 4 5 ° + a ; ) -t a n ( 4 5 ° -x ) = 2 t a n 2 x . 7 3 . t a n ( 4 5 ° -» ) + c o t ( 4 5 ° -¡ b ) = 2 s e c 2 x . „ . t a n 2 ( 4 5 ° + a ; ) - 1 . „ 7 4 . J - — ' -= s i n 2 a ; . t a n 2 ( 4 5 ° + x ) + 1 7 5 . c o s ( a ; + 4 5 ° ) = s e c 2 a ; _ t a n 2 a ; . c o s ( x — 4 5 ° ) 7 6 . t a n a ; = s i n a ; + s i n 2 a ; . 1 + c o s x + c o s 2 x „ „ 4 . s i n 2 a ; — s i n a ; 7 7 . t a n x — 1 — c o s a s + c o s 2 a ; 7 8 . £ 2 i l £ = 2 c o s 2 a ¡ - l . c o s a ; E X A M P L E S . 8 3 r r n 3 s i n x — s i n 3 x , , 7 9 . = t a n 3 x . c o s 3 x + 3 c o s x O A , „ c o f a ; — 3 c o t x 8 0 . c o t 3 a ; = — — 3 c o t 2 a ; - l 8 1 . 1 ~ c o s 3 a ; = ( l + 2 c o s a ; ) 2 . 1 — c O S X o r > s i n x + c o s x . o , o 8 2 . — ;— = t a n 2 x + s e c 2 x . c o s x — s i n a ; g g c o s 2 x + c o s 1 2 x c o s 7 a ; — c o s 3 a ; 2 s i n 4 a ; _ ^ q c o s 6 x + c o s 8 a ; c o s a ; — c o s 3 x s i n 2 x 8 4 . s i n 2 a ; s i n 2 ? / — s i n 2 ( a ; + y ) — s i n 2 ( a ; — y ) . 8 5 . t a n 5 0 ° + c o t 5 0 ° = 2 s e c 1 0 ° . 8 6 . s i n 3 x = 4 s i n x s i n ( 6 0 ° + x ) s i n ( 6 0 ° -x ) . 8 7 . c o t í - t a n - = 2 . 8 8 8 8 . t a n 4 f l = 4 t a n f l ( l - t a n 2 g ) _ l - 6 t a n 2 0 + t a n 4 0 8 9 . 2 c o s ^ = V 2 + V 2 . o 9 0 . ( 3 s i n 6 -4 s i n 3 0 ) 2 + ( 4 c o s 3 0 -3 c o s 6 ) 2 = 1 . 9 1 s i n 2 6 c o s 0 _ 6 ( l + e o s 2 0 ) ( l + c o s 0 ) ~ a n 2 1 2 1 9 2 . I f t a n 6 = - , a n d t a n < f > = — , p r o v e t a n ( 2 6 + < / > ) = - -7 1 1 2 9 5 . P r o v e t h a t t a n -a n d c o t -a r e t h e r o o t s o f t h e 2 2 e q u a t i o n a ? — 2 a ; c o s e c 0 + l = O . 8 4 P L A N E T R I G O N O M E T R Y . 9 4 . I f t a n 6 = -p r o v e t h a t a V a + b , j a — b 2 c o s 6 + b V c o s 2 0 9 5 . F i n d t h e v a l u e s o f ( 1 ) s i n 9 ° , ( 2 ) c o s 9 ° , ( 3 ) s i n 8 1 ° , ( 4 ) c o s 1 8 9 ° , ( 5 ) t a n 2 0 2 ^ ° , ( 6 ) t a n 9 7 £ ° . A n s . ( 1 ) 1 ( ^ 3 + ^ 5 -V 5 - V 5 ) , ( 2 ) i ( V 3 + V 5 + V 5 -V 5 ) , ( 3 ) s i n 8 1 ° = c o s 9 ° , ( 4 ) c o s 1 8 9 ° = - c o s 9 ° , ( 5 ) V 2 -1 , ( 6 ) _ ( V 3 + V 2 ) ( V 2 + 1 ) . 9 6 . I f A = 2 0 0 ° , p r o v e t h a t ( 1 ) 2 s i n -= + V l + s i n A + V l -s j n A . L i ( 2 ) t a n A = - ( 1 + V l + t a n 2 A ) ' . ' 2 t a n A 9 7 . I f A l i e s b e t w e e n 2 7 0 ° a n d 3 6 0 ° , p r o v e t h a t ( 1 ) 2 s i n A = + V l - s i n A -V l + s i n A . 2 A ( 2 ) t a n — = — c o t A + c o s e c A . 9 8 . I f A l i e s b e t w e e n 4 5 0 ° a n d 6 3 0 ° , p r o v e t h a t A ( 1 ) 2 s i n — = — V l + s i n A — V 1 — s i n A . 2 ( 2 ) 2 c o s — = _ V T + s i n A + V 1 — s i n A . 2 E X A M P L E S . 8 5 P r o v e t h e f o l l o w i n g s t a t e m e n t s , A , B , C b e i n g t h e a n g l e s o f a t r i a n g l e . 9 9 . s i n A - s i n B = t a n C t a n A - B 1 0 0 . s i n A + s i n B 2 2 s i n 3 B - s i n 3 C _ t a n 3 A c o s 3 C - c o s 3 B _ 2 1 A 1 A A , . B B , . C C 1 0 1 . s i n — c o s — k s i n — c o s _ ■+ s i n -c o s -2 2 2 2 2 2 o A B C = 2 c o s — c o s — c o s - -2 2 2 1 0 2 . c o s 2 — + c o s 2 — — c o s 2 - = 2 c o s — c o s — s i n — -2 2 2 2 2 2 1 0 3 . s i n A c o s A — s i n B c o s B + s ; n C c o s C = 2 c o s A s i n B c o s C . 1 0 4 . c o s 2 A - + - c o s 2 B + c o s 2 C = — 1 — 4 c o s A c o s B c o s C . 1 0 5 . s i n 2 A — s i n 2 B + s i n 2 C = 2 s i n A c o s B s i n C . 1 0 6 . t a n ? t a n — + t a n 5 t a n — + t a n — t a n ? = 1 . 2 2 2 2 2 2 P r o v e t h e f o l l o w i n g s t a t e m e n t s w h e n w e t a k e f o r s i n - 1 , c o s - 1 , e t c . , t h e i r l e a s t p o s i t i v e v a l u e . 1 0 7 . s i n - 1 \ = c o s - 1 — = c o t - 1 V 3 . 1 0 8 . 2 t a n - 1 ( c o s 2 6 ) = t a n - 1 ^ 1 0 9 . 4 t a n - 1 i - t a n - 1 — _ ( . i n - i / , c o t 2 f l — t a n 2 A N 2 3 9 4 , - + s i n - 1 5 1 7 8 5 2 1 1 0 . s i n - 1 - + s i n - ~ + s i n - 1 — = 8 6 P L A N E T R I G O N O M E T R Y . 1 1 1 . 1 1 2 . 1 1 3 . 1 1 4 . 1 1 5 . 1 1 6 . t a n - 1 V 5 ( 2 -V 3 ) -c o t - 1 V 5 ( 2 + V 3 ) = c o t ' 1 V 5 . s e c - 1 V 3 = 2 c o t - 1 V 2 . 2 c o t - 1 a ; = c o s e c - 1 j — 2 a ; t a n - . V 3 + V 2 _ j I ; V 3 - V 2 Y s i n - 1 + c o t - 1 3 = - -V 5 4 ' 3 = 3 j r 2 4 ' c o s - 1 — + 2 t a n 6 5 - , i ^ s i n - 1 ? -5 5 I f 0 = s i n - 1 - , a n d < j > = c o s - 1 -t h e n 6 + < f , = 9 0 ° 5 5 1 1 7 . 1 1 8 . P r o v e t h a t c o s ( 2 t a n - 1 a ; ) = 1 1 9 . 1 2 0 . - i 3 = ( i n s ' - , 5 , 1 - a r 1 2 1 . 1 2 2 . 1 2 3 . 1 2 4 . 1 2 5 . 1 + a ; 2 ' t a n - 1 - + c o s e c - 1 V l 0 = - . 2 4 2 j _ 1 2 _ i 5 1 3 3 t a n 1 c o s e c - = s i n 1 — 3 3 6 5 < 2 t a n - 1 l + c o s - 1 | = | ' s i n - 1 ( c o s x ) + c o s - 1 ( s i n y ) + x + y = i t . i 1 2 t a n - 1 - — 1 - t a n - 1 h t a n - 1 — = t i w . 1 + x t a n - 1 x ~ - + t a n - 1 ' s i n - 1 a ; — s i n - 1 ? / 1 - a ; 1 . i r = n i r A — 2 a ; - 1 4 = c o s - 1 ( x y ± \ / l — x 2 — y 2 + x 1 } / " ) . N A T U H E A N D U S E O F L O G A R I T H M S . 8 7 C H A P T E R I V . L O G A R I T H M S A N D L O G A R I T H M I C T A B L E S . — T R I G O N O M E T R I C T A B L E S . 6 2 . N a t u r e a n d U s e o f L o g a r i t h m s . — T h e n u m e r i c a l c a l c u l a t i o n s w h i c h o c c u r i n T r i g o n o m e t r y a r e v e r y m u c h a b b r e v i a t e d b y t h e a i d o f l o g a r i t h m s ; a n d t h u s i t i s n e c e s s a r y t o e x p l a i n t h e n a t u r e a n d u s e o f l o g a r i t h m s , a n d t h e m a n n e r o f c a l c u l a t i n g t h e m . T h e l o g a r i t h m o f a n u m b e r t o a g i v e n b a s e i s t h e e x p o n e n t o f t h e p o w e r t o w h i c h t h e b a s e m u s t b e r a i s e d t o g i v e t h e n u m b e r . T h u s , i f a x = m , x i s c a l l e d t h e " l o g a r i t h m o f m t o t h e b a s e a , " a n d i s u s u a l l y w r i t t e n x = l o g a m , t h e b a s e b e i n g p u t a s a s u f f i x . T h e r e l a t i o n b e t w e e n t h e b a s e , l o g a r i t h m , a n d n u m b e r i s e x p r e s s e d b y t h e e q u a t i o n , ( b a s e ) , o g = n u m b e r . T h u s , i f t h e b a s e o f a s y s t e m o f l o g a r i t h m s i s 2 , t h e n 3 i s t h e l o g a r i t h m o f t h e n u m b e r 8 , b e c a u s e 2 s — 8 . I f t h e b a s e b e 5 , t h e n 3 i s t h e l o g a r i t h m o f 1 2 5 , b e c a u s e 5 3 = 1 2 5 . 6 3 . P r o p e r t i e s o f L o g a r i t h m s . — T h e u s e o f l o g a r i t h m s d e p e n d s o n t h e f o l l o w i n g p r o p e r t i e s w h i c h a r e t r u e f o r a l l l o g a r i t h m s , w h a t e v e r m a y b e t h e b a s e . F r o m t h e d e f i n i t i o n i t f o l l o w s t h a t ( 1 ) l o g a a x = x , a n d c o n v e r s e l y ( 2 ) o l 0 8 a ™= m . T a k i n g t h e l o g a r i t h m s o f b o t h s i d e s o f t h e e q u a t i o n a x = m , w e h a v e l o g a a x = x = l o g m . C o n v e r s e l y , t a k i n g t h e e x p o n e n t i a l s o f b o t h s i d e s o f x = l o g a m t o b a s e a , w e h a v e a x = a l o Z a ™= m . a x = m a n d x = \ o g a m a r e t h u s s e e n t o b e e q u i v a l e n t , a n d t o e x p r e s s t h e s a m e r e l a t i o n b e t w e e n a n u m b e r , w i , a n d i t s l o g a r i t h m , x , t o b a s e a . 8 8 P L A N E T R I G O N O M E T R Y . ( 1 ) T h e l o g a r i t h m o f 1 i s z e r o . F o r a " = 1 , w h a t e v e r a m a y b e ; t h e r e f o r e l o g 1 = 0 . ( 2 ) T h e l o g a r i t h m o f t h e b a s e o f a n y s y s t e m i s u n i t y . F o r a } = a , w h a t e v e r a m a y b e ; t h e r e f o r e l o g a a = 1 . ( 3 ) T h e l o g a r i t h m o f z e r o i n a n y s y s t e m w h o s e b a s e i s g r e a t e r t h a n 1 i s m i n u s i n f i n i t y . F o r a ~ ° ° = — = -= 0 ; t h e r e f o r e l o g 0 = — a o . a " o o ( 4 ) T h e l o g a r i t h m o f a p r o d u c t i s e q u a l t o t h e s u m o f t h e l o g a r i t h m s o f i t s f a c t o r s . F o r l e t a ; = l o g a r a , a n d y — \ o g a n . . : m = a ' , a n d n = a ' . . : m n = a + . . : l o g a m n = x + y = l o g a m + l o g a n . S i m i l a r l y , \ o g a m n p = l o g a m + l o g a n + l o g a p , a n d s o o n f o r a n y n u m b e r o f f a c t o r s . T h u s , l o g 6 0 = l o g ( 3 x 4 x 5 ) , = l o g 3 + l o g 4 + l o g 5 . ( 5 ) T h e l o g a r i t h m o f a q u o t i e n t i s e q u a l t o t h e l o g a r i t h m o f t h e d i v i d e n d m i n u s t h e l o g a r i t h m o f t h e d i v i s o r . l o g a m , a n d y = l o g a n . a ' , a n d n = a " . a ' - ' . x — y = l o g a m — l o g a n . l o g 1 7 — l o g 5 . F o r l e t x = m : m _ n m i f i t T h u s , P R O P E R T I E S O F L O G A R I T H M S . 8 9 ( 6 ) T h e l o g a r i t h m o f a n y p o w e r o f a n u m b e r i s e q u a l t o t h e l o g a r i t h m o f t h e n u m b e r m u l t i p l i e d b y t h e e x p o n e n t o f t h e p o w e r . F o r l e t c c = l o g a m . . - . m = a z . . : m " = a " . . : \ o g a m p = p x = p l o g a m . ( 7 ) T h e l o g a r i t h m o f a n y r o o t o f a n u m b e r i s e q u a l t o t h e l o g a r i t h m o f t h e n u m b e r d i v i d e d b y t h e i n d e x o f t h e r o o t . F o r l e t x = \ o g a m . . : m = a " . 1 x . : m ' = a r . I x i . - . l o g ( m r ) = ~ = - l o g . m . r r I t f o l l o w s f r o m t h e s e p r o p o s i t i o n s t h a t b y m e a n s o f l o g a r i t h m s , t h e o p e r a t i o n s o f m u l t i p l i c a t i o n a n d d i v i s i o n a r e c h a n g e d i n t o t h o s e o f a d d i t i o n a n d s u b t r a c t i o n ; a n d t h e o p e r a t i o n s o f i n v o l u t i o n a n d e v o l u t i o n a r e c h a n g e d i n t o t h o s e o f m u l t i p l i c a t i o n a n d d i v i s i o n . 1 . S u p p o s e , f o r i n s t a n c e , i t i s r e q u i r e d t o f i n d t h e p r o d u c t o f 2 4 6 a n d 3 5 7 ; w e a d d t h e l o g a r i t h m s o f t h e f a c t o r s , a n d t h e s u m i s t h e l o g a r i t h m o f t h e p r o d u c t : t h u s , l o g 1 0 2 4 6 = 2 . 3 9 0 9 3 l o g 1 0 3 5 7 = 2 . 5 5 2 6 7 4 . 9 4 3 6 0 w h i c h i s t h e l o g a r i t h m o f 8 7 8 2 2 , t h e p r o d u c t r e q u i r e d . 2 . I f w e a r e r e q u i r e d t o d i v i d e 3 7 1 . 4 9 b y 5 2 . 3 7 6 , w e p r o c e e d t h u s : l o g 1 0 3 7 1 . 4 9 = 2 . 5 6 9 9 5 l o g 1 0 5 2 . 3 7 6 = 1 . 7 1 9 1 3 0 . 8 5 0 8 2 w h i c h i s t h e l o g a r i t h m o f 7 . 0 9 2 7 5 2 , t h e q u o t i e n t r e q u i r e d . 9 0 P L A N E T R I G O N O M E T l i V . 3 . I f w e h a v e t o f i n d t h e f o u r t h p o w e r o f 1 3 , w e p r o c e e d t h u s : l o g 1 0 1 3 = 1 . 1 1 3 9 4 4 4 . 4 5 5 7 6 w h i c h i s t h e l o g a r i t h m o f 2 8 5 6 1 , t h e n u m b e r r e q u i r e d . 4 . I f w e a r e t o f i n d t h e f i f t h r o o t o f 1 C 8 0 7 , w e p r o c e e d t h u s : 5 ) 4 . 2 2 5 4 9 -l o g , , , 1 6 8 0 7 , 0 . 8 4 5 0 9 8 w h i c h i s t h e l o g a r i t h m o f 7 , t h e r o o t r e q u i r e d . 5 . G i v e n l o g 1 0 2 = 0 . 3 0 1 0 3 ; f i n d l o g 1 0 1 2 8 , l o g 1 0 5 1 2 . A n a . 2 . 1 0 7 2 1 , 2 . 7 0 9 2 7 . 6 . G i v e n l o g 1 0 3 = 0 . 4 7 7 1 2 ; f i n d l o g 1 0 8 1 , l o g , „ 2 1 8 7 . A n a . 1 . 9 0 8 4 9 , 3 . 3 3 9 8 5 . 7 . G i v e n l o g 1 0 3 ; f i n d l o g 1 0 0 . 2 8 6 2 7 . 8 . F i n d t h e l o g a r i t h m s t o t h e b a s e a o f , J O 4 / -S r ~ r , - i a , « 3 , V a , v « " , a J . 9 . F i n d t h e l o g a r i t h m s t o t h e b a s e 2 o f 8 , 6 4 , £ , . 1 2 5 , . 0 1 5 6 2 5 , ^ 6 4 . A n a . 3 , 6 , - 1 , - 3 , - 6 , 2 . 1 0 . F i n d t h e l o g a r i t h m s t o b a s e 4 o f 8 , - ^ 1 6 , \ / . 5 > \ / . 0 1 5 6 2 5 . A n a . f , f , - \ , - 1 . E x p r e s s t h e f o l l o w i n g l o g a r i t h m s i n t e r m s o f l o g a , l o g b , a n d l o g c : 1 1 . l o g V ( « - 6 3 c ) 6 . A n s . 6 1 o g a + 9 1 o g 6 + 3 1 o g c . 1 2 . l o g - \ / a ? b 5 ( H . f l o g a + f l o g 6 + | d o g e . 1 3 . l o g - . l o g « . ( a - 1 b - 2 c - A y S Y S T E M S O F L O G A I U T U M S . 9 1 6 4 . C o m m o n S y s t e m o f L o g a r i t h m s . — T h e r e a r e t w o s y s t e m s o f l o g a r i t h m s i n u s e , v i z . , t h e N a p e r i a n s y s t e m a n d t h e c o m m o n s y s t e m . T h e N a p e r i a n s y s t e m i s u s e d f o r p u r e l y t h e o r e t i c i n v e s t i g a t i o n s ; i t s b a s e i s e = 2 . 7 1 8 2 8 1 8 . T h e c o m m o n s y s t e m f o f l o g a r i t h m s i s t h e s y s t e m t h a t i s u s e d i n a l l p r a c t i c a l c a l c u l a t i o n s ; i t s b a s e i s 1 0 . B y a s y s t e m o f l o g a r i t h m s t o t h e b a s e 1 0 , i s m e a n t a s u c c e s s i o n o f v a l u e s o f x w h i c h s a t i s f y t h e e q u a t i o n m = 1 0 , f o r a l l p o s i t i v e v a l u e s o f m , i n t e g r a l o r f r a c t i o n a l . T h u s , i f w e s u p p o s e m t o a s s u m e i n s u c c e s s i o n e v e r y v a l u e f r o m 0 t o c o , t h e c o r r e s p o n d i n g v a l u e s o f x w i l l f o r m a s y s t e m o f l o g a r i t h m s , t o t h e b a s e 1 0 . S u c h a s y s t e m i s f o r m e d b y m e a n s o f t h e s e r i e s o f l o g a r i t h m s o f t h e n a t u r a l n u m b e r s f r o m 1 t o 1 0 0 0 0 0 , w h i c h c o n s t i t u t e t h e l o g a r i t h m s r e g i s t e r e d i n o u r o r d i n a r y t a b l e s . N o w 1 0 ° = 1 , . - -l o g l = 0 ; 1 0 1 = 1 0 , . - . l o g 1 0 = 1 ; 1 0 2 = 1 0 0 , . - . l o g 1 0 0 _ 2 -a n d s o o n . 1 0 3 = 1 0 0 0 , . - . l o g 1 0 0 0 = 3 . A l s o , 1 0 - = t V = - ! , . - . l o g . l = - i ; i o - 2 = t U = - ° i . . - . l o g . 0 1 = - 2 ; i o - 3 = t 7 V i 7 = - o o i , . - . l o g . 0 0 1 = - 3 . a n d s o o n . H e n c e , i n t h e c o m m o n s y s t e m , t h e l o g a r i t h m o f a n y n u m b e r b e t w e e n 1 a n d 1 0 i s s o m e n u m b e r b e t w e e n 0 a n d 1 ; i . e . , 0 + a d e c i m a l ; S o c a l l e d f r o m I t s i n v e n t o r , B a r o n X a p i e r , a H c o t c h m a t h e m a t i c i a n , f F i r s t i n t r o d u c e d i n 1 6 1 5 b y B r i f f f f S , a c o n t e m p o r a r y o f N a p i e r . 9 2 P L A N E T R I G O N O M E T R Y . 1 0 a n d 1 0 0 i s s o m e n u m b e r b e t w e e n 1 a n d 2 ; i . e . , 1 + a d e c i m a l ; 1 0 0 a n d 1 0 0 0 i s s o m e n u m b e r b e t w e e n 2 a n d 3 ; i . e . , 2 + a d e c i m a l ; 1 a n d . 1 i s s o m e n u m b e r b e t w e e n 0 a n d — 1 ; i . e . , — 1 + a d e c i m a l ; . 1 a n d . 0 1 i s s o m e n u m b e r b e t w e e n — 1 a n d — 2 ; i . e . , — 2 + a d e c i m a l ; . 0 1 a n d . 0 0 1 i s s o m e n u m b e r b e t w e e n — 2 a n d — 3 ; i . e . , — 3 + a d e c i m a l ; a n d s o o n . I t t h u s a p p e a r s t h a t ( 1 ) T h e ( c o m m o n ) l o g a r i t h m o f a n y n u m b e r g r e a t e r t h a n 1 i s p o s i t i v e . ( 2 ) T h e l o g a r i t h m o f a n y p o s i t i v e n u m b e r l e s s t h a n 1 i s n e g a t i v e . ( 3 ) I n g e n e r a l , t h e c o m m o n l o g a r i t h m o f a n u m b e r c o n s i s t s o f t w o p a r t s , a n i n t e g r a l p a r t a n d a d e c i m a l p a r t . T h e i n t e g r a l p a r t o f a l o g a r i t h m i s c a l l e d t h e c h a r a c t e r i s t i c o f t h e l o g a r i t h m , a n d m a y b e e i t h e r p o s i t i v e o r n e g a t i v e . T h e d e c i m a l p a r t o f a l o g a r i t h m i s c a l l e d t h e m a n t i s s a o f t h e l o g a r i t h m , a n d i s a l w a y s k e p t p o s i t i v e . N o t e . — I t i s c o n v e n i e n t t o k e e p t h e d e c i m a l p a r t o f t h e l o g a r i t h m s a l w a y s p o s i t i v e , i n o r d e r t h a t n u m b e r s c o n s i s t i n g o f t h e s a m e d i g i t s i n t h e s a m e o r d e r m a y c o r r e s p o n d t o t h e s a m e m a n t i s s a . I t i s e v i d e n t f r o m t h e a b o v e e x a m p l e s t h a t t h e c h a r a c t e r i s t i c o f a l o g a r i t h m c a n a l w a y s b e o b t a i n e d b y t h e f o l l o w i n g r u l e : R u l e . — T h e c h a r a c t e r i s t i c o f t h e l o g a r i t h m o f a n u m b e r g r e a t e r t h a n u n i t y i s o n e l e s s t h a n t h e n u m b e r o f d i g i t s i n t h e w h o l e n u m b e r . T h e c h a r a c t e r i s t i c o f t h e l o g a r i t h m o f a n u m b e r l e s s t h a n u n i t y i s n e g a t i v e , a n d i s o n e m o r e t h a n t h e n u m b e r o f c i p h e r s i m m e d i a t e l y a f t e r t h e d e c i m a l p o i n t . R U L E S F O R T H E C H A R A C T E R I S T I C . 9 3 T h u s , t h e c h a r a c t e r i s t i c s o f t h e l o g a r i t h m s o f 1 2 3 4 , 1 2 3 . 4 , 1 . 2 3 4 , . 1 2 3 4 , . 0 0 0 0 1 2 3 4 , 1 2 3 4 0 , a r e r e s p e c t i v e l y , 3 , 2 , 0 , -1 , - 5 , 4 . N o t e . — W h e n t h e c h a r a c t e r i s t i c i s n e g a t i v e , t h e m i n u s s i g n i s w r i t t e n o v e r i t t o i n d i c a t e t h a t t h e c h a r a c t e r i s t i c a l o n e i s n e g a t i v e , t h e m a n t i s s a h e i n g a l w a y s p o s i t i v e . W r i t e d o w n t h e c h a r a c t e r i s t i c s o f t h e c o m m o n l o g a r i t h m s o f t h e f o l l o w i n g n u m b e r s : 1 . 1 7 6 0 1 , 3 6 1 . 1 , 4 . 0 1 , 7 2 3 0 0 0 , 2 9 . A n s . 4 , 2 , 0 , 5 , 1 . 2 . . 0 4 , . 0 0 0 0 6 1 2 , . 7 9 6 3 , . 0 0 1 2 0 1 , . 1 . A n s . - 2 , - 5 , - 1 , - 3 , - 1 . 3 . H o w m a n y d i g i t s a r e t h e r e i n t h e i n t e g r a l p a r t o f t h e n u m b e r s w h o s e c o m m o n l o g a r i t h m s a r e r e s p e c t i v e l y 3 . 4 6 1 , 0 . 3 0 2 0 3 , 5 . 4 7 1 2 3 , 2 . 6 7 1 0 1 ? 4 . G i v e n l o g 2 = 0 . 3 0 1 0 3 ; f i n d t h e n u m b e r o f d i g i t s i n t h e i n t e g r a l p a r t o f 8 1 0 , 2 n , 1 6 2 0 , 2 1 0 0 . A n s . 1 0 , 4 , 2 5 , 3 1 . 6 5 . C o m p a r i s o n o f T w o S y s t e m s o f L o g a r i t h m s . — G i v e n t h e l o g a r i t h m o f a n u m b e r t o b a s e a ; t o f i n d t h e l o g a r i t h m o f t h e s a m e n u m b e r t o b a s e b . L e t m b e a n y n u m b e r w h o s e l o g a r i t h m t o b a s e b i s r e q u i r e d . L e t a ; = l o g j m ; t h e n b ' = m . . : l o g a ( 6 " ) = l o g a m ; o r x l o g a b = l o g . m . . - - x = r ^ — . X l o g a m , l o g a o o r l o g s m = - & 2 — ( 1 ) l o g a 6 H e n c e , t o t r a n s f o r m t h e l o g a r i t h m o f a n u m b e r f r o m b a s e a t o b a s e b , w e m u l t i p l y i t b y — — — l o g . b 9 4 P L A N E T R I G O N O M E T R Y . T h i s c o n s t a n t m u l t i p l i e r i s c a l l e d t h e m o d u l u s o f l o g . b t h e s y s t e m o f w h i c h t h e b a s e i s b w i t h r e f e r e n c e t o t h e s y s t e m o f w h i c h t h e b a s e i s a . I f , t h e n , a l i s t o f l o g a r i t h m s t o s o m e b a s e e c a n b e m a d e , w e c a n d e d u c e f r o m i t a l i s t o f c o m m o n l o g a r i t h m s b y m u l t i p l y i n g e a c h l o g a r i t h m i n t h e g i v e n l i s t b y t h e m o d u l u s o f t h e c o m m o n s y s t e m — J 3 l o g e 1 0 P u t t i n g a f o r m i n ( 1 ) , w e h a v e l o g , a = = b y ( 2 ) o f A r t . 6 3 . l o g a 6 l o g a 6 . - . l o g , a x l o g a 6 = 1 . E X A M P L E S . 1 . S h o w h o w t o t r a n s f o r m l o g a r i t h m s w i t h b a s e 5 t o l o g a r i t h m s w i t h b a s e 1 2 5 . L e t m b e a n y n u m b e r , a n d l e t x b e i t s l o g a r i t h m t o b a s e 1 2 5 . T h e n m = 1 2 5 = ( 5 3 ) = 5 3 . . - . 3 x = \ o g 5 m . . : = l o g 1 2 5 m = £ l o g 5 m . T h u s , t h e l o g a r i t h m o f a n y n u m b e r t o b a s e 5 , d i v i d e d b y 3 ( i . e . , b y l o g s 1 2 5 ) , i s t h e l o g a r i t h m o f t h e s a m e n u m b e r t o t h e b a s e 1 2 5 . O t h e r w i s e b y t h e r u l e g i v e n i n ( 1 ) . T h u s , , l o g . m l o g . m 1 0 g - m = I o i i 2 5 = 3 . S h o w h o w t o t r a n s f o r m 2 . L o g a r i t h m s w i t h b a s e 2 t o l o g a r i t h m s w i t h b a s e 8 . A n s . D i v i d e e a c h l o g a r i t h m b y 3 . T A B L E S O F L O G A R I T H M S . 9 5 3 . L o g a r i t h m s w i t h b a s e 9 t o l o g a r i t h m s w i t h b a s e 3 . A n s . M u l t i p l y e a c h l o g a r i t h m b y 2 . 4 . F i n d l o g 2 8 , l o g 5 l , l o g 8 2 , l o g 7 l , l o g s 2 1 2 8 . A n a . 3 , 0 , i , 0 , f 6 6 . T a b l e s o f L o g a r i t h m s . — T h e c o m m o n l o g a r i t h m s o f a l l i n t e g e r s f r o m 1 t o 1 0 0 0 0 0 h a v e b e e n f o u n d a n d r e g i s t e r e d i n t a b l e s , w h i c h a r e t h e r e f o r e c a l l e d t a b u l a r l o g a r i t h m s . I n m o s t t a b l e s t h e y a r e g i v e n t o s i x p l a c e s o f d e c i m a l s , t h o u g h t h e y m a y b e c a l c u l a t e d t o v a r i o u s d e g r e e s o f a p p r o x i m a t i o n , s u c h a s f i v e , s i x , s e v e n , o r a h i g h e r n u m b e r o f d e c i m a l p l a c e s . T a b l e s o f l o g a r i t h m s t o s e v e n p l a c e s o f d e c i m a l s a r e i n c o m m o n u s e f o r a s t r o n o m i c a l a n d m a t h e m a t i c a l c a l c u l a t i o n s . T h e c o m m o n s y s t e m t o b a s e 1 0 i s t h e o n e i n p r a c t i c a l u s e , a n d i t h a s t w o g r e a t a d v a n t a g e s : ( 1 ) F r o m t h e r u l e ( A r t . 6 4 ) t h e c h a r a c t e r i s t i c s c a n b e w r i t t e n d o w n a t o n c e , s o t h a t o n l y t h e m a n t i s s a e h a v e t o b e g i v e n i n t h e t a b l e s . ( 2 ) T h e m a n t i s s a e a r e t h e s a m e f o r t h e l o g a r i t h m s o f a l l n u m b e r s w h i c h h a v e t h e s a m e s i g n i f i c a n t d i g i t s , i n t h e s a m e o r d e r , s o t h a t i t i s s u f f i c i e n t t o t a b u l a t e t h e i n a n t i s s a e o f t h e l o g a r i t h m s o f i n t e g e r s . F o r , s i n c e a l t e r i n g t h e p o s i t i o n o f t h e d e c i m a l p o i n t w i t h o u t c h a n g i n g t h e s e q u e n c e o f f i g u r e s m e r e l y m u l t i p l i e s o r d i v i d e s t h e n u m b e r b y a n i n t e g r a l p o w e r o f 1 0 , i t f o l l o w s t h a t i t s l o g a r i t h m w i l l b e i n c r e a s e d o r d i m i n i s h e d b y a n i n t e g e r ; i . e . , t h a t t h e m a n t i s s a o f t h e l o g a r i t h m r e m a i n s u n a l t e r e d . I n G e n e r a l . — I f N b e a n y n u m b e r , a n d p a n d q a n y i n t e g e r s , i t f o l l o w s t h a t N x 1 0 p a n d N 1 0 « a r e n u m b e r s w h o s e s i g n i f i c a n t d i g i t s a r e t h e s a m e a s t h o s e o f N . T h e n l o g ( N x 1 0 " ) = l o g N + p l o g l 0 = l o g N + p . ( 1 ) A l s o , l o g ( N - s - 1 0 « ) = l o g N - g > g l 0 = l o g N - g . ( 2 ) 9 6 P L A N E T R I G O N O M E T R Y . I n ( 1 ) t h e l o g a r i t h m o f N i s i n c r e a s e d b y a n i n t e g e r , a n d i n ( 2 ) i t i s d i m i n i s h e d b y a n i n t e g e r . T h a t i s , t h e s a m e m a n t i s s a s e r v e s f o r t h e l o g a r i t h m s o f a l l n u m b e r s , w h e t h e r g r e a t e r o r l e s s t h a n u n i t y , w h i c h h a v e t h e s a m e s i g n i f i c a n t d i g i t s , a n d d i f f e r o n l y i n t h e p o s i t i o n o f t h e d e c i m a l p o i n t . T h i s w i l l p e r h a p s b e b e t t e r u n d e r s t o o d i f w e t a k e a p a r t i c u l a r c a s e . F r o m a t a b l e o f l o g a r i t h m s w e f i n d t h e m a n t i s s a o f t h e l o g a r i t h m o f 7 8 7 t o b e 8 9 5 9 7 5 ; t h e r e f o r e , p r e f i x i n g t h e c h a r a c t e r i s t i c w i t h i t s a p p r o p r i a t e s i g n a c c o r d i n g t o t h e r u l e , w e h a v e l o g 7 8 7 N o w l o g 7 . 8 7 A l s o , l o g . 0 7 8 7 A l s o , l o g 7 8 7 0 0 N o t e 1 . — W e d o n o t w r i t e l o g 1 0 7 8 7 ; f o r b o l o n g a s w e a r e t r e a t i n g o f l o g a r i t h m s t o t h e p a r t i c u l a r b a s e 1 0 , w e m a y o m i t t h e s u f f i x . N o t e 2 . — S o m e t i m e s i n w o r k i n g w i t h n e g a t i v e l o g a r i t h m s , a n a r i t h m e t i c a r t i f i c e w i l l b e n e c e s s a r y t o m a k e t h e m a n t i s s a p o s i t i v e . F o r e x a m p l e , a r e s u l t s u c h a s — 2 . 6 9 8 9 7 , i n w h i c h t h e w h o l e e x p r e s s i o n i s n e g a t i v e , m a y b e t r a n s f o r m e d b y s u b t r a c t i n g 1 f r o m t h e c h a r a c t e r i s t i c , a n d a d d i n g 1 t o t h e m a n t i s s a . T h u s , -2 . 6 9 8 9 7 = -8 + ( 1 - . 6 9 8 9 7 ) = 3 . 3 0 1 0 3 . N o t e 3 . — W h e n t h e c h a r a c t e r i s t i c o f a l o g a r i t h m i s n e g a t i v e , i t i s o f t e n , e s p e c i a l l y i n A s t r o n o m y a n d G e o d e s y , f o r c o n v e n i e n c e , m a d e p o s i t i v e b y t h e a d d i t i o n o f 1 0 , w h i c h c a n l e a d t o n o e r r o r , i f w e a r e c a r e f u l t o s u b t r a c t 1 0 . T h u s , i n s t e a d o f t h e l o g a r i t h m 3 . 6 0 8 5 8 2 , w e m a y w r i t e 7 . 6 0 3 5 8 2 -1 0 . I n c a l c u l a t i o n s w i t h n e g a t i v e c h a r a c t e r i s t i c s w e f o l l o w t h e r u l e s o f A l g e b r a . = 2 . 8 9 5 9 7 5 . = l o g ~ 5 = l o g 7 8 7 - 2 = 0 . 8 9 5 9 7 5 . = l o s ( = l o g 7 8 7 -4 V i o o o o y 5 = 2 . 8 9 5 9 7 5 . = l o g ( 7 8 7 x 1 0 0 ) = l o g 7 8 7 + 2 = 4 . 8 9 5 9 7 5 . E X A M P L E S . 9 7 E X A M P L E S . 1 . A d d t o g e t h e r 5 . 2 1 4 3 1 . 3 1 4 2 5 . 9 0 6 8 7 . 4 3 5 3 A n s . 2 . F r o m 3 . 2 4 5 6 9 t a k e 5 . 6 2 4 9 3 1 . 6 2 0 7 6 t h e 1 c a r r i e d f r o m t h e l a s t s u b t r a c t i o n i n d e c i m a l p l a c e s c h a n g e s — 5 i n t o — 4 , a n d t h e n — 4 s u b t r a c t e d f r o m — 3 g i v e s 1 a s a r e s u l t . 3 . M u l t i p l y 2 . 1 5 2 8 b y 7 . 2 . 1 5 2 8 7 1 3 . 0 6 9 6 t h e 1 c a r r i e d f r o m t h e l a s t m u l t i p l i c a t i o n o f t h e d e c i m a l p l a c e s b e i n g a d d e d t o — 1 4 , a n d t h u s - g i v i n g — 1 3 a s a r e s u l t . N o t e 4 . — W h e n a l o g a r i t h m w i t h n e g a t i v e c h a r a c t e r i s t i c h a s t o b e d i v i d e d b y a n u m b e r w h i c h i s n o t a n e x a c t d i v i s o r o f t h e c h a r a c t e r i s t i c , w e p r o c e e d a s f o l l o w s i n o r d e r t o k e e p t h e c h a r a c t e r i s t i c i n t e g r a l . I n c r e a s e t h e c h a r a c t e r i s t i c n u m e r i c a l l y b y a n u m b e r w h i c h w i l l m a k e i t e x a c t l y d i v i s i b l e , a n d p r e f i x a n e q u a l p o s i t i v e n u m b e r t o t h e m a n t i s s a . 4 . D i v i d e 3 . 7 2 6 8 b y 5 . I n c r e a s e t h e n e g a t i v e c h a r a c t e r i s t i c s o t h a t i t m a y b e e x a c t l y d i v i s i b l e b y 5 ; t h u s 3 . 7 2 6 8 = 5 + 2 - 7 2 6 8 = f 5 l 5 3 5 5 G i v e n t h a t l o g 2 = . 3 0 1 0 3 , l o g 3 = . 4 7 7 1 2 , a n d l o g 7 = . 8 4 5 1 0 ; f i n d t h e v a l u e s o f 5 . l o g 6 , l o g 4 2 , l o g 1 6 . A n s . . 7 7 8 1 5 , 1 . 6 2 3 2 5 , 1 . 2 0 4 1 2 . 9 8 P L A N E T R I G O N O M E T R Y . 6 . l o g 4 9 , l o g 3 6 , l o g 6 3 . A n s . 1 . 6 9 0 2 0 , 1 . 5 5 6 3 0 , 1 . 7 9 9 3 4 . 7 . I o g 2 0 0 , l o g 6 0 0 , l o g 7 0 . 2 . 3 0 1 0 3 , 2 . 7 7 8 1 5 , 1 . 8 4 5 1 0 . 8 . I o g 6 0 , l o g . 0 3 , l o g 1 . 0 5 , l o g . 0 0 0 0 4 3 2 . N o t e . — T h e l o g a r i t h m o f 5 a n d i t s p o w e r s c a n a l w a y s b e o b t a i n e d f r o m l o g 2 . A n s . 1 . 7 7 8 1 5 , 2 . 4 7 7 1 2 , . 0 2 1 1 9 , 5 . 6 3 5 4 8 . 9 . G i v e n l o g 2 = . 3 0 1 0 3 ; f i n d l o g 1 2 8 , l o g 1 2 5 , a n d l o g 2 5 0 0 . A n s . 2 . 1 0 7 2 1 , 2 . 0 9 6 9 1 , 3 . 3 9 7 9 4 . G i v e n t h e l o g a r i t h m s o f 2 , 3 , a n d 7 , a s a b o v e ; f i n d t h e l o g a r i t h m s o f t h e f o l l o w i n g : 1 0 . 2 0 7 3 6 , 4 3 2 , 9 8 , 6 8 6 , 1 . 7 2 8 , . 3 3 6 . A n s . 4 . 3 1 6 7 2 , 2 . 6 3 5 4 8 , 1 . 9 9 1 2 2 , 2 . 8 3 6 3 2 , . 2 3 7 5 4 , 1 . 5 2 6 3 4 . 1 1 . V . 2 , ( . 0 3 ) , ( . 0 0 2 1 ) ( . 0 9 8 ) 3 , ( . 0 0 0 4 2 ) 5 , ( . 0 3 3 6 ) . A n s . 1 . 6 5 0 5 2 , 1 . 6 1 9 2 8 , 1 . 4 6 4 4 4 , 4 . 9 7 3 6 8 , 1 7 . 1 1 6 2 5 , 1 . 2 6 3 1 7 . 6 7 . U s e o f T a b l e s o f L o g a r i t h m s o f N u m b e r s . — I n o u r e x p l a n a t i o n s o f t h e u s e o f t a b l e s o f c o m m o n l o g a r i t h m s w e s h a l l u s e t a b l e s o f s e v e n , p l a c e s o f d e c i m a l s . f T h e s e t a b l e s a r e a r r a n g e d s o a s t o g i v e t h e m a n t i s s a e o f t h e l o g a r i t h m s o f t h e n a t u r a l m e m b e r s f r o m 1 t o 1 0 0 0 0 0 ; i . e . , o f n u m b e r s c o n t a i n i n g f r o m o n e t o f i v e d i g i t s . A t a b l e o f l o g a r i t h m s o f n u m b e r s c o r r e c t t o s e v e n d e c i m a l p l a c e s i s e x a c t f o r a l l t h e p r a c t i c a l p u r p o s e s o f A s t r o n o m y a n d G e o d e s y . F o r a n a c t u a l m e a s u r e m e n t o f a n y k i n d m u s t b e m a d e w i t h t h e g r e a t e s t c a r e , w i t h t h e m o s t a c c u r a t e i n s t r u m e n t s , b y t h e m o s t s k i l f u l o b s e r v e r s , i f i t i s t o a t t a i n t o a n y t h i n g l i k e t h e a c c u r a c y r e p r e s e n t e d b y s e v e n s i g n i f i c a n t f i g u r e s . T h e m e t h o d s b y w h i c h t h e s e t a b l e s a r e f o r m e d w i l l b e g i v e n i n C h a p . V I I I . t T h e s t u d e n t s h o u l d h e r e p r o v i d e h i m s e l f w i t h l o g a r i t h m i c a n d t r i g o n o m e t r i c t a b l e s o f s e v e n d e c i m a l p l a c e s . T h e m o s t c o n v e n i e n t s e v e n - f i g u r e t a b l e s u s e d i n t h i s c o u n t r y a r e S t a n l e y ' s , V e g a ' s , B r u h n s ' , e t c . I n t h e a p p e n d i x t o t h e E l e m e n t a r y T r i g o n o m e t r y a r e g i v e n f i v e - f i g u r e d t a b l e s , w h i c h a r e s u f f i c i e n t l y n e a r f o r m o s t p r a c t i c a l a p p l i c a t i o n s . U S E O F L O G A R I T H M I C T A B L E S . 9 9 I f t h e m e a s u r e o f a n y l e n g t h i s k n o w n a c c u r a t e l y t o s e v e n f i g u r e s , i t i s p r a c t i c a l l y e x a c t ; i . e . , i t i s k n o w n t o w i t h i n t h e l i m i t s o f o b s e r v a t i o n . I f t h e m e a s u r e o f a n y a n g l e i s k n o w n t o w i t h i n t h e t e n t h p a r t o f a s e c o n d , t h e g r e a t e s t a c c u r a c y p o s s i b l e , a t p r e s e n t , i n t h e m e a s u r e m e n t o f a n g l e s i s r e a c h e d . T h e t e n t h p a r t o f a s e c o n d i s a b o u t t h e t w o -m i l l i o n t h p a r t o f a r a d i a n . T h i s d e g r e e o f a c c u r a c y i s a t t a i n a b l e o n l y w i t h t h e l a r g e s t a n d b e s t i n s t r u m e n t s , a n d u n d e r t h e m o s t f a v o r a b l e c o n d i t i o n s . O n p a g e 1 0 1 i s a s p e c i m e n p a g e o f L o g a r i t h m i c T a b l e s . I t c o n s i s t s o f t h e m a n t i s s a e o f t h e l o g a r i t h m s , c o r r e c t t o s e v e n p l a c e s o f d e c i m a l s , o f a l l n u m b e r s b e t w e e n 6 2 5 0 0 a n d 6 3 0 0 9 . T h e f i g u r e s o f t h e n u m b e r a r e t h o s e i n t h e l e f t c o l u m n h e a d e d N , f o l l o w e d b y o n e i n l a r g e r t y p e a t t h e t o p o f t h e p a g e . T h e f i r s t t h r e e f i g u r e s o f t h e m a n t i s s a e 7 9 5 , 7 9 6 , 7 9 7 a n d t h e r e m a i n i n g f o u r a r e i n t h e s a m e h o r i z o n t a l l i n e w i t h t h e f i r s t f o u r f i g u r e s o f t h e n u m b e r , a n d i n t h e v e r t i c a l c o l u m n u n d e r t h e l a s t . L o g a r i t h m s a r e i n g e n e r a l i n c o m m e n s u r a b l e n u m b e r s . T h e i r v a l u e s c a n t h e r e f o r e o n l y b e g i v e n a p p r o x i m a t e l y . T h r o u g h o u t a l l a p p r o x i m a t e c a l c u l a t i o n s i t i s u s u a l t o t a k e f o r t h e l a s t f i g u r e w h i c h w e r e t a i n , t h a t f i g u r e w h i c h g i v e s t h e n e a r e s t a p p r o a c h t o t h e t r u e v a l u e . W h e n o n l y a c e r t a i n n u m b e r o f d e c i m a l p l a c e s i s r e q u i r e d , t h e g e n e r a l r u l e i s t h i s : S t r i k e o u t t h e r e s t o f t h e f i g u r e s , a n d i n c r e a s e t h e l a s t f i g u r e r e t a i n e d b y 1 i f t h e f i r s t f i g u r e s t r u c k o f f i s 5 o r g r e a t e r t h a n 5 . 6 8 . T o f i n d t h e L o g a r i t h m o f a G i v e n N u m b e r . — W h e n t h e g i v e n n u m b e r h a s n o t m o r e t h a n f i v e d i g i t s , w e h a v e m e r e l y t o t a k e t h e m a n t i s s a i m m e d i a t e l y f r o m t h e t a b l e , a n d p r e f i x t h e c h a r a c t e r i s t i c b y t h e r u l e ( A r t . 6 4 ) . T h u s , s u p p o s e w e r e q u i r e t h e l o g a r i t h m o f 6 2 5 4 1 . T h e t a b l e g i v e s . 7 9 6 1 6 4 8 a s t h e m a n t i s s a , a n d t h e c h a r a c t e r i s t i c i s 4 , b y t h e r u l e ; t h e r e f o r e l o g 6 2 5 4 J . = 4 . 7 9 6 1 6 4 8 . S i m i l a r l y , l o g . 0 0 6 2 8 1 = 3 . 7 9 8 0 2 8 8 . . ( A r t . 6 4 ) 1 0 0 P L A N E T R I G O N O M E T R Y . S u p p o s e , h o w e v e r , t h a t t h e g i v e n n u m b e r h a s m o r e t h a n f i v e d i g i t s . F o r e x a m p l e : S u p p o s e w e r e q u i r e t o f i n d l o g 6 2 7 6 1 . 6 . W e f i n d f r o m t h e t a b l e l o g 6 2 7 6 1 = 4 . 7 9 7 6 8 9 9 l o g 6 2 7 6 2 = 4 . 7 9 7 6 9 6 8 a n d d i f f . f o r 1 = 0 . 0 0 0 0 0 6 9 T h u s f o r a n i n c r e a s e o f 1 i n t h e n u m b e r t h e r e i s a n i n c r e a s e o f . 0 0 0 0 0 6 9 i n t h e l o g a r i t h m . H e n c e , a s s u m i n g t h a t t h e i n c r e a s e o f t h e l o g a r i t h m i s p r o p o r t i o n a l t o t h e i n c r e a s e o f t h e n u m b e r , t h e n a n i n c r e a s e i n t h e n u m b e r o f . 6 w i l l c o r r e s p o n d t o a n i n c r e a s e i n t h e l o g a r i t h m o f . 6 x . 0 0 0 0 0 6 9 = . 0 0 0 0 0 4 1 , t o t h e n e a r e s t s e v e n t h d e c i m a l p l a c e . H e n c e , l o g 6 2 7 6 1 = 4 . 7 9 7 6 8 9 9 d i f f . f o r . 6 = 4 1 . - . l o g 6 2 7 6 1 . 6 = 4 . 7 9 7 6 9 4 0 T h i s e x p l a i n s t h e u s e o f t h e c o l u m n o f p r o p o r t i o n a l p a r t s o n t h e e x t r e m e r i g h t o f t h e p a g e . I t w i l l b e s e e n t h a t t h e d i f f e r e n c e b e t w e e n t h e l o g a r i t h m s o f t w o c o n s e c u t i v e n u m b e r s i s n o t a l w a y s t h e s a m e ; f o r i n s t a n c e , t h o s e i n t h e u p p e r p a r t o f t h e p a g e b e f o r e u s d i f f e r b y . 0 0 0 0 0 7 0 , w h i l e t h o s e i n t h e m i d d l e a n d t h e l o w e r p a r t s d i f f e r b y . 0 0 0 0 0 6 9 a n d . 0 0 0 0 0 6 8 . U n d e r t h e c o l u m n w i t h t h e h e a d i n g 6 9 w e s e e t h e d i f f e r e n c e 4 1 c o r r e s p o n d i n g t o t h e f i g u r e 6 , w h i c h i m p l i e s t h a t w h e n t h e d i f f e r e n c e b e t w e e n t h e l o g a r i t h m s o f t w o c o n s e c u t i v e m e m b e r s i s . 0 0 0 0 0 6 9 , t h e i n c r e a s e i n t h e l o g a r i t h m c o r r e s p o n d i n g t o a n i n c r e a s e o f . 6 i n t h e n u m b e r i s . 0 0 0 0 0 4 1 ; f o r . 0 6 i t i s e v i d e n t l y . 0 0 0 0 0 0 4 , a n d s o o n . N o t e . — W e a s s u m e i n t i n s m e t h o d t h a t t h e I n c r e a s e i n a l o g a r i t h m i s p r o p o r t i o n a l t o t h e i n c r e a s e i n t h e n u m b e r . A l t h o u g h t h i s i s n o t s t r i c t l y t r u e , y e t i t i s i n m o s t c a s e s s u f f i c i e n t l y e x a c t f o r p r a c t i c a l p u r p o s e s . H a d w e t a k e n n w h o l e n u m b e r o r a d e c i m a l , t h e p r o c e s s w o u l d h a v e b e e n t h e s a m e . T A B L E O F L O G A R I T H M S ! ] \ ; l j t t N . O 1 1 2 3 I 5 - 7 . . 8 1 . 9 P . P . 6 2 5 0 7 9 5 8 8 0 0 8 8 7 0 , 8 9 3 9 9 0 0 9 9 0 7 8 9 1 4 8 9 2 1 7 9 2 S ? | 9 A 5 . 6 j ? t e < 3 --- 5 1 9 4 9 5 9 5 6 4 1 9 6 3 4 9 7 0 3 9 7 7 3 9 8 4 2 9 9 1 2 9 9 8 l ' o 0 5 1 0 1 2 0 5 2 7 9 6 0 1 9 0 0 2 5 9 0 3 2 9 0 3 9 8 0 4 6 8 0 5 3 7 , 0 6 0 6 0 6 7 6 0 7 4 5 0 8 1 5 5 3 0 8 8 4 0 9 5 4 1 0 2 3 1 0 9 3 1 1 6 2 1 2 3 2 1 3 0 1 1 3 7 0 1 4 4 0 1 5 0 9 5 4 1 5 7 9 1 6 4 8 1 7 1 8 1 7 8 7 1 8 5 7 1 9 2 6 1 9 9 5 2 0 6 5 2 1 3 4 2 2 0 4 5 5 2 2 7 3 2 3 4 3 2 4 1 2 2 4 8 1 2 5 5 1 2 6 2 0 2 6 9 0 2 7 5 9 2 8 2 9 2 8 9 8 5 6 2 9 6 7 3 0 3 7 3 1 0 6 3 1 7 6 3 2 4 5 3 3 1 4 3 3 8 4 3 4 5 3 3 5 2 3 1 3 5 9 2 7 0 5 7 3 6 6 2 3 7 3 1 3 8 0 0 3 8 7 0 3 9 3 9 4 0 0 9 4 0 7 8 4 1 4 7 4 2 1 7 , 4 2 8 6 1 7 . 0 5 8 4 3 5 6 4 4 2 5 4 4 9 4 4 5 6 4 4 6 3 3 4 7 0 3 4 7 7 2 4 8 4 1 4 9 1 1 4 9 8 0 2 1 4 . 0 5 9 5 0 5 0 5 1 1 9 5 1 8 8 5 2 5 S 5 3 2 7 5 3 9 6 5 4 6 6 5 5 3 5 5 6 0 5 5 6 7 4 3 2 1 . 0 6 2 6 0 7 % 5 7 4 3 5 8 1 3 5 8 8 2 5 9 5 1 6 0 2 1 6 0 9 0 6 1 6 0 6 2 2 9 6 2 9 8 6 3 6 8 4 2 8 . 0 6 1 6 4 3 7 6 5 0 6 ; 6 5 7 6 6 6 4 5 6 7 1 4 6 7 8 4 6 8 5 3 6 9 2 3 6 9 9 2 1 7 0 6 1 5 3 5 . 0 6 2 7 1 3 1 7 2 0 0 7 2 6 9 7 3 3 9 7 4 0 8 7 4 7 7 7 5 4 7 7 6 1 6 7 6 8 5 7 7 5 5 6 4 2 . 0 6 3 7 8 2 4 7 8 9 3 7 9 6 3 8 0 3 2 8 1 0 1 8 1 7 1 8 2 4 0 8 3 0 9 8 3 7 9 8 4 4 8 7 4 9 . 0 6 4 8 5 1 7 8 5 8 7 8 6 5 6 8 7 2 5 8 7 9 5 8 8 6 4 8 9 3 3 9 0 0 3 9 0 7 2 9 1 4 1 8 5 6 . 0 6 5 9 2 1 1 9 2 8 0 9 3 4 9 9 4 1 9 9 4 8 8 9 5 5 7 9 6 2 7 9 6 9 6 9 7 6 5 9 8 3 5 9 6 3 . 0 6 6 9 9 0 4 9 9 7 3 0 0 4 3 0 1 1 2 0 1 8 1 0 2 5 0 0 3 2 0 0 3 8 9 0 4 5 8 0 5 2 8 6 7 7 9 7 0 5 9 7 0 6 6 6 , 0 7 3 6 0 8 0 5 0 8 7 4 0 9 4 3 1 0 1 3 1 0 8 2 1 1 5 1 1 2 2 1 6 8 1 2 9 0 1 3 5 9 , 1 4 2 8 1 4 9 8 1 5 6 7 1 6 3 6 , 1 7 0 6 1 7 7 5 1 8 4 4 1 9 1 3 6 9 1 9 8 3 2 0 5 2 2 1 2 1 2 1 9 1 1 2 2 6 0 2 3 2 9 2 3 9 8 2 4 6 8 2 5 3 7 2 6 0 6 6 2 7 0 7 9 7 2 6 7 5 2 7 4 5 2 8 1 4 2 8 8 3 2 9 5 2 3 0 2 2 3 0 9 1 3 1 6 0 3 2 2 9 3 2 9 9 7 1 3 3 6 8 3 4 3 7 , 3 5 0 7 3 5 7 6 3 6 4 5 3 7 1 4 3 7 8 4 3 S 5 3 3 9 2 2 3 9 9 1 6 9 7 2 4 0 6 0 4 1 3 0 4 1 9 9 4 2 6 8 4 3 3 7 4 4 0 7 | 4 4 7 6 4 5 4 5 4 6 1 4 4 6 8 4 1 6 . 9 7 3 4 7 5 3 4 8 2 2 4 8 9 1 4 9 6 1 5 0 3 0 5 0 9 9 5 1 6 8 5 2 3 7 5 3 0 7 5 3 7 6 2 1 3 . 8 7 4 5 4 4 5 5 5 1 4 5 5 8 4 5 6 5 3 5 7 2 2 5 7 9 1 5 8 6 0 5 9 3 0 5 9 9 9 6 0 6 8 3 2 0 . 7 7 5 6 1 3 7 6 2 0 7 6 2 7 6 6 3 4 5 6 4 1 4 6 4 8 3 6 5 5 3 6 6 2 2 6 6 9 1 6 7 6 0 4 2 7 . 6 7 6 6 8 2 9 6 8 9 9 6 9 6 8 7 0 3 7 7 1 0 6 7 1 7 5 7 2 4 5 7 3 1 4 7 3 8 3 7 4 5 2 5 3 4 . 5 7 7 7 5 2 1 7 5 9 0 7 6 6 0 7 7 2 9 7 7 9 8 7 8 6 7 7 9 3 6 8 0 0 6 8 0 7 5 8 1 4 4 6 4 1 . 4 7 8 8 2 1 3 8 2 8 2 8 3 5 1 8 4 2 1 8 4 9 0 8 5 5 9 8 6 2 8 8 6 9 7 8 7 6 6 8 8 3 6 7 4 8 . 3 7 9 8 9 0 5 8 9 7 4 9 0 4 3 9 1 1 2 , 9 1 8 1 9 2 5 1 9 3 2 0 9 3 8 9 9 4 5 8 9 5 2 7 8 5 5 . 2 6 2 8 0 7 9 7 9 5 9 6 9 6 6 6 9 7 3 5 9 8 0 4 , 9 8 7 3 9 9 4 2 0 0 1 1 0 0 8 0 0 1 5 0 0 2 1 9 9 6 2 . 1 8 1 7 9 8 0 2 8 8 0 3 5 7 0 4 2 6 0 4 9 5 0 5 6 5 0 6 3 4 0 7 0 3 0 7 7 2 0 8 4 1 0 9 1 0 8 2 0 9 7 9 1 0 4 8 1 1 1 8 1 1 8 7 1 2 5 6 1 3 2 5 1 3 9 4 1 4 6 3 1 5 3 2 1 6 0 1 8 3 1 6 7 1 1 7 4 0 1 8 0 9 1 8 7 8 1 9 4 7 2 0 1 6 2 0 8 5 2 1 5 4 2 2 2 4 2 2 9 3 8 4 2 3 6 2 2 4 3 1 2 5 0 0 2 5 6 9 2 6 3 8 2 7 0 7 2 7 7 6 2 8 4 6 2 9 1 5 2 9 8 4 8 5 3 0 5 3 3 1 2 2 3 1 9 1 3 2 6 0 3 3 2 9 3 3 9 8 3 4 6 7 3 5 3 6 3 6 0 6 3 6 7 5 8 6 3 7 4 4 3 8 1 3 3 8 8 2 3 9 5 1 4 0 2 0 4 0 8 9 4 1 5 8 4 2 2 7 4 2 9 6 4 3 6 6 6 8 8 7 4 4 3 5 4 5 0 4 4 5 7 3 4 6 4 2 1 4 7 1 1 4 7 8 0 4 8 4 9 4 9 1 8 4 9 8 7 5 0 5 6 1 6 . 8 8 8 5 1 2 5 5 1 9 4 5 2 6 3 5 3 3 3 | 5 4 0 2 5 4 7 1 5 5 4 0 5 6 0 9 5 6 7 8 5 7 4 7 2 1 3 . 6 8 9 5 8 1 6 5 8 8 5 5 9 5 4 6 0 2 3 , 6 0 9 2 6 1 6 1 6 2 3 0 6 2 9 9 6 3 6 8 6 4 3 7 3 2 0 . 4 6 2 9 0 7 9 8 6 5 0 6 6 5 7 5 6 6 4 5 6 7 1 4 6 7 8 3 6 8 5 2 6 9 2 1 6 9 9 0 7 0 5 9 7 1 2 8 4 2 7 . 2 9 1 7 1 9 7 7 2 6 6 7 3 3 5 7 4 0 4 7 4 7 3 7 5 4 2 7 6 1 1 7 6 8 0 7 7 4 9 7 8 1 8 5 3 4 . 0 9 2 7 8 8 7 7 9 5 6 8 0 2 5 8 0 9 4 8 1 6 3 8 2 3 2 8 3 0 1 8 3 7 0 8 4 3 9 8 5 0 8 6 4 0 . 8 9 3 8 5 7 7 8 6 4 6 8 7 1 5 8 7 8 4 8 8 5 3 8 9 2 2 8 9 9 1 9 0 6 0 9 1 2 9 9 1 9 8 7 4 7 . 6 9 4 9 2 6 7 9 3 3 6 9 4 0 5 9 4 7 4 9 5 4 3 9 6 1 2 9 6 8 1 9 7 5 0 9 8 1 9 9 8 8 8 8 5 4 . 4 9 5 9 9 5 7 0 0 2 6 0 0 9 5 0 1 6 4 0 2 3 3 0 3 0 2 0 3 7 1 0 4 4 0 0 5 0 9 0 5 7 8 9 6 1 . 2 9 6 7 9 9 0 6 4 7 0 7 1 6 0 7 8 5 0 8 5 4 0 9 2 3 0 9 9 2 1 0 6 1 1 1 3 0 1 1 9 9 1 2 6 8 9 7 1 3 3 7 1 4 0 6 1 4 7 5 1 5 4 4 1 6 1 3 1 6 8 2 1 7 5 1 1 8 2 0 1 8 8 9 1 9 5 8 9 8 2 0 2 7 2 0 9 6 2 1 6 4 2 2 3 3 2 3 0 2 2 3 7 1 2 4 4 0 2 5 0 9 2 5 7 8 2 6 4 7 9 9 2 7 1 6 2 7 8 5 2 8 5 4 2 9 2 3 2 9 9 2 3 0 6 1 3 1 3 0 3 1 9 9 3 2 6 8 3 3 3 7 6 3 0 0 7 9 9 3 4 0 5 3 4 7 4 3 5 4 3 3 6 1 2 3 6 8 1 3 7 5 0 3 8 1 9 3 8 8 8 3 9 5 7 4 0 2 6 N . 0 1 2 3 4 5 6 7 8 9 P . P . i " Q 2 u " P L A N E T R I G O N O M E T R Y . « - - . ' t h u s , s u p p o s e w e r e q u i r e t o f i n d l o g 6 2 7 6 1 6 a u d l o g . 6 2 7 6 1 6 . T h e m a n t i s s a i s e x a c t l y t h e s a m e a s b e f o r e ( A r t . 6 6 ) , a n d t h e o n l y d i f f e r e n c e t o b e m a d e i n t h e f i n a l r e s u l t i s t o c h a n g e t h e c h a r a c t e r i s t i c a c c o r d i n g t o r u l e ( A r t . 6 4 ) . T h u s l o g 6 2 7 6 1 6 = 5 . 7 9 7 6 9 4 2 , a n d l o g . 6 2 7 6 1 6 = 1 . 7 9 7 6 9 4 2 . 6 9 . T o f i n d t h e N u m b e r c o r r e s p o n d i n g t o a G i v e n L o g a r i t h m . — I f t h e d e c i m a l p a r t o f t h e l o g a r i t h m i s f o u n d e x a c t l y i n t h e t a b l e , w e c a n t a k e o u t t h e c o r r e s p o n d i n g n u m b e r , a n d p u t t h e d e c i m a l p o i n t i n t h e n u m b e r , i n t h e p l a c e i n d i c a t e d b y t h e c h a r a c t e r i s t i c . T h u s i f w e h a v e t o f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 2 . 7 9 8 2 9 1 5 , w e l o o k i n t h e t a b l e f o r t h e m a n t i s s a . 7 9 8 2 9 1 5 , a n d w e f i n d i t s e t d o w n o p p o s i t e t h e n u m b e r 6 2 8 4 8 : a n d a s t h e c h a r a c t e r i s t i c i s 2 , t h e r e m u s t b e o n e c i p h e r b e f o r e t h e f i r s t s i g n i f i c a n t f i g u r e ( A r t . 6 4 ) . H e n c e 2 . 7 9 8 2 9 1 5 i s t h e l o g a r i t h m o f . 0 6 2 8 4 8 . N e x t , s u p p o s e t h a t t h e d e c i m a l p a r t o f t h e l o g a r i t h m ' i s n o t f o u n d e x a c t l y i n t h e t a b l e . F o r e x a m p l e , s u p p o s e w e h a v e t o f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 2 . 7 9 7 4 4 5 3 . W e f i n d f r o m t h e t a b l e l o g 6 2 7 2 6 = 4 . 7 9 7 4 4 7 6 l o g 0 2 7 2 5 = 4 . 7 9 7 4 4 0 7 d i f f . f o r l = . 0 0 0 0 0 0 9 T h u s f o r a d i f f e r e n c e o f 1 i n t h e n u m b e r s t h e r e i s a d i f f e r e n c e o f . 0 0 0 0 0 6 9 i n t h e l o g a r i t h m s . T h e e x c e s s o f t h e g i v e n m a n t i s s a a b o v e . 7 9 7 4 4 0 7 i s ( . 7 9 7 4 4 5 3 -. 7 9 7 4 4 0 7 ) o r . 0 0 0 0 0 4 6 . H e n c e , a s s u m i n g t h a t t h e i n c r e a s e o f t h e n u m b e r i s p r o p o r t i o n a l t o t h e i n c r e a s e o f t h e l o g a r i t h m , w e h a v e . 0 0 0 0 0 6 9 : . 0 0 0 0 0 4 6 : : 1 : n u m b e r t o b e a d d e d t o 6 2 7 . 2 5 . . - . n u m b e r t o b e a d d e d = ' f ) 0 0 0 0 4 r ' = 4 « = . 6 6 7 6 9 ) 4 6 . 0 ( . 6 6 6 . 0 0 0 0 0 0 9 6 9 4 1 4 . - . l o g 6 2 7 2 5 . 6 6 7 = 4 . 7 9 7 4 4 5 3 , ~ 4 6 0 a n d . - . l o g 6 2 7 . 2 5 6 6 7 = 2 . 7 9 7 4 4 5 3 ; 4 1 4 t h e r e f o r e n u m b e r r e q u i r e d i s 6 2 7 . 2 5 6 6 7 . 4 6 0 A R I T H M E T I C C O M P L E M E N T . 1 0 3 W e m i g h t h a v e s a v e d t h e l a b o r o f d i v i d i n g 4 6 b y 6 9 , b y u s i n g t h e t a b l e o f p r o p o r t i o n a l p a r t s a s f o l l o w s : g i v e n m a n t i s s a = . 7 9 7 4 4 5 3 m a n t i s s a o f 6 2 7 2 5 = . 7 9 7 4 4 0 7 d i f f . o f m a n t i s s a e = 4 6 p r o p o r t i o n a l p a r t f o r . 6 = 4 1 . 4 4 . 6 " " " . 0 6 = 4 . 1 4 . 4 6 " " " . 0 0 6 = . 4 1 4 a n d s o o n . . - . n u m b e r = 6 2 7 . 2 5 6 6 6 6 - . . . 6 9 a . A r i t h m e t i c C o m p l e m e n t . — B y t h e a r i t h m e t i c c o m p l e m e n t o f t h e l o g a r i t h m o f a n u m b e r , o r , b r i e f l y , t h e c o l o g a r i t h m o f t h e n u m b e r , i s m e a n t t h e r e m a i n d e r f o u n d b y s u b t r a c t i n g t h e l o g a r i t h m f r o m 1 0 . T o s u b t r a c t o n e l o g a r i t h m f r o m a n o t h e r i s t h e s a m e a s t o a d d t h e c o -l o g a r i t h m a n d t h e n s u b t r a c t 1 0 f r o m t h e r e s u l t . T h u s , a - b = a + ( 1 0 - 6 ) - 1 0 , w h e r e a a n d b a r e l o g a r i t h m s , a n d 1 0 — 6 i s t h e a r i t h m e t i c c o m p l e m e n t o f b . W h e n o n e l o g a r i t h m i s t o b e s u b t r a c t e d f r o m t h e s u m o f s e v e r a l o t h e r s , i t i s m o r e c o n v e n i e n t t o a d d i t s c o l o g a r i t h m t o t h e s u m , a n d r e j e c t 1 0 . T h e a d v a n t a g e o f u s i n g t h e c o l o g a r i t h m i s t h a t i t e n a b l e s u s t o e x h i b i t t h e w o r k i n a m o r e c o m p a c t f o r m . T h e c o l o g a r i t h m i s e a s i l y t a k e n f r o m t h e t a b l e m e n t a l l y b y s u b t r a c t i n g t h e l a s t s i g n i f i c a n t f i g u r e o n t h e r i g h t f r o m 1 0 , a n d a l l t h e o t h e r s f r o m 9 . 1 0 4 P L A N E T R I G O N O M E T R Y . E X A M P L E S . 1 . G i v e n f i n d 2 . G i v e n f i n d 3 . G i v e n f i n d 4 . G i v e n f i n d 5 . G i v e n l o g 5 2 5 0 2 = 4 . 7 2 0 1 7 5 8 , l o g 5 2 5 0 3 = 4 . 7 2 0 1 8 4 1 ; l o g 5 2 5 0 2 . 5 . l o g 3 . 0 0 4 2 = 0 . 4 7 7 7 2 8 8 , l o g 3 . 0 0 4 3 = 0 . 4 7 7 7 4 3 3 ; l o g 3 0 0 . 4 2 5 . l o g 7 . 6 5 4 3 = 0 . 8 8 3 9 0 5 5 , l o g 7 . 6 5 4 4 = 0 . 8 8 3 9 1 1 2 ; l o g 7 . 6 5 4 3 2 . l o g 6 . 4 3 7 1 = 0 . 8 0 8 6 9 0 3 , l o g 6 . 4 3 7 2 = 0 . 8 0 8 6 9 7 0 ; l o g 6 4 3 7 1 2 5 . A n s . 4 . 7 2 0 1 7 9 9 . 2 . 4 7 7 7 3 6 0 . l o g 1 2 9 5 4 = 4 . 1 1 2 4 0 3 9 , l o g 1 2 9 5 5 = 4 . 1 1 2 4 3 7 4 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 4 . 1 1 2 4 3 0 7 . 6 . G i v e n l o g 6 0 1 9 5 = 4 . 7 7 9 5 5 3 2 , l o g 6 0 1 9 6 = 4 . 7 7 9 5 6 0 4 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 2 . 7 7 9 5 5 6 1 . 7 . G i v e n l o g 3 . 7 0 4 0 = . 5 0 8 6 7 1 0 , l o g 3 . 7 0 4 1 = . 5 6 8 6 8 2 7 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s . 5 6 8 6 7 6 0 . 8 . G i v e n l o g 2 . 4 9 2 8 = . 3 9 6 6 8 7 4 , l o g 2 . 4 9 2 9 = . 3 9 6 7 0 4 9 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 6 . 3 9 6 6 9 3 8 . . 8 8 3 9 0 6 6 . 6 . 8 0 8 6 9 2 0 . 1 2 9 5 4 . 8 . 6 0 1 . 9 5 4 0 3 . 3 . 7 0 4 0 4 . 2 4 9 2 8 3 7 . N A T U R A L T R I G O N O M E T R I C F U N C T I O N S . 1 0 5 9 . G i v e n l o g 3 2 6 4 2 = 4 . 5 1 3 7 7 6 8 , l o g 3 2 6 4 3 = 4 . 5 1 3 7 9 0 1 ; l o g 3 2 6 4 2 . 5 . f i n d A n s . 4 . 5 1 3 7 8 3 5 . 1 0 . F i n d t h e l o g a r i t h m o f 6 2 6 5 4 3 2 6 . U s e s p e c i m e n p a g e . 7 . 7 9 6 9 5 1 0 . 1 1 . F i n d t h e n u m b e r w h o s e l o g a r i t h m i s 4 . 7 9 8 9 6 7 2 . 7 0 . U s e o f T r i g o n o m e t r i c T a b l e s . — T r i g o n o m e t r i c T a b l e s a r e o f t w o k i n d s , — T a b l e s o f N a t u r a l T r i g o n o m e t r i c F u n c t i o n s a n d T a b l e s o f L o g a r i t h m i c T r i g o n o m e t r i c F u n c t i o n s . A s t h e g r e a t e r p a r t o f t h e c o m p u t a t i o n s o f T r i g o n o m e t r y i s c a r r i e d o n b y l o g a r i t h m s , t h e l a t t e r t a b l e s a r e b y f a r t h e m o s t u s e -W e h a v e e x p l a i n e d i n A r t . 2 7 h o w t o f i n d t h e a c t u a l n u m e r i c a l v a l u e s o f c e r t a i n t r i g o n o m e t r i c f u n c t i o n s , e x a c t l y o r a p p r o x i m a t e l y . T h u s , s i n 3 0 ° = i ; t h a t i s , . 5 e x a c t l y . A l s o , t a n 6 0 ° = V 3 ; t h a t i s , 1 . 7 3 2 0 5 a p p r o x i m a t e l y . A t a b l e o f n a t u r a l t r i g o n o m e t r i c f u n c t i o n s g i v e s t h e i r a p p r o x i m a t e n u m e r i c a l v a l u e s f o r a n g l e s a t r e g u l a r i n t e r v a l s i n t h e f i r s t q u a d r a n t . I n s o m e t a b l e s t h e a n g l e s s u c c e e d e a c h o t h e r a t i n t e r v a l s o f 1 " , i n o t h e r s , a t i n t e r v a l s o f 1 0 " , b u t i n o r d i n a r y t a b l e s a t i n t e r v a l s o f 1 ' : a n d t h e v a l u e s o f t h e f u n c t i o n s a r e g i v e n c o r r e c t t o f i v e , s i x , a n d s e v e n p l a c e s . T h e f u n c t i o n s o f i n t e r m e d i a t e a n g l e s c a n b e f o u n d b y t h e p r i n c i p l e o f p r o p o r t i o n a l p a r t s a s a p p l i e d i n t h e t a b l e o f l o g a r i t h m s o f n u m b e r s ( A r t s . 6 8 a n d 6 9 ) . I t i s s u f f i c i e n t t o h a v e t a b l e s w h i c h g i v e t h e f u n c t i o n s o f a n g l e s o n l y i n t h e f i r s t q u a d r a n t , s i n c e t h e f u n c t i o n s o f a l l a n g l e s o f w h a t e v e r s i z e c a n b e r e d u c e d t o f u n c t i o n s o f a n g l e s l e s s t h a n 9 0 ° ( A r t . 3 5 ) . A n s . 6 2 9 4 5 . 8 7 6 . f u l . 1 0 6 P L A N E T R I G O N O M E T R Y . T L U s e o f T a b l e s o f N a t u r a l T r i g o n o m e t r i c F u n c t i o n s . — T h e s e t a b l e s , w h i c h c o n s i s t o f t h e a c t u a l n u m e r i c a l v a l u e s o f t h e t r i g o n o m e t r i c f u n c t i o n s , a r e c o m m o n l y c a l l e d t a b l e s o f n a t u r a l s i n e s , c o s i n e s , e t c . , s o a s t o d i s t i n g u i s h t h e m f r o m t h e t a b l e s o f t h e l o g a r i t h m s o f t h e s i n e s , c o s i n e s , e t c . W e s h a l l n o w e x p l a i n , f i r s t , h o w t o d e t e r m i n e t h e v a l u e o f a f u n c t i o n t h a t l i e s b e t w e e n t h e f u n c t i o n s o f t w o c o n s e c u t i v e a n g l e s g i v e n i n t h e t a b l e s ; a n d s e c o n d l y , h o w t o d e t e r m i n e t h e a n g l e t o w h i c h a g i v e n r a t i o c o r r e s p o n d s . 7 2 . T o f i n d t h e S i n e o f a G i v e n A n g l e . F i n d t h e s i n e o f 2 5 ° 1 4 ' 2 0 " , L a v i n g g i v e n f r o m t h e t a b l e s i n 2 5 ° 1 5 ' = . 4 2 6 5 6 8 7 s i n 2 5 ° 1 4 ' = . 4 2 6 3 0 5 6 d i f f . f o r 1 ' = . 0 0 0 2 6 3 1 L e t d = d i f f . f o r 2 0 " ; a n d a s s u m i n g t h a t a n i n c r e a s e i n t h e a n g l e i s p r o p o r t i o n a l t o a n i n c r e a s e i n t h e s i n e , w e h a v e 6 0 : 2 0 : : . 0 0 0 2 6 3 1 : d . . : d . - . s i n 2 5 ° 1 4 ' 2 0 " N o t e . — W e a s s u m e d h e r e t h a t a n i n c r e a s e i n t h e a n g l e i s p r o p o r t i o n a l t o t h e i n c r e a s e i n t h e c o r r e s p o n d i n g s i n e , w h i c h i s s u f f i c i e n t l y e x a c t f o r p r a c t i c a l p u r p o s e s , w i t h c e r t a i n e x c e p t i o n s . 7 3 . T o f i n d t h e C o s i n e o f a G i v e n A n g l e . F i n d t h e c o s i n e o f 4 4 ° 3 5 ' 2 5 " , h a v i n g g i v e n f r o m t h e t a b l e c o s 4 4 ° 3 5 ' = . 7 1 2 2 3 0 3 c o s 4 4 ° 3 6 ' = . 7 1 2 0 2 6 0 d i f f . f o r 1 ' = . 0 0 0 2 0 4 3 o b s e r v i n g t h a t t h e c o s i n e d e c r e a s e s a s t h e a n g l e i n c r e a s e s f r o m 0 ° t o 9 0 ° . . 2 0 x . 0 0 0 2 6 3 1 = . 0 0 0 0 8 7 7 . 6 0 = . 4 2 6 3 0 5 6 + . 0 0 0 0 8 7 7 = . 4 2 6 3 9 3 3 . E X A M P L E S . 1 0 7 L e t d = d e c r e a s e o f c o s i n e f o r 2 5 " ; t h e n 6 0 : 2 5 : : . 0 0 0 2 0 4 3 : d . . : d = — x . 0 0 0 2 0 4 3 = . 0 0 0 0 8 5 1 . 6 0 . - . c o s 4 4 ° 3 5 ' 2 5 " = . 7 1 2 2 3 0 3 -. 0 0 0 0 8 5 1 = 7 1 2 1 4 5 2 . S i m i l a r l y , w e m a y f i n d t h e v a l u e s o f t h e o t h e r t r i g o n o m e t r i c f u n c t i o n s , r e m e m b e r i n g t h a t , i n t h e f i r s t q u a d r a n t , t h e t a n g e n t a n d s e c a n t i n c r e a s e a n d t h e c o t a n g e n t a n d c o s e c a n t d e c r e a s e , a s t h e a n g l e i n c r e a s e s . 1 . G i v e n f i n d 2 . G i v e n f i n d 3 . G i v e n f i n d 4 . G i v e n f i n d 5 . G i v e n f i n d E X A M P L E S . s i n 4 4 ° 3 5 ' = . 7 0 1 9 4 5 9 , s i n 4 4 ° 3 6 ' = . 7 0 2 1 5 3 1 ; s i n 4 4 ° 3 5 ' 2 5 " . s i n 4 2 ° 1 5 ' = . 6 7 2 3 6 6 8 , s i n 4 2 ° 1 0 ' = . 6 7 2 5 8 2 1 ; s i n 4 2 ° 1 5 ' 1 6 " . s i n 4 3 ° 2 3 ' = . 6 8 6 8 7 6 1 , s i n 4 3 ° 2 2 ' = . 6 8 6 6 6 4 7 ; s i n 4 3 ° 2 2 ' 5 0 " . s i n 3 1 ° 6 ' = . 5 1 6 5 3 3 3 , s i n 3 1 ° 7 ' = . 5 1 6 7 8 2 4 ; s i n 3 1 ° 6 ' 2 5 " . c o s 7 4 ° 4 5 ' = . 4 2 6 5 6 8 7 , c o s 7 4 ° 4 6 ' = . 4 2 6 3 0 5 6 ; c o s 7 4 ° 4 5 ' 4 0 " . A n a . . 7 0 2 0 3 2 2 . . 6 7 2 4 2 4 2 . . 6 8 6 8 4 0 8 . . 5 1 6 6 3 7 1 . . 4 2 6 3 9 3 3 . 1 0 8 P L A N E T R I G O N O M E T R Y . 6 . G i v e n c o s 4 1 ° 1 3 ' = . 7 5 2 2 2 3 3 , c o s 4 1 ° 1 4 ' = . 7 5 2 0 3 1 6 ; c o s 4 1 ° 1 3 ' 2 6 " . f i n d A n s . . 7 5 2 1 4 0 3 . 7 . G i v e n c o s 4 7 ° 3 8 ' = . 6 c o s 4 7 ° 3 9 ' = . 6 c o s 4 7 ° 3 8 ' 3 0 " . . 6 7 3 8 7 2 7 , . 6 7 3 6 5 7 7 ; f i n d . 6 7 3 7 6 5 2 . 7 4 . T o f i n d t h e A n g l e w h o s e S i n e i s G i v e n . F i n d t h e a n g l e w h o s e s i n e i s . 5 0 8 2 7 8 4 , h a v i n g g i v e n f r o m s i n 3 0 ° 3 3 ' = . 5 0 8 2 9 0 1 s i n 3 0 ° 3 2 ' = . 5 0 8 0 3 9 6 d i f f . f o r 1 ' = . 0 0 0 2 5 0 5 g i v e n s i n e = . 5 0 8 2 7 8 4 s i n 3 0 ° 3 2 ' = . 5 0 8 0 3 9 6 d i f f . = . 0 0 0 2 3 8 8 L e t d = d i f f . b e t w e e n 3 0 ° 3 2 ' a n d r e q u i r e d a n g l e ; t h e n . 0 0 0 2 5 0 5 : . 0 0 0 2 3 8 8 : : 6 0 : d . d _ 2 3 8 8 x 6 0 _ 6 5 5 2 2 5 0 5 1 6 7 = 5 7 . 2 n e a r l y . . - . r e q u i r e d a n g l e = 3 0 ° 3 2 ' 5 7 " . 2 . 7 5 . T o f i n d t h e A n g l e w h o s e C o s i n e i s G i v e n . F i n d t h e a n g l e w h o s e c o s i n e i s . 4 0 4 3 2 8 1 , h a v i n g g i v e n f r o m t h e t a b l e c o g 0 g o 9 < = . 4 0 4 3 4 3 6 c o s 6 6 ° 1 0 ' = . 4 0 4 0 7 7 5 d i f f . f o r 1 ' = . 0 0 0 2 6 6 1 c o s 6 6 ° 9 ' = . 4 0 4 3 4 3 6 g i v e n c o s i n e = . 4 0 4 3 2 8 1 t h e t a b l e d i f f . = . 0 0 0 0 1 5 5 E X A M P L E S . 1 0 9 L e t d — d i f f . b e t w e e n 6 6 ° 9 ' a n d r e q u i r e d a n g l e ; t h e n . 0 0 0 2 6 6 1 : . 0 0 0 0 1 5 5 : : 6 0 : d . . . . d = 1 5 5 6 ° = 3 . 5 . 2 6 6 1 R e q u i r e d a n g l e i s g r e a t e r t h a n 6 6 ° 9 ' b e c a u s e i t s c o s i n e i . ; l e s s t h a n c o s 6 6 ° 9 ' . . - . r e q u i r e d a n g l e = 6 6 ° 9 ' 3 " . 5 . E X A M P L E S . 1 . G i v e n s i n 4 4 ° 1 2 ' = . 6 9 7 1 6 5 1 , s i n 4 4 ° 1 1 ' = . 6 9 6 9 5 6 5 ; f i n d t h e a n g l e w h o s e s i n e i s . 6 9 7 0 8 8 6 . A n a . 4 4 ° 1 1 ' 3 8 " . 2 . G i v e j s i n 4 8 ° 4 7 ' = . 7 5 2 2 2 3 3 , s i n 4 8 ° 4 6 ' = . 7 5 2 0 3 1 6 ; f i n d t h e a n g l e w h o s e s i n e i s . 7 5 2 1 4 0 . 4 8 ° 4 6 ' 3 4 " . 3 . G i v e n s i n 2 4 ° 1 1 ' = . 4 0 9 6 5 7 7 , s i n 2 4 ° 1 2 ' = . 4 0 9 9 2 3 0 ; f i n d t h e a n g l e w h o s e s i n e i s . 4 0 9 7 5 5 9 . 2 4 ° 1 1 ' 2 2 " . 2 . 4 . G i v e n e o s 3 2 ° 3 1 ' = . 8 4 3 2 3 5 1 , c o s 3 2 ° 3 2 ' = . 8 4 3 0 7 8 7 ; f i n d t h e a n g l e w h o s e c o s i n e i s . 8 4 3 2 . 3 2 ° 3 1 ' 1 3 " . 5 . 5 . G i v e n c o s 4 4 ° 1 1 ' = . 7 1 7 1 1 3 4 , c o s 4 4 ° 1 2 ' = . 7 1 6 9 1 0 6 ; f i n d t h e a n g l e w h o s e c o s i n e i s . 7 1 6 9 8 4 8 . 4 4 ° 1 1 ' 3 8 " . 6 . G i v e n c o s 7 0 ° 3 2 ' = . 3 3 3 2 5 8 4 , c o s 7 0 ° 3 1 ' = . 3 3 3 5 3 2 6 ; f i n d t h e a n g l e w h o s e c o s i n e i s . 3 3 3 3 3 3 3 . 7 0 ° 3 1 ' 4 3 " . 6 . 1 1 0 P L A N E T R I G O N O M E T R Y . 7 6 . U s e o f T a b l e s o f L o g a r i t h m i c T r i g o n o m e t r i c F u n c t i o n s . — S i n c e t h e s i n e s , c o s i n e s , t a n g e n t s , e t c . , o f a n g l e s a r e n u m b e r s , w e m a y u s e t h e l o g a r i t h m s o f t h e s e n u m b e r s i n n u m e r i c a l c a l c u l a t i o n s i n w h i c h t r i g o n o m e t r i c f u n c t i o n s a r e i n v o l v e d ; a n d t h e s e l o g a r i t h m s a r e i n p r a c t i c e m u c h m o r e u s e f u l t h a n t h e n u m b e r s t h e m s e l v e s , a s w i t h t h e i r a s s i s t a n c e w e a r e a b l e t o a b b r e v i a t e g r e a t l y o u r c a l c u l a t i o n s ; t h i s i s e s p e c i a l l y t h e c a s e , a s w e s h a l l s e e h e r e a f t e r , i n t h e s o l u t i o n o f t r i a n g l e s . I n o r d e r t o a v o i d t h e t r o u b l e o f r e f e r r i n g t w i c e t o t a b l e s — f i r s t t o t h e t a b l e o f n a t u r a l f u n c t i o n s f o r t h e v a l u e o f t h e f u n c t i o n , a n d t h e n t o a t a b l e o f l o g a r i t h m s f o r t h e l o g a r i t h m o f t h a t f u n c t i o n — t h e l o g a r i t h m s o f t h e t r i g o n o m e t r i c f u n c t i o n s h a v e b e e n c a l c u l a t e d a n d a r r a n g e d i n t a b l e s , f o r m i n g t a b l e s o f t h e l o g a r i t h m s o f t h e s i n e s , l o g a r i t h m s o f t h e c o s i n e s , e t c . ; t h e s e t a b l e s a r e c a l l e d t a b l e s o f l o g a r i t h m i c s i n e s , l o g a r i t h m i c c o s i n g , e t c . S i n c e t h e s i n e s a n d c o s i n e s o f a l l a n g l e s a n d t h e t a n g e n t s o f a n g l e s l e s s t h a n 4 5 ° a r e l e s s t h a n u n i t y , t h e l o g a r i t h m s o f t h e s e f u n c t i o n s a r e n e g a t i v e . T o a v o i d t h e i n c o n v e n i e n c e o f u s i n g n e g a t i v e c h a r a c t e r i s t i c s , 1 0 i s a d d e d t o t h e l o g a r i t h m s o f a l l t h e f u n c t i o n s b e f o r e t h e y a r e e n t e r e d i n t h e t a b l e . T h e l o g a r i t h m s s o i n c r e a s e d a r e c a l l e d t h e t a b u l a r l o g a r i t h m s o f t h e s i n e , c o s i n e , e t c . T h u s , t h e t a b u l a r l o g a r i t h m i c s i n e o f 3 0 ° i s 1 0 + l o g s i n 3 0 ° = 1 0 + l o g ^ = 1 0 -l o g 2 = 9 . 6 9 8 9 7 0 0 . L i I n c a l c u l a t i o n s w e h a v e t o r e m e m b e r a n d a l l o w f o r t h i s i n c r e a s e o f t h e t r u e l o g a r i t h m s . W h e n t h e v a l u e o f a n y o n e o f t h e t a b u l a r l o g a r i t h m s i s g i v e n , w e m u s t t a k e a w a y 1 0 f r o m i t t o o b t a i n t h e t r u e v a l u e o f t h e l o g a r i t h m . T h u s i n t h e t a b l e s w e f i n d l o g s i n 3 1 ° 1 5 ' = 9 . 7 1 4 9 7 7 6 . T h e r e f o r e t h e t r u e v a l u e o f t h e l o g a r i t h m o f t h e s i n e o f 3 1 ° 1 5 ' i s 9 . 7 1 4 9 7 7 6 -1 0 = 1 . 7 1 4 9 7 7 6 . S i m i l a r l y w i t h t h e l o g a r i t h m s o f o t h e r f u n c t i o n s . T A B L E S O F L O G A R I T H M I C F U N C T I O N S . I l l N o t e . — E n g l i s h a u t h o r s u s u a l l y d e n o t e t h e s e t a b u l a r l o g a r i t h m s b y t h e l e t t e r L T h u s , L s i n A d e n o t e s t h e t a b u l a r l o g a r i t h m o f t b e s i n e o f A . F r e n c h a u t h o r s u s e t h e l o g a r i t h m s o f t h e t a b l e s d i m i n i s h e d b y 1 0 . T h u s , T h e T a b l e s c o n t a i n t h e t a b u l a r l o g s o f t h e f u n c t i o n s o f a l l a n g l e s i n t h e f i r s t q u a d r a n t a t i n t e r v a l s o f 1 ' ; a n d f r o m t h e s e t h e l o g a r i t h m i c f u n c t i o n s o f a l l o t h e r a n g l e s c a n b e f o u n d . S i n c e e v e r y a n g l e b e t w e e n 4 5 ° a n d 9 0 ° i s t h e c o m p l e m e n t o f a n o t h e r a n g l e b e t w e e n 4 5 ° a n d 0 ° , e v e r y s i n e , t a n g e n t , e t c . , o f a n a n g l e l e s s t h a n 4 5 ° i s t h e c o s i n e , c o t a n g e n t , e t c . , o f a n o t h e r a n g l e g r e a t e r t h a n 4 5 ° ( A r t . 1 6 ) . H e n c e t h e d e g r e e s a t t h e t o p o f t h e t a b l e s a r e g e n e r a l l y m a r k e d f r o m 0 ° t o 4 5 ° , a n d t h o s e a t t h e b o t t o m f r o m 4 5 ° t o 9 0 ° , w h i l e t h e m i n u t e s a r e m a r k e d b o t h i n t h e f i r s t c o l u m n a t t h e l e f t , a n d i n t h e l a s t c o l u m n a t t h e r i g h t . E v e r y n u m b e r t h e r e f o r e i n e a c h c o l u m n , e x c e p t t h o s e m a r k e d d i f f . , s t a n d s f o r t w o f u n c t i o n s — t h e o n e n a m e d a t t h e t o p o f t h e c o l u m n , a n d t h e c o m p l e m e n t a l f u n c t i o n n a m e d a t t h e b o t t o m o f t h e c o l u m n . I n l o o k i n g f o r a f u n c t i o n o f a n a n g l e , i f i t b e l e s s t h a n 4 5 ° , t h e d e g r e e s a r e f o u n d a t t h e t o p , a n d t h e m i n u t e s a t t h e l e f t - h a n d s i d e . I f g r e a t e r t h a n 4 5 ° , t h e d e g r e e s a r e f o u n d a t t h e f o o t , a n d t h e m i n u t e s a t t h e r i g h t - h a n d s i d e . O n p a g e 1 1 3 i s a s p e c i m e n p a g e o f M a t h e m a t i c a l T a b l e s . I t g i v e s t h e t a b u l a r l o g a r i t h m i c f M i c t i o n s o f a l l a n g l e s b e t w e e n 3 8 ° a n d 3 9 ° , a n d a l s o o f t h o s e b e t w e e n 5 1 ° a n d 5 2 ° , b o t h i n c l u s i v e , a t i n t e r v a l s o f 1 ' . T h e n a m e s o f t h e f u n c t i o n s f o r 3 8 ° a r e p r i n t e d a t t h e t o p o f t h e p a g e , a n d t h o s e f o r 5 1 ° a t t h e f o o t . T h e c o l u m n o f m i n u t e s f o r 3 8 ° i s o n t h e l e f t , t h a t f o r 5 1 ° i s o n t h e r i g h t . T h u s w e f i n d l o g s i n A = 1 . 8 5 9 8 2 1 3 , i n s t e a d o f 9 . 8 5 9 8 2 1 3 . l o g s i n 3 8 ° 2 9 ' l o g c o s 3 8 ° 4 5 ' l o g t a n 5 1 ° 1 8 ' 9 . 7 9 3 9 9 0 7 . 9 . 8 9 2 0 3 0 3 . 1 0 . 0 9 6 2 8 5 6 . M a n y t a b l e s a r e c a l c u l a t e d f o r a n g l e s a t i n t e r v a l s o f 1 0 " . 1 1 2 P L A N E T R I G O N O M E T R Y . 7 7 . T o f i n d t h e L o g a r i t h m i c S i n e o f a G i v e n A n g l e . F i n d l o g s i n 3 8 ° 5 2 ' 4 6 " . W e h a v e f r o m p a g e 1 1 3 l o g s i n 3 8 ° 5 3 ' = 9 . 7 9 7 7 7 7 5 l o g s i n 3 8 ° 5 2 ' = 9 . 7 9 7 6 2 0 8 c l i f f , f o r V = . 0 0 0 1 5 6 7 L e t d = d i f f . f o r 4 6 " , a n d a s s u m i n g t h a t t h e c h a n g e i n t h e l o g s i n e i s p r o p o r t i o n a l t o t h e c h a n g e i n t h e a n g l e , w e h a v e 6 0 : 4 6 : : . 0 0 0 1 5 6 7 : d . . . . d = 4 g X - 0 0 0 1 5 ^ 0 0 0 1 2 0 1 6 0 . - . l o g s i n 3 8 ° 5 2 ' 4 6 " = 9 . 7 9 7 6 2 0 8 + . 0 0 0 1 2 0 1 = 9 . 7 9 7 7 4 0 9 . 7 8 . T o f i n d t h e L o g a r i t h m i c C o s i n e o f a G i v e n A n g l e . F i n d l o g c o s 8 3 ° 2 7 ' 2 3 " , h a v i n g g i v e n f r o m t h e t a b l e l o g c o s 8 3 ° 2 7 ' = 9 . 0 5 7 1 7 2 3 l o g c o s 8 3 ° 2 8 ' = 9 . 0 5 6 0 7 0 6 d i f f . f o r 1 ' = . 0 0 1 1 0 1 7 L e t d = d e c r e a s e o f l o g c o s i n e f o r 2 3 " ; t h e n 6 0 : 2 3 : : . 0 0 1 1 0 1 7 : d . , 2 3 x . 0 0 1 1 0 1 7 A A , w 0 ( > 0 i . - . d = = . 0 0 0 4 2 2 3 , n e a r l y . 6 0 , J . : l o g c o s 8 3 ° 2 7 ' 2 3 " = 9 . 0 5 7 1 7 2 3 -. 0 0 0 4 2 2 3 = 9 . 0 5 6 7 5 0 0 . E X A M P L E S . 1 . G i v e n l o g s i n 6 ° 3 3 ' = 9 . 0 5 7 1 7 2 3 , l o g s i n 6 ° 3 2 ' = 9 . 0 5 6 0 7 0 6 ; f i n d l o g s i n 6 ° 3 2 ' 3 7 " . A n s . 9 . 0 5 6 7 5 . 3 8 D e g . 1 1 3 T A B L E O F L O G A R I T H M S . ; S i n e . o 9 . 7 8 9 3 4 2 0 I 9 . 7 8 9 5 0 3 6 2 9 . 7 8 9 6 6 5 2 3 9 . 7 8 9 8 2 6 6 4 9 . 7 8 9 9 8 8 0 5 1 9 - 7 9 ° i 4 9 3 6 9 . 7 9 0 3 1 0 4 7 9 . 7 9 0 4 7 1 5 8 9 . 7 9 0 6 3 2 5 9 9 . 7 9 0 7 9 3 3 1 0 9 . 7 9 0 9 5 4 1 1 1 9 . 7 9 1 1 1 4 8 1 3 9 . 7 9 1 2 7 5 4 1 3 9 . 7 9 I 4 3 5 9 1 4 9 . 7 9 1 5 9 6 3 I S ^ 7 9 1 7 5 6 6 1 6 9 . 7 9 1 9 1 6 8 1 7 9 . 7 9 2 0 7 6 9 1 8 9 . 7 9 2 2 3 6 9 1 9 9 . 7 9 2 3 9 6 8 3 0 J 1 ; 7 9 2 5 S 6 6 _ 3 1 9 . 7 9 2 7 1 6 3 3 2 9 . 7 9 2 8 7 6 0 3 3 9 . 7 9 3 0 3 5 5 4 9 . 7 9 3 1 9 4 9 » S 9 - 7 9 3 3 5 4 3 2 6 9 - 7 9 3 5 ' 3 5 2 7 9 . 7 9 3 6 7 2 7 2 8 9 - 7 9 3 8 3 I 7 3 9 9 - 7 9 3 9 9 0 7 3 0 9 . 7 9 4 1 4 9 6 3 1 9 . 7 9 4 3 0 8 3 3 a 9 . 7 9 4 4 6 7 0 3 3 9 . 7 9 4 6 2 5 6 3 4 9 . 7 9 4 7 8 4 1 3 5 9 . 7 9 4 9 4 2 5 3 6 9 . 7 9 5 1 0 0 8 3 7 9 . 7 9 5 2 5 9 0 3 8 9 - 7 9 5 4 I 7 I 3 9 9 - 7 9 5 5 7 5 1 4 0 9 - 7 9 5 7 3 3 ° 4 1 9 . 7 9 5 8 9 0 9 4 2 9 . 7 9 6 0 4 8 6 4 3 9 . 7 9 6 2 0 6 2 4 4 9 . 7 9 6 3 6 3 8 4 5 9 . 7 9 6 5 2 1 2 4 6 9 . 7 9 6 6 7 8 6 4 7 9 . 7 9 6 8 3 5 9 4 8 9 . 7 9 6 9 9 3 0 4 9 9 . 7 9 7 1 5 0 1 5 ° 9 - 7 9 7 3 ° 7 S i 9 - 7 9 7 4 6 4 0 S 9 . 7 9 7 6 2 0 8 5 3 9 - 7 9 7 7 7 7 5 5 4 9 - 7 9 7 9 3 4 1 5 5 9 . 7 9 8 0 9 0 6 5 6 9 . 7 9 8 2 4 7 0 5 7 9 . 7 9 8 4 0 3 4 5 8 9 - 7 9 8 5 5 9 6 5 9 9 . 7 9 8 7 1 5 8 6 0 9 . 7 9 8 8 7 1 8 r C o s i n o . C o s i n e . D i f f . 1 6 1 6 1 6 1 6 1 6 1 4 1 6 1 4 1 6 1 3 1 6 1 1 1 6 1 1 1 6 1 0 1 6 0 8 1 6 0 8 1 6 0 7 1 6 0 6 1 6 0 5 1 6 0 4 1 6 0 3 1 6 0 2 1 6 0 1 1 6 0 0 1 5 9 9 1 5 9 S 1 5 9 7 1 5 9 7 1 5 9 5 1 5 9 4 1 5 9 4 1 5 9 2 1 5 9 2 1 5 9 0 1 5 9 0 1 5 8 9 1 5 8 7 1 5 8 7 1 5 8 6 1 5 8 5 1 5 8 4 1 5 8 3 1 5 8 2 1 5 8 1 1 5 8 0 1 5 7 9 1 5 7 9 1 5 7 7 1 5 7 6 1 5 7 6 ' 5 7 4 1 5 7 4 1 5 7 3 1 5 7 1 I S 7 I 1 5 7 0 5 6 9 1 5 6 8 1 5 6 7 1 5 6 6 I S 6 5 1 5 6 4 1 5 6 4 1 5 6 2 1 5 6 2 1 5 6 0 D j f f T T a n g . 9 . 8 9 2 8 0 9 8 9 . 8 9 3 0 7 0 2 9 . 8 9 3 3 3 0 6 9 . 8 9 3 5 9 0 9 9 . 8 9 3 8 5 1 1 9 . 8 9 4 1 1 1 4 9 . 8 9 4 3 7 1 5 9 . 8 9 4 6 3 1 7 9 . 8 9 4 8 9 1 8 9 - 8 9 S I 5 I 9 9 . 8 9 5 4 1 1 9 ^ 9 . 8 9 5 6 7 1 9 9 . 8 9 5 9 3 1 9 9 . 8 9 6 1 9 1 8 9 . 8 9 6 4 5 1 7 9 . 8 9 6 7 1 1 6 9 . 8 9 6 9 7 1 4 9 . 8 9 7 2 3 1 2 9 . 8 9 7 4 9 1 0 9 - 8 9 7 7 5 0 7 9 . 8 9 8 0 1 0 4 9 . 8 9 8 2 7 0 0 9 . 8 9 8 5 2 9 6 9 . 8 9 8 7 8 9 2 9 . 8 9 9 0 4 8 7 9 . 8 9 9 3 0 8 2 9 . 8 9 9 5 6 7 7 9 . 8 9 9 8 2 7 1 9 . 9 0 0 0 8 6 5 9 . 9 0 0 3 4 5 9 9 . 9 0 0 6 0 5 2 9 . 9 0 0 8 6 4 5 9 . 9 0 1 1 2 3 7 9 . 9 0 1 3 8 3 0 9 . 9 0 1 6 4 2 2 9 , 9 0 1 9 0 1 3 9 . 9 0 2 1 6 0 4 9 . 9 0 2 4 1 9 5 9 . 9 0 2 6 7 8 6 9 . 9 0 2 9 3 7 6 9 . 9 0 3 1 9 6 6 9 - 9 0 3 4 5 5 5 9 . 9 0 3 7 1 4 4 9 . 9 0 3 9 7 3 3 9 . 9 0 4 2 3 2 1 9 . 9 0 4 4 9 1 0 9 . 9 0 4 7 4 9 7 9 . 9 0 5 0 0 8 5 9 . 9 0 5 2 6 7 2 9 - 9 0 5 5 2 5 9 9 . 9 0 5 7 8 4 5 D i f f . 9 . 9 0 6 0 4 3 1 9 . 9 0 6 3 0 1 7 9 . 9 0 6 5 6 0 3 9 . 9 0 6 8 1 8 8 9 . 9 0 7 0 7 7 3 9 9 0 7 3 3 5 7 9 . 9 0 7 5 9 4 1 9 . 9 0 7 8 5 2 5 9 . 9 0 8 1 1 0 9 9 . 9 0 8 3 6 9 2 C o t a n g , 2 6 0 4 2 6 0 4 2 6 0 3 2 6 0 2 2 6 0 3 2 6 0 1 2 6 0 2 2 6 0 1 2 6 0 1 2 6 0 0 2 6 0 0 2 6 0 0 2 5 9 9 2 5 9 9 2 5 9 9 2 5 9 8 2 5 9 8 2 5 9 8 2 5 9 7 2 5 9 7 2 5 9 6 2 5 9 6 2 5 9 6 2 5 9 5 2 5 9 5 2 5 9 5 2 5 9 4 2 5 9 4 2 5 9 4 2 5 9 3 2 5 9 3 2 5 9 2 2 5 9 3 2 5 9 2 2 5 9 1 2 5 9 1 2 5 9 1 2 5 9 1 2 5 9 0 2 5 9 0 2 5 8 9 2 5 8 9 I 2 5 8 9 2 5 8 8 2 5 8 9 2 5 8 7 2 5 8 8 2 5 8 7 2 5 8 7 2 5 8 6 2 5 8 6 2 5 8 6 2 5 8 6 2 5 8 5 2 5 8 5 2 5 8 4 2 5 8 4 2 5 8 4 2 5 8 4 2 5 8 3 O o t a n g , 0 . 1 0 7 1 9 0 2 0 . 1 0 6 9 2 9 8 : o . 1 0 6 6 6 9 4 0 . 1 0 6 4 0 9 1 0 . 1 0 6 1 4 8 9 0 . 1 0 5 8 8 8 6 0 . 1 0 5 6 2 8 5 : o . i o 5 3 6 8 3 0 . 1 0 5 1 0 8 2 0 . 1 0 4 8 4 8 1 0 . 1 0 4 5 8 8 1 0 . 1 0 4 3 2 8 1 0 . 1 0 4 0 6 8 1 0 . 1 0 3 8 0 8 2 0 . 1 0 3 5 4 8 3 0 . 1 0 3 2 S 8 4 0 . 1 0 3 0 2 8 6 0 . 1 0 2 7 6 8 8 0 . 1 0 2 5 0 9 0 0 . 1 0 2 2 4 9 3 0 . 1 0 1 9 8 9 6 0 . 1 0 1 7 3 0 0 0 . 1 0 1 4 7 0 4 0 . 1 0 1 2 1 0 8 0 . 1 0 0 9 5 1 3 0 . 1 0 0 6 9 1 8 0 . 1 0 0 4 3 2 3 0 . 1 0 0 1 7 2 9 0 . 0 9 9 9 1 3 5 0 . 0 9 9 6 5 4 1 0 . 0 9 9 3 9 4 8 D i f f . 0 . 0 9 9 1 3 5 5 0 . 0 9 8 8 7 6 3 0 . 0 9 8 6 1 7 0 0 . 0 9 8 3 5 7 8 0 . 0 9 8 0 9 8 7 0 . 0 9 7 8 3 9 6 0 . 0 9 7 5 8 0 5 0 . 0 9 7 3 2 1 4 0 . 0 9 7 0 6 2 4 0 . 0 9 6 8 0 3 4 0 . 0 9 6 5 4 4 5 0 . 0 9 6 2 8 5 6 0 . 0 9 6 0 2 6 7 0 . 0 9 5 7 6 7 9 0 . 0 9 5 5 0 9 0 0 . 0 9 5 2 5 0 3 0 . 0 9 4 9 9 1 5 0 . 0 9 4 7 3 2 8 0 . 0 9 4 4 7 4 1 0 . 0 9 4 2 1 5 5 0 . 0 9 3 9 5 6 9 0 . 0 9 3 6 9 8 3 0 . 0 9 3 4 3 9 7 0 . 0 9 3 1 8 1 2 0 . 0 9 2 9 2 2 7 0 . 0 9 2 6 6 4 3 0 . 0 9 2 4 0 5 9 0 . 0 9 2 1 4 7 5 0 . 0 9 1 8 8 9 1 0 . 0 9 1 6 3 0 8 D i f f . T a n g . 9 8 7 9 8 8 9 8 8 9 8 9 9 9 0 9 9 ° 9 9 1 9 9 2 9 9 2 9 9 2 9 9 3 9 9 4 9 9 5 9 9 5 9 9 S 9 9 7 9 9 6 9 9 8 9 9 8 9 9 8 1 0 0 0 9 9 9 0 0 1 0 0 1 0 0 1 0 0 3 0 0 2 0 0 4 0 0 4 0 0 4 0 0 5 0 0 6 0 0 7 0 0 7 0 0 7 0 0 8 0 0 9 0 1 0 0 1 0 0 1 0 0 1 1 0 1 2 0 1 3 0 1 3 0 1 3 0 1 4 0 1 5 0 1 6 0 1 6 0 1 6 ; o i 8 : o i 7 0 1 9 0 1 9 : o 2 0 0 2 0 0 2 1 0 2 1 0 3 2 0 3 3 9 . 8 9 6 5 3 2 I 6 0 9 . 8 9 6 4 3 3 4 5 9 9 . 8 9 6 3 3 4 6 5 8 9 . 8 9 6 2 3 5 8 5 7 9 . 8 9 6 1 3 6 9 j 5 6 9 . 8 9 6 0 3 7 9 I 5 5 5 4 9 . 8 9 5 9 3 8 9 9 . 8 9 5 8 3 9 8 9 . 8 9 5 7 4 0 6 9 . 8 9 5 6 4 1 4 9 . 8 9 5 5 4 2 2 9 . 8 9 5 4 4 2 9 9 - 8 9 5 3 4 3 5 9 . 8 9 5 2 4 4 0 9 . 8 9 5 1 4 4 5 9 . 8 9 5 0 4 5 0 9 . 8 9 4 9 4 5 3 9 . 8 9 4 8 4 5 7 9 . 8 9 4 7 4 5 9 9 . 8 9 4 6 4 6 1 9 . 8 9 4 5 4 6 3 9 . 8 9 4 4 4 6 3 9 . 8 9 4 3 4 6 4 9 . 8 9 4 2 4 6 3 9 . 8 9 4 1 4 6 2 9 . 8 9 4 0 4 6 1 D i f f . 9 . 8 9 3 9 4 5 8 9 . 8 9 3 8 4 5 6 9 - 8 9 3 7 4 5 2 9 . 8 9 3 6 4 4 8 9 . 8 9 3 5 4 4 4 9 . 8 9 3 4 4 3 9 9 . 8 9 3 3 4 3 3 9 . 8 9 3 2 4 2 6 2 7 9 . 8 9 3 1 4 1 9 2 6 9 . 8 9 3 0 4 1 2 2 5 9 . 8 9 2 9 4 0 4 9 . 8 9 2 8 3 9 5 9 . 8 9 2 7 3 8 5 9 . 8 9 2 6 3 7 5 9 - 8 9 2 5 3 6 5 9 . 8 9 2 4 3 5 4 9 . 8 9 2 3 3 4 2 9 . 8 9 2 2 3 2 9 9 . 8 9 2 1 3 1 6 9 . 8 9 2 0 3 0 3 9 . 8 9 1 9 2 8 9 9 . 8 9 1 8 2 7 4 9 . 8 9 1 7 2 5 8 9 . 8 9 1 6 2 4 2 9 . 8 9 1 5 2 2 6 j 1 0 9 . 8 9 1 4 2 0 8 9 . 8 9 1 3 1 9 1 9 . 8 9 1 2 1 7 2 9 . 8 9 1 1 1 5 3 9 . 8 9 1 0 1 3 3 9 . 8 9 0 9 1 1 3 9 . 8 9 0 8 0 9 2 9 . 8 9 0 7 0 7 1 9 . 8 9 0 6 0 4 9 9 . 8 9 0 5 0 2 6 S i n e . 5 1 D e p . 1 1 4 P L A N E T R I G O N O M E T R Y . 2 . G i v e n f i n d 3 . G i v e n f i n d 4 . G i v e n l o g s i n 5 5 ° 3 3 l o g s i n 5 5 ° 3 4 l o g s i n 5 5 ° 3 3 l o g c o s 3 7 ° 2 8 l o g c o s 3 7 ° 2 9 l o g c o s 3 7 ° 2 8 l o g c o s 4 4 ° 3 5 l o g c o s 4 4 ° 3 5 f i n d l o g c o s 4 4 ° 3 5 S e e f o o t - n o t e o f A r t . 7 6 . 5 . G i v e n l o g c o s 5 5 ° 1 1 l o g c o s 5 5 ° 1 2 l o g c o s 5 5 ° 1 1 l o g t a n 2 7 ° 1 3 l o g t a n 2 7 ° 1 4 l o g t a n 2 7 ° 1 3 f i n d 6 . G i v e n f i n d = 9 . 9 1 6 2 5 3 9 , = 9 . 9 1 6 3 4 0 6 ; 5 4 " . A n a . 9 . 9 1 6 3 3 1 9 . = 9 . 8 9 9 6 6 0 4 , = 9 . 8 9 9 5 6 3 6 ; 3 6 " . 2 0 " = 9 . 8 5 2 5 7 8 9 , 3 0 " = 9 . 8 5 2 5 5 8 2 ; 2 5 " . 7 . = 9 . 7 5 6 5 9 9 9 , = 9 . 7 5 6 4 1 8 2 ; 1 2 " . = 9 . 7 1 1 2 1 4 8 , = 9 . 7 1 1 5 2 5 4 ; 4 5 " . 9 . 8 9 9 6 0 2 3 . 9 . 8 5 2 5 6 7 1 . 9 . 7 5 6 5 6 3 6 . 9 . 7 1 1 4 4 7 7 . 7 9 . T o f i n d t h e A n g l e w h o s e L o g a r i t h m i c S i n e i s G i v e n . F i n d t h e a n g l e w h o s e l o g s i n e i s 8 . 8 7 8 5 9 4 0 , h a v i n g g i v e n f r o m t h e t a b l e l o g s i n 4 ° 2 1 ' = 8 . 8 7 9 9 4 9 3 l o g s i n 4 ° 2 0 ' = 8 . 8 7 8 2 8 5 4 d i f f . f o r 1 ' = . 0 0 1 6 6 3 9 g i v e n l o g s i n e = 8 . 8 7 8 5 9 4 0 l o g s i n 4 ° 2 0 ' = 8 . 8 7 8 2 8 5 4 d i f f . = . 0 0 0 3 0 8 6 L e t d = d i f f . b e t w e e n 4 ° 2 0 ' a n d r e q u i r e d a n g l e ; t h e n . 0 0 1 6 6 3 9 : . 0 0 0 3 0 8 6 : : 6 0 : d . , 3 0 8 6 x 6 0 „ . , . - . d = — = 2 4 , n e a r l y . 1 6 6 3 9 J . - . r e q u i r e d a n g l e = 4 ° 2 0 ' 2 4 " . E X A M P L E S . 1 1 5 8 0 . T o f i n d t h e A n g l e w h o s e L o g a r i t h m i c C o s i n e i s G i v e n . F i n d t h e a n g l e w h o s e l o g c o s i n e i s 9 . 8 9 3 4 3 4 2 . ^ V e h a v e f r o m p a g e 1 1 3 l o g c o s 3 8 ° 3 1 ' = 9 . 8 9 3 4 4 3 9 l o g c o s 3 8 ° 3 2 ' = 9 . 8 9 3 3 4 3 3 d i f f . f o r 1 ' = . 0 0 0 1 0 0 6 l o g c o s 3 8 ° 3 1 ' = 9 . 8 9 3 4 4 3 9 g i v e n l o g c o s i n e = 9 . 8 9 3 4 3 4 2 d i f f . = . 0 0 0 0 0 9 7 L e t d = d i f f . b e t w e e n 3 8 ° 3 1 ' a n d r e q u i r e d a n g l e ; t h e n . 0 0 0 1 0 0 6 : . 0 0 0 0 0 9 7 : : 6 0 : d . . d = . 0 0 0 0 0 9 7 9 7 _ x _ 6 0 = 5 „ 8 . 0 0 0 1 0 0 6 1 0 0 6 . - . r e q u i r e d a n g l e = 3 8 ° 3 1 ' 5 " . 8 . N o t e . — I n u s i n g b o t h t h e t a b l e s o f t h e n a t u r a l s i n e s , c o s i n e s , e t c . , a n d t h e t a b l e s o f t h e l o g a r i t h m i c n i n e , c o s i n e s , e t c . , t h e s t u d e n t w i l l r e m e m b e r t h a t , i n t h e f i r s t q u a d r a n t , a s t h e a n g l e i n c r e a s e s , t h e s i n e , t a n g e n t , a n d s e c a n t i n c r e a s e , b u t t h e c o s i n e , c o t a n g e n t , a n d c o s e c a n t d e c r e a s e . E X A M P L E S . 1 . G i v e n l o g s i n 1 4 ° 2 4 ' = 9 . 3 9 5 6 5 8 1 , l o g s i n 1 4 ° 2 . 7 = 9 . 3 9 6 1 4 9 9 ; f i n d t h e a n g l e w h o s e l o g s i n e i s 9 . 3 9 5 0 4 4 9 . A n a . 1 4 ° 2 4 ' 3 5 " . 2 . G i v e n l o g s i n 7 1 ° 4 0 ' = 9 . 9 7 7 3 7 7 2 , l o g s i n 7 1 ° 4 1 ' = 9 . 9 7 7 4 1 9 1 ; f i r d t h e a n g l e w h o s e l o g s i n e i s 9 . 9 7 7 3 8 9 7 . 7 1 ° 4 0 ' 1 8 " . 3 . G i v e n l o g c o s 2 8 ° 1 7 ' = 9 . 9 4 4 7 8 6 2 , l o g c o s 2 8 ° 1 6 ' = 9 . 9 4 4 8 5 4 1 ; f i n d t h e a n g l e w h o s e l o g c o s i n e i s 9 . 9 4 4 8 2 3 0 . 2 8 ° 1 6 ' 2 7 " . 5 . 1 1 6 P L A N E T R I G O N O M E T R Y . 4 . G i v e n l o g c o s 8 0 ° 5 3 ' = 9 . 1 9 9 8 7 9 3 , l o g c o s 8 0 ° 5 2 ' 5 0 " = 9 . 2 0 0 0 1 0 5 j f i n d t h e a n g l e w h o s e l o g c o s i n e i s 9 . 2 0 0 0 0 0 0 . A n s . 8 0 ° 5 2 ' 5 1 " . 5 . G i v e n l o g t a n 3 5 ° 4 ' = 9 . 8 4 6 3 0 1 8 , l o g t a n 3 5 ° 5 ' = 9 . 8 4 6 5 7 0 5 ; f i n d t h e a n g l e w h o s e l o g t a n g e n t i s 9 . 8 4 6 4 0 2 8 . 3 5 ° 4 ' 2 3 " . 6 . G i v e n l o g s i n 4 4 ° 3 5 ' 3 0 " = 9 . 8 4 6 3 6 7 8 , l o g s i n 4 4 ° 3 5 ' 2 0 " = 9 . 8 4 6 3 4 6 4 ; f i n d t h e a n g l e w h o s e l o g s i n e i s 9 . 8 4 6 3 5 8 6 . 4 4 ° 3 5 ' 2 5 " . 7 . 7 . F i n d t h e a n g l e b y p a g e 1 1 3 w h o s e l o g t a n g e n t i s 1 0 . 1 0 1 8 5 4 2 . A n s . 5 1 ° 3 9 ' 2 8 " . 7 . 8 1 . A n g l e s n e a r t h e L i m i t s o f t h e Q u a d r a n t . — I t w a s a s s u m e d i n A r t s . 7 2 - 8 0 t h a t , i n g e n e r a l , t h e d i f f e r e n c e s o f t h e t r i g o n o m e t r i c f u n c t i o n s , b o t h n a t u r a l a n d l o g a r i t h m i c , a r e a p p r o x i m a t e l y p r o p o r t i o n a l t o t h e d i f f e r e n c e s o f t h e i r c o r r e s p o n d i n g i i n g l e s , w i t h c e r t a i n e x c e p t i o n s . T h e e x c e p t i o n a l c a s e s a r e a s f o l l o w s : ( 1 ) N a t u r a l f u n c t i o n s . — F o r t h e s i n e t h e d i f f e r e n c e s a r e i n s e n s i b l e f o r a n g l e s n e a r 9 0 ° ; f o r t h e c o s i n e t h e y a r e i n s e n s i b l e f o r a n g l e s n e a r 0 ° . F o r t h e t a n g e n t t h e d i f f e r e n c e s a r e i r r e g u l a r f o r a n g l e s n e a r 9 0 ° ; f o r t h e c o t a n g e n t t h e y a r e i r r e g u l a r f o r a n g l e s n e a r 0 ° . ( 2 ) L o g a r i t h m i c f u n c t i o n s . — T h e p r i n c i p l e o f p r o p o r t i o n a l p a r t s f a i l s b o t h f o r a n g l e s n e a r 0 ° a n d a n g l e s n e a r 9 0 ° . F o r t h e l o g s i n e a n d t h e l o g c o s e c a n t t h e d i f f e r e n c e s a r e i r r e g u l a r f o r a n g l e s n e a r 0 ° , a n d i n s e n s i b l e f o r a n g l e s n e a r 9 0 ° . F o r t h e l o g c o s i n e a n d t h e l o g s e c a n t t h e d i f f e r e n c e s a r e i n s e n s i b l e f o r a n g l e s n e a r 0 ° , a n d i r r e g u l a r f o r a n g l e s n e a r 9 0 ° . F o r t h e l o g t a n g e n t a n d t h e , l o g c o t a n g e n t t h e d i f f e r e n c e s a r e i r r e g u l a r f o r a n g l e s n e a r 0 ° a n d a n g l e s n e a r 9 0 ° . E X A M P L E S . 1 1 7 I t f o l l o w s , t h e r e f o r e , t h a t a n g l e s n e a r 0 ° a n d a n g l e s n e a r 9 0 ° c a n n o t b e f o u n d w i t h e x a c t n e s s f r o m t h e i r l o g t r i g o n o m e t r i c f u n c t i o n s . T h e s e d i f f i c u l t i e s m a y b e m e t i n t h r e e w a y s . ( 1 ) F o r a n a n g l e n e a r 0 ° u s e t h e p r i n c i p l e t h a t t h e s i n e s a n d t a n g e n t s o f s m a l l a n g l e s a r e a p p r o x i m a t e l y p r o p o r t i o n a l t o t h e a n g l e s t h e m s e l v e s . ( S e e A r t . 1 3 0 . ) ( 2 ) F o r a n a n g l e n e a r 9 0 ° u s e t h e h a l f a n g l e ( A r t . 9 9 ) . ( 3 ) I n u s i n g t h e p r o p o r t i o n a l p a r t s , f i n d t w o , t h r e e , o r m o r e o r d e r s o f d i f f e r e n c e s ( A l g . , A r t . 1 9 7 ) . S p e c i a l t a b l e s a r e e m p l o y e d f o r a n g l e s n e a r t h e l i m i t s o f t h e q u a d r a n t . E X A M P L E S . 1 . G i v e n l o g 1 0 7 = . 8 4 5 0 9 8 0 , f i n d l o g M 3 4 3 , l o g , „ 2 4 0 1 , a n d l o g 1 0 1 6 . 8 0 7 . A n s . 2 . 5 3 5 2 9 4 0 , 3 . 3 8 0 3 9 2 0 , 1 . 2 2 5 4 9 0 0 . 2 . F i n d t h e l o g a r i t h m s t o t h e b a s e 3 o f 9 , 8 1 , \ , . 1 , f a . A n s . 2 , 4 , -1 , -3 , -2 , -4 . 3 . F i n d t h e v a l u e o f l o g 2 8 , l o g 2 . 5 , l o g 3 2 4 3 , l o g 5 ( . 0 4 ) , l o g 1 0 1 0 0 0 , l o g , „ . 0 0 1 . A n s . 3 , -1 , 5 , -2 , 3 , -3 . 4 . F i n d t h e v a l u e o f l o g 0 a , l o g b - Z / ¥ 2 , l o g 8 2 , l o g ^ 3 , l o g ^ O . A n s . , | , \ , \ , \ . G i v e n l o g j 0 2 = . 3 0 1 0 3 0 0 , l o g 1 0 3 = . 4 7 7 1 2 1 3 , a n d l o g , „ 7 = . 8 4 5 0 9 8 0 , f i n d t h e v a l u e s o f t h e f o l l o w i n g : 5 . l o g 1 0 3 5 , l o g 1 0 1 5 0 , l o g i „ . 2 . A n s . 1 . 5 4 4 0 6 8 , 2 . 1 7 6 0 9 1 3 , 1 . 3 0 1 0 3 . 6 . l o g 1 0 3 . 5 , l o g , „ 7 . 2 9 , l o g 1 0 . 0 8 1 . A n s . . 5 4 4 0 6 8 0 , . 8 6 2 7 2 7 8 , 2 . 9 0 8 4 8 5 2 . 7 . l o g 1 0 | , l o g 1 0 3 5 , l o g 1 0 ^ . A n s . . 3 6 7 9 7 6 7 , 2 . 3 8 5 6 0 6 5 , . 0 7 8 0 2 7 8 . 1 1 8 P L A N E T R I G O N O M E T R Y . 8 . W r i t e d o w n t h e i n t e g r a l p a r t o f t h e c o m m o n l o g a r i t h m s o f 7 9 6 3 , . 1 , 2 . 6 1 , 7 9 . 6 3 4 1 , 1 . 0 0 0 6 , . 0 0 0 0 0 0 7 9 . A n s . 3 , -1 , 0 , 1 , 0 , -7 . 9 . G i v e t h e p o s i t i o n o f t h e f i r s t s i g n i f i c a n t f i g u r e i n t h e n u m b e r s w h o s e l o g a r i t h m s a r e 2 . 4 6 1 2 3 1 0 , 1 . 2 7 9 3 4 0 0 , 6 . 1 7 6 3 2 4 1 . 1 0 . G i v e t h e p o s i t i o n o f t h e f i r s t s i g n i f i c a n t f i g u r e i n t h e n u m b e r s w h o s e l o g a r i t h m s a r e 4 . 2 9 9 0 7 1 3 , . 3 0 4 0 5 9 5 , 2 . 5 8 6 0 2 4 4 , 3 . 1 7 6 0 9 1 3 , 1 . 3 1 8 0 6 3 3 , . 4 9 8 0 3 4 7 . A n s . t e n t h o u s a n d s , u n i t s , h u n d r e d s , 3 r d d e c . p l . , 1 s t d e c . p l . , u n i t s . 1 1 . G i v e n l o g 7 = . 8 4 5 0 9 8 0 , f i n d t h e n u m b e r o f d i g i t s i n t h e i n t e g r a l p a r t o f 7 1 0 , 4 9 6 , 3 4 3 l ° f i , ( V 0 ) 2 0 , ( 4 . 9 ) 1 2 , ( 3 . 4 3 ) 1 0 . A n s . 9 , 1 1 , 8 5 , 4 , 9 , 6 . 1 2 . F i n d t h e p o s i t i o n o f t h e f i r s t s i g n i f i c a n t f i g u r e i n t h e n u m e r i c a l v a l u e o f 2 0 7 , ( . 0 2 ) 7 , ( . 0 0 7 ) 2 , ( 3 . 4 3 ) " , ( . 0 3 4 3 ) 8 , ( . 0 3 4 3 ) ^ . A n s . t e n t h i n t e g r a l p l . , 1 2 t h d e c . p l . , 5 t h d e c . p l . , u n i t s , 1 2 t h d e c . p l . , 1 s t d e c . p l . S h o w h o w t o t r a n s f o r m 1 3 . C o m m o n l o g a r i t h m s t o l o g a r i t h m s w i t h b a s e 2 . A n s . D i v i d e e a c h l o g a r i t h m b y . 3 0 1 0 3 . 1 4 . L o g a r i t h m s w i t h b a s e 3 t o c o m m o n l o g a r i t h m s . A n s . M u l t i p l y e a c h l o g b y . 4 7 7 1 2 1 3 . 1 5 . G i v e n l o g 1 0 2 = . 3 0 1 0 3 0 0 , f i n d l o g 2 1 0 . 3 . 3 2 1 9 0 . 1 6 . G i v e n l o g , „ 7 = . 8 4 5 0 9 8 0 , f i n d l o g i 1 0 . 1 . 1 8 3 . 1 7 . G i v e n l o g 1 0 2 = . 3 0 1 0 3 0 0 , f i n d l o g 8 1 0 . 1 . 1 0 7 3 0 . 1 8 . T h e m a n t i s s a o f t h e l o g o f 8 5 7 6 2 i s 9 3 3 2 9 4 9 ; f i n d ( 1 ) t h e l o g o f a / . 0 0 8 5 7 6 2 , a n d ( 2 ) t h e n u m b e r o f f i g u r e s i n ( 8 5 7 6 2 ) 1 1 , w h e n i t i s m u l t i p l i e d o u t . A n s . ( 1 ) 1 . 8 1 2 1 1 7 7 , ( 2 ) 5 5 . E X A M P L E S . 1 1 9 1 9 . W h a t a r e t h e c h a r a c t e r i s t i c s o f t h e l o g a r i t h m s o f 3 7 4 2 t o t h e b a s e s 3 , 6 , 1 0 , a n d 1 2 r e s p e c t i v e l y ? A n s . 7 , 4 , 3 , 3 . 2 0 . P r o v e t h a t 7 l o g + 6 l o g f + 5 l o g £ + l o g f | = l o g 3 . 1 A n s . 2 + l o g j 0 7 3 . 7 1 4 5 . 2 1 . G i v e n l o g 1 0 7 , f i n d l o g 7 4 9 0 . 2 2 . F r o m 5 . 3 4 2 9 t a k e 3 . 6 2 8 4 . 2 3 . D i v i d e 1 3 . 2 6 1 5 b y 8 . 2 . 4 0 7 0 . 2 4 . P r o v e t h a t 6 l o g § + 4 l o g T % + 2 l o g ? / ' = < > -2 5 . F i n d l o g J 2 9 7 V 1 l } $ t o t h e b a s e 3 V 1 1 . 1 . 8 . G i v e n l o g 2 = . 3 0 1 0 3 0 0 , l o g 3 = . 4 7 7 1 2 1 3 . 2 6 . F i n d l o g 2 1 6 , 6 4 8 0 , 5 4 0 0 , f . A n s . 2 . 3 3 4 4 5 3 9 , 3 . 8 1 1 5 7 5 2 , 3 . 7 3 2 3 9 3 9 , 1 . 6 4 7 8 1 7 4 . 2 7 . F i n d l o g . 0 3 , 6 ~ , ( 5 ) ~ . A n s . 2 . 4 7 7 1 2 1 3 , 1 . 7 4 0 6 1 6 2 , 1 . 6 3 6 5 0 0 6 . 2 8 . F i n d l o g . 1 8 , l o g 2 . 4 , l o g ^ . A n s . 1 . 2 5 5 2 7 2 6 , . 3 8 0 2 1 1 3 , 1 . 2 7 3 0 0 1 3 . 2 9 . F i n d l o g ( 6 . 2 5 ) l o g 4 V . 0 ' 0 5 . 3 0 . G i v e n . 1 1 3 6 9 7 1 , 1 . 4 5 1 5 4 . f i n d 3 1 . G i v e n f i n d 3 2 . G i v e n f i n d l o g 5 6 3 2 1 = 4 . 7 5 0 6 7 0 4 , l o g 5 6 3 2 2 = 4 . 7 5 0 6 7 8 1 ; l o g 5 6 3 2 1 4 7 . 6 . 7 5 0 6 7 4 0 . l o g 5 3 4 0 3 = 4 . 7 2 7 5 6 5 7 , l o g 5 3 4 0 2 = 4 . 7 2 7 5 5 7 5 ; l o g 5 3 4 0 2 3 4 . 6 . 7 2 7 5 6 0 3 . l o g 5 6 4 1 2 = 4 . 7 5 1 3 7 1 5 , l o g 5 6 4 1 3 = 4 . 7 5 1 3 7 9 2 ; l o g 5 6 4 . 1 2 3 . 2 . 7 5 1 3 7 3 8 . 1 2 0 P L A N E T R I G O N O M E T R Y . 3 3 . G i v e n f i n d 3 4 . G i v e n f i n d 3 5 . G i v e n f i n d 3 6 . G i v e n f i n d 3 7 . G i v e n f i n d 3 8 . G i v e n f i n d 3 9 . G i v e n f i n d t h e n u m b e r 4 0 . G i v e n f i n d t h e n u m b e r 4 1 . G i v e n l o g 8 7 3 0 4 = 4 . 9 4 1 3 3 2 5 , l o g 8 7 3 0 5 = 4 . 9 4 1 3 3 7 5 ; l o g . 0 0 0 8 7 3 0 4 1 0 . A n s . 4 . 9 4 1 3 3 3 3 . l o g 3 7 2 4 5 = 4 . 5 7 1 0 0 8 0 , l o g 3 7 2 4 6 = 4 . 5 7 1 0 7 9 0 ; l o g 3 . 7 2 4 5 6 . . 5 7 1 0 7 5 0 . l o g 3 2 0 2 5 = 4 . 5 0 5 4 8 9 1 , l o g 3 2 0 2 6 = 4 . 5 0 5 5 0 2 7 ; l o g 3 2 . 0 2 5 6 1 3 . 1 . 5 0 5 4 9 7 4 . l o g 0 5 9 3 1 = 4 . 8 1 9 0 8 9 7 , l o g 0 5 9 3 2 = 4 . 8 1 9 0 9 0 2 ; l o g . 0 0 0 0 0 0 5 9 3 1 7 1 . 6 . 8 1 9 0 9 4 3 . l o g 2 5 8 1 9 = 4 . 4 1 1 9 3 9 4 , l o g 2 5 8 2 0 = 4 . 4 1 1 9 5 6 2 ; l o g 2 . 5 8 1 9 2 6 . . 4 1 1 9 4 3 8 . l o g 2 3 4 5 4 = 4 . 3 7 0 2 1 0 9 , l o g 2 3 4 5 3 = 4 . 3 7 0 1 9 8 4 ; l o g 2 3 4 5 3 4 8 7 . 7 . 3 7 0 2 0 7 4 . l o g 4 5 7 4 0 = 4 . 0 0 0 2 9 6 2 , l o g 4 5 7 4 1 = 4 . 0 0 0 3 0 5 7 ; w h o s e l o g a r i t h m i s 4 . 6 6 0 2 9 8 7 . 4 5 7 4 0 . 2 0 . l o g 4 3 9 0 5 = 4 . 6 4 3 1 0 7 1 , l o g 4 3 9 0 0 = 4 . 0 4 3 1 1 7 0 ; w h o s e l o g a r i t h m i s 4 . 0 4 3 1 1 5 0 . . 0 0 0 4 3 9 0 5 8 . 5 . 0 8 9 1 5 8 . l o g 5 6 8 9 1 = 4 . 7 5 5 0 4 3 6 , l o g 5 6 8 9 2 = 4 . 7 5 5 0 5 1 2 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s . 7 5 5 0 4 8 0 . E X A M P L E S . 1 2 1 4 2 . G i v e n l o g 3 4 5 7 2 = 4 . 5 3 8 7 2 4 5 , l o g 3 4 5 7 3 = 4 . 5 3 8 7 3 7 1 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 2 . 5 3 8 7 3 5 9 . A n s . 3 4 5 . 7 2 9 1 . 4 3 . G i v e n l o g 1 0 9 0 5 = 4 . 0 3 7 C 2 5 7 , l o g 1 0 9 0 6 = 4 . 0 3 7 6 6 5 5 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 3 . 0 3 7 6 3 7 1 . 1 0 9 0 . 5 2 8 6 . 4 4 . G i v e n l o g 2 5 7 2 5 = 4 . 4 1 0 3 5 5 4 , l o g 2 5 7 2 6 = 4 . 4 1 0 3 7 2 3 ; f i n d t h e n u m b e r w h o s e l o g a r i t h m i s 7 . 4 1 0 3 7 2 0 . A n s . . 0 0 0 0 0 0 2 5 7 2 5 9 8 2 . I n t h e f o l l o w i n g s i x e x a m p l e s t h e s t u d e n t m u s t t a k e h i s l o g a r i t h m s f r o m t h e t a b l e s . 4 5 . R e q u i r e d t h e p r o d u c t o f 3 6 7 0 . 2 5 7 a n d 1 2 . 6 1 1 5 8 , b y l o g a r i t h m s . A n s . 4 6 2 8 7 . 7 4 . 4 6 . R e q u i r e d t h e q u o t i e n t o f . 1 2 3 4 5 6 7 b y 5 4 . 8 7 6 4 5 , b y l o g a r i t h m s . A n s . . 0 0 2 2 4 9 7 2 1 . 4 7 . R e q u i r e d t h e c u b e o f . 3 1 8 0 2 3 6 , b y l o g a r i t h m s . 4 8 . R e q u i r e d t h e c u b e r o o t o f . 3 6 6 3 2 6 5 , b y l o g a r i t h m s . A n s . . 0 3 2 1 6 4 5 8 . A n s . . 7 1 5 5 2 1 6 . 4 9 . R e q u i r e d t h e e l e v e n t h r o o t o f 6 3 . 7 4 2 . 5 0 . R e q u i r e d t h e f i f t h r o o t o f . 0 7 . 1 . 4 5 8 9 4 . . 5 8 7 5 2 . 5 1 . G i v e n s i n 4 2 ° 2 1 ' = . 6 s i n 4 2 ° 2 2 ' = . 6 s i n 4 2 ° 2 1 ' 3 0 " . 6 7 3 6 5 7 7 , 6 7 3 8 7 2 7 ; f i n d . 6 7 3 7 6 5 2 . 5 2 . G i v e n s i n 6 7 ° 2 2 ' = . 9 2 2 9 8 6 5 , s i n 6 7 ° 2 3 ' = . 9 2 3 0 9 8 4 ; s i n 6 7 ° 2 2 ' 4 8 " . 5 . f i n d . 9 2 3 0 7 6 9 . 1 2 2 P L A N E T R I G O N O M E T R Y . 5 3 . G i v e n s i n 7 , " I T = . 1 2 6 7 7 6 1 , s i n 7 ° 1 8 ' = . 1 2 7 0 6 4 6 ; f i n d s i n 7 ° 1 7 ' 2 5 " . A n s . . 1 2 6 8 9 6 3 . 5 4 . G i v e n c o s 2 1 ° 2 7 ' = . 9 3 0 7 3 7 0 , c o s 2 1 ° 2 8 ' = . 9 3 0 6 3 0 6 ; f i n d c o s 2 1 ° 2 7 ' 4 5 " . . 9 3 0 6 5 7 2 . 5 5 . G i v e n c o s 3 4 ° 1 2 ' = . 8 2 7 0 8 0 6 , c o s 3 4 ° 1 3 ' = . 8 2 6 9 1 7 0 ; f i n d c o s 3 4 ° 1 2 ' 1 9 " . 6 . . 8 2 7 0 2 7 2 . 5 6 . G i v e n s i n 4 1 ° 4 8 ' = . 6 6 6 5 3 2 5 , s i n 4 1 ° 4 9 ' = . 6 6 6 7 4 9 3 ; f i n d t h e a n g l e w h o s e s i n e i s . 6 6 6 6 6 6 6 . 4 1 ° 4 8 ' 3 7 " . 5 7 . G i v e n s i n 7 3 ° 4 4 ' = . 9 5 9 9 6 8 4 , s i n 7 3 ° 4 5 ' = . 9 6 0 0 4 9 9 ; f i n d t h e a n g l e w h o s e s i n e i s . 9 6 . 7 3 ° 4 4 ' 2 3 " . 2 . 5 8 . G i v e n c o s 7 5 ° 3 2 ' = . 2 4 9 8 1 6 7 , c o s 7 5 ° 3 1 ' = . 2 5 0 0 9 8 4 ; f i n d t h e a n g l e w h o s e c o s i n e i s . 2 5 . 7 5 " 3 1 ' 2 1 " . 5 9 . G i v e n c o s 5 3 ° 7 ' = . 6 0 0 1 8 7 6 , c o s 5 3 ° 8 ' = . 5 9 9 9 5 4 9 ; f i n d t h e a n g l e w h o s e c o s i n e i s . 6 . 5 3 ° 7 ' 4 8 " . 4 . 6 0 . G i v e n l o g s i n 4 5 ° 1 6 ' = 9 . 8 5 1 4 9 6 9 , l o g s i n 4 5 ° 1 7 ' = 9 . 8 5 1 6 2 2 0 ; f i n d l o g s i n 4 5 ° 1 6 ' 3 0 " . 9 . 8 5 1 5 5 9 4 . 6 1 . G i v e n l o g s i n 3 8 ° 2 4 ' = 9 . 7 9 3 1 9 4 9 , l o g s i n 3 8 ° 2 5 ' = 9 . 7 9 3 3 5 4 3 ; f i n d l o g s i n 3 8 ° 2 4 ' 2 7 " . 9 . 7 9 3 2 6 6 6 . E X A M P L E S . 1 2 3 = 9 . 7 2 9 8 1 9 7 , = 9 . 7 3 0 0 1 8 2 ; 3 6 " . A m . 9 . 7 2 9 9 3 8 8 . 6 2 . G i v e n f i n d . 6 3 . G i v e n f i n d 6 4 . G i v e n f i n d 6 5 . G i v e n f i n d 6 6 . G i v e n f i n d 6 7 . G i v e n f i n d 6 8 . G i v e n f i n d 6 9 . G i v e n f i n d 7 0 . G i v e n f i n d l o g s i n 3 2 ° 2 8 ' l o g s i n 3 2 ° 2 9 ' l o g s i n 3 2 ° 2 8 ' l o g s i n 1 7 ° 1 ' l o g s i n 1 7 ° 0 ' l o g s i n 1 7 ° 0 ' l o g s i n 2 6 ° 2 4 ' l o g s i n 2 6 ° 2 5 ' l o g s i n 2 6 ° 2 4 ' l o g c o s 1 7 ° 3 1 ' l o g c o s 1 7 ° 3 2 ' l o g c o s 1 7 ° 3 1 ' l o g t a n 2 1 ° 1 7 ' l o g t a n 2 1 ° 1 8 ' l o g t a n 2 1 ° 1 7 ' l o g t a n 2 7 ° 2 6 ' l o g t a n 2 7 ° 2 7 ' l o g t a n 2 7 ° 2 6 ' l o g c o t 7 2 ° 1 5 ' l o g c o t 7 2 ° 1 6 ' l o g c o t 7 2 ° 1 5 ' l o g c o t 3 6 ° 1 8 ' l o g c o t 3 6 ° 1 9 ' l o g c o t 3 6 ° 1 8 ' l o g c o t 5 1 ° 1 7 ' l o g c o t 5 1 ° 1 8 ' . l o g c o t 5 1 ° 1 7 = 9 . 4 6 6 3 4 8 3 . = 9 . 4 6 5 9 3 5 3 ; 1 2 " . = 9 . 6 4 8 0 0 3 8 . = 9 . 6 4 8 2 5 8 2 ; 1 2 " . = 9 . 9 7 9 3 7 9 6 , = 9 . 9 7 9 3 3 9 8 ; 2 5 " . 2 . = 9 . 5 9 0 5 6 1 7 , = 9 . 5 9 0 9 3 5 1 ; 1 2 " . = 9 . 7 1 5 2 4 1 9 , . = 9 . 7 1 5 5 5 0 8 ; 4 2 " . = 9 . 5 0 5 2 8 9 1 , = 9 . 5 0 4 8 5 3 8 ; 3 5 " . = 1 0 . 1 3 3 9 6 5 0 , = 1 0 . 1 3 3 7 0 0 3 ; 2 0 " . = 9 . 9 0 3 9 7 3 3 , = 9 . 9 0 3 7 1 4 4 ; ' 3 2 " . 9 . 4 6 6 0 1 7 9 . 9 . 6 4 8 0 5 4 7 . 9 . 9 7 9 3 6 2 9 . 9 . 5 9 0 6 3 6 4 . 9 . 7 1 5 4 5 8 1 . 9 . 5 0 5 0 3 5 2 . 1 0 . 1 3 3 8 7 6 8 . 9 . 9 0 3 8 3 5 2 . 1 2 4 P L A N E T R I G O N O M E T R Y . 7 1 . G i v e n l o g s i n 1 6 ° 1 9 ' = 9 . 4 4 8 6 2 2 7 , l o g s i n 1 6 ° 2 0 ' = 9 . 4 4 9 0 5 4 0 ; f i n d t h e a n g l e w h o s e l o g s i n e i s 9 . 4 4 8 8 1 0 5 . A n s . 1 6 ° 1 9 ' 2 6 " . 7 2 . G i v e n l o g s i n 6 ° 5 3 ' = 9 . 0 7 8 6 3 1 0 , l o g s i n 6 ° 5 3 ' 1 0 " = 9 . 0 7 8 8 0 5 4 ; f i n d t h e a n g l e w h o s e l o g s i n e i s 9 . 0 7 8 7 7 4 3 . 6 ° 5 3 ' 8 " . 7 3 . G i v e n l o g c o s 2 2 ° 2 8 ' 2 0 " = 9 . 9 6 5 7 0 2 5 , l o g c o s 2 2 ° 2 8 ' 1 0 " = 9 . 9 6 5 7 1 1 2 ; f i n d t h e a n g l e w h o s e l o g c o s i n e i s 9 . 9 6 5 7 0 5 6 . 2 2 ° 2 8 ' 1 6 " . I n t h e f o l l o w i n g e x a m p l e s t h e t a b l e s a r e t o b e u s e d : 7 4 . F i n d l o g t a n 5 5 ° 3 7 ' 5 3 " . A n s . 1 0 . 1 6 5 0 0 1 1 . 7 5 -F i n d l o g s i n 7 3 ° 2 0 ' 1 5 " . 7 . 9 . 9 8 1 3 7 0 7 . 7 6 . F i n d l o g c o s 5 5 ° 1 1 ' 1 2 " . 9 . 7 5 6 5 6 3 6 . 7 7 . F i n d l o g t a n 1 6 ° 0 ' 2 7 " . 9 . 4 5 7 7 1 0 9 . 7 8 . F i n d l o g s e c 1 6 ° 0 ' 2 7 " . 1 0 . 0 1 7 1 7 4 7 . 7 9 . F i n d t h e a n g l e w h o s e l o g c o s i n e i s 9 . 9 7 1 3 3 8 3 . A n s . 2 0 ° 3 5 ' 1 6 " . 8 0 . F i n d t h e a n g l e w h o s e l o g c o s i n e i s 9 . 9 1 6 5 6 4 6 . A n s . 3 4 ° 2 3 ' 2 5 " . 8 1 . F i n d l o g c o s 3 4 ° 2 4 ' 2 6 " . 9 . 9 1 6 4 7 6 2 . 8 2 . F i n d l o g c o s 3 7 ° 1 9 ' 4 7 " . 9 . 9 0 0 4 5 4 0 . 8 3 . F i n d l o g s i n 3 7 ° 1 9 ' 4 7 " . 9 . 7 8 2 7 5 9 9 . 8 4 . F i n d l o g t a n 3 7 ° 1 9 ' 4 7 " . 9 . 8 8 2 3 0 5 9 . 8 5 . F i n d l o g s i n 3 2 ° 1 8 ' 2 4 " . 6 . 9 . 7 2 7 9 0 9 6 . 8 6 . F i n d l o g c o s 3 2 ° 1 8 ' 2 4 " . 6 . 9 . 9 2 6 9 5 8 5 . 8 7 . F i u d l o g t a n 3 2 ° 1 8 ' 2 4 " . 6 . 9 . 8 0 0 9 5 1 1 . E X A M P L E S . P r o v e t h e f o l l o w i n g b y t h e u s e o f l o g a r i t h m s 8 8 . ( 7 - 0 1 4 ) ' -1 9 9 4 2 2 0 7 . ( 7 . 0 1 4 ) 3 + 1 g 9 x 0 0 0 0 8 0 7 5 = - 0 0 0 2 8 8 m ^ 8 0 - s - ^ 0 0 0 0 0 1 1 2 6 P L A N E T R I G O N O M E T R Y . C H A P T E R V . S O L U T I O N O F T R I G O N O M E T R I C E Q U A T I O N S . 8 2 . A T r i g o n o m e t r i c E q u a t i o n i s a n e q u a t i o n i n w h i c h t h e u n k n o w n q u a n t i t i e s i n v o l v e t r i g o n o m e t r i c f u n c t i o n s . T h e s o l u t i o n o f a t r i g o n o m e t r i c e q u a t i o n i s t h e p r o c e s s o f f i n d i n g t h e v a l u e s o f t h e u n k n o w n q u a n t i t y w h i c h s a t i s f y t h e e q u a t i o n . A s i n A l g e b r a , w e m a y h a v e t w o o r m o r e s i m u l t a n e o u s e q u a t i o n s , t h e n u m b e r o f a n g l e s i n v o l v e d b e i n g e q u a l t o t h e n u m b e r o f e q u a t i o n s . E X A M P L E S . 1 . S o l v e s i n 0 = - . T h i s i s a t r i g o n o m e t r i c e q u a t i o n . T o s o l v e i t w e m u s t f i n d s o m e a n g l e w h o s e s i n e i s - . W e k n o w t h a t s i n 3 0 ° = b 2 2 T h e r e f o r e , i f 3 0 ° b e p u t f o r 0 , t h e e q u a t i o n i s s a t i s f i e d . . - . 0 = 3 0 ° i s a s o l u t i o n o f t h e e q u a t i o n . . - . e = ? n r + ( - l ) " | ( A r t . 3 8 ) 2 . S o l v e c o s 0 + s e c 0 = - -2 T h e u s u a l m e t h o d o f s o l u t i o n i s t o e x p r e s s a l l t h e f u n c t i o n s i n t e r m s o f o n e o f t h e m . T h u s , w e p u t f o r s e c 0 , a n d g e t c O S 0 c o s 0 + - ] - = ^ . c o s 0 2 T R I G O N O M E T R I C E Q U A T I O N S . 1 2 7 T h i s i s a n e q u a t i o n i n w h i c h 6 , a n d t h e r e f o r e c o s 0 , i s u n k n o w n . W e p r o c e e d t o s o l v e t h e e q u a t i o n a l g e b r a i c a l l y j u s t a s w e s h o u l d i f x o c c u p i e d t h e p l a c e o f c o s 0 , t h u s : c o s 2 0 - - c o s 0 = - l . 2 . - . c O s 0 = f ± ? 4 4 = 2 o r - -2 T h e v a l u e 2 i s i n a d m i s s i b l e , f o r t h e r e i s n o a n g l e w h o s e c o s i n e i s n u m e r i c a l l y g r e a t e r t h a n 1 ( A r t . 2 1 ) . . - . c O S 0 = . 2 B u t c o s 6 0 ° = 2 . - . c o s 0 = c o s 6 0 ° . T h e r e f o r e o n e v a l u e o f 6 w h i c h s a t i s f i e s t h e e q u a t i o n i s 6 0 ° . 3 . S o l v e c o s e c 6 -c o t J 0 + 1 = 0 . W e h a v e c o s e c 6 -( c o s e c 2 0 -1 ) + 1 = 0 . . ( A r t . 2 3 ) c o s e c 2 0 — c o s e c 6 = 2 . . - . c o s e c 0 = - ± -= 2 o r -1 . B u t c o s e c 3 0 ° = 2 . . - . c o s e c 6 = c o s e c 3 0 ° . T h e r e f o r e 3 0 ° i s o n e v a l u e o f 6 w h i c h s a t i s f i e s t h e e q u a t i o n . F i n d a v a l u e o f 6 w h i c h w i l l s a t i s f y t h e f o l l o w i n g e q u a t i o n s : 4 . c o s 6 = c o s 2 6 . - A n s . 3 t t . 5 . 2 c o s 0 = s e c 0 . 4 5 ° . 1 2 8 P L A N E T R I G O N O M E T R Y . 6 . 4 s i n 0 — 3 c o s e c 0 = 0 . A n s . 6 0 ° . 7 . 4 c o s 0 = 3 s e c 6 . 3 0 ° . 8 . 3 s i n 0 - 2 c o s 2 0 = 0 . 3 0 ° . 9 . V 2 s i n 0 = t a n 0 . 0 ° o r 4 5 ° . 1 0 . t a n 0 = 3 c o t 0 . 6 0 ° . 1 1 . t a n 0 + 3 c o t 0 = 4 . 4 5 ° . i T 2 i T 1 2 . c o s 6 + c o s 3 0 + c o s 5 6 = 0 . 2 , T ' 1 3 . s i n ( 0 -< £ ) = £ , c o s ( 0 + < £ ) = 0 . 0 = 6 0 ° , < j > = 3 0 ° . 8 3 . S o l v e t h e e q u a t i o n s ( 1 ) ( 2 ) w h e r e a a n d b a r e g i v e n , a n d t h e v a l u e s o f m a n d a r e r e q u i r e d . D i v i d i n g ( 1 ) b y ( 2 ) , w e g e t t a n d > = - , w h i c h g i v e s t w o v a l u e s o f < f > , d i f f e r i n g b y 1 8 0 ° , a n d t h e r e f o r e t w o v a l u e s o f m a l s o f r o m e i t h e r o f t h e e q u a t i o n s s i n < j > c o s < j > T h e t w o v a l u e s o f m w i l l b e e q u a l n u m e r i c a l l y w i t h o p p o s i t e s i g n s . I n p r a c t i c e , m i s a l m o s t a l w a y s p o s i t i v e b y t h e c o n d i t i o n s o f t h e p r o b l e m . A c c o r d i n g l y , s i n < j > h a s t h e s i g n o f a , a n d c o s < f > t h e s i g n o f b , a n d h e n c e < f > m u s t b e t a k e n i n t h e q u a d r a n t d e n o t e d b y t h e s e s i g n s . T h e s e c a s e s m a y b e c o n s i d e r e d a s f o l l o w s : ( 1 ) S i n < j > a n d c o s < f > b o t h p o s i t i v e . T h i s r e q u i r e s t h a t t h e a n g l e < j > b e t a k e n i n t h e f i r s t q u a d r a n t , b e c a u s e s i n < j > a n d c o s < j > a r e b o t h p o s i t i v e i n n o o t h e r q u a d r a n t . T R I G O N O M E T R I C E Q U A T I O N S . 1 2 9 ( 2 ) S i n < j > p o s i t i v e a n d c o s n e g a t i v e . T h i s r e q u i r e s t h a t < f > b e t a k e n i n t h e s e c o n d q u a d r a n t , b e c a u s e o n l y i n t h i s q u a d r a n t i s s i n < f > p o s i t i v e a n d c o s < £ n e g a t i v e f o r t h e s a m e a n g l e . ( 3 ) S i n < j > a n d c o s < j > b o t h n e g a t i v e . T h i s r e q u i r e s t h a t < f > b e t a k e n i n t h e t h i r d q u a d r a n t , b e c a u s e o n l y i n t h i s q u a d r a n t a r e s i n < f > a n d c o s < j > b o t h n e g a t i v e f o r t h e s a m e a n g l e . ( 4 ) S i n < j > n e g a t i v e a n d c o s < f > p o s i t i v e . T h i s r e q u i r e s t h a t < f > b e t a k e n i n t h e f o u r t h q u a d r a n t , b e c a u s e o n l y i n t h i s q u a d r a n t i s s i n n e g a t i v e a n d c o s < j > p o s i t i v e f o r t h e s a m e a n g l e . E x . 1 . S o l v e t h e e q u a t i o n s m s i n < f > = 3 3 2 . 7 6 , a n d m c o s < j > = 2 9 0 . 0 8 , f o r m a n d < j > . l o g m s i n < £ = 2 . 5 2 2 1 3 l o g m c o s = 2 . 4 6 2 5 2 l o g t a n < f , = 0 . 0 5 9 6 1 . - . = 4 8 ° 5 5 ' . 2 . l o g m s i n = 2 . 5 2 2 1 3 l o g s i n < f > = 9 . S 7 7 2 5 l o g m = 2 . 6 4 4 8 8 . - . m = 4 4 1 . 4 5 . E x . 2 . S o l v e m s i n < f > = — 7 2 . 6 3 1 , a n d m c o s < £ = 3 8 . 4 1 2 . A n s . < t > = 1 1 7 ° 5 2 ' . 3 , m = -8 2 . 1 6 4 . 8 4 . S o l v e t h e e q u a t i o n a s i n < f > + b c o s < £ = c ( 1 ) o , 6 , a n d c b e i n g g i v e n , a n d < f > r e q u i r e d . F i n d i n t h e t a b l e s t h e a n g l e w h o s e t a n g e n t i s -; l e t i t b e B . T h e n _ = t a n B , a n d ( 1 ) b e c o m e s a a ( s i n t f > + t a n B c o s < £ ) = c ; o r f l / s i n c o s B + c o s < f r s i n g \ _ , ^ c o s / ? / , o r s i n ( < 4 + / 3 ) = - c o s / J = - s i n f i . . . . ( 2 ) a 6 1 3 0 P L A N E T R I G O N O M E T R Y . T h e r e w i l l b e t w o s o l u t i o n s f r o m t h e t w o v a l u e s o f < j > + / 3 g i v e n i n ( 2 ) . F i n d f r o m t h e t a b l e s t h e v a l u e o f c o s / ? . N e x t f i n d f r o m t h e t a b l e s t h e m a g n i t u d e o f t h e a n g l e a w h o s e s i n e = -a c o s / ? , a n d w e g e t s i n ( < £ 4 - p ) = s i n a , . - . < £ + / 8 = n i r + ( - l ) n a . . ( A r t . 3 8 ) . - . < £ = - 0 + n i r + ( - l ) " a , w h e r e n i s z e r o o r a n y p o s i t i v e o r n e g a t i v e i n t e g e r . I n o r d e r t h a t t h e s o l u t i o n m a y b e p o s s i b l e , i t i s n e c e s s a r y t o h a v e - c o s / ? = , o r < 1 . a N o t e . — T h i s e x a m p l e m i g h t h a v e b e e n s o l v e d b y s q u a r i n g b o t h s i d e s o f t h e e q u a t i o n ; b u t i n s o l v i n g t r i g o n o m e t r i c e q u a t i o n s , i t i s i m p o r t a n t , i f p o s s i b l e , t o a v o i d s q u a r i n g b o t h s i d e s o f t h e e q u a t i o n . T h u s , s o l v e c o s 9 = k s i n 9 ( 3 ) I f w e s q u a r e b o t h s i d e s w e g e t c o s 2 9 = V s i n 9 = -c o s 2 9 ) . Z - 2 I . . - . c o s s 9 = — - — ; o r c o s 9 = ± — - — ( 4 1 1 + » ^ k N o w i f a b e t h e l e a s t a n g l e w h o s e c o s i n e = — , w e g e t f r o m ( 4 ) - J l + k 8 = n i r + a ( 5 ) B u t ( 3 ) m a y b e w r i t t e n c o t 9 = k . . - . 9 = n i r + a ( 6 ) ( 6 ) i s t h e c o m p l e t e s o l u t i o n o f t h e g i v e n e q u a t i o n ( 3 ) , w h i l e ( 5 ) i s t h e s o l u t i o n o f b o t h c o s 9 = k s i n 9 , a n d a l s o o f c o s 8 = — k s i n 9 . T h e r e f o r e b y s q u a r i n g b o t h m e m b e r s o f a n e q u a t i o n w e o b t a i n s o l u t i o n s w h i c h d o n o t b e l o n g t o t h e g i v e n e q u a t i o n . E X A M P L E S . 1 . S o l v e 0 . 7 4 6 6 8 9 8 s i n < j > -1 . 0 4 9 8 c o s < f > = -0 . 4 3 1 6 8 9 , w h e n < 1 8 0 ° . l o g b = 0 . 0 2 1 1 2 - l o g a = 1 . 8 7 3 1 4 l o g t a n / J = 0 . 1 4 7 9 8 -. - . / 3 = 1 2 5 ° 2 5 ' 2 0 " . T h e m i n u s s i g n i s w r i t t e n t h u s t o d e n o t e t h a t i t b e l o n g s t o t h e n a t u r a l n u m b e r a n d d o e s n o t a f f e c t t h e l o g a r i t h m . S o m e t i m e s t h e l e t t e r n i s w r i t t e n i n s t e a d o f t h e m i n u s s i g n , t o d e n o t e t h e s a m e t h i n g . T R I G O N O M E T R I C E Q U A T I O N S . 1 3 1 l o g s i n y 3 = 9 . 9 1 1 1 1 l o g c = 1 . 6 3 5 1 7 -c o l o g 6 = 9 . 9 7 8 8 8 -l o g s i n ( < £ + / 3 ) = 9 . 5 2 5 1 6 + . - . < f , + 1 3 = 1 9 ° 3 4 ' 4 0 " o r 1 6 0 ° 2 5 ' 2 0 " . . - . < t > = -1 0 5 ° 5 0 ' 4 0 " o r 3 5 ° 0 ' 0 " . 2 . S o l v e -2 3 . 8 s i n + 1 9 . 3 c o s i f , = 1 7 . 5 ( < £ < 1 8 0 ° ) . A n s . < f > = 4 ° 1 2 ' . 7 o r -1 0 6 ° 7 ' . 9 . 3 . S o l v e 2 s i n 6 + 2 c o s 0 = V 2 . ^ 4 n s . -- + n i r + ( - l ) " - -4 6 4 . " s i n 0 + V 3 c o s 0 = l . -- + n + ( -1 ) " - . 3 6 5 . " s i n 0 -c o s 0 = 1 . - + n i r + ( - l ) n | i r . 6 . " V 3 s i n 0 - c o s 0 = V 2 . £ t t + n i r + ( -l ) " ^ i r . 8 5 . S o l v e t h e e q u a t i o n s i n ( a + x ) = m s i n x ( 1 ) i n w h i c h a a n d m a r e g i v e n . F r o m ( 1 ) w e h a v e s i n ( a + x ) + s i n x _ m + 1 s i n ( a + x ) — s i n x m — 1 t a n / z + ^ j V . . ( A r t . 4 6 ) t a n ? . - . t a n ( a ; + £ a ) = 2 L ± l t a n £ . . . . ( 3 ) m — 1 2 w h i c h d e t e r m i n e s x + 1 « , a n d t h e r e f o r e a ; . I f w e i n t r o d u c e a n a u x i l i a r y a n g l e , t h e c a l c u l a t i o n q £ e q u a t i o n ( 3 ) i s f a c i l i t a t e d . 1 3 2 P L A N E T R I G O N O M E T R Y . T h u s , l e t m = t a n < £ ; t h e n w e h a v e b y [ ( 1 4 ) o f A r t . 6 1 ] m + 1 t a n d > + 1 t / , , ^ — = y = c o t ( < A — 4 o ) , m -1 t a n < £ - l K ^ " w h i c h i n ( 3 ) g i v e s t a n ^ a ; + 1 ^ = c o t ( < £ -4 5 ° ) t a n £ a . ( 4 ) T h i s , w i t h t a n < f > = m , g i v e s t h e l o g a r i t h m i c s o l u t i o n . T h e l o g a r i t h m i c s o l u t i o n o f t h e e q u a t i o n s i n ( « — x ) = m s i n a ; i s f o u n d i n t h e s a m e m a n n e r t o b e t a n < f > = m , a n d t a n — = c o t ( < j > + 4 5 ° ) t a n ^ > w h i c h t h e s t u d e n t m a y s h o w . E x a m p l e . — S o l v e s i n ( 1 0 6 ° + x ) = -1 . 2 6 3 s i n a ; ( a ; < 1 8 0 ° ) . l o g t a n = l o g m = l o g ( -1 . 2 6 3 ) = 0 . 1 0 1 4 0 -. . - . = 1 2 8 ° 2 2 ' . 3 . ^ . - 4 5 ° = 8 3 ° 2 2 ' . 3 ; l o g c o t ( < £ -4 5 ° ) = 9 . 0 6 5 2 3 $ a = 5 3 ° O ' . O , l o g t a n £ « = 1 0 . 1 2 2 8 9 l o g t a n ^ a ; + ^ = 9 . 1 8 8 1 2 a ; + - = 8 ° 4 6 ' . 0 o r 1 8 8 ° 4 6 ' . 2 . - . x = - 4 4 ° 1 4 ' o r 1 3 5 ° 4 6 ' . 8 6 . S o l v e t h e e q u a t i o n t a n ( r c + x ) = m t & n x ( 1 ) i n w h i c h a a n d m a r e g i v e n . T R I G O N O M E T R I C E Q U A T I O N S . 1 3 3 F r o m ( 1 ) w e h a v e t a n ( « + x ) + t a n x _ m + 1 t a n ( a + x ) — t a n x m — 1 s i n ( a + 2 a ; ) [ ( 2 1 ) o f A r t . 6 1 ] . - . s i n ( a + 2 a ; ) = ? ^ - ^ s i n a ( 2 ) m — 1 = c o t ( < £ - 4 5 ° ) s i n a . ( A r t . 8 5 ) w h e r e t a n < j > = m . E x a m p l e . — S o l v e t a n ( 2 3 ° 1 6 ' + x ) = . 2 9 6 t a n a ; , l o g t a n = l o g m = l o g ( . 2 9 6 ) = 1 . 4 7 1 2 9 . . - . = 1 6 ° 2 9 ' . 3 . < j > -4 5 ° = -2 8 ° 3 0 ' . 7 ; l o g c o t ( < £ -4 5 ° ) = 1 0 . 2 6 5 0 2 -« = 2 3 ° 1 6 ' . l , l o g s i n a = 9 . 5 9 6 6 1 l o g s i n ( a + 2 x ) = 9 . 8 6 1 6 3 -e t + 2 x = 2 2 6 ° 3 8 ' . 9 o r 3 1 3 0 2 1 ' . 1 . . - . x = 1 0 1 ° 4 1 ' . 5 o r 1 4 5 ° 2 ' . 6 . 8 7 . S o l v e t h e e q u a t i o n t a n ( a + a ; ) , t a n a ; = m ( 1 ) i n w h i c h a a n d m a r e g i v e n . F r o m ( 1 ) w e h a v e 1 + t a n ( a + x ) t a n x _ 1 + m 1 — t a n ( « + a ; ) t a n x 1 — m c o s a c o s ( a + 2 x ) . : c o s ( a + 2 a ; ) = c o s a 1 + m ( E x . 4 o f A r t . 4 7 ) = t a n ( 4 5 ° -< £ ) c o s « [ ( 1 6 ) o f A r t . 6 1 ] w h e r e t a n < j > = m . 1 3 4 P L A N E T R I G O N O M E T R Y . E x a m p l e . — S o l v e t a n ( 6 5 ° + ) t a n a ; = 1 . 5 1 9 0 ( « < 1 8 0 ° ) . l o g t a n $ = l o g t o = 0 . 1 8 1 7 3 . . - . < £ = 5 0 ° 3 9 ' 9 " ; 4 5 ° -< t > = -1 1 ° 3 9 ' 9 " ; l o g t a n ( 4 5 ° - < £ ) = 9 . 3 1 4 3 4 -a = 6 5 ° 0 ' 0 " ; l o g c o s « = 9 . 6 2 5 9 5 l o g c o s ( a + 2 x ) = 8 . 9 4 0 2 9 -a + 2 x = 9 5 ° o r 2 6 5 ° . . - . 2 x = 3 0 ° o r 2 0 0 ° . . - . x = 1 5 ° o r 1 0 0 ° . 8 8 . S o l v e t h e e q u a t i o n s t o s i n ( 6 + x ) = a ( 1 ) m s i n ( < j > + x ) = b ( 2 ) f o r m a n d x , t h e o t h e r f o u r q u a n t i t i e s , 6 , < f > , a , b , b e i n g k n o w n . E x p a n d i n g ( 1 ) a n d ( 2 ) b y ( A r t . 4 4 ) , w e g e t t o s i n 6 c o s x + t o c o s 0 s i n x = a ( 3 ) t o s i n < £ c o s x + m c o s s i n a ; = 6 ( 4 ) M u l t i p l y i n g ( 3 ) b y s i n < f > a n d ( 4 ) b y s i n 6 , a n d s u b t r a c t i n g t h e l a t t e r f r o m t h e f o r m e r , w e h a v e t o s i n x ( s i n < j > c o s 6 — c o s < j > s i n 6 ) = a s i n — 6 s i n 0 . a s i n < A — 6 s i n 0 , r x . - . t o s i n a ; = — ~ ( o ) s i n ( < £ - 0 ) v ' T o f i n d t h e v a l u e o f t o c o s x , m u l t i p l y ( 3 ) a n d ( 4 ) b y c o s < j > a n d c o s 6 , r e s p e c t i v e l y , a n d s u b t r a c t t h e f o r m e r f r o m t h e l a t t e r . T h u s m c o s a ; ( s i n < f > c o s 6 — c o s < j > s i n 6 ) = b c o s 6 — a c o s < £ . 6 c o s 6 — a c o s < / > . - . m c o s x = ~ ( 6 ) s i n ( < £ - 0 ) v ' H a v i n g o b t a i n e d t h e v a l u e s o f m s i n x a n d t o c o s x f r o m ( 5 ) a n d ( 6 ) , t o a n d a ; c a n b e c a l c u l a t e d b y A r t . 8 3 . T R I G O N O M E T R I C E Q U A T I O N S . 1 3 5 E X A M P L E S . 1 . S o l v e t o c o s ( 6 + x ) = a , a n d t o s i n ( + a ; ) = b , f o r m s i n x a n d t o c o s x . . . b i ; o s 6 — a s m t b A n s . m s i n x = ~ , c o s ( 0 -< f > ) b s i n 0 + a c o s < £ t o c o s a ; = 1 - ~ . c o s ( ( ? - < £ ) 2 . S o l v e m c o s ( 6 + x ) = a , a n d m c o s ( < f > — x ) = b , f o r m s i n a ; a n d t o c o s x . A . b c o s 6 — a c o s < & y l n s . m s i n a ; = ~ , s i n ( 0 + < » ) 6 s i n 0 + a s i n < A t o c o s x — s m ( 0 4 - 0 ) 8 9 . S o l v e t h e e q u a t i o n x c o s a + y s m a = m ( 1 ) x s i n a — y c o s a = n ( 2 ) f o r x a n d y . M u l t i p l y i n g ( 1 ) b y c o s r a a n d ( 2 ) b y s i n a , a n d a d d i n g , w e g e t a ; = t o c o s a + n s i n a . T o f i n d t h e v a l u e o f y , m u l t i p l y ( 1 ) b y s i n a a n d ( 2 ) b y c o s a , a n d s u b t r a c t t h e l a t t e r f r o m t h e f o r m e r . T h u s y = m s i n a — n c o s a . E x a m p l e . — S o l v e x s i n a + y c o s « = a , x c o s a — y s i n « = b . 9 0 . T o a d a p t F o r m u l a e t o L o g a r i t h m i c C o m p u t a t i o n . — A s c a l c u l a t i o n s a r e p e r f o r m e d p r i n c i p a l l y b y m e a n s o f l o g a r i t h m s , a n d a s w e a r e n o t a b l e b y l o g a r i t h m s d i r e c t l y t o a d d a n d s u b t r a c t q u a n t i t i e s , i t b e c o m e s n e c e s s a r y t o k n o w h o w t o t r a n s f o r m s u m s a n d d i f f e r e n c e s i n t o p r o d u c t s A d d i t i o n a n d S u b t r a c t i o n T a b l e s a r e p u b l i s h e d , b y m e a n s o f w b i c h t h e l o g a r i t h m o f t h e s u m o r d i f f e r e n c e o f t w o n u m b e r s m a y b e o b t a i n e d . ( S e e T a f e l n d e r A d d i t i o n s , u n d S u b t r a c t i o n s , L o g a r i t h m e n f u r s i e b e n S t e l l e n , v o n J . Z e c b , B e r l i n . ) 1 3 6 P L A N E T R I G O N O M E T R Y . a n d q u o t i e n t s . A n e x p r e s s i o n i n t h e f o r m o f a p r o d u c t o r q u o t i e n t i s s a i d t o b e a d a p t e d t o l o g a r i t h m i c c o m p u t a t i o n . A n a n g l e , i n t r o d u c e d i n t o a n e x p r e s s i o n i n o r d e r t o a d a p t i t t o l o g a r i t h m i c c o m p u t a t i o n , i s c a l l e d a S u b s i d i a r y A n g l e . S u c h a n a n g l e w a s i n t r o d u c e d i n t o e a c h o f t h e A r t s . 8 4 , 8 5 , 8 6 , a n d 8 7 . T h e f o l l o w i n g a r e f u r t h e r e x a m p l e s o f t h e u s e o f s u b s i d i a r y a n g l e s : 1 . T r a n s f o r m a c o s 6 ± b s i n 6 i n t o a p r o d u c t , s o a s t o a d a p t i t t o l o g a r i t h m i c c o m p u t a t i o n . P u t - = t a n < f > ; t h u s a a c o s 6 ± b s i n 0 = a ( c o s 6 ± -s i n 6 ] \ « / = a ( c o s 6 ± t a n < f > s i n 6 ) = - ^ — c o s ( 0 T < £ ) . c o s < f > 2 . S i m i l a r l y , a s i n ^ ± 6 c o s 0 = — 2 -s i n ( 0 ± < £ ) . c o s < £ 3 . T r a n s f o r m a ± b i n t o a p r o d u c t , a + b — a { \ + - \ = a ( 1 + t a n 2 < £ ) = a s e c 2 < £ , i f - = t a n 2 < £ . a i — b = a ^ l — = a c o s 2 < f > , i f - = s i n > . a T h e f u n d a m e n t a l f o r m u l a e c o s ( a ; ± y ) a n d s i n ( a r ± y ) ( A r t . 4 2 ) a f f o r d e x a m p l e s o f o n e t e r m e q u a l t o t h e s u m o r d i f f e r e n c e o f t w o t e r m s ; h e n c e w e m a y t r a n s f o r m a n e x p r e s s i o n a c o s 0 J r b s i n 0 i n t o a n e q u i v a l e n t p r o d u c t , b y c o n f o r m i n g i t t o t h e f o r m u l a ? j u s t m e n t i o n e d . T h u s , c o m p a r i n g t h e i d e n t i t y , m c o s < f > c o s 9 ± m s i n s i n 9 = m c o s ( < p t 9 ) o r m c o s ( 9 T < f > ) , w i t h a c o s 9 ± 6 s i n 9 , w e w i l l h a v e a c o s 0 ± 6 s i n 0 = m c o s ( 0 T < £ ) i f w e a s s u m e a = m c o s < f > a n d 6 = m s i n < t > ; i . e . ( A r t . 8 3 ) , i f t a n < f l > = - a n d m = — — — -a s a b o v e . S e e A r t . 8 4 . » c 0 s s i n A D A P T A T I O N T O L O G A R I T H M I C C O M P U T A T I O N . . : l o g ( a + f t ) = l o g a + 2 1 o g s e c < £ ; a n d l o g ( a — b ) = l o g a + 2 l o g c o s < f > . 4 . T r a n s f o r m 1 2 3 9 . 3 s i n 6 — 7 2 4 . 6 c o s 0 t o a p r o d u c t . l o g b = l o g ( -7 2 4 . 6 ) = 2 . 8 6 0 1 0 -l o g a = l o g ( 1 2 3 9 . 3 ) = 3 . 0 9 3 1 8 l o g t a n < £ = 9 . 7 6 6 9 2 -. - . = -3 0 ° 1 8 ' . 8 . l o g a = 3 . 0 9 3 1 8 l o g c o s < £ = 9 . 9 3 6 1 5 l o g - ^ — = 3 . 1 5 7 0 3 c o s < £ 1 4 3 5 . 6 . c o s < £ . - . 1 2 3 9 . 3 s i n 6 -7 2 4 . 6 c o s 0 = 1 4 3 5 . 6 s i n ( 0 -3 0 ° 1 8 ' . 8 ) . 9 1 . S o l v e t h e e q u a t i o n s r c o s t j > c o s 6 = a r c o s < £ s i n 0 = 6 r s i n < £ = c f o r r , < f > , a n d 6 . D i v i d i n g ( 2 ) b y ( 1 ) , w e h a v e t a n 6 = - > a f r o m w h i c h w e o b t a i n 6 . F r o m ( 1 ) a n d ( 2 ) w e h a v e a h r c o s < b ; c o s 6 s i n ^ f r o m w h i c h w e o b t a i n r c o s < £ . F r o m ( 3 ) a n d ( 4 ) w e o b t a i n r a n d < j > ( A r t . 8 3 ) . 1 3 8 P L A N E T R I G O N O M E T R Y . E X A M P L E S . 1 . S o l v e r c o s c o s 6 = — 5 3 . 9 5 3 , r c o s < j > s i n 0 = 1 9 7 . 2 0 7 , r s i n < £ = -3 9 . 0 6 2 , f o r r , < f > , 6 . l o g b = 2 . 2 9 4 9 3 l o g o = 1 . 7 3 2 0 1 -. - . < £ = - 1 0 ° 4 9 ' . l o g r c o s < # > = 2 . 3 1 0 6 0 l o g c o s < # . = 9 . 9 9 2 2 1 l o g t a n 0 = 0 . 5 6 2 9 2 -. - . 0 = 1 0 5 ° 1 8 ' . 0 . l o g r = 2 . 3 1 8 3 9 . - . r = 2 0 8 . 1 6 . l o g s i n 6 = 9 . 9 8 4 3 3 l o g r c o s < £ = 2 . 3 1 0 6 0 l o g r s i n < f r = 1 . 5 9 1 7 5 -l o g t a n < £ = 9 . 2 8 1 1 5 -9 2 . T r i g o n o m e t r i c E l i m i n a t i o n . — S e v e r a l s i m u l t a n e o u s e q u a t i o n s m a y b e g i v e n , a s i n A l g e b r a , b y t h e c o m b i n a t i o n o f w h i c h c e r t a i n q u a n t i t i e s m a y b e e l i m i n a t e d , a n d a r e s u l t o b t a i n e d i n v o l v i n g t h e r e m a i n i n g q u a n t i t i e s . T r i g o n o m e t r i c e l i m i n a t i o n o c c u r s c h i e f l y i n t h e a p p l i c a t i o n o f T r i g o n o m e t r y t o t h e h i g h e r b r a n c h e s o f M a t h e m a t i c s , a s , f o r e x a m p l e , i n P h y s i c a l A s t r o n o m y , M e c h a n i c s , A n a l y t i c G e o m e t r y , e t c . A s n o s p e c i a l r u l e s c a n b e g i v e n , w e i l l u s t r a t e t h e p r o c e s s b y a f e w e x a m p l e s . E X A M P L E S . 1 . E l i m i n a t e < j > f r o m t h e e q u a t i o n s x = a c o s < f > , y = b s i n < j > . F r o m t h e g i v e n e q u a t i o n s w e h a v e w h i c h i n g i v e s s i n < £ , T R I G O N O M E T R I C E L I M I N A T I O N S . 2 . E l i m i n a t e < f > f r o m t h e e q u a t i o n s a c o s < j > + b s i n = c , b c o s ( f > + c s i n < j > = a . S o l v i n g t h e s e e q u a t i o n s f o r , s i n < f > a n d c o s < j > , w e h a v e b e — a 2 s i n < £ = e o s < £ = b 2 — a c c 2 — a b a c — b 2 , w h i c h i n c o s 2 + s i n 2 = 1 , g i v e s ( b e — a 2 ) 2 + ( c 2 — a b ) 2 = ( a c — b 2 ) 2 . 3 . E l i m i n a t e f r o m t h e e q u a t i o n s y c o s — x s i n = a c o s 2 < £ , y s i n + a ; c o s < £ = 2 a s i n 2 < £ . S o l v e f o r x a n d y , t h e n a d d a n d s u b t r a c t , a n d w e g e t x + y = a ( s i n ^ > + c o s < £ ) ( 1 + s i n 2 < £ ) , x — y = a ( s i n < £ — c o s < £ ) ( 1 — s i n 2 < £ ) . . - . ( x + 2 / ) 2 = a 2 ( l + s i n 2 ^ ) 3 , . ( x - 2 / ) 2 = a 2 ( l - s i n 2 < £ ) 3 . . . . ( x + y ) + ( x - y ) = 2 < $ . 4 . E l i m i n a t e a a n d f r o m t h e e q u a t i o n s a = s i n a c o s / 3 s i n 0 + c o s a c o s 0 . . . 6 = s i n a c o s / J c o s 0 — c o s a s i n 0 . . . c = s i n « s i n / J s i n 0 S q u a r i n g ( 1 ) a n d ( 2 ) , a n d a d d i n g , w e g e t a 2 + b 2 = s i n 2 « c o s 2 f 3 + c o s 2 a r 2 1 4 0 P L A N E T R I G O N O M E T R Y . A d d i n g ( 4 ) a n d ( 5 ) , w e h a v e 5 . E l i m i n a t e f r o m t h e e q u a t i o n s o s i n < f > + b c o s < f > = c > a c o s — b s i n = d . A n s . a 2 + f t 2 = c 2 + d 2 . 6 . E l i m i n a t e 6 f r o m t h e e q u a t i o n s m = c o s e c 6 — s i n 6 , n = s e c 6 — c o s 6 . m W ( w ' + » ' ) = l . 7 . E l i m i n a t e 0 a n d f r o m t h e e q u a t i o n s s i n 6 + s i n < f > = a , c o s 0 + c o s < t > = b , c o s ( 0 - < £ ) = c . a 2 + 6 2 - 2 c = 2 . 8 . E l i m i n a t e x a n d y f r o m t h e e q u a t i o n s t a n x 4 - t a n y = a , c o t x + c o t y = b , x + y = c . c o t c = - — — a 6 9 . E l i m i n a t e < £ f r o m t h e e q u a t i o n s x = c o s 2 + c o s < f > , y = s i n 2 < j > + s i n < £ . ^ n s . 2 x = ( x 2 + y 2 ) 2 - 3 ( x + y 2 ) . E X A M P L E S . S o l v e t h e f o l l o w i n g e q u a t i o n s : 1 . t a n 6 + c o t 6 = 2 . A n s . 4 5 ° . 2 . 2 s i n 2 0 + V 2 c o s 6 = 2 . 9 0 ° , o r 4 5 ° . 3 . 3 t a n 2 0 - 4 s i n 2 0 = l . 4 5 ° . 4 . 2 s i n 0 + V 2 s i n 0 = 2 . 4 5 ° . E X A M P L E S . 1 4 1 5 . c o s 2 0 - V 3 c o s 0 + f = 0 . ^ 4 n s . 3 0 ° . 6 . s i n 5 0 = 1 6 s i n 5 0 . n v , o r n i r ± - -6 7 . s i n 9 0 — s i n 0 = s i n 4 f t \ m r , o r f n - ± ~1 5 8 . 2 s i n 0 = t a n f t n j r , o r 2 r i i r ± - -9 . 6 c o t 2 0 — 4 c o s 2 0 = 1 . 3 1 0 . t a n 0 + t a n ( 0 -4 5 ° ) = 2 . 1 1 . c o s 0 + V 3 s i n 0 = V 2 . 2 m r ± - .3 1 2 . t a n ( 0 + 4 5 ° ) = l + s i n 2 0 . n i T O r 7 1 i T . 4 1 3 . ( c o t 0 - t a n 0 ) 2 ( 2 + V 3 ) = 4 ( 2 - V 3 ) . £ r ± £ . 2 4 1 4 c o s e c 6 c a t 6 = 2 V 3 . 2 n ^ ± ? .6 1 5 . c o s e c 0 4 - c o t 0 = V 3 \ 2 n i T + ^ - i T . 1 6 . s i n - = c o s e c 0 — c o t f t 2 M i r . 1 7 . s i n 5 0 c o s 3 6 = s i n 9 0 c o s 7 f t ^ + ( _ 1 ) | . 1 8 . s i n 2 0 + c o s 2 2 0 = J . n t t ± ^ , o r J l t t ± t ^ t t . 1 9 . V 3 s i n 0 — c o s 0 = V 2 . 2 0 . t a n 0 + c o t 0 = 4 . 2 1 . s i n ( 0 + < £ ) = c o s ( 0 - ) = 0 = T , 2 4 ' V 1 2 2 2 . S o l v e m s i n < f > = 1 . J 9 7 4 3 , a n d m c o s = 6 . 0 0 2 4 . A n a . ( < £ < 1 8 0 ° ) . 1 4 2 P L A N E T R I G O N O M E T R Y . 2 3 . S o l v e m s i n < £ = - 0 . 3 0 7 6 2 5 8 , a n d m c o s < £ = 0 . 4 2 7 8 7 3 5 . ( m p o s i t i v e . ) A n s . < ¿ = 3 2 4 ° 1 7 ' 6 " . 6 , m = 0 . 5 2 6 9 8 . 2 4 . S o l v e m s i n < ¿ > = 0 . 0 8 2 1 9 , a n d m c o s = 0 . 1 2 8 8 . 2 5 . S o l v e m s i n < f > = 1 9 4 . 6 8 3 , a n d m c o s = 8 4 6 0 . 7 . 2 6 . I f a s i n 0 + b c o s 0 = c , a n d a c o s 0 + 6 s i n 0 = c s i n 9 c o s 0 , s h o w t h a t s i n 2 0 ( c ' 2 — a 2 — 6 2 ) = 2 a 6 . S o l v e t h e f o l l o w i n g e q u a t i o n s : 2 7 . V 2 s i n 0 + V 2 c o s 0 = V 3 . A n s . -1 + r n r + ( -1 ) " | -2 8 . 2 s i n x + 5 c o s x = 2 . S u g . [ 2 . 5 = t a n 6 8 ° 1 2 ' ] . A n s . x = -6 8 ° 1 2 ' + n l 8 0 ° + ( -1 ) " ( 2 1 ° 4 8 ' ) . 2 9 . 3 c o s x -8 s i n x = 3 . S u g . [ 2 . 6 = t a n 6 9 ° 2 6 ' 3 0 " ] . A n s . a ; = -6 9 ° 2 6 ' 3 0 " + 2 n 1 8 0 ° ± ( 6 9 ° 2 6 ' 3 0 " ) . 3 0 . 4 s i n x — 1 5 c o s x = A . S u g . [ 3 . 7 5 = t a n 7 5 ° 4 ' ] . A n s . x = 7 5 ° 4 ' + n 1 8 0 ° + ( -1 ) " ( 1 4 ° 5 6 ' ) . 3 1 . c o s ( « - + - a ; ) = s i n ( a + x ) + V 2 c o s ¡ 3 . A n s . x = — a — J + 2 r n r ± / 3 . 3 2 . c o s 0 + c o s 3 0 + c o s 5 0 = 0 . . I n s . £ ( 2 n + l ) i r , o r ^ ( 3 ? i ± 1 ) t t . 3 3 . s i n 5 0 = s i n 3 0 + s i n 0 = 3 - 4 s i n 2 0 . A n s . r n r ± - , o r | ( 2 n + 1 ) " " -o 3 4 . 2 s i n ! 3 0 + s i n 2 6 0 = 2 . 3 5 . a ( c o s 2 0 - l ) + 2 6 ( c o s 0 + l ) = 0 . . ¿ I n s . ( 2 n + l ) i r , o r c o s - 1 - ^ ^ E X A M P L E S . 1 4 3 3 6 . S o l v e m s i n ( 0 + x ) = a c o s f 3 , a n d m c o s ( 6 — x ) = a s i n / J , f o r m s i n a ; a n d m c o s x . ( A r t . 6 7 . ) A n s . m s i n a ; = = a R o s ( / ? j l ! ? 2 c o s 2 0 a s i n ( / 3 — 0 ) m c o s a ; = ^ c o s 2 0 3 7 . S o l v e m c o s ( 0 + < £ ) = 3 . 7 9 , a n d m c o s ( 0 — < \ > ) = 2 . 0 6 , f o r m a n d 0 , w h e n < £ = 3 1 ° 2 7 ' . 4 . ( A r t . 6 7 . ) 3 8 . S o l v e r c o s < £ c o s 0 = 1 . 2 7 1 , r c o s s i n 0 = — 0 . 9 8 1 , r s i n = 0 . 8 9 0 , f o r r , < £ , 0 . ( A r t . 7 0 . ) 3 9 . S o l v e r c o s < j > c o s 0 = — 2 , r c o s < f > s i n 0 = + 3 , r s i n = — 4 , f o r r , < £ , 0 . 4 0 . S o l v e r s i n < j > s i n 0 = 1 9 . 7 6 5 , r s i n c o s 0 = — 7 . 1 9 2 , r c o s < t > = 1 2 . 1 2 4 , f o r r , < j > , 0 . 4 1 . S o l v e c o s ( 2 a ; + 3 y ) = ^ , c o s ( 3 a ; + 2 y ) = | V 3 . , 4 n s . X = | » i r ± 1 ^ r i r ± ^ i r , y = | n ± \ i r ± ^ j r . 4 2 . S o l v e c o s 3 0 + c o s 5 0 + V 2 ( c o s 0 + s i n 0 ) c o s 0 = O . ^ n s . 4 0 ± 0 = 2 n j r ± f j r , o r £ ( 2 n + l ) j r . 4 3 . S o l v e c o s 3 0 + s i n 3 0 = c o s 0 + s i n 0 . . 4 n s . s i n 0 = 0 , o r t a n 0 = — 1 ± V 2 . 4 4 . S o l v e 3 s i n 0 + c o s 0 = 2 x , s i n 0 + 2 c o s 0 = x . A n s . 0 = 7 1 ° 3 4 ' , s e = $ V l O . 1 4 4 P L A N E T R I G O N O M E T R Y . 4 5 . S o l v e 1 . 2 6 8 s i n < f > = 0 . 9 4 8 + m s i n ( 2 5 ° 2 7 ' . 2 ) , 1 . 2 6 8 c o s < t > = 0 . 2 8 1 + m c o s ( 2 5 ° 2 7 ' . 2 ) . A n s . 4 > = 6 0 ° 5 3 ' . 8 , m = 0 . 3 7 2 . 4 6 . T r a n s f o r m a f + + 2 4 — 2 ? / V — 2 z l i . - 2 — 2 a r f y 2 i n t o a p r o d u c t . A n s . — ( x + y + z ) ( y + z — x ) ( z + x — y ) ( x - { - y — z ) . 4 7 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s m s i n 2 6 = n s i n 0 , p c o s 2 6 = q c o s f t . / I n s . m 2 + p 2 = n 2 + ? 2 -4 8 . E l i m i n a t e 0 a n d f r o m t h e e q u a t i o n s x = a c o s ™0 c o s ™ < j > , y = b c o s ™0 s i n ™ < £ , a = c s i n ™ f t ^ n s -© - + ( ! ) " + ( ^ = 1 -4 9 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s a s i n 0 + 6 c o s 6 = h , a c o s 0 — 6 s i n 0 = k . A n s . a 2 + 6 2 = h 2 + k 2 . 5 0 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s a t a n 0 + 6 s e c 0 = c , a ' c o t 0 + 6 ' c o s e c 0 = c ' . y l n s . ( a ' 6 + c 6 ' ) 2 + ( a 6 ' + c ' 6 ) 2 = ( c c ' -a a ' ) 2 . 5 1 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s x = 2 a c o s 0 c o s 2 0 — a c o s 0 , y ~ 2 a c o s 0 s i n 2 0 — a s i n 0 . . 4 n s . a ^ + y 2 = a 2 . 5 2 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s a ; = a c o s 0 + 6 c o s 2 0 , a n d y = a s i n 0 + 6 s i n 2 0 . . 4 n s . a 2 [ ( x + 6 ) 2 + , y 2 ] = [ a ; 2 + y 2 -6 2 ] 2 . 5 3 . E l i m i n a t e a a n d / J f r o m t h e e q u a t i o n s 6 + c c o s a = u c o s ( a — 0 ) , 6 + c c o s / ? = m c o s ( £ — 0 ) , a — / ? = 2 < £ ; a n d s h o w t h a t « 2 — 2 m c c o s 0 + c 2 = 6 2 s e c 2 < j > . E X A M P L E S I N E L I M I N A T I O N . 1 4 5 5 4 . E l i m i n a t e 6 a n d < j > f r o m t h e e q u a t i o n s x c o s 6 + y s i n 6 = a , b s i n ( 6 + < £ ) = a s i n < £ , a ; c o s ( 0 + 2 < £ ) — y s i n ( 0 + 2 < £ ) = a . ^ 4 / i s . a r + 1 / 2 = a _ 5 5 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s x _ s e c 2 6 — c o s 2 f l a s e c 2 0 4 - c o s 2 0 , 2 5 = „ . ! - u £ V . b 2 = s e c 2 0 + c o s 2 6 . a J ^ 6 2 5 6 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s ( a + 6 ) t a n ( 0 -< f > ) = ( a — b ) t a n ( 6 + a c o s 2 < j > + 6 c o s 2 6 = c . A n s . 6 2 = c 2 + a 2 — 2 « c c o s 2 < £ . 5 7 . E l i m i n a t e 0 f r o m t h e e q u a t i o n s a a r ^ ~ , — 2 c o s 2 0 . s i n 2 0 1 z s i n t f — y c o s O = \ > x + y 2 , — t ~ + — y T V + y 2 , 4 n s . — — l . 5 8 . E l i m i n a t e 6 a n d f r o m t h e e q u a t i o n s a 2 c o s 2 0 — 6 2 c o s 2 < £ = c 2 , a c o s f l + 6 c o s < £ = r , a t a n 6 = b t a n < £ . A n s . r ' + c 2 ) 2 J _ [ ( r ' - c 2 ) 2 _ ' 5 9 . E l i m i n a t e < £ f r o m t h e e q u a t i o n s n s i n 6 — m c o s 0 = 2 m s i n < f > , n s i n 2 0 — m c o s 2 < f > = n . ^ 4 n s . ( n s i n 0 + m c o s 0 ) 2 = 2 m ( m + » ) . 6 0 . E l i m i n a t e a f r o m t h e e q u a t i o n s a ; t a n ( a — / ? ) = 2 / t a n ( « + / ? ) , ( x — c o s 2 a - + - ( x + y ) c o s 2 / J = z . A n s . z 2 + 4 a / = 2 z ( x + y ) c o s 2 / 3 . 1 4 6 P L A N E T R I G O N O M E T R Y . C H A P T E R V I . R E L A T I O N S B E T W E E N T H E S I D E S O F A T R I A N G L E A N D T H E F U N C T I O N S O F I T S A N G L E S . 9 3 . F o r m u l a e . — I n t h i s c h a p t e r w e s h a l l d e d u c e f o r m u l a e w h i c h e x p r e s s c e r t a i n r e l a t i o n s b e t w e e n t h e s i d e s o f a t r i a n g l e a n d t h e f u n c t i o n s o f i t s a n g l e s . T h e s e r e l a t i o n s w i l l b e a p p l i e d i n t h e n e x t c h a p t e r t o t h e s o l u t i o n o f t r i a n g l e s . O n e o f t h e p r i n c i p a l o b j e c t s o f T r i g o n o m e t r y , a s i t s n a m e i m p l i e s ( A r t . 1 ) , i s t o e s t a b l i s h c e r t a i n r e l a t i o n s b e t w e e n t h e s i d e s a n d a n g l e s o f t r i a n g l e s , s o t h a t w h e n s o m e o f t h e s e a r e k n o w n t h e r e s t m a y b e d e t e r m i n e d . R I G H T T R I A N G L E S . 9 4 . L e t A B C b e a t r i a n g l e , r i g h t - a n g l e d a t C . t h e a n g l e s o f t h e t r i a n g l e b y t h e l e t t e r s A , B , C , a n d t h e l e n g t h s o f t h e s i d e s r e s p e c t i v e l y o p p o s i t e t h e s e a n g l e s , b y t h e l e t t e r s a , b , c . T h e n w e h a v e ( A r t . 1 4 ) t h e f o l l o w i n g r e l a t i o n s : a = c s i n A = c c o s B : b = c s i n B = c c o s A : A 6 b t a n A = b c o t B a t a n B = a c o t A c = f t s e c A = a s e c B = 6 c o s e c B = a c o s e c A . ( 3 ) w h i c h m a y b e e x p r e s s e d i n t h e f o l l o w i n g g e n e r a l t h e o r e m s : T h e s t u d e n t m u s t r e m e m b e r t h a t a , b , c , a r e n u m b e r s e x p r e s s i n g t h e l e n g t h s o f t h e s i d e s i n t e r m s o f s o m e u n i t o f l e n g t h , s u c h a s a f o o t o r a m i l e . T h e u n i t m a y b e w h a t e v e r w e p l e a s e , b u t m u s t b e t h e s a m e f o r a l l t h e s i d e s . O B L I Q U E T R I A N G L E S . 1 4 7 I . I n a r i g h t t r i a n g l e e a c h s i d e i s e q u a l t o t h e p r o d u c t o f t h e h y p o t e n u s e i n t o t h e s i n e o f t h e o p p o s i t e a n g l e o r t h e c o s i n e o f t h e a d j a c e n t a n g l e . I I . I n a r i g h t t r i a n g l e e a c h s i d e i s e q u a l t o t h e p r o d u c t o f t h e o t h e r s i d e i n t o t h e t a n g e n t o f t h e a n g l e a d j a c e n t t o t h a t o t h e r s i d e , o r t h e c o t a n g e n t o f t h e a n g l e a d j a c e n t t o i t s e l f . I I I . I n a r i g h t t r i a n g l e t h e h y p o t e n u s e i s e q u a l t o t h e p r o d u c t o f a s i d e i n t o t h e s e c a n t o f i t s a d j a c e n t a n g l e , o r t h e c o s e c a n t o f i t s o p p o s i t e a n g l e . E X A M P L E S . I n a r i g h t t r i a n g l e A B C , i n w h i c h C i s a r i g h t a n g l e , p r o v e t h e f o l l o w i n g : 1 . t a n B = c o t A + c o s C . 3 . c o s 2 A + c o s 2 B = 0 . 5 . c o s e c 2 B = — + 2 b 2 a 2 . s i n 2 A = s i n 2 B . 2 a b c 2 ' b ! - a 2 7 . t a n 2 A : 2 a b 6 2 - a 2 4 . s i n 2 A = 6 . c o s 2 A = 8 . s i n 3 A = c 2 3 a b 2 — a 3 O B L I Q U E T R I A N G L E S . 9 5 . L a w o f S i n e s . — I n a n y t r i a n g l e t h e s i d e s a r e p r o p o r t i o n a l t o t h e s i n e s o f t h e o p p o s i t e a n g l e s . L e t A B C b e a n y t r i a n g l e . D r a w C D p e r p e n d i c u l a r t o A B . W e h a v e , t h e n , i n b o t h f i g u r e s C D - = a s i n B = 6 s i n A . ( A r t . 9 4 ) . - . a s i n B = b s i n A . a _ b s i n A s i n B S i m i l a r l y , b y d r a w i n g a p e r p e n d i c u l a r f r o m A o r B t o t h e o p p o s i t e s i d e , w e m a y p r o v e t h a t 1 4 8 P L A N E T R I G O N O M E T R Y . o r s i n B s i n C a b , a n d s i n 0 s i n A s i n A s i n B s i n C a : b : c = s i n A : s i n B : s i n C . 9 6 . L a w o f C o s i n e s . — I n a n y t r i a n g l e t h e s q u a r e o f a n y s i d e i s e q u a l t o t h e s u m o f t h e s q u a r e s o f t h e o t h e r t w o s i d e s m i n u s t w i c e t h e p r o d u c t o f t h e s e s i d e s a n d t h e c o s i n e o f t h e i n c l u d e d a n g l e . I n a n a c u t e - a n g l e d t r i a n g l e ( s e e f i r s t f i g u r e ) w e h a v e ( G e o m . , B o o k I I I . , P r o p . 2 6 ) B C 2 = A C 2 + A B 2 - 2 A B x A D , o r a 2 = 6 2 + c 2 - 2 c - A D . B u t A D = b c o s A . . - . a , = o 2 + c J - 2 o c c o s A . I n a n o b t u s e - a n g l e d t r i a n g l e ( s e e s e c o n d f i g u r e ) w e h a v e ( G e o m . , B o o k I I I . , P r o p . 2 7 ) o r B u t S i m i l a r l y , B ( ? = A C ? + A B 2 + 2 A B x A D , 6 2 + c 2 + 2 c - A D . b c o s C A D = — b c o s A . 6 2 + c 2 _ 2 6 c c o s A . c 2 + a 2 — 2 c a c o s B , a + t f - 2 a b c o s C . a ' A D - . a 2 6 2 < ? N o t k . — W h e n o n e e q u a t i o n i n t h e s o l u t i o n o f t r i a n g l e s h a s b e e n o b t a i n e d , t h e o t h e r t w o m a y g e n e r a l l y b e o b t a i n e d b y a d v a n c i n g t h e l e t t e r s s o t h a t a b e c o m e s b , b b e c o m e s c , a n d c b e c o m e s a ; t h e o r d e r i s a b c , b c a , c a b . I t i s o b v i o u s t h a t t h e f o r m u l a s t h u s o b t a i n e d a r e t r u e , s i n c e t h e n a m i n g o f t h e s i d e s m a k e s n o d i f f e r e n c e , p r o v i d e d t h e r i g h t o r d e r i s m a i n t a i n e d . O B L I Q U E T R I A N G L E S . 1 4 9 9 7 . L a w o f T a n g e n t s . — I n a n y t r i a n g l e t h e s u m o f a n y t w o s i d e s i s t o t h e i r d i f f e r e n c e a s t h e t a n g e n t o f h a l f t h e s u m o f t h e o p p o s i t e a n g l e s i s t o t h e t a n g e n t o f h a l f t h e i r d i f f e r e n c e . B y A r t . 9 5 , a : b = s i n A : s i n B . B y c o m p o s i t i o n a n d d i v i s i o n , a + b _ s i n A + s i n B a — b s i n A — s i n B t a n i ( A + B ) . ^ f A g l ( 1 ) t a n | ( A - B ) , y y ' K ' a -- i i b + c t a n A ( B + C ) / O N S i m i l a r l y . - 2 — = - — ) u ' ( 2 ) b — c t a n £ ( H — C ) c + a _ t a n ^ ( C + A ) ^ c — a t a n £ ( C — A ) S i n c e t a n £ ( A + B ) = t a n ( 9 0 ° -£ C ) = u o t ^ C , t h e r e s u l t i n ( 1 ) m a y b e w r i t t e n a + b _ c o t a - 6 ~ t a n £ ( A - B ) a n d s i m i l a r e x p r e s s i o n s f o r ( 2 ) a n d ( 3 ) . 9 8 . T o s h o w t h a t i n a n y t r i a n g l e c — < i c o s 1 ! 4 b c o s A . I n a n a c u t e - a n g l e d t r i a n g l e ( f i r s t f i g u r e o f A r t . 9 6 ) w e h a v e c = D B + D A = a c o s B + b c o s A . I n a n o b t u s e - a n g l e d t r i a n g l e ( s e c o n d f i g u r e o f A r t . 9 6 ) w e h a v e c = D B - D A = a c o s B — 6 c o s C A D . . - . c = a c o s B + b c o s A . S i m i l a r l y , b = c c o s A a c o s C , a = b c o s C + c c o s B . 1 5 0 P L A N E T R I G O N O M E T R Y . E X A M P L E S . 1 . I n t h e t r i a n g l e A B C p r o v e ( 1 ) a + 6 : c = c o s \ ( A — B ) : s i n - | C , a n d ( 2 ) o — 6 : c = s i n \ ( A — B ) : c o s \ C . 2 . I f A D b i s e c t s t h e a n g l e A o f t h e t r i a n g l e A B C , p r o v e B D : D C = s i n C : s i n B . 3 . I f A D ' b i s e c t s t h e e x t e r n a l v e r t i c a l a n g l e A , p r o v e B D ' : C D ' = s i n C : s i n B . . - a 1 2 c o s i A c o s A ( B — C ) 4 . H e n c e p r o v e — = 2 ^ ' - , v D C a s i n B , a n d a l s o J _ = 2 s i n | A s i u j ( C -B ) . . D ' C a s i n B 9 9 . T o e x p r e s s t h e S i n e , t h e C o s i n e , a n d t h e T a n g e n t o f H a l f a n A n g l e o f a T r i a n g l e i n T e r m s o f t h e S i d e s . I . B y A r t . 9 6 w e h a v e c o s A = 6 " + c 2 ~ a 2 = l -2 s i n 2 -----( A r t . 4 9 ) 2 6 c ^ 2 K ' . : 2 s i n 2 A = i _ ^ + c 2 - t t 2 2 2 6 c _ q 2 - ( 6 _ C ) 2 2 f e e _ 4 - 6 — c ) ( a — 6 + c ) ~ 2 b c ~ L e t a + 6 + c = = 2 s ; t h e n a + 6 — c = 2 ( s — c ) , a n d a — 6 - + - c = 2 ( s — 6 ) . „ . „ A 2 ( s - c \ 2 ( s - b } O B L I Q U E T M A N G L E S . 1 5 1 S i m i l a r l y , s i n ^ = ^ . ( L ~ £ ) ( ? _ = . « ) . ( 2 ) -i -V ^ P ^ m I I . e o s A = 2 c o s 2 ^ - l ( A r t . 4 9 ) . - . 2 c o s j A = l + ? ' ! ± c 2 ^ 2 2 2 6 c ^ ( 6 + c ) ! _ g ! 2 6 c ( g + 6 + c ) ( 6 + c — a ) 2 6 c = 2 s - 2 ( s - a ) 2 6 c - - - - f - x ^ < « > S i m i l a r l y , c o s | = J s - i i ^ l ( 5 ) 2 V a c -l ' ^ 3 » I I I . D i v i d i n g ( 1 ) b y ( 4 ) , w e g e t t a n A ( « ^ ) ( 7 ) S i m i l a r l y , t a n | = ( 8 ) t a n g = K s - a ) ( a - b ) ( 9 ) 2 \ s ( s - c ) W S i n c e a n y a n g l e o f a t r i a n g l e i s < 1 8 0 ° , t h e h a l f a n g l e i s < 9 0 ° ; t h e r e f o r e t h e p o s i t i v e s i g n m u s t b e g i v e n t o t h e r a d i c a l s w h i c h o c c u r i n t h i s a r t i c l e . 1 5 2 P L A N E T R I G O N O M E T R Y . 1 0 0 . T o e x p r e s s t h e S i n e o f a n A n g l e i n T e r m s o f t h e S i d e s . A A s i n A = 2 s i n — c o s — ( A r t . 4 9 ) ' - -= 2 J ( » - f t ) " ( - c ) > ( - « ) \ b e M b e ( A r t . 9 9 ) 2 . . s i n A = — V s ( s — a ) ( s — b ) ( s — c ) . b e S i m i l a r l y , s i n B = — V « ( s — a ) ( s — b ) ( s — c ) , a c 2 • s i u C = — V s ( s — a ) ( s — b ) ( s — c ) . a o C o r . s i n A = — V 2 6 V + 2 c W + 2 o 2 6 ! -a 4 - 6 ^ c 4 , 2 6 c a n d s i m i l a r e x p r e s s i o n s f o r s i n B , s i n C . E X A M P L E S . I n a n y t r i a n g l e A B C p r o v e t h e f o l l o w i n g s t a t e m e n t s : 1 . a ( o c o s C — c c o s B ) = b 2 — c 2 . 2 . ( 6 + c ) c o s A + ( c + a ) c o s B + ( a + 6 ) c o s C = a + 6 + c . o s i n A + 2 s i n B _ s i n C a + 2 6 ~ ~ c ^ s i n 2 A — m s i n 2 B _ s i n 2 C a 3 — m b 2 c 2 5 . a c o s A + o c o s B — c c o s C = 2 c c o s A c o s B . ^ c o s A c o s B c o s C _ „ s i n B s i n C s i n C s i n A s i n A s i n B 7 . a s i n ( B -C ) + b s i n ( C -A ) + c s i n ( A -B ) = 0 . 8 . t a n £ A t a n £ B = s ^ ; . s 9 . t a n £ A - ; - t a n £ B = ( s — b ) - i - ( s — c ) . A R E A O F A T R I A N G L E . 1 5 3 1 0 1 . E x p r e s s i o n s f o r t h e A r e a o f a T r i a n g l e . ( 1 ) G i v e n t w o s i d e s a n d t h e i r i n c l u d e d a n g l e . L e t S d e n o t e t h e a r e a o f t h e t r i a n g l e A B C . T h e n b y G e o m e t r y , 2 S = c x C D . B u t i n e i t h e r f i g u r e , b y A r t . 9 4 , C D = b s i n A . . - . S » = ^ 6 c s i n A . S i m i l a r l y , S = £ a c s i n B , S = £ a b s i n C . ( 2 ) G i v e n o n e s i d e a n d t h e a n g l e s . S i n c e w h i c h i s a : b = s i n A : s i n B ( A r t . 9 5 ) , a s i n B s i n A S = ^ a b s i n C , g i v e s a 2 s i n B s i n C S i m i l a r l y , S S = 2 s i n A f t 2 s i n A s i n C c 2 s i n A s i n B 2 s i n B 2 s i n C ( 3 ) G i v e n t h e t h r e e s i d e s . 2 s i n A = -V s ( s -a ) ( s - 6 ) ( -c ) ( A r t . 1 0 0 ) b e S u b s t i t u t i n g i n S = \ b e s i n A , w e g e t S = V s ( s -a ) ( s -6 ) ( s -c ) . 1 5 4 P L A N E T R I G O N O M E T R Y . 1 0 2 . I n s c r i b e d C i r c l e . — T o f i n d t h e r a d i u s o f t h e i n s c r i b e d c i r c l e o f a t r i a n g l e . q L e t A B C b e a t r i a n g l e , O t h e c e n t r e o f t h e i n s c r i b e d c i r c l e , a n d r i t s r a d i u s . D r a w r a d i i t o t h e p o i n t s o f c o n t a c t D , E , F ; a n d j o i n O A , O B , O C . T h e n A c D S = a r e a o f A B C = A A O B + A B O C + A C O A a + b + c . . . r = S = H s ^ ) ( s - b ) ( s - c ) s \ s ( A r t . 9 9 ) ( A r t . 1 0 1 ) 1 0 3 . C i r c u m s c r i b e d C i r c l e . — T o f i n d t h e r a d i u s o f t h e c i r c u m s c r i b e d c i r c l e o f a t r i a n g l e i n t e r m s o f t h e s i d e s o f t h e t r i a n g l e . L e t O b e t h e c e n t r e o f t h e c i r c l e d e s c r i b e d a b o u t t h e t r i a n g l e A B C , a n d R i t s r a d i u s . T h r o u g h O d r a w t h e d i a m e t e r C D a n d j o i n B D . T h e n Z B D C = Z B A C = Z A . B C = - . E = 2 R s i n A , o r a = 2 R s i n A . a b c B u t s i n A = 2 s i n A 2 S b e 2 s i n B 2 s i n C . -( 1 ) ( A r t . 1 0 1 ) ( 2 ) R A D I I O F T H E E S C R I B E D C I R C L E S . 1 5 5 1 0 4 . E s c r i b e d C i r c l e . — T o f i n d t h e r a d i i o f t h e e s c r i b e d c i r c l e s o f a t r i a n g l e . A c i r c l e , w h i c h t o u c h e s o n e s i d e o f a t r i a n g l e a n d t h e o t h e r t w o s i d e s p r o d u c e d , i s c a l l e d a n e s c r i b e d c i r c l e o f t h e t r i a n g l e . L e t O b e t h e c e n t r e o f t h e e s c r i b e d c i r c l e w h i c h t o u c h e s t h e s i d e B C a n d t h e o t h e r s i d e s p r o d u c e d , a t t h e p o i n t s D , E , a n d F , r e s p e c t i v e l y , a n d l e t t h e r a d i u s o f t h i s c i r c l e b e r v W e t h e n h a v e f r o m t h e f i g u r e A A B C = A A O B + A A O C -A B O C , -g _ I 6 r i a r i " 2 2 2 = \ r 1 ( b + c — a ) = r l ( s — a ) . 8 ( A r t . 9 9 ) --( 1 ) s — a S i m i l a r l y i t m a y b e p r o v e d t h a t i f r 2 , r 3 a r e t h e r a d i i o f t h e c i r c l e s t o u c h i n g A C a n d A B r e s p e c t i v e l y , S S r 2 = s - b , s — c 1 0 5 . T o f i n d t h e D i s t a n c e b e t w e e n t h e C e n t r e s o f t h e I n s c r i b e d a n d C i r c u m s c r i b e d C i r c l e s o f a T r i a n g l e . L e t I a n d O b e t h e i n c e n t r e a n d c i r c u m c e n t r e , r e s p e c t i v e l y , o f t h e t r i a n g l e A B C , I A a n d 1 C b i s e c t t h e a n g l e s B A C a n d B C A ; O f t e n c u l l e d t h e i n c e n t r e a n d c i r c u m c e n t r e o f a t r i a n g l e . T V i / \ A m . / V \ \ / \ A 1 5 6 P L A N E T R I G O N O M E T R Y . t h e r e f o r e t h e a r c B D i s e q u a l t o t h e a r c D C , a n d D O H b i s e c t s B C a t r i g h t a n g l e s . D r a w I M p e r p e n d i c u l a r t o A C . T h e n Z D I C = A - i ^ = B C D + B C I = D C I . . - . D I = D C = 2 R s i n A -2 A A A l s o . A I = I M c o s e c — = r c o s e c — , 2 2 . - . D I - A I = 2 R r = E I - I F ; t h a t i s , ( R + O 1 ) ( R -O 1 ) = 2 R r . . - . O T = R 2 - 2 R r . E X A M P L E S . 1 . T h e s i d e s o f a t r i a n g l e a r e 1 8 , 2 4 , 3 0 ; f i n d t h e r a d i i o f i t s i n s c r i b e d , e s c r i b e d , a n d c i r c u m s c r i b e d c i r c l e s . A n a . 6 , 1 2 , 1 8 , 3 6 , 1 5 . 2 . P r o v e t h a t t h e a r e a o f t h e t r i a n g l e A B C i s 1 c 2 2 c o t A + c o t B 3 . F i n d t h e a r e a o f t h e t r i a n g l e A B C w h e n ( 1 ) a = 4 , b = 1 0 f t . , C = 3 0 ° . A n s . 1 0 s q . f t . ( 2 ) 6 = 5 , c = 2 0 i n c h e s , A = 6 0 ° . 4 3 . 3 s q . i n . ( 3 ) a = 1 3 , b = 1 4 , c = 1 5 c h a i n s . 8 4 s q . c h a i n s . A X . 1 1 1 1 4 . P r o v e - = — i r r z r 2 r 3 5 . P r o v e r = B s i n ° . c o s ^ A E X A M P L E S . 1 5 7 6 . P r o v e t h a t t h e a r e a o f t h e t r i a n g l e A B C i s r e p r e s e n t e d b y e a c h o f t h e t h r e e e x p r e s s i o n s : 2 R 2 s i n A s i n B s i n C , r s , a n d R r ( s i n A + s i n B + s i n C ) . 7 . I f A = 6 0 ° , a = s / 3 , b = V 2 , p r o v e t h a t t h e a r e a = i ( 3 + V 3 ) . 8 . P r o v e R ( s i n A + s i n B + s i n C ) = s . 9 . P r o v e t h a t t h e b i s e c t o r s o f t h e a n g l e s A , B , C , o f a t r i a n g l e a r e , r e s p e c t i v e l y , e q u a l t o A B C 2 6 c c o s — 2 c a c o s — 2 a b c o s — 2 2 2 6 + c , c + a , a + 6 1 0 6 . T o f i n d t h e A r e a o f a C y c l i c Q u a d r i l a t e r a l . L e t A B C D b e t h e q u a d r i l a t e r a l , a n d a , 6 , c , a n d d i t s s i d e s . J o i n B D . T h e n , a r e a o f f i g u r e = S = \ a d s i n A + £ b e s i n C = ^ ( o d + 6 c ) s i n A . . . ( 1 ) N o w i n A A B D , B D 2 = a 2 + d 2 -2 a d c o s A , a n d i n A C B D , 5 1 ? = b 2 + c 2 -2 6 c c o s C = 6 2 + c 2 - 2 6 c c o s A . a 2 -6 2 -c 2 + d 2 . - . c o s A = - — 2 ( a d + b e ) r r a 2 _ 6 2 _ , s i n A = - » 1 — \ | _ 2 ( a d + c 2 + c T | 2 b e ) = V ( 2 a d + 2 6 c ) 2 -( a 2 -6 » -c 2 + d 2 ) 2 2 ( a r f + 6 c ) S e e G e o m e t r y , A r t . 2 5 1 . 1 5 8 P L A N E T R I G O N O M E T R Y . _ ^ [ l a + d y - ( b ~ -c ) ^ T r ( f t + c ) Y - ~ ( a -d ) 2 ] 2 ( a d + 6 c ) V ( o + d + 6 — c ) ( a ~ + d ^ 6 + e ) ( 6 + c + a — d ) ( 6 + c — a + d ) ^ 2 ( a d + 6 c ) = 2 V ( 8 -a ) ( a -b ) ( s -c ) ( s -d ) a d + 6 c ( w h e r e 2 s = a + 6 + c + d ) . S u b s t i t u t i n g i n ( 1 ) , w e h a v e S = V ( a -a ) ( a -6 ) ( s -c ) ( s -d ) . T h e m o r e i m p o r t a n t f o r m u l a e p r o v e d i n t h i s c h a p t e r a r e s u m m e d u p a s f o l l o w s : 1 . - A -= ^ 5 = - ^ t ; ( A r t . 9 5 ) s i n A s i n B s i n C 2 . a 2 = 6 2 + c 2 -2 6 c c o s A ( A r t . 9 6 ) 3 . i ± j = t a p j ( A + B ) ( A r t . 9 7 ) a -b t a n ^ ( A - B ) v ' 4 . s i n £ A = J ( « - f t ) ( a - c ) ( A r t . 9 9 ) \ 6 c 5 . c o s + A = - v / ^ - ^ — 5 ^ -3 v 6 c 7 . s i n A = -V s ( s - a ) ( s - 6 ) ( s -c ) . . ( A r t . 1 0 0 ) 6 c = _ L V 2 6 V + 2 c 2 a 2 + 2 a 2 6 2 -a 4 -6 4 -c 4 . 2 6 c 8 . A r e a o f A = V s ( s -a ) ( s - b ) ( s -c ) . . ( A r t . 1 0 1 ) E X A M P L E S . 1 5 9 9 . A r e a o f A = f ( a + b + c ) = r s . . . . ( A r t . 1 0 2 ) 2 1 0 . 1 1 . R = ^ ( A r t . 1 0 3 ) 4 o E X A M P L E S . I n a r i g h t t r i a n g l e A B C , i n w h i c h C i s t h e r i g h t a n g l e , p r o v e t h e f o l l o w i n g : 1 . c o s 2 B = s i n 2 A - s i D 2 B . s i n 2 A + s i n 2 B 2 . s i n 2 — = 2 2 c / A , . A \ ! a + c 3 . i c o s — h s i n -) = — X _ r . V 2 2 ; c . « A 6 + c 4 . c o s 2 — = — - — 2 2 c 5 . s i n ( A - B ) + c o s 2 A = 0 . e a — b , A — B 6 . t a n a + b 2 7 . s i n ( A - B ) + s i n ( 2 A + C ) = 0 . 8 . t a n A = - 2 — T 6 + c 9 . ( s i n A - s i n B ) 2 + ( c o s A + c o s B ) J = 2 . 1 0 I a ^ + j a — b _ 2 s i n A \ a _ 6 \ a + 6 ~ ~ V c o s 2 B I n a n y t r i a n g l e A B C , p r o v e t h e f o l l o w i n g s t a t e m e n t s : 1 1 . ( a + b ) s i n - = c c o s - ^ — ^ ' 2 2 1 6 0 P L A N E T R I G O N O M E T R Y . 1 2 . ( b — c ) c o s — = C T s i n ^ ~ ^ -1 3 . a ( 6 2 + c 2 ) c o s A + 6 ( c 2 + a 2 ) c o s B + c ( a 2 + 6 ! ) c o s C = 3 a 6 c . a — 6 c o s B — c o s A 1 4 . 1 5 . c 1 + c o s C 6 + c c o s B + c o s C a 1 — c o s A / t — r — — i— ^ 6 2 s i n C + c 2 s i n B 1 6 . V 6 c s i n B s i n C = — b + c 1 7 . a + 6 + c = ( 6 + c ) c o s A + ( c + a ) c o s B + ( a + 6 ) c o s C . 1 8 . b + c — a = ( b + c ) c o s A — ( c — a ) c o s B + ( a — 6 ) c o s C . 1 9 . a c o s ( A + B + C ) -b c o s ( B + A ) -c c o s ( A + C ) = 0 . 2 q c o s A _ ^ c o s B _ | _ c o s C _ a 2 + b 2 + c 2 a 6 c 2 a 6 c a s i n C 2 1 . t a n A = b — a c o s C 2 2 . 6 c o s 2 — + c c o s 2 — . = s . 2 2 o o i B , C 6 + c — a 2 3 . t a n — t a n - = — ' ■ 2 2 ò + c + a A B 2 4 . t a n — ( 6 + c — a ) = t a n — ( c + a — b ) . 2 2 2 5 . c 2 « = ( a + 6 ) 2 s i n 2 ^ + ( a - 6 ) 2 c o s 2 ^ -2 6 . c ( c o s A + c o s B ) = 2 ( « + 6 ) s i n 2 - -2 7 . c ( c o s A - c o s B ) = 2 ( 6 - a ) c o s 2 ^ -2 8 . t a n B - f - t a n C = ( a 2 + 6 2 -c 2 ) ( a 2 -b 2 + c 2 ) . E X A M P L E S . 1 6 1 2 9 . a 1 + 6 2 + c 2 = 2 ( a b c o s C + 6 c c o s A + c a c o s B ) . 3 0 . c o s 2 — - s - c o s 2 ? = ( s — a ) - : - 6 ( s — 6 ) . 2 2 3 1 . 6 s i n 2 — + c s i n 2 — = s — a . 2 2 3 2 . I f p i s t h e l e n g t h o f t h e p e r p e n d i c u l a r f r o m A o n B C , s i n A = ^ £ . b e 3 3 . I f A = 3 B , t h e n s i n B = | ^ ' 3 6 - o 2 \ 6 3 4 . I t v 6 c s i n B s i n C = , t h e n B = C . b + c o k B C A 3 5 . a c o s — c o s c o s e c — = s . 2 2 2 3 6 . I f c o s A = a n d c o s B = | | , t h e n c o s C = — 3 7 . I f s i n 2 B + s i n 2 C = s i n 2 A , t h e n A = 9 0 ° . 3 8 . I f D i s t h e m i d d l e p o i n t o f B C , p r o v e t h a t 4 A D 2 = 2 6 2 + 2 c 2 - a 2 . 3 9 . I f a = 2 6 , a n d A = 3 B , p r o v e t h a t C = 6 0 ° . 4 0 . I f D , E , F , a r e t h e m i d d l e p o i n t s o f t h e s i d e s , B C , C A , A B , p r o v e 4 ( A D ' + B E 2 + C T ) = 3 ( a 2 + 6 2 + c 2 ) . 4 1 . I f a , 6 , c , t h e s i d e s o f a t r i a n g l e , a r e i n a r i t h m e t i c p r o g r e s s i o n , p r o v e A C 1 t a n — t a n — = — 2 2 3 i 0 J f t a n A — t a n B c — 6 . , , . « A 0 4 ^ 2 . I t — -p r o v e t h a t A = 6 0 . t a n A + t a n B c 4 3 . I f c o s B = „ s - ~ , p r o v e t h a t B = C . 2 s i n C 1 6 2 P L A N E T R I G O N O M E T R Y . 4 4 . I f a 1 = W -b e + c 2 , p r o v e t h a t A = 6 0 ° . 4 5 . I f t h e s i d e s o f a t r i a n g l e a r e a , b , a n d V a 2 - f a b + b - , p r o v e t h a t i t s g r e a t e s t a n g l e i s 1 2 0 ° . 4 6 . P r o v e t h a t t h e v e r t i c a l a n g l e o f a n y t r i a n g l e i s d i v i d e d b y t h e m e d i a n w h i c h b i s e c t s t h e b a s e , i n t o s e g m e n t s w h o s e s i n e s a r e i n v e r s e l y p r o p o r t i o n a l t o t h e a d j a c e n t s i d e s . 4 7 . I f A D b e t h e m e d i a n t h a t b i s e c t s B C , p r o v e ( 1 ) ( 6 2 -c 2 ) t a n A D B = 2 6 c s i n A , a n d ( 2 ) c o t B A D + c o t D A C = 4 c o t A + c o t B + c o t C . 4 8 . F i n d t h e a r e a o f t h e t r i a n g l e A B C w h e n a = 6 2 5 , b = 5 0 5 , c = 9 0 4 y a r d s . A n s . 1 5 1 8 7 2 s q . y a r d s . 4 9 . F i n d t h e r a d i i o f t h e i n s c r i b e d a n d e a c h o f t h e e s c r i b e d c i r c l e s o f t h e t r i a n g l e A B C w h e n a = 1 3 , 6 = 1 4 , c = 1 5 . A n s . 4 ; 1 0 . 5 ; 1 2 ; 1 4 . 5 0 . P r o v e t h e a r e a S = | a 2 s i n B s i n C c o s e c A . 5 1 . " " " " = V r ? w v 5 2 . " " " " = 2 a b c / c o s ~ c o s - c o s C « + 6 + c V 2 2 2 j 5 3 . P r o v e t h a t t h e l e n g t h s o f t h e s i d e s o f t h e p e d a l t r i a n g l e , t h a t i s , t h e t r i a n g l e f o r m e d b y j o i n i n g t h e f e e t o f t h e p e r p e n d i c u l a r s , a r e a c o s A , 6 c o s B , c c o s C , r e s p e c t i v e l y . 5 4 . P r o v e t h a t t h e a n g l e s o f t h e p e d a l t r i a n g l e a r e , r e s p e c t i v e l y , i r — 2 A , i t — 2 B , i r — 2 C . 5 5 . P r o v e r ^ r 3 = r 3 c o t 2 — c o t 2 — c o t 2 5 . 2 2 2 5 6 . P r o v e r , c o s — = a c o s ? c o s — -2 2 2 5 7 . P r o v e t h a t t h e a r e a o f t h e i n c i r c l e : a r e a o f t h e t r i -, t A , B . C a n g l e : : t t : c o t — c o t — c o t — . 2 2 2 E X A M P L E S . 1 6 3 P r o v e t h e f o l l o w i n g s t a t e m e n t s : 5 8 . I f a , b , c , a r e i n A . P . , t h e n a c = 6 r R . 5 9 . I f t h e a l t i t u d e o f a n i s o s c e l e s t r i a n g l e i s e q u a l t o t h e b a s e , R i s f i v e - e i g h t h s o f t h e b a s e . 6 0 . 6 c = 4 R 2 ( c o s A - f c o s B c o s C ) . 6 1 . I f C i s a r i g h t a n g l e , 2 r + 2 R = a + 6 . 6 2 . r 2 r 3 + + j v 2 = 5 2 . 6 3 . ! + i + ! = J _ b e c a a b 2 r R 6 4 . r t + r 2 = c c o t ^ -a K A . B -C 6 5 . r c o s — = a s i n — s i n -. 2 2 2 6 6 . I f p u p 2 , p 3 b e t h e d i s t a n c e s t o t h e s i d e s f r o m t h e c i r c u m c e n t r e , t h e n 2 l - | - — j f . c — a ^ c P i P P i ^ I h P l P s 6 7 . T h e r a d i u s R o f t h e c i r c u m c i r c l e _ 1 3 / a b c 2 V s i n A s i n B s i n C 6 8 . S = - s i n 2 B + - s i n 2 A . 4 4 6 9 . J L - + 1 + - J L -s — a s — 0 s — c s b 7 0 . a 6 c r = 4 R ( s - a ) ( s - 6 ) ( s - c ) . 7 1 . T h e d i s t a n c e s b e t w e e n t h e c e n t r e s o f t h e i n s c r i b e d A a n d e s c r i b e d c i r c l e s o f t h e t r i a n g l e A B C a r e 4 R s i n — , 4 R s i n ? , 4 R s i n — -2 2 7 2 . I f A i s a r i g h t a n g l e , r 2 + r s = a . 1 6 4 P L A N E T R I G O N O M E T R Y . 7 3 . I n a n e q u i l a t e r a l t r i a n g l e 3 R = 6 r = 2 r r 7 4 . I f r , r „ r 2 , r s d e n o t e t h e r a d i i o f t h e i n s c r i b e d a n d e s c r i b e d c i r c l e s o f a t r i a n g l e , 7 5 . T h e s i d e s o f a t r i a n g l e a r e i n a r i t h m e t i c p r o g r e s s i o n , a n d i t s a r e a i s t o t h a t o f a n e q u i l a t e r a l t r i a n g l e o f t h e s a m e p e r i m e t e r a s 3 i s t o 5 . F i n d t h e r a t i o o f t h e s i d e s a n d t h e v a l u e o f t h e l a r g e s t a n g l e . A n s . A s 7 , 5 , 3 ; 1 2 0 ° . 7 6 . I f a n e q u i l a t e r a l t r i a n g l e b e d e s c r i b e d w i t h i t s a n g u l a r p o i n t s o n t h e s i d e s o f a g i v e n r i g h t i s o s c e l e s t r i a n g l e , a n d o n e s i d e p a r a l l e l t o t h e h y p o t e n u s e , i t s a r e a w i l l b e 2 a 2 s i n 2 1 5 ° s i n 6 0 ° , w h e r e a i s a s i d e o f t h e g i v e n t r i a n g l e . 7 7 . I f h b e t h e d i f f e r e n c e b e t w e e n t h e s i d e s c o n t a i n i n g t h e r i g h t a n g l e o f a r i g h t t r i a n g l e , a n d S i t s a r e a , t h e d i a m e t e r o f t h e c i r c u m s c r i b i n g c i r c l e = y / h ? + 4 S . 7 8 . T h r e e c i r c l e s t o u c h o n e a n o t h e r e x t e r n a l l y : p r o v e t h a t t h e s q u a r e o f t h e a r e a o f t h e t r i a n g l e f o r m e d b y j o i n i n g t h e i r c e n t r e s i s e q u a l t o t h e p r o d u c t o f t h e s u m a n d p r o d u c t o f t h e i r r a d i i . 7 9 . O n t h e s i d e s o f a n y t r i a n g l e e q u i l a t e r a l t r i a n g l e s a r e d e s c r i b e d e x t e r n a l l y , a n d t h e i r c e n t r e s a r e j o i n e d : p r o v e t h a t t h e t r i a n g l e t h u s f o r m e d i s e q u i l a t e r a l . 8 0 . I f O 1 ; O 2 , O 3 a r e t h e c e n t r e s o f t h e e s c r i b e d c i r c l e s o f a t r i a n g l e , t h e n t h e a r e a o f t h e t r i a n g l e O ^ O 2 O s = a r e a o f 8 1 . I f t h e c e n t r e s o f t h e t h r e e e s c r i b e d c i r c l e s o f a t r i a n g l e a r e j o i n e d , t h e n t h e a r e a o f t h e t r i a n g l e t h u s f o r m e d i s w h e r e r i s t h e r a d i u s o f t h e i n s c r i b e d c i r c l e o f t h e 2 r o r i g i n a l t r i a n g l e . t r i a n g l e A B C 1 + b + c — a a + c — b a + b — c S O L U T I O N O F T R I A N G L E S . 1 6 5 C H A P T E R V I I . S O L U T I O N O F T R I A N G L E S . 1 0 7 . T r i a n g l e s . — I n e v e r y t r i a n g l e t h e r e a r e s i x e l e m e n t s , t h e t h r e e s i d e s a n d t h e t h r e e a n g l e s . W h e n a n y t h r e e e l e m e n t s a r e g i v e n , o n e a t l e a s t o f t h e t h r e e b e i n g a s i d e , t h e o t h e r t h r e e c a n b e c a l c u l a t e d . T h e p r o c e s s o f d e t e r m i n i n g t h e u n k n o w n e l e m e n t s f r o m t h e k n o w n i s c a l l e d t h e s o l u t i o n o f t r i a n g l e s . N o t e . — I f t h e t h r e e a n g l e s o n l y o f a t r i a n g l e a r e g i v e n , i t i s i m p o s s i b l e t o d e t e r m i n e t h e s i d e s , f o r t h e r e i s a n I n f i n i t e n u m b e r o f t r i a n g l e s t h a t a r e e q u i a n g u l a r t o o n e a n o t h e r . T r i a n g l e s a r e d i v i d e d i n T r i g o n o m e t r y i n t o r i g h t a n d o b l i q u e . W e s h a l l c o m m e n c e w i t h r i g h t t r i a n g l e s , a n d s h a l l s u p p o s e C t h e r i g h t a n g l e . R I G H T T R I A N G L E S . 1 0 8 . T h e r e a r e F o u r C a s e s o f R i g h t T r i a n g l e s . I . G i v e n o n e s i d e a n d t h e h y p o t e n u s e . I I . G i v e n a n a c u t e a n g l e a n d t h e h y p o t e n u s e . I I I . G i v e n o n e s i d e a n d a n a c u t e a n g l e . I V . G i v e n t h e t w o s i d e s . L e t A B C b e a t r i a n g l e , r i g h t - a n g l e d a t C , a n d l e t a , b , a n d c , a s b e f o r e , b e t h e s i d e s o p p o s i t e t h e a n g l e s A , B , a n d C , r e s p e c t i v e l y . T h e f o r m u l a e f o r t h e s o l u t i o n o f r i g h t t r i a n g l e s a r e ( 1 ) , ( 2 ) , ( 3 ) o f A r t . 9 4 . 1 6 6 P L A N E T R I G O N O M E T R Y . 1 0 9 . C a s e I . — G i v e n a s i d e a n d t h e h y p o t e n u s e , a s a a n d c ; t o f i n d A , B , b . W e h a v e s i n A = . . . l o g s i n A = l o g a — l o g c , f r o m w h i c h A i s d e t e r m i n e d ; t h e n B = 9 0 ° — A . L a s t l y , b = c c o s A . . - . l o g b = l o g c + l o g c o s A . T h u s A , B , a n d b a r e d e t e r m i n e d . E x . 1 . G i v e n a = 5 3 6 , c = 9 4 1 ; f i n d A , B , 6 . S o l u t i o n b y N a t u r a l F u n c t i o n s . W e h a v e s i n A = a 5 3 6 = . 5 6 9 6 0 7 . c 9 4 1 F r o m a t a b l e o f n a t u r a l s i n e s w e f i n d t h a t A = 3 4 ° 4 3 ' 2 2 " . . - . B = 5 5 ° 1 6 ' 3 8 " . L a s t l y , b = c c o s A = 9 4 1 x . 8 2 1 9 1 8 = 7 7 3 . 4 2 5 . l o g s i n A = l o g a — l o g c l o g a = 2 . 7 2 9 1 6 4 8 l o g c = 2 . 9 7 3 5 8 9 6 l o g s i n A = 9 . 7 5 5 5 7 5 2 . - . A = 3 4 ° 4 3 ' 2 2 " . . - . B = 5 5 ° 1 6 ' 3 8 " . L o g a r i t h m i c S o l u t i o n . l o g b = l o g c + l o g c o s A . l o g c = 2 . 9 7 3 5 8 9 6 l o g c o s A = 9 . 9 1 4 8 2 8 3 l o g 6 = 2 . 8 8 8 4 1 7 9 . - . 6 = 7 7 3 . 4 2 4 . O u r t w o m e t h o d s o f c a l c u l a t i o n g i v e r e s u l t s w h i c h d o n o t q u i t e a g r e e . T h e d i s c r e p a n c i e s a r i s e f r o m t h e d e f e c t s o f t h e t a b l e s . T e n i s a d d e d s o a s t o g e t t h e t a b u l a r l o g a r i t h m s ( A r t . 7 6 ) . R I G H T T R I A N G L E S . 1 6 7 T h e p r o c e s s o f s o l u t i o n b y n a t u r a l s i n e s , c o s i n e s , e t c . , c a n b e u s e d t o a d v a n t a g e o n l y i n c a s e s i n w h i c h t h e m e a s u r e s o f t h e s i d e s a r e s m a l l n u m b e r s . W e m i g h t h a v e d e t e r m i n e d b t h u s : b = V ( c — a ) ( c + a ) ; o r t h u s : b = a t a n B . N o t e . — I t i s g e n e r a l l y b e t t e r t o c o m p u t e a l l t h e r e q u i r e d p a r t s f r o m t h e g i v e n o n e s , s o t h a t I f a n e r r o r i s m a d e i n d e t e r m i n i n g o n e p a r t , t h a t e r r o r w i l l n o t a f f e c t l h « c o m p u t a t i o n o f t h e o t h e r p a r t s . T o t e s t t h e a c c u r a c y o f t h e w o r k , c o m p u t e t h e s a m e p a r t s b y d i f f e r e n t f o r m u l a s . E x . 2 . G i v e n a = 2 1 , c = 2 9 ; f i n d A , B , b . A n s . A = 4 6 ° 2 3 ' 5 0 " , B = 4 3 ° 3 6 ' 1 0 " , 6 = 2 0 . N o t e . — I n t h e s e e x a m p l e s t h e s t u d e n t m u s t f i n d t h e n e c e s s a r y l o g a r i t h m s f r o m t h e t a b l e s . 1 1 0 . C a s e I I . — G i v e n a n a c u t e a n g l e a n d t h e h y p o t e n u s e , a s A a n d c ; t o f i n d B , a , b . W e h a v e B = 9 0 ° -A . A l s o a = c s i n A , a n d b = c c o s A . . - . l o g a = l o g c + l o g s i n A ; a n d l o g b — l o g c + l o g c o s A . T h u s B , a , a n d b a r e d e t e r m i n e d . E x . 1 . G i v e n A = 5 4 ° 2 8 ' , c = 1 2 5 ; f i n d B , a , b . B = 9 0 ° -A = 3 5 ° 3 2 ' . S o l u t i o n b y N a t u r a l F u n c t i o n s . W e h a v e a = c s i n A , a n d b = c c o s A . U s i n g a t a b l e o f n a t u r a l s i n e s , w e h a v e a = 1 2 5 x . 8 1 3 7 7 8 = 1 0 1 . 7 2 2 , a n d b = 1 2 5 x . 5 8 1 1 7 7 = 7 2 . 6 4 7 . 1 6 8 P L A N E T R I G O N O M E T R Y . L o g a r i t h m i c S o l u t i o n . l o g a = l o g c + l o g s i n A . l o g c = 2 . 0 9 6 9 1 0 0 l o g s i n A = 9 . 9 1 0 5 0 5 7 l o g a = 2 . 0 0 7 4 1 5 7 . - . a = 1 0 1 . 7 2 2 . l o g 6 = l o g c + l o g c o s A . l o g c = 2 . 0 9 6 9 1 0 0 l o g c o s A = 9 . 7 6 4 3 0 8 0 l o g 6 = 1 . 8 6 1 2 1 8 0 . - . 6 = 7 2 . 6 4 7 . E x . 2 . G i v e n A = 3 7 ° 1 0 ' , c = 8 7 6 2 ; f i n d a a n d 6 . A n s . 5 2 9 3 . 4 ; 6 9 8 2 . 3 . 1 1 1 . C a s e I I I . — G i v e n a s i d e a n d a n a c u t e a n g l e , a s A a n d a ; t o f i n d B , 6 , c . W e h a v e A l s o B = 9 0 ° -A . 6 = — — — , a n d c = -a n d t a n A s i n A l o g b = l o g a — l o g t a n A , l o g c = l o g a — l o g s i n A . E x . 1 . G i v e n A = 3 2 ° 1 5 ' 2 4 " , a = 5 4 7 2 . 5 ; f i n d B , 6 , c . S o l u t i o n . B = 9 0 ° -A = 5 7 ° 4 4 ' 3 6 " . l o g b = l o g a — l o g t a n A . l o g a = 3 . 7 3 8 1 8 5 8 l o g t a n A = 9 . 8 0 0 1 0 9 0 l o g 6 = 3 . 9 3 8 0 7 6 8 . - . 6 = 8 6 7 1 . 1 5 2 . l o g c = l o g a — l o g s i n A . l o g a = 3 . 7 3 8 1 8 5 8 l o g s i n A = 9 . 7 2 7 3 0 7 6 l o g c = 4 . 0 1 0 8 7 8 2 . - . c = 1 0 2 5 3 . 6 4 . E x . 2 . G i v e n A = 3 4 ° 1 8 ' , a = 2 3 7 . 6 ; f i n d B , 6 , c . A n s . B = 5 5 ° 4 2 ' ; 6 = 3 4 8 . 3 1 ; c = 4 2 1 . 6 3 . T e n i s r e j e c t e d b e c a u s e t h e t a b u l a r l o g a r i t h m i c f u n c t i o n s a r e t o o l a r g e b y t e n ( A r t r e ) . R I G H T T R I A N G L E S . 1 6 9 1 1 2 . C a s e I V . — G i v e n t h e t w o s i d e s , a s a a n d b ; t o f i n d A , B , c . W e h a v e t a n A = -; t h e n B = 9 0 ° -A . 6 A l s o c = a c o s e c A = .s i n A . - . l o g t a n A = l o g a — l o g b , a n d l o g c = l o g a — l o g s i n A . E x . G i v e n a = 2 2 6 6 . 3 5 , 6 = 5 4 3 9 . 2 4 ; f i n d A , B , c . S o l u t i o n . l o g t a n A = l o g a — l o g b . l o g o = 3 . 3 5 5 3 2 7 0 l o g 6 = 3 . 7 3 5 5 3 8 2 l o g t a n A = 9 . 6 1 9 7 8 8 8 . - . A = 2 2 ° 3 7 ' 1 2 " . . - . B = 6 7 ° 2 2 ' 4 8 " . l o g c = l o g a — l o g s i n A . l o g a = 3 . 3 5 5 3 2 7 0 l o g s i n A = 9 . 5 8 5 0 2 6 6 l o g c = 3 . 7 7 0 3 0 0 4 . - . c = 5 8 9 2 . 5 1 . N o t e . — I n t h i s e x a m p l e w e m i g h t h a v e f o u n d c b y m e a n s o f t h e f o r m u l a c = S a 2 + b ' l \ b u t w e w o u l d h a v e h a d t o g o t h r o u g h t h e p r o c e s s o f s q u a r i n g t h e v a l u e s o f a a n d 6 . I f t h e s e v a l u e s a r e s i m p l e n u m b e r s , i t i s o f t e n e a s i e r t o f i n d c i n t h i s w a y ; b u t t h i s v a l u e o f c i s n o t a d a p t e d t o l o g a r i t h m s . A f o r m u l a w h i c h c o n s i s t s e n t i r e l y o f f a c t o r s i s a l w a y s p r e f e r r e d t o o n e w h i c h c o n s i s t s o f t e r m s , w h e n a n y o f t h o s e t e r m s c o n t a i n a n y p o w e r o f t h e q u a n t i t i e s i n v o l v e d . 1 1 3 . W h e n a S i d e a n d t h e H y p o t e n u s e a r e n e a r l y E q u a l . — W h e n a s i d e a n d t h e h y p o t e n u s e a r e g i v e n , a s a a n d c i n C a s e I . , a n d a r e n e a r l y e q u a l i n v a l u e , t h e a n g l e A i s v e r y n e a r 9 0 ° , a n d c a n n o t b e d e t e r m i n e d w i t h m u c h a c c u r a c y f r o m t h e t a b l e s , b e c a u s e t h e s i n e s o f a n g l e s n e a r 9 0 ° d i f f e r v e r y l i t t l e f r o m o n e a n o t h e r ( A r t . 8 1 ) . I t i s t h e r e f o r e d e s i r a b l e , i n t h i s c a s e , t o f i n d B f i r s t , b y e i t h e r o f t h e f o l l o w i n g f o r m u l a e : B s i n 2 V I — c o s B , . . C A , 2 ^ ^ . 4 ^ ™ 1 7 0 P L A N E T R I G O N O M E T R Y . B / 1 - c o s B ( A r t . 5 0 ) 2 \ l + c o s B V ' . I^ ( 2 ) T h e n b = c c o s A ( 3 ) o r = V ( o + a ) ( c - a ) ( 4 ) E x . 1 . G i v e n a = 4 6 0 2 . 2 1 0 5 9 , c = 4 6 0 2 . 8 3 6 ; f i n d B . c -a = 0 . 6 2 5 4 1 , l o g ( c -a ) = 1 . 7 9 6 1 6 4 8 2 c = 9 2 0 5 . 6 7 2 , l o g 2 c = 3 . 9 6 4 0 5 5 5 2 ) 5 . 8 3 2 1 0 9 3 l o g s i n 5 = 7 . 9 1 6 0 5 4 7 6 2 B = 5 6 ' 4 0 " . 3 6 . . - . ? = 2 8 ' 2 0 " . 1 8 . 2 N o t e . — T h e c h a r a c t e r i s t i c 6 i s i n c r e a s e d n u m e r i c a l l y t o 6 t o m a k e i t d i v i s i b l e b y 2 ( s e e N o t e 4 o f A r t . 6 6 ) . T e n i s t h e n a d d e d t o t h e c h a r a c t e r i s t i c 3 , m a k i n g i t 7 , s u a s t o a g r e e w i t h t h e T a b l e s ( A r t . 7 6 ) . T h e r e i s a s l i g h t e r r o r i n t h e a b o r e v a l u e o f B o n a c c o u n t o f t h e i r r e g u l a r d i f f e r e n c e s o f t h e l o g s i n e s f o r a n g l e s n e a r 0 ° ( A r t . 8 1 ) . A m o r e a c c u r a t e v a l u e m a y b e f o u n d b y t h e p r i n c i p l e t h a t t h e s i n e s o f s m a l l a n g l e s a r e a p p r o x i m a t e l y p r o p o r t i o n a l t o t h e a n g l e s ( A r t . 1 3 0 ) . E X A M P L E S . T h e f o l l o w i n g r i g h t t r i a n g l e s m u s t b e s o l v e d b y l o g a r i t h m s . 1 . G i v e n a = 0 0 , c = 1 0 0 ; f i n d A , B , b . A n s . A = 3 6 ° 5 2 ' ; B = 5 3 ° 8 ' ; b = 8 0 . 2 . G i v e n a = 1 3 7 . 0 6 , c = 2 4 0 ; f i n d A , B , b . A n s . A = ; ! ; - , ° ; B = 5 5 ° ; b = 1 9 ( 5 . 5 9 . R I G H T T R I A N G L E S . 1 7 1 1 3 . G i v e n a = 1 4 7 , c = 1 8 4 ; f i n d A , B , 6 . A n s . A = 5 3 ° 1 ' 3 6 " j B = 3 6 ° 5 8 ' 2 5 " ; 6 = 1 1 0 . 6 7 . 4 . G i v e n a = 1 0 0 , c = 2 0 0 ; f i n d A , B , 6 . A n s . A = 3 0 ° ; B = 6 0 ° ; b = 1 0 0 V S . 5 . G i v e n A = 4 0 ° , c = 1 0 0 ; f i n d B , a , b . A n s . B = 5 0 ° ; « = 6 4 . 2 7 9 ; 6 = 7 6 . 6 0 4 . 6 . G i v e n A = 3 0 ° , c = 1 5 0 ; f i n d B , a , b . A n s . B = 6 0 ° ; a = 7 5 ; 6 = 7 5 V 3 . 7 . G i v e n A = 3 2 ° , c = 1 7 6 0 ; f i n d B , a , b . A n s . B = 5 8 ° ; a = 9 3 2 . 6 6 ; b = 1 4 9 2 . 5 7 . 8 . G i v e n A = 3 5 ° 1 6 ' 2 5 " , c = 6 7 2 . 3 4 1 2 ; f i n d B , a , b . A n s . B = 5 4 ° 4 3 ' 3 5 " ; a = 3 8 8 . 2 6 ; 6 = 5 4 8 . 9 . 9 . G i v e n A = 7 5 ° , a = 8 0 ; f i n d B , 6 , c . A n s . B = 1 5 ° ; 6 = 8 0 ( 2 - V 3 ) ; c = 8 0 ( V 6 -V 2 ) . 1 0 . G i v e n A = 3 6 ° , a = 5 2 0 ; f i n d B , b , c . A n s . B = 5 4 ° ; 6 = 7 1 5 . 7 2 ; c = 8 8 4 . 6 8 . 1 1 . G i v e n A = 3 4 ° 1 5 ' , a = 8 4 3 . 2 ; f i n d B , 6 , c . A n s . B = 5 5 ° 4 5 ' ; c = 1 4 9 8 . 2 . 1 2 . G i v e n A = 6 7 ° 3 7 ' 1 5 " , 6 = 2 5 4 . 7 3 ; f i n d B , a , c . A n s . B = 2 2 ° 2 2 ' 4 5 " ; a = 6 1 8 . 6 6 ; c = 6 6 9 . 0 5 . 1 3 . G i v e n a = 7 5 , 6 = 7 5 ; f i n d A , B , c . A n s . A = 4 5 ° = B ; c . = 7 5 V 2 . 1 4 . G i v e n a = 2 1 , 6 = 2 0 ; f i n d A , B , c . A n s . A = 4 6 ° 2 3 ' 5 0 " ; c = 2 9 . 1 5 . G i v e n a = 3 0 0 . 4 3 , 6 = 5 0 0 ; f i n d A , B , c . A n s . A = 3 1 ° ; B = 5 9 ° ; c = 5 8 3 . 3 1 . 1 6 . G i v e n a = 4 8 4 5 , 6 = 4 7 4 2 ; f i n d A , B , c . A n s . A = 4 5 ° 3 6 ' 5 6 " . 1 7 2 P L A N E T R I G O N O M E T R Y . O B L I Q U E T R I A N G L E S . 1 1 4 . T h e r e a r e F o u r C a s e s o f O b l i q u e T r i a n g l e s . I . G i v e n a s i d e a n d t w o a n g l e s . I I . O i v e n t w o s i d e s a n d t h e a n g l e o p p o s i t e o n e o f t h e m . I I I . O i v e n t w o s i d e s a n d t h e i n c l u d e d a n g l e . I V . O i v e n t h e t h r e e s i d e s . T h e f o r m u l a e f o r t h e s o l u t i o n o f o b l i q u e t r i a n g l e s w i l l b e t a k e n f r o m C h a p . V I . S p e c i a l a t t e n t i o n m u s t b e g i v e n t o t h e f o l l o w i n g t h r e e , p r o v e d i n A r t s . 9 5 , 9 6 , 9 7 . a b c ( 1 ) T h e S i n e - r u l e , s i n A s i n B s i n C ( 2 ) T h e C o s i n e - r u l e , c o s A = 6 2 + c 2 ~ a 2 -2 6 c ( 3 ) T h e T a n g e n t - r u l e , t a n A -B = l ^ c o t - -2 a + b 2 1 1 5 . C a s e I . O i v e n a s i d e a n d t w o a n g l e s , a s a , B , C ; f i n d A , b , c . ( 1 ) A = 1 8 0 ° - ( B + C ) . . - . A i s f o u n d . , Q s b a , a s i n B s i n B s i n A s i n A c a a s i n C s i n C s i n A s i n A T h e s e d e t e r m i n e b a n d c . E x . 1 . G i v e n a = 7 0 1 2 . 6 , B = 3 8 ° 1 2 ' 4 8 " , C = 6 0 ° ; f i n d A , b , c . S o l u t i o n . A = 1 8 0 ° -( B + C ) = 8 1 ° 4 7 ' 1 2 " . l o g a = 3 . 8 4 5 8 7 2 9 l o g s i n B = 9 . 9 7 1 4 0 3 8 c o l o g s i n A = 0 . 0 0 4 4 7 7 5 l o g b = a 8 2 1 7 5 4 2 . - . 6 = 6 6 3 3 . 6 7 . l o g a = 3 . 8 4 5 8 7 2 9 l o g s i n C = 9 . 9 3 7 5 3 0 6 c o l o g s i n A = 0 . 0 0 4 4 7 7 5 l o g c = 3 . 7 8 7 8 8 1 0 . - . c = 6 1 3 5 . 9 4 . S e e A r t . 6 9 . O B L I Q U E T R I A N G L E S . 1 7 3 E x . 2 . G i v e n a = 1 0 0 0 , B = 4 5 ° , C = 1 2 7 ° 1 9 ' ; f i n d A , b , c . A r t s . A = 7 ° 4 1 ' ; 6 = 5 2 8 8 . 8 ; c = 5 9 4 8 . 5 . 1 1 6 . C a s e I I . G i v e n t w o s i d e s a n d t h e a n g l e o p p o s i t e o n e o f t h e m , a s a , b , A ; f i n d B , C , c . ( 1 ) s i n B = - s 1 n — ; t h u s B i s f o u n d . ( 2 ) C = 1 8 0 ° -( A + B ) ; t h u s C i s f o u n d . ( 3 ) c = -s 1 n — ; t h u s c i s f o u n d . s i n A T h i s i s u s u a l l y k n o w n a s t h e a m b i g u o u s c a s e , a s s h o w n i n g e o m e t r y ( B . I I . , P r o p . 3 1 ) . T h e a m b i g u i t y i s f o u n d i n t h e e q u a t i o n -t , b s i n A s i n B = a S i n c e t h e a n g l e i s d e t e r m i n e d b y i t s s i n e , i t a d m i t s o f t w o v a l u e s , w h i c h a r e s u p p l e m e n t s o f e a c h o t h e r ( A r t . 2 9 ) . T h e r e f o r e , e i t h e r v a l u e o f B m a y b e t a k e n , u n l e s s e x c l u d e d b y t h e c o n d i t i o n s o f t h e p r o b l e m . I . I f a < b s i n A , s i n B > 1 , w h i c h i s i m p o s s i b l e ; a n d t h e r e f o r e t h e r e i s n o t r i a n g l e w i t h t h e g i v e n p a r t s . I I . I f a = b s i n A , s i n B = 1 , a n d B = 9 0 ° ; t h e r e f o r e t h e r e i s o n e t r i a n g l e — a r i g h t t r i a n g l e — w i t h t h e g i v e n p a r t s . I I I . I f a > b s i n A , a n d < b , s i n B < 1 ; h e n c e t h e r e a r e t w o v a l u e s o f B , o n e b e i n g t h e s u p p l e m e n t o f t h e o t h e r , i . e . , o n e a c u t e , t h e o t h e r o b t u s e , a n d b o t h a r e a d m i s s i b l e ; t h e r e f o r e t h e r e a r e t w o t r i a n g l e s w i t h t h e g i v e n p a r t s . I V . I f a > 6 , t h e n A > B , a n d s i n c e A i s g i v e n , B m u s t b e a c u t e ; t h u s t h e r e i s o n l y o n e t r i a n g l e w i t h t h e g i v e n p a r t s . 1 7 4 P L A N E T R I G O N O M E T R Y . T h e s e f o u r c a s e s m a y b e i l l u s t r a t e d g e o m e t r i c a l l y . D r a w A , t h e g i v e n a n g l e . M a k e A C = b ; d r a w t h e p e r p e n d i c u l a r C D , w h i c h = b s i n A . W i t h c e n t r e C a n d r a d i u s a , d e s c r i b e a c i r c l e . I . I f a < 6 s i n A , t h e c i r c l e w i l l n o t m e e t A X , a n d t h e r e f o r e n o t r i a n g l e c a n b e f o r m e d w i t h t h e g i v e n p a r t s . I I . I f a = b s i n A , t h e c i r c l e t o u c h e s A X i n B ' ; t h e r e f o r e t h e r e i s o n e t r i a n g l e , r i g h t - a n g l e d a t B . I I I . I f a > 6 s i n A , a n d < b , t h e c i r c l e c u t s A X i n t w o p o i n t s B a n d B ' , o n t h e s a m e s i d e o f A ; t h u s t h e r e B s a r e t w o t r i a n g l e s A B C a n d A B ' C , e a c h h a v i n g t h e g i v e n p a r t s , t h e a n g l e s A B C , A B ' C b e i n g s u p p l e m e n t a r y . I V . I f a > b , t h e c i r c l e c u t s A X o n o p p o s i t e s i d e s o f A , a n d o n l y t h e t r i a n g l e A B C h a s t h e g i v e n p a r t s , b e c a u s e t h e a n g l e B ' A C o f t h e t r i a n g l e A B ' C i s n o t t h e g i v e n a n g l e A , b u t i t s s u p p l e m e n t . T h e s e r e s u l t s m a y b e s t a t e d a s f o l l o w s : a < b s i n A , a • = b s i n A , a > b s i n A a n d < b , a > b , n o s o l u t i o n . o n e s o l u t i o n ( r i g h t t r i a n g l e ) , t w o s o l u t i o n s , o n e s o l u t i o n . T h e s e r e s u l t s m a y b e o b t a i n e d a l g e b r a i c a l l y t h u s : W e h a v e : 6 2 + c 2 — 2 b e c o s A ( A r t . 9 6 ) c = b c o s A ± V a 2 — b 2 s i n 2 A , O B L I Q U E T R I A N G L E S , 1 7 5 g i v i n g t w o r o o t s , r e a l a n d u n e q u a l , e q u a l o r i m a g i n a r y , a c c o r d i n g a s a > , = , o r < b s i n A . A d i s c u s s i o n o f t h e s e t w o v a l u e s o f c g i v e s t h e s a m e r e s u l t s a s a r e f o u n d i n t h e a b o v e f o u r c a s e s . W e l e a v e t h e d i s c u s s i o n a s a n e x e r c i s e f o r t h e s t u d e n t . N o t e . — W h e n t w o s i d e s a n d t h e a n g l e o p p o s i t e t h e g r e a t e r a r e g i v e n , t h e r e c a n b e n o a m b i g u i t y , f o r t h e a n g l e o p p o s i t e t h e l e s s m u s t b e a c u t e . W h e n t h e g i v e n a n g l e i s a r i g h t a n g l e o r o b t u s e , t h e o t h e r t w o a n g l e s a r e b o t h a c u t e , a n d t h e r e c a n b e n o a m b i g u i t y . I n t h e s o l u t i o n o f t r i a n g l e s t h e r e c a n b e n o a m b i g u i t y , e x c e p t w h e n a n a n g l e i s d e t e r m i n e d b y t h e s i n e o r c o s e c a n t , a n d i n n o c a s e w h a t e v e r w h e n t h e t r i a n g l e h a s a r i g h t a n g l e . E x . 1 . G i v e n a = 7 , 6 = 8 , A = 2 7 ° 4 7 ' 4 5 " ; f i n d B , C , a S o l u t i o n . l o g b = 0 . 9 0 3 0 9 0 0 l o g s i n A = 9 . 6 6 8 6 8 6 0 c o l o g a = 9 . 1 5 4 9 0 2 0 l o g a ; l o g s i n C : c o l o g s i n A : l o g c : 0 . 8 4 5 0 9 8 0 : 9 . 9 3 7 5 3 0 6 : 0 . 3 3 1 3 1 4 0 : 1 . 1 1 3 9 4 2 6 1 3 . l o g s i n B = 9 . 7 2 6 6 7 8 0 . - . B = 3 2 ° 1 2 ' 1 5 " , o r l 4 7 ° 4 7 ' 4 5 " . . - . c = . - . C = 1 2 0 ° , o r 4 ° 2 4 ' 3 0 " . T a k i n g t h e s e c o n d v a l u e o f C a s f o l l o w s : l o g a = 0 . 8 4 5 0 9 8 0 l o g s i n C = 8 . 8 8 5 7 2 3 2 c o l o g s i n A = 0 . 3 3 1 3 1 4 0 l o g c = 0 . 0 6 2 1 3 5 2 . - . c = 1 . 1 5 3 8 . T h u s , t h e r e a r e t w o s o l u t i o n s . S e e C a s e I I I . E x . 2 . G i v e n a = 3 1 . 2 3 9 , 6 = 4 9 . 5 0 5 3 , A = 3 2 ° 1 8 ' ; f i n d B , C , c . A n s . B = 5 6 ° 5 6 ' 5 6 " . 3 , o r 1 2 3 ° 3 ' 3 " . 7 ; C = 9 0 ° 4 5 ' 3 " . 7 , o r 2 4 ° 3 8 ' 5 6 " . 3 ; c = 5 8 . 4 5 6 , o r 2 4 . 3 8 2 . 1 7 6 P L A N E T R I G O N O M E T R Y . 1 1 7 . C a s e I I I . — G i v e n t w o s i d e s a n d t h e i n c l u d e d a n g l e , a s a , b , 0 ; f i n d A , B , c . ( 1 ) t a n ^ - B a - b , C = c o t — 2 a + b 2 ( A r t . 1 1 4 ) H e n c e — — — i s k n o w n , a n d — — ^ = 9 0 ° — — -2 , 2 2 A a n d B a r e f o u n d . ( 2 ) a s i n C b s i n C c = — -— - — , o r s i n A s i n B a n d t h u s c i s f o u n d a n d t h e t r i a n g l e s o l v e d . I n s i m p l e c a s e s t h e t h i r d s i d e c m a y b e f o u n d d i r e c t l y b y t h e f o r m u l a c = V a 2 + b 2 -2 a b c o s C . . . ( A r t . 9 6 ) o r t h e f o r m u l a m a y b e a d a p t e d t o l o g a r i t h m i c c a l c u l a t i o n b y t h e u s e o f a s u b s i d i a r y a n g l e ( A r t . 9 0 ) . E x . 1 . G i v e n a = 2 3 4 . 7 , b = 1 8 5 . 4 , C = 8 4 ° 3 6 ' ; f i n d A , B , 6 . S o l u t i o n . a b : a — b -a + b : C " 2 : 2 3 4 . 7 : 1 8 5 . 4 : 4 9 . 3 : 4 2 0 . 1 - = 4 2 ° 1 8 ' . l o g ( a - 6 ) = 1 . 6 9 2 8 4 6 9 c o l o g ( a + b ) = 7 . 3 7 6 6 4 7 3 l o g c o t C = 1 0 . 0 4 0 9 9 2 0 l o g t a n 9 . 1 1 0 4 8 6 2 2 A - B = 7 ° 2 0 ' 5 6 " . A + B = 4 7 ° 4 2 ' . A = 5 5 ° 2 ' 5 6 " , B = 4 0 ° 2 1 ' 4 " , C = 8 4 ° 3 6 ' , c = 2 8 5 . 0 7 4 5 . l o g b -2 . 2 6 8 1 0 9 7 l o g s i n 0 = 9 . 9 9 8 0 6 8 3 c o l o g s i n B = 0 . 1 8 8 7 8 0 4 l o g c = 2 . 4 5 4 9 5 8 4 O B L I Q U E T R I A N G L E S . 1 7 7 E x . 2 . G i v e n a = . 0 6 2 3 8 7 , 6 = . 0 2 3 4 7 5 , C = 1 1 0 ° 3 2 ' ; f i n d A , B , c . A n s . A = 5 2 ° 1 0 ' 3 3 " ; B = 1 7 ° 1 7 ' 2 7 " ; c = . 0 7 3 9 6 3 5 . 1 1 8 . C a s e I V . G i v e n t h e t h r e e s i d e s , a s a , b , c ; f i n d A , B , C . T h e s o l u t i o n i n t h i s c a s e m a y b e p e r f o r m e d b y t h e f o r m u l a e o f A r t . 9 9 . B y m e a n s o f t h e s e f o r m u l a s w e m a y c o m p u t e t w o o f t h e a n g l e s , a n d f i n d t h e t h i r d b y s u b t r a c t i n g t h e i r s u m f r o m 1 8 0 ° . B u t i n p r a c t i c e i t i s b e t t e r t o c o m p u t e t h e t h r e e a n g l e s i n d e p e n d e n t l y , a n d c h e c k t h e a c c u r a c y o f t h e w o r k b y t a k i n g t h e i r s u m . I f o n l y o n e a n g l e i s t o b e f o u n d , t h e f o r m u l a e f o r t h e s i n e s o r c o s i n e s m a y b e u s e d . I f a l l t h e a n g l e s a r e t o b e f o u n d , t h e t a n g e n t f o r m u l a e a r e t h e m o s t c o n v e n i e n t , b e c a u s e t h e n w e r e q u i r e o n l y t h e l o g a r i t h m s o f t h e s a m e f o u r q u a n t i t i e s , s , s — a , s — b , s — c , t o f i n d a l l t h e a n g l e s ; w h e r e a s t h e s i n e a n d c o s i n e f o r m u l a e r e q u i r e i n a d d i t i o n t h e l o g s o f a , b , c . T h e t a n g e n t f o r m u l a e ( A r t . 9 9 ) m a y b e r e d u c e d a s f o l l o w s : . - . t a n A 2 s — a r ( A r t . 1 0 2 ) S i m i l a r l y , r 2 s - b t a n — = r 2 s - c N o t e . — T h e q u a n t i t y r i s t h e r a d i u s o f t h e i n s c r i b e d c i r c l e ( A r t . 1 0 2 ) . 1 7 8 P L A N E T R I G O N O M E T R Y . 1 E x . 1 . G i v e n a = 1 3 , b = 1 4 , c = 1 5 ; f i n d A , B , C . S o l u t i o n . a = 1 3 l o g ( s - a ) = . 9 0 3 0 9 0 0 6 = 1 4 l o g ( s - 6 ) = . 8 4 5 0 9 8 0 c = 1 5 l o g ( s - c ) = . 7 7 8 1 5 1 3 2 s = 4 2 c o l o g s = 8 . 6 7 7 7 8 0 7 s = 2 1 . l o g r 2 = 1 . 2 0 4 1 2 0 0 s — a = 8 , l o g r = . 6 0 2 0 6 0 0 . s -6 = 7 , . - . l o g t a n : | = 9 . 6 9 8 9 7 0 0 . s — c = 6 . . : l o g t a n | = 9 . 7 5 6 9 6 2 0 . . - . l o g t a n | = 9 . 8 2 3 9 0 8 7 . . - . A = 5 3 ° 7 ' 4 8 " . 3 8 ; B = 5 9 ° 2 9 ' 2 3 " . 1 8 ; C = 6 7 ° 2 2 ' 4 8 " . 4 4 . W i t h o u t t h e u s e o f l o g a r i t h m s , t h e a n g l e s m a y b e f o u n d b y t h e c o s i n e f o r m u l a e ( A r t . 9 6 ) . T h e s e m a y s o m e t i m e s b e u s e d w i t h a d v a n t a g e , w h e n t h e g i v e n l e n g t h s o f a , b , c e a c h c o n t a i n l e s s t h a n t h r e e d i g i t s . E x . 2 . F i n d t h e g r e a t e s t a n g l e i n t h e t r i a n g l e w h o s e s i d e s a r e 1 3 , 1 4 , 1 5 . L e t a = 1 5 , b = 1 4 , c = 1 3 . T h e n A i s t h e g r e a t e s t a n g l e . r p , . t f + t f - a 1 1 4 ! + 1 3 J - 1 5 2 T h e n c o s A = — — = ' . . .— — 2 6 c 2 x 1 4 x 1 3 = 1 5 J = . 3 8 4 6 1 5 = c o s 6 7 ° 2 3 ' , n e a r l y ( b y t h e t a b l e o f n a t u r a l s i n e s ) . . - . t h e g r e a t e s t a n g l e i s 6 7 ° 2 3 ' . E X A M P L E S . 1 7 9 E X A M P L E S . 1 . G i v e n a = 2 5 4 , B = 1 6 ° , C = 6 4 ° ; f i n d 6 = 7 1 . 0 9 1 9 . 2 . G i v e n c = 3 3 8 . 6 5 , A = 5 3 ° 2 4 ' , B = 6 6 ° 2 7 ' ; f i n d a = 3 1 3 . 4 6 . . 3 . G i v e n c = 3 8 , A = 4 8 ° , B = 5 4 ° ; f i n d a = 2 8 . 8 7 , 6 = 3 1 . 4 3 . 4 . G i v e n a = 7 0 1 2 . 5 , B = 3 8 ° 1 2 ' 4 8 " , C = 6 0 ° ; f i n d b a n d c . A n s . b = 4 3 8 2 . 8 2 ; c = 6 1 3 5 . 9 4 . 5 . G i v e n « = 5 2 8 , 6 = 2 5 2 , A = 1 2 4 ° 3 4 ' ; f i n d B a n d C . A n s . B = 2 3 ° 8 ' 3 3 " ; C = 3 2 ° 1 7 ' 2 7 " . 6 . G i v r a a = 1 7 0 . 6 , 6 = 1 4 0 . 5 , B = 4 0 ° ; f i n d A a n d C . A n s . A = 5 1 ° 1 8 ' 2 1 " , o r 1 2 8 ° 4 1 ' 3 9 " ; C = 8 8 ° 4 1 ' 3 9 " , o r 1 1 ° 1 8 ' 2 1 " . 7 . G i v e n a = 9 7 , 6 = 1 1 9 , A = 5 0 ° ; f i n d B a n d C . A n s . B = 7 0 ° 0 ' 5 6 " , o r 1 0 9 ° 5 9 ' 4 " ; C = 5 9 ° 5 9 ' 4 " , o r 2 0 ° 0 ' 5 6 " . 8 . G i v e n a = 7 , 6 = 8 , A = 2 7 ° 4 7 ' 4 5 " ; f i n d B , C , c . A n s . B = 3 2 ° 1 2 ' 1 5 " , o r 1 4 7 ° 4 7 ' 4 5 " ; 0 = 1 2 0 ° , o r 4 ° 2 4 ' 3 0 " ; c = 1 3 , o r 1 . 1 5 3 8 5 . 9 . G i v e n 6 = 5 5 , c = 4 5 , A = 6 ° ; f i n d B a n d C . A n s . B = 1 4 9 ° 2 0 ' 3 1 " ; C = 2 4 ° 3 9 ' 2 9 " . 1 0 . G i v e n 6 = 1 3 1 , c = 7 2 , A = 4 0 ° ; f i n d B a n d C . A n s . B = 1 0 8 ° 3 6 ' 3 0 " ; C = 3 1 ° 2 3 ' 3 0 " . 1 1 . G i v e n a = 3 5 , 6 = 2 1 , C = 5 0 ° ; f i n d A a n d B . A n s . A = 9 3 ° 1 1 ' 4 9 " ; B = 3 6 ° 4 8 ' 1 1 " . 1 2 . G i v e n a = 6 0 1 , 6 = 2 8 9 , C = 1 0 0 ° 1 9 ' 6 " ; f i n d A a n d B . A n s . A = 5 6 ° 8 ' 4 2 " ; B = 2 3 ° 3 2 ' 1 2 " , 1 8 0 P L A N E T R I G O N O M E T R Y . 1 3 . G i v e n a = 2 2 2 , 6 = 3 1 8 , c = 4 0 6 ; f i n d A = 3 2 ° 5 7 ' 8 " . 1 4 . G i v e n a = 2 7 5 . 3 5 , 6 = 1 8 9 . 2 8 , c = 3 0 1 . 4 7 ; f i n d A , B , C . A n s . A = 6 3 ° 3 0 ' 5 7 " ; B = 3 7 ° 5 8 ' 2 0 " ; C = 7 8 ° 3 0 ' 4 3 " . 1 5 . G i v e n a = 5 2 3 8 , 6 = 5 6 6 2 , c = 9 3 8 4 ; f i n d A a n d B . A n s . A = 2 9 ° 1 7 ' 1 6 " ; B = 3 1 ° 5 5 ' 3 1 " . 1 6 . G i v e n a = 3 1 7 , b = 5 3 3 , c = 5 1 0 ; f i n d A , B , C . A n s . A = 3 5 ° 1 8 ' 0 " ; B = 7 6 ° 1 8 ' 5 2 " ; C = 6 8 ° 2 3 ' 8 " . 1 1 9 . A r e a o f a T r i a n g l e ( A r t . 1 0 1 ) . E X A M P L E S . F i n d t h e a r e a : 1 . G i v e n a = 1 1 6 . 0 8 2 , 6 = 1 0 0 , C = 1 1 8 ° 1 5 ' 4 1 " . A n s . 5 1 1 2 . 2 5 . 2 . G f i v e n a = 8 , 6 = 5 , C = 6 0 ° . 1 7 . 3 2 0 5 . 3 . G i v e n 6 = 2 1 . 5 , c = 3 0 . 4 5 6 , A = 4 1 ° 2 2 ' . 2 1 6 . 3 7 2 . 4 . G i v e n a = 7 2 . 3 , A = 5 2 ° 3 5 ' , B = 6 3 ° 1 7 ' . 2 6 4 4 . 9 4 . 5 . G i v e n 6 = 1 0 0 , A = 7 6 ° 3 8 ' 1 3 " , C = 4 0 ° 5 ' . 3 5 0 6 . 8 1 5 . 6 . G i v e n a = 3 1 . 3 2 5 , B = 1 3 ° 5 7 ' 2 " , A = 5 3 ° 1 1 ' 1 8 " . A n s . 1 3 5 . 3 5 4 5 . 7 . G i v e n a = . 5 8 2 , 6 = . 6 0 1 , c = . 4 2 7 . . 1 1 7 6 5 5 . 8 . G i v e n a = 4 0 8 , 6 = 4 1 , c = 4 0 1 . 8 1 6 0 . 9 . G i v e n a = . 9 , 6 = 1 . 2 , c = 1 . 5 . . 5 4 . 1 0 . G i v e n a = 2 1 , 6 = 2 0 , c = 2 9 . 2 1 0 . 1 1 . G i v e n a = 2 4 , 6 = 3 0 , c = 1 8 . 2 1 6 . 1 2 . G i v e n a = 6 3 . 8 9 , 6 = 1 3 8 . 2 4 , c = 1 2 1 . 1 5 . 3 8 6 9 . 2 . H E I G H T O F A N O B J E C T . 1 8 1 M E A S U R E M E N T O F H E I G H T S A N D D I S T A N C E S . 1 2 0 . D e f i n i t i o n s . — O n e o f t h e m o s t i m p o r t a n t a p p l i c a t i o n s o f T r i g o n o m e t r y i s t h e d e t e r m i n a t i o n o f t h e h e i g h t s a n d d i s t a n c e s o f o b j e c t s w h i c h c a n n o t b e a c t u a l l y m e a s u r e d . T h e a c t u a l m e a s u r e m e n t , w i t h s c i e n t i f i c a c c u r a c y , o f a l i n e o f a n y c o n s i d e r a b l e l e n g t h , i s a v e r y l o n g a n d d i f f i c u l t o p e r a t i o n . B u t t h e a c c u r a t e m e a s u r e m e n t o f a n a n g l e , w i t h p r o p e r i n s t r u m e n t s , c a n b e m a d e w i t h c o m p a r a t i v e e a s e a n d r a p i d i t y . B y t h e a i d o f t h e S o l u t i o n o f T r i a n g l e s w e c a n d e t e r m i n e : ( 1 ) T h e d i s t a n c e b e t w e e n p o i n t s w h i c h a r e i n a c c e s s i b l e . ( 2 ) T h e m a g n i t u d e o f a n g l e s w h i c h c a n n o t b e p r a c t i c a l l y o b s e r v e d . ( 3 ) T h e r e l a t i v e h e i g h t s o f d i s t a n t a n d i n a c c e s s i b l e p o i n t s . A v e r t i c a l l i n e i s t h e l i n e a s s u m e d b y a p l u m m e t w h e n f r e e l y s u s p e n d e d b y a c o r d , a n d a l l o w e d t o c o m e t o r e s t . A v e r t i c a l p l a n e i s a n y p l a n e c o n t a i n i n g a v e r t i c a l l i n e . A h o r i z o n t a l p l a n e i s a p l a n e p e r p e n d i c u l a r t o a v e r t i c a l l i n e . A v e r t i c a l a n g l e i s o n e l y i n g i n a v e r t i c a l p l a n e . A h o r i z o n t a l a n g l e i s o n e l y i n g i n a h o r i z o n t a l p l a n e . A n a n g l e o f e l e v a t i o n i s a v e r t i c a l a n g l e h a v i n g o n e s i d e h o r i z o n t a l a n d t h e o t h e r a s c e n d i n g . A n a n g l e o f d e p r e s s i o n i s a v e r t i c a l a n g l e h a v i n g o n e s i d e h o r i z o n t a l a n d t h e o t h e r d e s c e n d i n g . B y d i s t a n c e i s m e a n t t h e h o r i z o n t a l d i s t a n c e , u n l e s s o t h e r w i s e n a m e d . B y h e i g h t i s m e a n t t h e v e r t i c a l h e i g h t a b o v e o r b e l o w t h e h o r i z o n t a l p l a n e o f t h e o b s e r v e r . F o r a d e s c r i p t i o n o f t h e r e q u i s i t e i n s t r u m e n t s , a n d t h e m e t h o d o f u s i n g t h e m , t h e s t u d e n t i s r e f e r r e d t o b o o k s o n p r a c t i c a l s u r v e y i n g . S e e J o h n s o n ' s S u r v e y i n g . G i l l e s p i e ' s S u r v e y i n g , C l a r k e ' s G e o d e s y , G o r e ' s G e o d e s y , e t c . 1 8 2 P L A N E T R I G O N O M E T R Y . 1 2 1 . T o f i n d t h e H e i g h t o f a n O b j e c t s t a n d i n g o n a H o r i z o n t a l P l a n e , t h e B a s e o f t h e O b j e c t b e i n g A c c e s s i b l e . L e t B C b e a v e r t i c a l o b j e c t , s u c h a s a c h u r c h s p i r e o r a t o w e r . F r o m t h e b a s e C m e a s u r e a h o r i z o n t a l l i n e C A . A t t h e p o i n t A m e a s u r e t h e a n g l e o f e l e v a t i o n C A B . W e c a n t h e n d e t e r m i n e t h e h e i g h t o f t h e o b j e c t B C ; f o r B C = A C t a n C A B . E X A M P L E S . 1 . I f A C = 1 0 0 f e e t a n d C A B = 6 0 ° , f i n d B C . A n s . 1 7 3 . 2 f e e t . 2 . I f A C = 1 2 5 f e e t a n d C A B = 5 2 ° 3 4 ' , f i n d B C . A n s . 1 6 3 . 3 f e e t . 3 . A C , t h e b r e a d t h o f a r i v e r , i s 1 0 0 f e e t . A t t h e p o i n t A , o n o n e b a n k , t h e a n g l e o f e l e v a t i o n o f B , t h e t o p o f a t r e e o n t h e o t h e r b a n k d i r e c t l y o p p o s i t e , i s 2 5 ° 3 7 ' ; f i n d t h e h e i g h t o f t h e t r e e . A n s . 4 7 . 9 f e e t . 1 2 2 . T o f i n d t h e H e i g h t a n d D i s t a n c e o f a n I n a c c e s s i b l e O b j e c t o n a H o r i z o n t a l P l a n e . L e t C D b e t h e o b j e c t , w h o s e b a s e D i s i n a c c e s s i b l e ; a n d l e t i t b e r e q u i r e d / t o f i n d t h e h e i g h t C D , a n d i t s h o r i z o n -t a l d i s t a n c e f r o m A , t h e n e a r e s t a c c e s -B - s i b l e p o i n t . ( 1 ) A t A i n t h e h o r i z o n t a l l i n e B A D o b s e r v e t h e A D A C = a ; m e a s u r e A B = a , a n d a t B o b s e r v e t h e Z D B C = / ? . T h e n C A = a s i n / ? ( A r t . 9 5 ) s m ( a -/ J ) v H E I G H T O F A N O B J E C T . 1 8 3 C D = C A s i n « : a s i n a s i n B s i n ( a — / J ) a n d . - r . . r . a s i n B c o s a A D = A C c o s a = — — " s i n ( « — B ) ( 2 ) W h e n t h e l i n e B A c a n n o t b e m e a s u r e d d i r e c t l y t o w a r d t h e o b j e c t . A t A o b s e r v e t h e v e r t i c a l Z C A D = a , a n d t h e h o r i z o n t a l Z D A B = B ; m e a s u r e A B = a , a n d a t B o b s e r v e t h e Z D B A = y . T h e n A D = a s i n -s i n O S + y ) C D = A D t a n a _ a s i n y t a n a s i n ( / 8 + y ) E X A M P L E S . 1 . A r i v e r 3 0 0 f e e t w i d e r u n s a t t h e f o o t o f a t o w e r , w h i c h s u b t e n d s a n a n g l e o f 2 2 ° 3 0 ' a t t h e e d g e o f t h e r e m o t e b a n k ; f i n d t h e h e i g h t o f t h e t o w e r . A r t s . 1 2 4 . 2 6 f e e t . 2 . A t 3 6 0 f e e t f r o m t h e f o o t o f a s t e e p l e t h e e l e v a t i o n i s h a l f w h a t i t i s a t 1 3 5 f e e t ; f i n d i t s h e i g h t . A n s . 1 8 0 f e e t . 3 . A p e r s o n s t a n d i n g o n t h e b a n k o f a r i v e r o b s e r v e s t h e a n g l e s u b t e n d e d b y a t r e e o n t h e o p p o s i t e b a n k t o b e 6 0 ° , a n d w h e n h e r e t i r e s 4 0 f e e t f r o m t h e r i v e r , s b a n k h e f i n d s t h e a n g l e t o b e 3 0 ° ; f i n d t h e h e i g h t o f t h e t r e e a n d t h e b r e a d t h o f t h e r i v e r . A n s . 2 0 V 3 ; 2 0 . 4 . W h a t i s t h e h e i g h t o f a h i l l w h o s e a n g l e o f e l e v a t i o n , t a k e n a t t h e b o t t o m , w a s 4 6 ° , a n d 1 0 0 y a r d s f a r t h e r o f f , o n a l e v e l w i t h t h e b o t t o m , t h e a n g l e w a s 3 1 ° ? A n s . 1 4 3 . 1 4 y a r d s . 1 8 4 P L A N E T R I G O N O M E T R Y . 1 2 3 . T o f i n d t h e H e i g h t o f a n I n a c c e s s i b l e O b j e c t s i t u a t e d a b o v e a H o r i z o n t a l P l a n e , a n d i t s H e i g h t a b o v e t h e P l a n e . L e t C D b e t h e o b j e c t , a n d l e t A a n d B b e t w o p o i n t s i n t h e h o r i z o n t a l p l a n e , a n d i n t h e s a m e v e r t i c a l p l a n e w i t h C D . A t A , i n t h e h o r i z o n t a l l i n e B A E , o b s e r v e t h e A C A E = a , a n d D A E = y ; m e a s u r e A B = a , a n d a t B o b s e r v e t h e Z C B E = 0 . T h e n A l s o , C E = A E = D E = C D = a s i n a s i n / ? ( A r t 1 2 2 ) s i n ( a — / 3 ) a c o s a s i n f t . . . . ( A r t . 1 2 2 ) s i n ( a — ( 3 ) a c o s a s i n ( 3 t a n y s i n ( a — / 3 ) a s l n ) 8 j s i n a -c o s a t a n y \ s i n ( a — ( 3 ) a s i n / 3 s i n ( a — y ) c o s y s i n ( « — / } ) E X A M P L E S . 1 . A m a n 6 f e e t h i g h s t a n d s a t a d i s t a n c e o f 4 f e e t 9 i n c h e s f r o m a l a m p - p o s t , a n d i t i s o b s e r v e d t h a t h i s s h a d o w i s 1 9 f e e t l o n g : f i n d t h e h e i g h t o f t h e l a m p . A n s . 1 \ f e e t . 2 . A f l a g s t a f f , 2 5 f e e t h i g h , s t a n d s o n t h e t o p o f a c l i f f , a n d f r o m a p o i n t o n t h e s e a s h o r e t h e a n g l e s o f e l e v a t i o n o f t h e h i g h e s t a n d l o w e s t p o i n t s o f t h e f l a g s t a f f a r e o b s e r v e d t o b e 4 7 ° 1 2 ' a n d 4 5 ° 1 3 ' r e s p e c t i v e l y : f i n d t h e h e i g h t o f t h e c l i f f . A n s . 3 4 8 f e e t . 3 . A c a s t l e s t a n d i n g o n t h e t o p o f a c l i f f i s o b s e r v e d f r o m t w o s t a t i o n s a t s e a , w h i c h a r e i n l i n e w i t h i t ; t h e i r H E I G H T O F A N O B J E C T . 1 8 5 d i s t a n c e i s a q u a r t e r o f a m i l e : t h e e l e v a t i o n o f t h e t o p o f t h e c a s t l e , s e e n f r o m t h e r e m o t e s t a t i o n , i s 1 6 ° 2 8 ' ; t h e e l e v a t i o n s o f t h e t o p a n d b o t t o m , s e e n f r o m t h e n e a r s t a t i o n , a r e 5 2 ° 2 4 ' a n d 4 8 ° 3 8 ' r e s p e c t i v e l y : ( 1 ) w h a t i s i t s h e i g h t , a n d ( 2 ) w h a t i t s e l e v a t i o n a b o v e t h e s e a ? A n s . ( 1 ) 6 0 . 8 2 f e e t ; ( 2 ) 4 4 5 . 2 3 f e e t . 1 2 4 . T o f i n d t h e D i s t a n c e o f a n O b j e c t o n a H o r i z o n t a l P l a n e , f r o m O b s e r v a t i o n s m a d e a t T w o P o i n t s i n t h e S a m e V e r t i c a l L i n e , a b o v e t h e P l a n e . L e t t h e p o i n t s o f o b s e r v a t i o n A a n d B b e i n t h e s a m e v e r t i c a l l i n e , a n d a t a g i v e n d i s t a n c e f r o m e a c h o t h e r ; l e t C b e t h e p o i n t o b s e r v e d , w h o s e h o r i z o n t a l d i s t a n c e C D a n d v e r t i c a l d i s t a n c e A D a r e r e q u i r e d . M e a s u r e t h e a n g l e s o f d e p r e s s i o n , 6 B C , a A C , e q u a l t o a a n d / J r e s p e c t i v e l y , a n d d e n o t e A B b y a . T h e n B D = C D t a n « , A D = C D t a n j 3 . . : a = C D ( t a n a — t a n a c o s a c o s / ? s i n ( « -0 ) , a c o s a s i n / ? s i n ( a — f t ) a n d C D : A D : E X A M P L E S . 1 . F r o m t h e t o p o f a h o u s e , a n d f r o m a w i n d o w 3 0 f e e t b e l o w t h e t o p , t h e a n g l e s o f d e p r e s s i o n o f a n o b j e c t o n t h e g r o u n d a r e 1 5 ° 4 0 ' a n d 1 0 ° : f i n d ( 1 ) t h e h o r i z o n t a l d i s t a n c e o f t h e o b j e c t , a n d ( 2 ) t h e h e i g h t o f t h e h o u s e . A n s . ( 1 ) 2 8 8 . 1 f e e t ; ( 2 ) 8 0 . 8 f e e t . 2 . F r o m t h e t o p a n d b o t t o m o f a c a s t l e , w h i c h i s 6 8 f e e t h i g h , t h e d e p r e s s i o n s o f a s h i p a t s e a a r e o b s e r v e d t o b e 1 6 ° 2 8 ' a n d 1 4 ° : f i n d i t s d i s t a n c e . A n s . 5 7 0 . 2 y a r d s . 1 8 6 P L A N E T B I G O N O M E T R Y . 1 2 5 . T o f i n d t h e D i s t a n t b e t w e e n T w o I n a c c e s s i b l e O b j e c t s o n a H o r i z o n t a l P l a n e . L e t C a n d D b e t h e t w o i n a c c e s s i b l e o b j e c t s . M e a s u r e a b a s e l i n e A B , f r o m w h o s e e x t r e m i t i e s C a n d D a r e v i s i b l e . A t A o b s e r v e t h e a n g l e s C A D , D A B ; a n d a t B o b s e r v e t h e a n g l e s C B A a n d C B D . T h e n , i n t h e t r i a n g l e A B C , w e k n o w t w o a n g l e s a n d t h e s i d e A B . . - . A C m a y b e f o u n d . I n t h e t r i a n g l e A B D w e k n o w t w o a n g l e s a n d t h e s i d e A B . . - . A D m a y b e f o u n d . L a s t l y , i n t h e t r i a n g l e A C D , A C a n d A D h a v e b e e n d e t e r m i n e d , a n d t h e i n c l u d e d a n g l e C A D h a s b e e n m e a s u r e d ; a n d t h u s C D c a n b e f o u n d . E X A M P L E S . 1 . L e t A B = 1 0 0 0 y a r d s , t h e a n g l e s B A C , B A D = 7 6 ° 3 0 ' a n d 4 4 ° 1 0 ' , r e s p e c t i v e l y ; a n d t h e a n g l e s A B D , A B C = 8 1 ° 1 2 ' a n d 4 6 ° 5 ' , r e s p e c t i v e l y : f i n d t h e d i s t a n c e b e t w e e n C a n d D . A n s . 6 6 9 . 8 y a r d s . 2 . A a n d B a r e t w o t r e e s o n o n e s i d e o f a r i v e r ; a t t w o s t a t i o n s P a n d Q o n t h e o t h e r s i d e o b s e r v a t i o n s a r e t a k e n , a n d i t i s f o u n d t h a t t h e a n g l e s A P B , B P Q , A Q P a r e e a c h e q u a l t o 3 0 ° , a n d t h a t t h e a n g l e A Q B i s e q u a l t o 6 0 ° . I f P Q = a , s h o w t h a t A B = 2 V 2 1 . 6 1 2 6 . T h e D i p o f t h e H o r i z o n . — S i n c e t h e s u r f a c e o f t h e e a r t h i s s p h e r i c a l , i t i s o b v i o u s t h a t a n o b j e c t o n i t w i l l b e v i s i b l e o n l y f o r a c e r t a i n d i s t a n c e d e p e n d i n g o n i t s h e i g h t ; a n d , c o n v e r s e l y , t h a t a t a c e r t a i n h e i g h t a b o v e t h e g r o u n d t h e v i s i b l e h o r i z o n w i l l b e l i m i t e d . T H E D I P O F T H E H O R I Z O N . 1 8 7 L e t 0 b e t h e c e n t r e o f t h e e a r t h , P a p o i n t a b o v e t h e s u r f a c e , P D a t a n g e n t t o t h e s u r f a c e a t D . T h e n D i s a p o i n t o n t h e t e r r e s t r i a l h o r i z o n ; a n d C P D , w h i c h i s t h e a n g l e o f d e p r e s s i o n o f t h e m o s t d i s t a n t p o i n t o n t h e h o r i z o n s e e n f r o m P , i s c a l l e d t h e d i p o f t h e h o r i z o n a t P . T h e a n g l e D O P i s e q u a l t o i t . D e n o t e t h e a n g l e C P D b y 6 , t h e h e i g h t A P b y h , a n d t h e r a d i u s O D b y r . T h e n B h = O P -O A = r s e c 6 -r _ r ( 1 — c o s 6 ) c o s 6 h c o s 6 . : r = 1 — c O S 0 t > t i i a h s i n 6 P D = r t a n 6 = 1 — c o s 6 = f t c o t ( ^ . . ( A r t . 4 8 , E x . 8 ) A l s o P D ! = P A x P B = 7 i ( H 2 r ) . . . ( G e o m . ) S i n c e , i n a l l c a s e s w h i c h c a n o c c u r i n p r a c t i c e , h i s v e r y s m a l l c o m p a r e d w i t h 2 r , w e h a v e a p p r o x i m a t e l y P D 2 = 2 A r . L e t n = t h e n u m b e r o f m i l e s i n P D , h = t h e f e e t i n P A , a n d r = 4 0 0 0 m i l e s n e a r l y . T h e n f t = T ' D 2 = ( 5 2 8 0 n ) 2 2 r 8 0 0 0 x 5 2 8 0 _ 5 2 8 0 n 2 = 5 . 2 8 i _ 2 , 8 0 0 0 8 n 3 n ' I t w i l l b e n o t i c e d t h a t n i s a n u m b e r m e r e l y , a n d t h a t t h e r e s u l t w i l l b e i n f e e t , s i n c e t h e m i l e s h a v e b e e n r e d u c e d t o f e e t . 1 8 8 P L A N E T R I G O N O M E T R Y . T h a t i s , t h e h e i g h t a t w h i c h o b j e c t s c a n b e s e e n v a r i e s a s t h e s q u a r e o f t h e d i s t a n c e . T h u s , i f n = 1 m i l e , w e h a v e h = | f e e t = 8 i n c h e s ; i f n = 2 m i l e s , h = § - 2 2 = | f e e t , e t c . , e t c . T h u s i t a p p e a r s t h a t a n o b j e c t l e s s t h a n 8 i n c h e s a b o v e t h e s u r f a c e o f s t i l l w a t e r w i l l b e i n v i s i b l e t o a n e y e o n t h e s u r f a c e a t t h e d i s t a n c e o f a m i l e . E x a m p l e . F r o m a b a l l o o n , a t a n e l e v a t i o n o f 4 m i l e s , t h e d i p o f t h e s e a - h o r i z o n i s o b s e r v e d t o b e 2 ° 3 3 ' 4 0 " : f i n d ( 1 ) t h e d i a m e t e r o f t h e e a r t h , a n d ( 2 ) t h e d i s t a n c e o f t h e h o r i z o n f r o m t h e b a l l o o n . A n s . ( 1 ) 8 0 0 1 . 2 4 m i l e s ; ( 2 ) 1 7 8 . 9 4 4 m i l e s . 1 2 7 . P r o b l e m o f P o t h e n o t o r o f S n e l l i u s . — T o d e t e r m i n e a p o i n t i n t h e p l a n e o f a g i v e n t r i a n g l e , a t w h i c h t h e s i d e s o f t h e t r i a n g l e s u b t e n d g i v e n a n g l e s . L e t A B C b e t h e g i v e n t r i a n g l e , a n d P t h e r e q u i r e d p o i n t . J o i n P w i t h A , B , C . L e t t h e g i v e n a n g l e s A P C , B P C b e d e n o t e d b y a , / } , a n d t h e u n k n o w n a n g l e s P A C , P B C b y x , y r e s p e c t i v e l y ; t h e n a a n d / ? a r e k n o w n ; a n d w h e n x a n d y a r e f o u n d , t h e p o s i t i o n o f P c a n b e d e t e r m i n e d , f o r t h e d i s t a n c e s P A a n d P B c a n b e f o u n d b y s o l v i n g t h e t r i a n g l e s P A C , P B C . W e h a v e x - \ - y — 2 i r — a — j 3 — C . . . . A l s o 6 s i D x = O s i n y = p a s i n a s i n f } A s s u m e a n a u x i l i a r y a n g l e < j > s u c h t h a t t a n 4 , = ™^ ; b s i n / ? t h e n t h e v a l u e o f < f > c a n b e f o u n d f r o m t h e t a b l e s . ( 1 ) E X A M P L E S . 1 8 9 T h u s , — — -t a n < j > . s i n y s i n x — s i n ? / t a n A — 1 , , -. . r 0 \ . - . ^ -" — - = t a n ( < j > — 4 5 ) s i n x + s i n y t a n + 1 [ ( 1 4 ) o f A r t . 6 1 ] , . - . t a n £ ( a ; — y ) = t a n £ ( a ; + y ) t a n ( ^ > — 4 5 ° ) [ ( 1 3 ) o f A r t . 6 1 ] . = t a n ( 4 5 ° -< f > ) t a n \ ( « + / ? + C ) . ( 2 ) t h u s f r o m ( 1 ) a n d ( 2 ) x a n d y a r e f o u n d . E X A M P L E S . S o l v e t h e f o l l o w i n g r i g h t t r i a n g l e s : 1 . G i v e n a = 5 1 . 3 0 3 , c = 1 5 0 ; f i n d A = 2 0 ° , B = 7 0 ° , 6 = 1 4 0 . 9 5 . 2 . G i v e n a = 1 5 7 . 3 3 , c = 2 5 0 ; f i n d A = 3 9 ° , B = 5 1 ° , 6 = 1 9 4 . 2 8 . 3 . G i v e n a = 1 0 4 , c = 1 8 5 ; f i n d A = 3 4 ° 1 2 ' 1 9 " . 6 , B = 5 5 ° 4 7 ' 4 0 " . 4 , 6 = 1 5 3 . 4 . G i v e n a = 3 0 4 , c = 4 2 5 ; f i n d A = 4 5 ° 4 0 ' 2 " . 3 , B = 4 4 ° 1 9 ' 5 7 " . 7 , 6 = 2 9 7 . 5 . G i v e n 6 = 3 , c = 5 ; f i n d A = 5 3 ° 7 ' 4 8 " . 4 , B = 3 6 ° 5 2 ' l l " . 6 , a = 4 . 6 . G i v e n 6 = 1 5 , c = 1 7 ; f i n d A = 2 8 ° 4 ' 2 0 " . 9 , B = 6 1 ° 5 5 ' 3 9 " . l , a = 8 . 7 . G i v e n 6 = 2 1 , c = 2 9 ; f i n d A = 4 3 ° 3 6 ' 1 0 " . l , B = 4 6 ° 2 3 ' 4 9 " . 9 , o = 2 0 . 8 . G i v e n 6 = 7 , c = 2 5 ; f i n d A = 7 3 ° 4 4 ' 2 3 " . 3 , B = 1 6 ° 1 5 ' 3 6 " . 7 , a = 2 4 . 1 9 0 P L A N E T R I G O N O M E T R Y . 9 . G i v e n 6 = 3 3 , f i n d A = 5 9 ° 2 9 ' 2 3 " . 2 , 1 0 . G i v e n c = 6 2 5 , f i n d a = 4 3 4 . 1 6 1 , 1 1 . G i v e n c = 3 0 0 , f i n d a = 2 3 6 . 4 0 3 , 1 2 . G i v e n c = 1 3 , f i n d B = 2 2 ° 3 7 ' 1 1 " . 5 , 1 3 . G i v e n A = 7 7 ° 1 9 ' 1 0 " . 6 , f i n d B = 1 2 ° 4 0 ' 4 9 " . 4 , 1 4 . G i v e n B = 4 8 ° 5 3 ' 1 6 " . 5 , f i n d A : = 4 1 ° 6 ' 4 3 " . 5 , 1 5 . G i v e n B = 6 4 ° 0 ' 3 8 " . 8 , f i n d A = 2 5 ° 5 9 ' 2 1 " . 2 , 1 6 . G i v e n A = 7 7 ° 1 9 ' 1 0 " . 6 , f i n d B = 1 2 ° 4 0 ' 4 9 " . 4 , 1 7 . G i v e n A = 8 7 ° 1 2 ' 2 0 " . 3 , f i n d B = 2 ° 4 7 ' 3 9 " . 7 , 1 8 . G i v e n A = 3 2 ° 3 1 ' 1 3 " . 5 , f i n d B = 5 7 ° 2 8 ' 4 6 " . 5 , 1 9 . G i v e n A = 8 2 ° 4 1 ' 4 4 " , f i n d B = 7 ° 1 8 ' 1 6 " , 2 0 . G i v e n A = 7 5 ° 2 3 ' 1 8 " . 5 , f i n d B = 1 4 ° 3 6 ' 4 1 " . 5 , 2 1 . G i v e n B = 8 7 ° 4 9 ' 1 0 " , f i n d A = 2 ° 1 0 ' 5 0 " , c = 6 5 ; B = 3 0 ° 3 0 ' 3 6 " . 8 , a = 5 6 . A = 4 4 ° ; 6 = 4 4 9 . 5 8 7 . A = 5 2 ° ; 6 = 1 8 4 . 6 9 8 . A = 6 7 ° 2 2 ' 4 8 " . 5 > a = 1 2 6 = 5 . c = 4 1 ; a = 4 0 , 6 = 9 . c = 7 3 ; a = 4 8 , 6 = 5 5 . c = 8 9 ; a = 3 9 , 6 = 8 0 . a = 4 0 ; 6 = 9 , c = 4 1 . a = 8 4 0 ; 6 = 4 1 , c = 8 4 1 . a = 3 3 6 ; 6 = 5 2 7 , c = 6 2 5 . a = 1 1 0 0 ; 6 = 1 4 1 , c = 1 1 0 9 . b = 1 9 5 ; a = 7 4 8 , c = 7 7 3 . 6 = 4 2 5 3 6 . 3 7 ; a = 1 6 1 9 . 6 2 6 , c = 4 2 5 6 7 . 2 . E X A M P L E S . 1 9 1 2 2 . G i v e n A = 8 8 ° 5 9 ' 6 = 2 . 2 3 4 8 7 5 ; f i n d B = 1 ° V , a = 1 2 5 . 9 3 6 5 , c = 1 2 5 . 9 5 6 3 . 2 3 . G i v e n A = 3 5 ° 1 6 ' 2 5 " , a = 3 8 8 . 2 6 4 7 ; f i n d B = 5 4 ° 4 3 ' 3 5 , " 6 = 5 4 8 . 9 0 1 8 , c = 6 7 2 . 3 4 1 2 . 2 4 . G i v e n a = 7 6 9 4 . 5 , 6 = 8 4 7 1 ; f i n d A = 4 2 ° 1 5 ' , B = 4 7 ° 4 5 ' , c = 1 1 4 4 4 . 2 5 . G i v e n a = 7 3 6 , 6 = 2 7 3 ; f i n d A = 6 9 ° 3 8 ' 5 6 " . 3 , B = 2 0 ° 2 1 ' 3 " . 7 , c = 7 8 5 . 2 6 . G i v e n a = 2 0 0 , 6 = 6 0 9 ; f i n d A = 1 8 ° 1 0 ' 5 0 " , B = 7 1 ° 4 9 ' 1 0 " , c = 6 4 1 . 2 7 . G i v e n a = 2 7 6 , 6 = 4 9 3 ; f i n d A = 2 9 ° 1 4 ' 3 0 " . 3 , B = 6 0 ° 4 5 ' 2 9 . " 7 , c = 5 6 5 . 2 8 . G i v e n a = 3 9 6 , 6 = 4 0 3 ; f i n d A = 4 4 ° 2 9 ' 5 3 " , B = 4 5 ° 3 0 ' 7 " , c = 5 6 5 . 2 9 . G i v e n a = 2 7 8 . 3 , 6 = 3 1 4 . 6 ; f i n d A = 4 1 ° 3 0 ' , B = 4 8 ° 3 0 ' , c = 4 2 0 . 3 0 . G i v e n a = 3 7 2 , 6 = 4 2 3 . 9 2 4 ; f i n d A = 4 1 ° 1 6 ' 2 " . 7 , B = 4 8 ° 4 3 ' 5 7 " . 3 , c = 5 0 4 . 3 1 . G i v e n a = 5 2 6 . 2 , 6 = 4 1 4 . 7 4 5 ; S o l v e t h e f o l l o w i n g o b l i q u e t r i a n g l e s : f i n d A = 5 1 ° 4 5 ' 1 8 " . 7 , B = 3 8 ° 1 4 ' 4 1 " . 3 , c = 6 7 0 . 3 2 . G i v e n B = 5 0 ° 3 0 ' , 0 = 1 2 2 ° 9 ' , a = 9 0 ; f i n d A = 7 ° 2 1 ' , 6 = 5 4 2 . 8 5 0 , c = 5 9 5 . 6 3 8 . 3 3 . G i v e n A = 8 2 ° 2 0 ' , B = 4 3 ° 2 0 ' , a = 4 7 9 ; f i n d C = 5 4 ° 2 0 ' , 6 = 3 3 1 . 6 5 7 , c = 3 9 2 . 4 7 3 . 3 4 . G i v e n A = 7 9 ° 5 9 ' , B = 4 4 ° 4 1 ' , a = 7 9 5 ; f i n d C = 5 1 ° 2 0 ' , 6 = 5 6 7 . 8 8 8 , c = 6 6 3 . 9 8 6 . 1 9 2 P L A N E T R I G O N O M E T R Y . 3 5 . G i v e n B = 3 7 ° 5 8 ' , C = 6 5 ° 2 ' , a = 9 9 9 ; f i n d A = 7 7 ° 0 ' , 6 = 6 3 0 . 7 7 1 , c = 8 2 9 . 4 8 0 . 3 6 . G i v e n A = 7 0 ° 5 5 ' , C = 5 2 ° 9 ' , a = 6 4 1 2 ; f i n d B = 5 6 ° 5 6 ' , 6 = 5 6 8 6 . 0 0 , c = 5 3 5 7 . 5 0 . 3 7 . G i v e n A = 4 8 ° 2 0 ' , B = 8 1 ° 2 ' 1 6 " , b = 5 . 7 5 ; f i n d C = 5 0 ° 3 7 ' 4 4 " , a = 4 . 3 4 8 5 , c = 4 . 5 . 3 8 . G i v e n A = 7 2 ° 4 ' , B = 4 1 ° 5 6 ' 1 8 " , c = 2 4 ; f i n d C = 6 5 ° 5 9 ' 4 2 " , a = 2 4 . 9 9 5 , b = 1 7 . 5 5 9 . 3 9 . G i v e n A = 4 3 ° 3 6 ' 1 0 " . l , C = 1 2 4 ° 5 8 ' 3 3 " . 6 , b = 2 9 ; f i n d B = 1 1 ° 2 5 ' 1 6 " . 3 , a = 1 0 1 , c = 1 2 0 . 4 0 . G i v e n A = 6 9 ° 5 9 ' 2 " . 5 , C = 7 0 ° 4 2 ' 3 0 " , b = 1 4 9 ; f i n d B = 3 9 ° 1 8 ' 2 7 " . 5 , a = 2 2 1 , c = 2 2 2 . 4 1 . G i v e n A = 2 1 ° 1 4 ' 2 5 " , a = 3 4 5 , b = 6 9 5 ; f i n d B = 4 6 ° 5 2 ' 1 0 " , C = 1 1 1 ° 5 3 ' 2 5 " , c = 8 8 3 . 6 5 . o r B ' = 1 3 3 ° 7 ' 5 0 " , C ' = 2 5 ° 3 7 ' 4 5 " , c ' = 4 1 1 . 9 2 . 4 2 . G i v e n A = 4 1 ° 1 3 ' 0 " , a = 7 7 . 0 4 > 6 = 9 1 . 0 6 ; f i n d B = 5 1 ° 9 ' 6 " , C = 8 7 ° 3 7 ' 5 4 " , c = 1 1 6 . 8 2 , o r B ' = 1 2 8 ° 5 0 ' 5 4 " , C ' = 9 ° 5 6 ' 6 " , c ' = 2 0 . 1 7 2 . 4 3 . G i v e n A = 2 1 ° 1 4 ' 2 5 " , 3 0 9 , 6 = 3 6 0 ; f i n d B = 2 4 ° 5 1 ' 5 4 " , c = 1 3 3 ° 4 7 ' 4 1 " , c = 6 1 5 . 6 7 , o r B ' = 1 5 5 ° 2 ' 6 " , C ' = 3 ° 4 3 ' 2 9 " , c ' = 5 5 . 4 1 . 4 4 . G i v e n B = 6 8 ° 1 0 ' 2 4 " , a = 8 3 . 8 5 6 , 6 = 8 3 . 1 5 3 ; f i n d A = 6 5 ° 5 ' 1 0 " , C = 4 5 ° 4 4 ' 2 6 " , c = 6 5 . 6 9 6 . 4 5 . G i v e n B = 6 0 ° 0 ' 3 2 " , a = 2 7 . 5 4 8 , b = 3 5 . 0 5 5 ; f i n d A = 4 2 ° 5 3 ' 3 4 " , C = 7 7 ° 5 ' 5 4 " , c = 3 9 . 4 5 3 . 4 6 . G i v e n A = 6 0 ° , a = 1 2 0 , 6 = 8 0 ; f i n d B = 3 5 ° 1 5 ' 5 2 " , C = 8 4 ° 4 4 ' 8 " , c = 1 3 7 . 9 7 9 6 . E X A M P L E S . 1 9 3 4 7 . G i v e n A = 5 0 ° , « = 1 1 9 , 6 = 9 7 ; f i n d B = 3 8 ° 3 8 ' 2 4 " , C = 9 1 ° 2 1 ' 3 6 " , c = 1 5 5 . 3 . 4 8 . G i v e n C = 6 5 ° 5 9 ' , a = 2 5 , c = 2 4 ; f i n d A = 7 2 ° 4 ' 4 8 " , B = 4 1 ° 5 6 ' 1 2 " , 6 = 1 7 . 5 6 , o r A ' = 1 0 7 ° 5 5 ' 1 2 " , B ' = 6 ° 5 ' 4 8 " . 4 9 . G i v e n A = 1 8 ° 5 5 ' 2 8 " . 7 , a = 1 3 , 6 = 3 7 ; f i n d B = 6 7 ° 2 2 ' 4 8 " . l , o r B ' = 1 1 2 ° 3 7 ' l l " . 9 . 5 0 . G i v e n C = 1 5 ° 1 1 ' 2 1 " , a = 2 3 2 , 6 = 2 2 9 ; f i n d A = 8 5 ° 1 1 ' 5 8 " , B = 7 9 ° 3 6 ' 4 0 " , c = 6 1 . 5 1 . G i v e n C = 1 2 6 ° 1 2 ' 1 4 " , a = 5 1 3 2 , 6 = 3 4 7 6 ; f i n d A = 3 2 ° 2 8 ' 1 9 " , B = 2 1 ° 1 9 ' 2 7 " , c = 7 7 1 3 . 3 . 5 2 . G i v e n C = 5 5 ° 1 2 ' 3 " , a = 2 0 . 7 1 , 6 = 1 8 . 8 7 ; f i n d A = 6 7 ° 2 8 ' 5 1 " . o , B = 5 7 ° 1 9 ' 5 " . 5 , c = 1 8 . 4 1 . 5 3 . G i v e n C = 1 2 ° 3 5 ' 8 " , a = 8 . 5 4 , 6 = 6 . 3 9 ; f i n d A = 1 3 6 ° 1 5 ' 4 8 " , B = 3 1 ° 9 ' 4 " , c = 2 . 6 9 . 5 4 . G i v e n C = 3 4 ° 9 ' 1 6 " , o = 3 1 8 4 , 6 = 9 1 7 ; f i n d A = 1 3 3 ° 5 1 ' 3 4 " , B = 1 1 ° 5 9 ' 1 0 " , c = 2 4 7 9 . 2 . 5 5 . G i v e n C = 3 2 ° 1 0 ' 5 3 " . 8 , a = 1 0 1 , 6 = 2 9 ; f i n d A = 1 3 6 ° 2 3 ' 4 9 " . 9 , B = 1 1 ° 2 5 ' 1 6 " . 3 , c = 7 8 . 5 6 . G i v e n C = 9 6 ° 5 7 ' 2 0 " . l , « = 4 0 1 , 6 = 4 1 ; f i n d A = 7 7 ° 1 9 ' 1 0 " . 6 , B = 5 ° 4 3 ' 2 9 " . 2 , c = 4 0 8 . 5 7 . G i v e n C = 3 0 ° 4 0 ' 3 5 " , a = 2 2 1 , 6 = 1 4 9 ; f i n d A = 1 1 0 ° 0 ' 5 7 " . 5 , B = 3 9 ° 1 8 ' 2 7 " . 5 , c = 1 2 0 . 5 8 . G i v e n C = 6 6 ° 5 9 ' 2 5 " . 4 , a = 1 0 9 , 6 = 6 1 ; f i n d A = 7 9 ° 3 6 ' 4 0 " , B = 3 3 ° 2 3 ' 5 4 " . 6 , c = 1 0 2 . 5 9 . G i v e n C = 1 3 1 ° 2 4 ' 4 4 " , a = 2 2 9 , 6 = 1 0 9 ; f i n d A = 3 3 ° 2 3 ' 5 4 " . 6 , B = 1 5 ° 1 1 ' 2 1 " . 4 , c = 3 1 2 . 1 9 4 P L A N E T R I G O N O M E T R Y . 6 0 . G i v e n 0 = : 1 0 4 ° 3 ' 5 1 " , a = 2 4 1 , 6 = 1 6 9 ; f i n d A = : 4 5 ° 4 6 ' 1 6 " . 5 , B = 3 0 ° 9 ' 5 2 " . 5 , c = 3 3 2 . 9 7 . 6 1 . G i v e n a = 2 8 9 , 6 = 6 0 1 , c = : 7 1 2 ; f i n d A = 2 3 ° 3 2 ' 1 2 " , B = 5 6 ° 8 ' 4 2 " , C = 1 0 0 ° 1 9 ' 6 " . 6 2 . G i v e n a = 1 7 , 6 = 1 1 3 , c = 1 2 0 ; f i n d A = 7 ° 3 7 ' 4 2 " , B = 6 1 ° 5 5 ' 3 8 " , C = 1 1 0 ° 2 6 ' 4 0 " . 6 3 . G i v e n a = 1 5 . 4 7 , 6 = 1 7 . 3 9 , c = 2 2 . 8 8 ; f i n d A = 4 2 ° 3 0 ' 4 4 " , B = 4 9 ° 2 5 ' 4 9 " , C = 8 8 ° 3 ' 2 7 " . 6 4 . G i v e n a = 5 1 3 4 , 6 = 7 2 6 8 , c = 9 3 1 3 ; f i n d A = 3 3 ° 1 5 ' 3 9 " , B = 5 0 ° 5 6 ' 0 " , C = 9 5 ° 4 8 ' 2 1 " . 6 5 . G i v e n a = 9 9 , 6 = 1 0 1 , c = 1 5 8 ; ' f i n d A = 3 7 ° 2 2 ' 1 9 " , B = 3 8 ° 1 5 ' 4 1 " , C = 1 0 4 ° 2 2 ' 0 " . 6 6 . G i v e n a = 1 1 , 6 = 1 3 , c = 1 6 ; f i n d A = 4 3 ° 2 ' 5 6 " , B = 5 3 ° 4 6 ' 4 4 " , C = 8 3 ° 1 0 ' 2 0 " . 6 7 . G i v e n a = 2 5 , 6 = 2 6 , c = 2 7 ; f i n d A = 5 6 ° 1 5 ' 4 " , B = 5 9 ° 5 1 ' 1 0 " , C = 6 3 ° 5 3 ' 4 6 " . 6 8 . G i v e n o = 1 9 7 , 6 = 5 3 , c = 2 4 0 ; f i n d A = 3 1 ° 5 3 ' 2 6 " . 8 , B = 8 ° 1 0 ' 1 6 " . 4 , C = 1 3 9 ° 5 6 ' 1 6 " . 8 . 6 9 . G i v e n a = 5 0 9 , 6 = 2 2 1 , c = 4 8 0 ; f i n d A = 8 4 ° 3 2 ' 5 0 " . 5 , B = 2 5 ° 3 6 ' 3 0 " . 7 , C = 6 9 ° 5 0 ' 3 8 " . 8 . 7 0 . G i v e n a = 5 3 3 , 6 = 3 1 7 , c = 5 1 0 ; f i n d A = 7 6 ° 1 8 ' 5 2 " , B = 3 5 ° 1 8 ' 0 " . 9 , C = 6 8 ° 2 3 ' 7 " . l . 7 1 . G i v e n a = 5 6 5 , 6 = 4 4 5 , c = 6 0 6 ; f i n d A = 6 2 ° 5 1 ' 3 2 " . 9 , B = 4 4 ° 2 9 ' 5 3 " , C = 7 2 ° 3 8 ' 3 4 " . l . 7 2 . G i v e n a = 1 0 , 6 = 1 2 , c = 1 4 ; f i n d A = 4 4 ° 2 4 ' 5 5 " . 2 , B = 5 7 ° 7 ' 1 8 " , C = 7 8 ° 2 7 ' 4 7 " . 7 3 . G i v e n a = . 8 7 0 6 , 6 = . 0 9 1 6 , c = . 7 9 0 2 ; f i n d A = 1 4 9 ° 4 9 ' 0 " . 4 , B = 3 ° 1 ' 5 6 " . 2 , C = 2 7 ° 9 ' 3 " . 4 . E X A M P L E S . 1 9 5 F i n d t h e a r e a : 7 4 . G i v e n a = 1 0 , 6 = 1 2 , C = 6 0 ° . A n s . 3 0 V 3 . 7 5 . " a = 4 0 , 6 = 6 0 , C = 3 0 ° . 6 0 0 . 7 6 . " 6 = 7 , c = 5 V 2 , A = 1 3 5 ° . 1 7 . 7 7 . " a = 3 2 . 5 , 6 = 5 6 . 3 , C = 4 7 ° 5 ' 3 0 " . 6 7 0 . 7 8 . " 6 = 1 4 9 , A = 7 0 ° 4 2 ' 3 0 " , B = 3 9 ° 1 8 ' 2 8 " . 1 5 5 4 0 . 7 9 . " c = 8 . 0 2 5 , B = 1 0 0 ° 5 ' 2 3 " , C = 3 1 ° 6 ' 1 2 " . 4 6 . 1 7 7 . 8 0 . " a = 5 , 6 = 6 , c = 7 . 1 2 . 8 1 . " a = 6 2 5 , 6 = 5 0 5 , c = 9 0 4 . 1 5 1 8 7 2 . 8 2 . " a = 4 0 9 , 6 = 1 6 9 , c = 5 1 0 . 3 0 6 0 0 . 8 3 . " a = 5 7 7 , 6 = 7 3 , c = 5 2 0 . 1 2 4 8 0 . 8 4 . " a = 5 2 . 5 3 , 6 = 4 8 . 7 6 , c = 4 4 . 9 8 . 1 0 1 6 . 9 4 8 7 . 8 5 . " a = 1 3 , 6 = 1 4 , c = 1 5 . 8 4 . 8 6 . " a = 2 4 2 y a r d s , 6 = 1 2 1 2 y a r d s , c = 1 4 5 0 y a r d s . A n s . 6 a c r e s . 8 7 . " a = 7 . 1 5 2 , 6 = 8 . 2 6 3 , c = 9 . 3 7 5 . 2 8 . 4 7 7 1 7 . 8 8 . T h e s i d e s o f a t r i a n g l e a r e a s 2 : 3 : 4 : s h o w t h a t t h e i d i i o f t h e e s c r i b e d c i r c l e s a r e a s 1 . 8 9 . T h e a r e a o f a t r i a n g l e i s a n a c r e ; t w o o f i t s s i d e s a r e 1 2 7 y a r d s a n d 1 5 0 y a r d s : f i n d t h e a n g l e b e t w e e n t h e m . A n s . 3 0 ° 3 2 ' 2 3 " . 9 0 . T h e a d j a c e n t s i d e s o f a p a r a l l e l o g r a m a r e 5 a n d 8 , a n d t h e y i n c l u d e a n a n g l e o f 6 0 ° : f i n d ( 1 ) t h e t w o d i a g o n a l s , a n d ( 2 ) t h e a r e a . A n s . ( 1 ) 7 , V l 2 9 ; ( 2 ) 2 0 V 3 . 9 1 . T w o a n g l e s o f a t r i a n g u l a r f i e l d a r e 2 2 ° a n d 4 5 ° , a n d t h e l e n g t h o f t h e s i d e o p p o s i t e t h e l a t t e r i s a f u r l o n g . S h o w t h a t t h e f i e l d c o n t a i n s 2 \ a c r e s . 1 9 6 P L A N E T R I G O N O M E T R Y . H E I G H T S A N D D I S T A N C E S . 9 2 . A t a p o i n t 2 0 0 f e e t i n a h o r i z o n t a l l i n e f r o m t h e f o o t o f a t o w e r , t h e a n g l e o f e l e v a t i o n o f t h e t o p o f t h e t o w e r i s o b s e r v e d t o b e 6 0 ° : f i n d t h e h e i g h t o f t h e t o w e r . A n s . 3 4 6 f e e t . 9 3 . F r o m t h e t o p o f a v e r t i c a l c l i f f , t h e a n g l e o f d e p r e s s i o n o f a p o i n t o n t h e s h o r e 1 5 0 f e e t f r o m t h e b a s e o f t h e c l i f f , i s o b s e r v e d t o b e 3 0 ° : f i n d t h e h e i g h t o f t h e c l i f f . A n s . 8 6 . 6 f e e t . 9 4 . F r o m t h e t o p o f a t o w e r 1 1 7 f e e t h i g h , t h e a n g l e o f d e p r e s s i o n o f t h e t o p o f a h o u s e 3 7 f e e t h i g h i s o b s e r v e d t o b e 3 0 ° : h o w f a r i s t h e t o p o f t h e h o u s e f r o m t h e t o w e r ? A n s . 1 3 8 . 5 f e e t . 9 5 . T h e s h a d o w o f a t o w e r i n t h e s u n l i g h t i s o b s e r v e d t o b e 1 0 0 f e e t l o n g , a n d a t t h e s a m e t i m e t h e s h a d o w o f a l a m p - p o s t 9 f e e t h i g h i s o b s e r v e d t o b e 3 V 3 f e e t l o n g : f i n d t h e a n g l e o f e l e v a t i o n o f t h e s u n , a n d h e i g h t o f t h e t o w e r . A n s . 6 0 ° ; 1 7 3 . 2 f e e t . 9 6 . A f l a g s t a f f 2 5 f e e t h i g h s t a n d s o n t h e t o p o f a h o u s e ; f r o m a p o i n t o n t h e p l a i n o n w h i c h t h e h o u s e s t a n d s , t h e a n g l e s o f e l e v a t i o n o f t h e t o p a n d b o t t o m o f t h e f l a g s t a f f a r e o b s e r v e d t o b e 6 0 ° a n d 4 5 ° r e s p e c t i v e l y : f i n d t h e h e i g h t o f t h e h o u s e a b o v e t h e p o i n t o f o b s e r v a t i o n . A n s . 3 4 . 1 5 f e e t . 9 7 . F r o m t h e t o p o f a c l i f f 1 0 0 f e e t h i g h , t h e a n g l e s o f d e p r e s s i o n o f t w o s h i p s a t s e a a r e o b s e r v e d t o b e 4 5 ° a n d 3 0 ° r e s p e c t i v e l y ; i f t h e l i n e j o i n i n g t h e s h i p s p o i n t s d i r e c t l y t o t h e f o o t o f t h e c l i f f , f i n d t h e d i s t a n c e b e t w e e n t h e s h i p s . A n s . 7 3 . 2 . 9 8 . A t o w e r 1 0 0 f e e t h i g h s t a n d s o n t h e t o p o f a c l i f f ; f r o m a p o i n t o n t h e s a n d a t t h e f o o t o f t h e c l i f f t h e a n g l e s E X A M P L E S . 1 9 7 o f e l e v a t i o n o f t h e t o p a n d b o t t o m o f t h e t o w e r a r e o b s e r v e d t o b e 7 5 ° a n d 6 0 ° r e s p e c t i v e l y : f i n d t h e h e i g h t o f t h e c l i f f . - 4 m s . 8 6 . 6 f e e t . 9 9 . A m a n w a l k i n g a s t r a i g h t r o a d o b s e r v e s a t o n e m i l e s t o n e a h o u s e i n a d i r e c t i o n m a k i n g a n a n g l e o f 3 0 ° w i t h t h e r o a d , a n d a t t h e n e x t m i l e s t o n e t h e a n g l e i s 6 0 ° : h o w f a r i s t h e h o u s e _ f r o m t h e r o a d ? A n s . 1 5 2 4 y d s . 1 0 0 . A m a n s t a n d s a t a p o i n t A o n t h e b a n k A B o f a s t r a i g h t r i v e r a n d o b s e r v e s t h a t t h e l i n e j o i n i n g A t o a p o s t C o n t h e o p p o s i t e b a n k m a k e s w i t h A B a n a n g l e o f 3 0 ° . H e t h e n g o e s 4 0 0 y a r d s a l o n g t h e b a n k t o B a n d f i n d s t h a t B C m a k e s w i t h B A a n a n g l e o f 6 0 ° : f i n d t h e b r e a d t h o f t h e r i v e r . . 4 n s . 1 7 3 . 2 y a r d s . 1 0 1 . F r o m t h e t o p o f a h i l l t h e a n g l e s o f d e p r e s s i o n o f t h e t o p a n d b o t t o m o f a f l a g s t a f f 2 5 f e e t h i g h a t t h e f o o t o f t h e h i l l a r e o b s e r v e d t o b e 4 5 ° 1 3 ' a n d 4 7 ° 1 2 ' r e s p e c t i v e l y : f i n d t h e h e i g h t o f t h e h i l l . A n s . 3 7 3 f e e t . 1 0 2 . F r o m e a c h o f t w o s t a t i o n s , e a s t a n d w e s t o f e a c h o t h e r , t h e a l t i t u d e o f a b a l l o o n i s o b s e r v e d t o b e 4 5 ° , a n d i t s b e a r i n g s t o b e r e s p e c t i v e l y N . W . a n d N . E . ; i f t h e s t a t i o n s b e 1 m i l e a p a r t , f i n d t h e h e i g h t o f t h e b a l l o o n . A n s . 3 7 3 3 f e e t . 1 0 3 . T h e a n g l e o f e l e v a t i o n o f a b a l l o o n f r o m a s t a t i o n d u e s o u t h o f i t i s 6 0 ° , a n d f r o m a n o t h e r s t a t i o n d u e w e s t o f t h e f o r m e r a n d d i s t a n t a m i l e f r o m i t i s 4 5 ° : f i n d t h e h e i g h t o f t h e b a l l o o n . A n s . 6 4 6 8 f e e t . 1 0 4 . F i n d t h e h e i g h t o f a h i l l , t h e a n g l e o f e l e v a t i o n a t i t s f o o t b e i n g 6 0 ° , a n d a t a p o i n t 5 0 0 y a r d s f r o m t h e f o o t a l o n g a h o r i z o n t a l p l a n e 3 0 ° . A n s . 2 5 0 V 3 y a r d s . 1 0 5 . A t o w e r 5 1 f e e t h i g h h a s a m a r k a t a h e i g h t o f 2 5 f e e t f r o m t h e g r o u n d : f i n d a t w h a t d i s t a n c e f r o m t h e f o o t t h e t w o p a r t s s u b t e n d e q u a l a n g l e s . 1 9 8 P L A N E T R I G O N O M E T R Y . 1 0 6 . T h e a n g l e s o f a t r i a n g l e a r e a s 1 : 2 : 3 , a n d t h e p e r p e n d i c u l a r f r o m t h e g r e a t e s t a n g l e o n t h e o p p o s i t e s i d e i s 3 0 y a r d s : f i n d t h e s i d e s . A n s . 2 0 V 3 , 6 0 , 4 0 V 3 . 1 0 7 . A t t w o p o i n t s A , B , a n o b j e c t D E , s i t u a t e d i n t h e s a m e v e r t i c a l l i n e C E , s u b t e n d s t h e s a m e a n g l e « ; i f A C , B C b e i n t h e s a m e r i g h t l i n e , a n d e q u a l t o a a n d b , r e s p e c t i v e l y , p r o v e D E = ( a + 6 ) t a n a . 1 0 8 . E r o m a s t a t i o n B a t t h e f o o t o f a n i n c l i n e d p l a n e B C t h e a n g l e o f e l e v a t i o n o f t h e s u m m i t A o f a m o u n t a i n i s 6 0 ° , t h e i n c l i n a t i o n o f B C i s 3 0 ° , t h e a n g l e B C A 1 3 5 ° , a n d t h e l e n g t h o f B C i s 1 0 0 0 y a r d s : f i n d t h e h e i g h t o f A o v e r B . A n s . 5 0 0 ( 3 + V 3 ) y a r d s . 1 0 9 . A r i g h t t r i a n g l e r e s t s o n i t s h y p o t e n u s e , t h e l e n g t h o f w h i c h i s 1 0 0 f e e t ; o n e o f t h e a n g l e s i s 3 6 ° , a n d t h e i n c l i n a t i o n o f t h e p l a n e o f t h e t r i a n g l e t o t h e h o r i z o n i s 6 0 ° : f i n d t h e h e i g h t o f t h e v e r t e x a b o v e t h e g r o u n d . A n s . 2 5 V 3 c o s 1 8 ° . 1 1 0 . A s t a t i o n a t A i s d u e w e s t o f a r a i l w a y t r a i n a t B ; a f t e r t r a v e l i n g N . W . 6 m i l e s , t h e b e a r i n g o f A f r o m t h e t r a i n i s S . 2 2 J ° W . : f i n d t h e d i s t a n c e A B . A n s . 6 m i l e s . 1 1 1 . T h e a n g l e s o f d e p r e s s i o n o f t h e t o p a n d b o t t o m o f a c o l u m n o b s e r v e d f r o m a t o w e r 1 0 8 f e e t h i g h a r e 3 0 ° a n d 6 0 ° r e s p e c t i v e l y : f i n d t h e h e i g h t o f t h e c o l u m n . A n s . 7 2 f e e t . 1 1 2 . A t t h e f o o t o f a m o u n t a i n t h e e l e v a t i o n o f i t s s u m m i t i s f o u n d t o b e 4 5 ° . A f t e r a s c e n d i n g f o r o n e m i l e , a t a s l o p e o f 1 5 ° , t o w a r d s t h e s u m m i t , i t s e l e v a t i o n i s f o u n d t o b e 6 0 ° : f i n d t h e h e i g h t o f t h e m o u n t a i n . A n s . m i l e s . V 2 1 1 3 . A a n d B a r e t w o s t a t i o n s o n a h i l l s i d e . T h e i n c l i n a t i o n o f t h e h i l l t o t h e h o r i z o n i s 3 0 ° . T h e d i s t a n c e b e t w e e n A a n d B i s 5 0 0 y a r d s . C i s t h e s u m m i t o f a n o t h e r h i l l i n E X A M P L E S . 1 9 9 t h e s a m e v e r t i c a l p l a n e a s A a n d B , o n a l e v e l w i t h A , b u t a t B i t s e l e v a t i o n a b o v e t h e h o r i z o n i s 1 5 ° : f i n d t h e d i s t a n c e b e t w e e n A a n d C . A n s . 5 0 0 ( V 3 + 1 ) . 1 1 4 . F r o m t h e t o p o f a c l i f f t h e a n g l e s o f d e p r e s s i o n o f t h e t o p a n d b o t t o m o f a l i g h t h o u s e 9 7 . 2 5 f e e t h i g h a r e o b s e r v e d t o b e 2 3 ° 1 7 ' a n d 2 4 ° 1 9 ' r e s p e c t i v e l y : h o w m u c h h i g h e r i s t h e c l i f f t h a n t h e l i g h t h o u s e ? A n s . 1 9 4 2 f e e t . 1 1 5 . T h e a n g l e o f e l e v a t i o n o f a b a l l o o n f r o m a s t a t i o n d u e s o u t h o f i t i s 4 7 ° 1 8 ' 3 0 " , a n d f r o m a n o t h e r s t a t i o n d u e w e s t o f t h e f o r m e r , a n d d i s t a n t 6 7 1 . 3 8 f e e t f r o m i t , t h e e l e v a t i o n i s 4 1 ° 1 4 ' : f i n d t h e h e i g h t o f t h e b a l l o o n . A n s . 1 0 0 0 f e e t . 1 1 6 . A p e r s o n s t a n d i n g o n t h e b a n k o f a r i v e r o b s e r v e s t h e e l e v a t i o n o f t h e t o p o f a t r e e o n t h e o p p o s i t e b a n k t o b e 5 1 ° ; a n d w h e n h e r e t i r e s 3 0 f e e t f r o m t h e r i v e r , s b a n k h e o b s e r v e s t h e e l e v a t i o n t o b e 4 6 ° : f i n d t h e b r e a d t h o f t h e r i v e r . A n s . 1 5 5 . 8 2 3 f e e t . 1 1 7 . F r o m t h e t o p o f a h i l l I o b s e r v e t w o m i l e s t o n e s o n t h e l e v e l g r o u n d i n a s t r a i g h t l i n e b e f o r e m e , a n d I f i n d t h e i r a n g l e s o f d e p r e s s i o n t o b e r e s p e c t i v e l y 5 ° a n d 1 5 ° : f i n d t h e h e i g h t o f t h e h i l l . A n s . 2 2 8 . 6 3 0 7 y a r d s . 1 1 8 . A t o w e r i s s i t u a t e d o n t h e t o p o f a h i l l w h o s e a n g l e o f i n c l i n a t i o n t o t h e h o r i z o n i s 3 0 ° . T h e a n g l e s u b t e n d e d b y t h e t o w e r a t t h e f o o t o f t h e h i l l i s f o u n d b y a n o b s e r v e r t o b e 1 5 ° ; a n d o n a s c e n d i n g 4 8 5 f e e t u p t h e h i l l t h e t o w e r i s f o u n d t o s u b t e n d a n a n g l e o f 3 0 ° : f i n d ( 1 ) t h e h e i g h t o f t h e t o w e r , a n d ( 2 ) t h e d i s t a n c e o f i t s b a s e f r o m t h e f o o t o f t h e h i l l . A n s . ( 1 ) 2 8 0 . 0 1 5 ; ( 2 ) 7 6 5 . 0 1 5 f e e t . 1 1 9 . T h e a n g l e o f e l e v a t i o n o f a t o w e r a t a p l a c e A d u e s o u t h o f i t i s 3 0 ° ; a n d a t a p l a c e B , d u e w e s t o f A , a n d a t a d i s t a n c e a f r o m i t , t h e e l e v a t i o n i s 1 8 ° : s h o w t h a t t h e h e i g h t o f t h e t o w e r i s — a ^ 2 + 2 V 5 2 0 0 P L A N E T R I G O N O M E T R Y . 1 2 0 . O n t h e b a n k o f a r i v e r t h e r e i s a c o l u m n 2 0 0 f e e t h i g h s u p p o r t i n g a s t a t u e 3 0 f e e t h i g h . T h e s t a t u e t o a n o b s e r v e r o n t h e o p p o s i t e b a n k s u b t e n d s a n e q u a l a n g l e w i t h a m a n 6 f e e t h i g h s t a n d i n g a t t h e b a s e o f t h e c o l u m n : f i n d t h e b r e a d t h o f t h e r i v e r . A n s . 1 0 V 1 1 5 f e e t . 1 2 1 . A m a n w a l k i n g a l o n g a s t r a i g h t r o a d a t t h e r a t e o f 3 m i l e s a n h o u r , s e e s i n f r o n t o f h i m , a t a n e l e v a t i o n o f 6 0 ° , a b a l l o o n w h i c h i s t r a v e l l i n g h o r i z o n t a l l y i n t h e s a m e d i r e c t i o n a t t h e r a t e o f 6 m i l e s a n h o u r ; t e n m i n u t e s a f t e r h e o b s e r v e s t h a t t h e e l e v a t i o n i s 3 0 ° : p r o v e t h a t t h e h e i g h t o f t h e b a l l o o n a b o v e t h e r o a d i s 4 4 0 V 3 y a r d s . 1 2 2 . A n o b s e r v e r i n a b a l l o o n o b s e r v e s t h e a n g l e o f d e p r e s s i o n o f a n o b j e c t o n t h e g r o u n d , d u e s o u t h , t o b e 3 5 ° 3 0 ' . T h e b a l l o o n d r i f t s d u e e a s t , a t t h e s a m e e l e v a t i o n , f o r 2 \ m i l e s , w h e n t h e a n g l e o f d e p r e s s i o n o f t h e s a m e o b j e c t i s o b s e r v e d t o b e 2 3 ° 1 4 ' : f i n d t h e h e i g h t o f t h e b a l l o o n . A n s . 1 . 3 4 3 9 4 m i l e s . 1 2 3 . A c o l u m n , o n a p e d e s t a l 2 0 f e e t h i g h , s u b t e n d s a n a n g l e 4 5 ° t o a p e r s o n o n t h e g r o u n d ; o n a p p r o a c h i n g 2 0 f e e t , i t a g a i n s u b t e n d s a n a n g l e 4 5 ° : s h o w t h a t t h e h e i g h t o f t h e c o l u m n i s 1 0 0 f e e t . I 1 2 4 . A t o w e r 5 1 f e e t h i g h h a s a m a r k 2 5 f e e t f r o m t h e g r o u n d : f i n d a t w h a t d i s t a n c e t h e t w o p a r t s s u b t e n d e q u a l a n g l e s t o a n e y e 5 f e e t f r o m t h e g r o u n d . A n s . 1 0 0 f e e t . 1 2 5 . F r o m t h e e x t r e m i t i e s o f a s e a - w a l l , 3 0 0 f e e t l o n g , t h e b e a r i n g s o f a b o a t a t s e a w e r e o b s e r v e d t o b e N . 2 3 ° 3 0 ' E . , a n d N . 3 5 ° 1 5 ' W . : f i n d t h e d i s t a n c e o f t h e b o a t f r o m t h e s e a - w a l l . A n s . 2 6 2 . 8 2 f e e t . 1 2 6 . A B C i s a t r i a n g l e o n a h o r i z o n t a l p l a n e , o n w h i c h s t a n d s a t o w e r C D , w h o s e e l e v a t i o n a t A i s 5 0 ° 3 ' 2 " ; A B i s 1 0 0 . 6 2 f e e t , a n d B C a n d A C m a k e w i t h A B a n g l e s 4 0 ° 3 5 ' 1 7 " a n d 9 ° 5 9 ' 5 0 " r e s p e c t i v e l y : f i n d C D . A n s . 1 0 1 . 1 6 6 f e e t . E X A M P L E S . 2 0 1 5 , 1 » 1 2 7 . T h e a n g l e o f e l e v a t i o n o f a t o w e r a t a d i s t a n c e o f 2 0 y a r d s f r o m i t s f o o t i s t h r e e t i m e s a s g r e a t a s t h e a n g l e ^ o f e l e v a t i o n 1 0 0 y a r d s f r o m t h e s a m e p o i n t : s h o w t h a t t h e h e i g h t o f t h e t o w e r i s f e e t . 1 2 8 . A m a n s t a n d i n g a t a p o i n t A , d u e s o u t h o f a t o w e r b u i l t o n a h o r i z o n t a l p l a i n , o b s e r v e s t h e a l t i t u d e o f t h e t o w e r t o b e 6 0 ° . H e t h e n w a l k s t o a p o i n t B d u e w e s t f r o m A a n d o b s e r v e s t h e a l t i t u d e t o b e 4 5 ° , a n d t h e n a t t h e p o i n t C i n A B p r o d u c e d h e o b s e r v e s t h e a l t i t u d e t o b e 3 0 ° : p r o v e t t h a t A B = B C . & ° m V 1 C , _ v . / > 1 2 9 . T h e a n g l e o f e l e v a t i o n o f a b a l l o o n , w h i c h i s a s c e n d -\ h i n g u n i f o r m l y a n d v e r t i c a l l y , w h e n i t i s o n e m i l e h i g h i s o b s e r v e d t o b e 3 5 ° 2 0 ' ; 2 0 m i n u t e s l a t e r t h e e l e v a t i o n i s o b s e r v e d t o b e 5 5 ° 4 0 ' : h o w f a s t i s t h e b a l l o o n m o v i n g ? A n s . 3 ( s i n 2 0 ° 2 0 ' ) ( s e c 5 5 ° 4 0 ' ) ( c o s e c 3 5 ° 2 0 ' ) m i l e s p e r h o u r . 1 3 0 . T h e a n g l e o f e l e v a t i o n o f t h e t o p o f a s t e e p l e a t a p l a c e d u e s o u t h o f i t i s 4 5 ° , a n d a t a n o t h e r p l a c e d u e w e s t o f t h e f o r m e r s t a t i o n a n d d i s t a n t 1 0 0 f e e t f r o m i t t h e e l e v a t i o n i s 1 5 ° : s h o w t h a t t h e h e i g h t o f t h e s t e e p l e i s 5 0 ( 3 - 3 ~ ) f e e t . 1 3 1 . A t o w e r s t a n d s a t t h e f o o t o f a n i n c l i n e d p l a n e w h o s e i n c l i n a t i o n t o t h e h o r i z o n i s 9 ° ; a l i n e i s m e a s u r e d u p t h e i n c l i n e f r o m t h e f o o t o f t h e t o w e r , o f 1 0 0 f e e t i n l e n g t h . A t t h e u p p e r e x t r e m i t y o f t h i s l i n e t h e t o w e r s u b t e n d s a n a n g l e o f 5 4 ° : f i n d t h e h e i g h t o f t h e t o w e r . A n s . 1 1 4 . 4 f e e t . 1 3 2 . T h e a l t i t u d e o f a c e r t a i n r o c k i s o b s e r v e d t o b e 4 7 ° , a n d a f t e r w a l k i n g 1 0 0 0 f e e t t o w a r d s t h e r o c k , u p a s l o p e i n c l i n e d a t a n a n g l e o f 3 2 ° t o t h e h o r i z o n , t h e o b s e r v e r f i n d s t h a t t h e a l t i t u d e i s 7 7 ° : p r o v e t h a t t h e v e r t i c a l h e i g h t o f t h e r o c k a b o v e t h e f i r s t p o i n t o f o b s e r v a t i o n i s 1 0 3 4 f e e t . 2 0 2 P L A N E T R I G O N O M E T R Y . 1 3 3 . F r o m a w i n d o w i t i s o b s e r v e d t h a t t h e a n g l e o f e l e v a t i o n o f t h e t o p o f a h o u s e o n t h e o p p o s i t e s i d e o f t h e s t r e e t i s 2 9 ° , a n d t h e a n g l e o f d e p r e s s i o n o f t h e b o t t o m o f t h e h o u s e i s 5 6 ° : f i n d t h e h e i g h t o f t h e h o u s e , s u p p o s i n g t h e b r e a d t h o f t h e s t r e e t t o b e 8 0 f e e t . A n s . 1 6 2 . 9 5 f e e t . 1 3 4 . A a n d B a r e t w o p o s i t i o n s o n o p p o s i t e s i d e s o f a m o u n t a i n ; C i s a p o i n t v i s i b l e f r o m A a n d B ; A C a n d B C a r e 1 0 m i l e s a n d 8 m i l e s r e s p e c t i v e l y , a n d t h e a n g l e B C A i s 6 0 ° : p r o v e t h a t t h e d i s t a n c e b e t w e e n A a n d B i s 9 . 1 6 5 m i l e s . 1 3 5 . P a n d Q a r e t w o i n a c c e s s i b l e o b j e c t s ; a s t r a i g h t l i n e A B , i n t h e s a m e p l a n e a s P a n d Q , i s m e a s u r e d , a n d f o u n d t o b e 2 8 0 y a r d s ; t h e a n g l e P A B i s 9 5 ° , t h e a n g l e Q A B i s 4 7 | ° , t h e a n g l e Q B A i s 1 1 0 ° , a n d t h e a n g l e P B A i s 5 2 ° 2 Q ' : f i n d t h e l e n g t h o f P Q . A n s . 5 0 9 . 7 7 y a r d s . 1 3 6 . T w o h i l l s e a c h 2 6 4 f e e t h i g h a r e j u s t v i s i b l e f r o m e a c h o t h e r o v e r t h e s e a : h o w f a r a r e t h e y a p a r t ? ( T a k e t h e r a d i u s o f t h e e a r t h = 4 0 0 0 m i l e s . ) A n s . 4 0 m i l e s . 1 3 7 . A s h i p s a i l i n g o u t o f h a r b o r i s w a t c h e d b y a n o b s e r v e r f r o m t h e s h o r e ; a n d a t t h e i n s t a n t s h e d i s a p p e a r s b e l o w t h e h o r i z o n h e a s c e n d s t o a h e i g h t o f 2 0 f e e t , a n d t h u s k e e p s h e r i n s i g h t 4 0 m i n u t e s l o n g e r : f i n d t h e r a t e a t w h i c h t h e s h i p i s s a i l i n g , a s s u m i n g t h e e a r t h , s r a d i u s t o b e 4 0 0 0 m i l e s , a n d n e g l e c t i n g t h e h e i g h t o f t h e o b s e r v e r . A n s . 4 0 V 3 3 0 f e e t p e r m i n u t e . 1 3 8 . F r o m t h e t o p o f t h e m a s t o f a s h i p 6 4 f e e t a b o v e t h e l e v e l o f t h e s e a t h e l i g h t o f a d i s t a n t l i g h t h o u s e i s j u s t s e e n i n t h e h o r i z o n ; a n d a f t e r t h e s h i p h a s s a i l e d d i r e c t l y t o w a r d s t h e l i g h t f o r 3 0 m i n u t e s i t i s s e e n f r o m t h e d e c k o f t h e s h i p , w h i c h i s 1 6 f e e t a b o v e t h e s e a : f i n d t h e r a t e a t w h i c h t h e s h i p i s s a i l i n g . ( T a k e r a d i u s = 4 0 0 0 m i l e s . ) A n s . 8 V f f m i l e s p e r h o u r . E X A M P L E S . 2 0 3 1 3 9 . A , B , C , a r e t h r e e o b j e c t s a t k n o w n d i s t a n c e s a p a r t ; n a m e l y , A B = 1 0 5 6 y a r d s , A C = 9 2 4 y a r d s , B C = 1 7 1 6 y a r d s . A n o b s e r v e r p l a c e s h i m s e l f a t a s t a t i o n P f r o m w h i c h C a p p e a r s d i r e c t l y i n f r o n t o f A , a n d o b s e r v e s t h e a n g l e C P B t o b e 1 4 ° 2 4 ' : f i n d t h e d i s t a n c e C P . A n s . 2 1 0 9 . 8 2 4 y a r d s . 1 4 0 . A , B , C , a r e t h r e e o b j e c t s s u c h t h a t A B : = 3 2 0 y a r d s , A C = 6 0 0 y a r d s , a n d B C = 4 3 5 y a r d s . F r o m a s t a t i o n P i t i s o b s e r v e d t h a t A P B = 1 5 ° , a n d B P C = 3 0 ° : f i n d t h e d i s t a n c e s o f P f r o m A , B , a n d C ; t h e p o i n t B b e i n g n e a r e s t t o P , a n d t h e a n g l e A P C b e i n g t h e . s u m o f t h e a n g l e s A P B a n d B P C . A n s . P A = 7 7 7 , P B = 5 0 2 , P C = 7 9 0 . 2 0 4 P L A N E T R I G O N O M E T R Y . C H A P T E R V I I I . C O N S T R U O T I O N O P L 0 G A R I T H M I 0 A N D T R I G 0 N 0 M E T R I 0 T A B L E S . 1 2 8 . L o g a r i t h m i c a n d T r i g o n o m e t r i c T a b l e s . — I n C h a p t e r s I V . , V . , a n d V I I . , i t w a s s h o w n h o w t o u s e l o g a r i t h m i c a n d t r i g o n o m e t r i c t a b l e s ; i t w i l l n o w b e s h o w n h o w t o c a l c u l a t e s u c h t a b l e s . A l t h o u g h t h e t r i g o n o m e t r i c f u n c t i o n s a r e s e l d o m c a p a b l e o f b e i n g e x p r e s s e d e x a c t l y , y e t t h e y c a n b e f o u n d a p p r o x i m a t e l y f o r a n y a n g l e ; a n d t h e c a l c u l a t i o n s m a y b e c a r r i e d t o a n y a s s i g n e d d e g r e e o f a c c u r a c y . W e s h a l l f i r s t s h o w h o w t o c a l c u l a t e l o g a r i t h m i c t a b l e s , a n d s h a l l r e p e a t h e r e s u b s t a n t i a l l y A r t s . 2 0 8 , 2 0 9 , 2 1 0 , f r o m t h e C o l l e g e A l g e b r a . 1 2 9 . E x p o n e n t i a l S e r i e s . — T o e x p a n d e " i n a s e r i e s o f a s c e n d i n g p o w e r s o f x . B y t h e B i n o m i a l T h e o r e m , + n x ( n x — 1 ) ( n x — 2 ) 1 ( 1 ) S i m i l a r l y , = 1 + 1 - r + - - . ( 2 ) L O G A R I T H M I C S E R I E S . 2 0 5 a n d t h e r e f o r e s e r i e s ( 1 ) i s e q u a l t o s e r i e s ( 2 ) h o w e v e r g r e a t n m a y b e . H e n c e i f n b e i n d e f i n i t e l y i n c r e a s e d , w e h a v e f r o m ( 1 ) a n d ( 2 ) l + + l + I + i + - - K 1 + 1 + i + H H T h e s e r i e s i n t h e p a r e n t h e s i s i s u s u a l l y d e n o t e d b y e ; h e n c e e = 1 + x + ( 3 ) [ Z [ 3 | 4 w h i c h i s t h e e x p a n s i o n o f e x i n p o w e r s o f x . T h i s r e s u l t i s c a l l e d t h e E x p o n e n t i a l T h e o r e m . I f w e p u t x = l , w e h a v e f r o m ( 3 ) F r o m t h i s s e r i e s w e m a y r e a d i l y c o m p u t e t h e a p p r o x i m a t e v a l u e o f e t o a n y r e q u i r e d d e g r e e o f a c c u r a c y . T h i s c o n s t a n t v a l u e e i s c a l l e d t h e N a p i e r i a n b a s e ( A r t . 6 4 ) . T o t e n p l a c e s o f d e c i m a l s i t i s f o u n d t o b e 2 . 7 1 8 2 8 1 8 2 8 4 . C o r . L e t a = e ° ; t h e n c = l o g e a , a n d a " = e " . S u b s t i t u t i n g i n ( 3 ) , w e h a v e e - = 1 + c a ; + _ + _ + . . . o r a ^ l + x l o g t a + ^ § ^ + ^ ^ . + . . . ( 4 ) w h i c h i s t h e e x p a n s i o n o f a ' i n p o w e r s o f x . 1 3 0 . L o g a r i t h m i c S e r i e s . — T o e x p a n d l o g e ( l + a ; ) i n a s e r i e s o f a s c e n d i n g p o w e r s o f x . B y t h e B i n o m i a l T h e o r e m , a = ( l + a - i y = l + x ( a - l ) + x ( x - V ( « - l ) 2 x ( x - l ) ( x - 2 ) ( 1 ) < + . . . [ 3 V ' 2 0 6 P L A N E T R I G O N O M E T R Y . + t e r m s i n v o l v i n g x 2 , X s , e t c . C o m p a r i n g t h i s v a l u e o f a w i t h t h a t g i v e n i n ( 4 ) o f A r t . 1 2 9 , a n d e q u a t i n g t h e c o e f f i c i e n t s o f x , w e h a v e l o g e a = a - l - | ( « - l ) 2 + H a - l ) 3 - i ( « - l ) 4 + -P u t a = 1 + x ; t h e n ™» 2 n J l f y t A l o g . ( l + x ) = x - | + + ( 3 ) T h i s i s t h e L o g a r i t h m i c S e r i e s ; b u t u n l e s s x b e v e r y s m a l l , t h e t e r m s d i m i n i s h s o s l o w l y t h a t a l a r g e n u m b e r o f t h e m w i l l h a v e t o b e t a k e n ; a n d h e n c e t h e s e r i e s i s o f l i t t l e p r a c t i c a l u s e f o r n u m e r i c a l c a l c u l a t i o n . I f x > 1 , t h e s e r i e s i s a l t o g e t h e r u n s u i t a b l e . W e s h a l l t h e r e f o r e d e d u c e s o m e m o r e c o n v e n i e n t f o r m u l a e . C h a n g i n g x i n t o — x , ( 1 ) b e c o m e s / » 2 o n S n / A l o g . ( l - « ) = - ! e - - | -( 2 ) S u b t r a c t i n g ( 2 ) f r o m ( 1 ) , w e h a v e l 0 g « r ^ = 2 ( ! e + 3 - + f + ! - + - ) ---< 3 ) P u t — -" — 1 — x n 2 n + 1 , a n d ( 3 ) b e c o m e s l o g ! L ± l = 2 f 1 . I . I + 1 + . . b e n | _ 2 m + 1 3 ( 2 n + l ) 3 5 ( 2 n + l ) 5 o r l o g e ( n + 1 ) = l o g „ n + 2 — 1 h + - - . -( 4 ) g e \ 2 n + l 3 ( 2 m + 1 ) 3 5 ( 2 » + l ) » _ k ' T h i s s e r i e s i s r a p i d l y c o n v e r g e n t , a n d g i v e s t h e l o g a r i t h m o f e i t h e r o f t w o c o n s e c u t i v e n u m b e r s t o a n y e x t e n t w h e n t h e l o g a r i t h m o f t h e o t h e r n u m b e r i s k n o w n . C O M P U T A T I O N O F L O G A R I T H M S . 2 0 7 1 3 1 . C o m p u t a t i o n o f L o g a r i t h m s . — L o g a r i t h m s t o t h e b a s e e a r e c a l l e d N a p i e r i a n L o g a r i t h m s ( A r t . 6 4 ) . T h e y a r e a l s o c a l l e d n a t u r a l l o g a r i t h m s , b e c a u s e t h e y a r e t h e f i r s t l o g a r i t h m s w h i c h o c c u r i n t h e i n v e s t i g a t i o n o f a m e t h o d o f c a l c u l a t i n g l o g a r i t h m s . L o g a r i t h m s t o t h e b a s e 1 0 a r e c a l l e d c o m m o n l o g a r i t h m s . W h e n l o g a r i t h m s a r e u s e d i n t h e o r e t i c a l i n v e s t i g a t i o n s , t h e b a s e e i s a l w a y s u n d e r s t o o d , j u s t a s i n a l l p r a c t i c a l c a l c u l a t i o n s t h e b a s e 1 0 i s i n v a r i a b l y e m p l o y e d . I t i s o n l y n e c e s s a r y t o c o m p u t e t h e l o g a r i t h m s o f p r i m e n u m b e r s f r o m t h e s e r i e s , s i n c e t h e l o g a r i t h m o f a c o m p o s i t e n u m b e r m a y b e o b t a i n e d b y a d d i n g t o g e t h e r t h e l o g a r i t h m s o f i t s c o m p o n e n t f a c t o r s . T h e l o g a r i t h m o f 1 = 0 . P u t t i n g n = 1 , 2 , 4 , 6 , e t c . , s u c c e s s i v e l y , i n ( 4 ) o f A r t . 1 3 0 , w e o b t a i n t h e f o l l o w i n g N a p i e r i a n L o g a r i t h m s : l o g . 2 = 2 \ ~ + — — h — H — - — I— — + . - . b | _ 3 3 - 3 3 5 - 3 5 7 - 3 i 9 - 3 9 l o g . 8 - k g . 2 + 2 [ i + 5 L . + J ^ + ^ + . . : l o g . 4 = 2 1 o g . 2 l o g . 5 = ^ 4 + 2 [ 9 1 + 3 ^ + ^ L + ^ + . . ; l o g . 6 = l o g . 2 + l o g . 3 l o g . 7 - l o f c 6 + 2 [ i + 5 ^ + s l 5 + . . : l o g . 8 = 3 1 o g . 2 l o g . 9 = 2 1 o g . 3 l o g . l 0 = l o g e o + l o g . 2 A n d s o o n . T h e n u m b e r o f t e r m s o f t h e s e r i e s w h i c h i t i s n e c e s s a r y t o i n c l u d e d i m i n i s h e s a s n i n c r e a s e s . T h u s , i n c o m p u t i n g = 0 . 6 9 3 1 4 7 1 8 . = 1 . 0 9 8 6 1 2 2 8 . = 1 . 3 8 6 2 9 4 3 6 . = 1 . 6 0 9 4 3 7 9 0 . = 1 . 7 9 1 7 5 9 4 6 . = 1 . 9 4 5 9 0 9 9 6 . = 2 . 0 7 9 4 4 1 5 4 . = 2 . 1 9 7 2 2 4 5 6 . = 2 . 3 0 2 5 8 5 0 9 . 2 0 8 P L A N E T R I G O N O M E T R Y . t h e l o g a r i t h m o f 1 0 1 , t h e f i r s t t e r m o f t h e s e r i e s g i v e s t h e r e s u l t t r u e t o s e v e n d e c i m a l p l a c e s . B y c h a n g i n g b t o 1 0 a n d a t o e i n ( 1 ) o f A r t . 6 5 , w e h a v e l o g l o m = l 2 S ^ L = l Q g . m = . 4 3 4 2 9 4 4 8 l o g . m , 6 l o g e 1 0 2 . 3 0 2 5 8 5 0 9 b o r c o m m o n l o g m = N a p i e r i a n l o g m x . 4 3 4 2 9 4 4 8 . T h e n u m b e r . 4 3 4 2 9 4 4 8 i s c a l l e d t h e m o d u l u s o f t h e c o m m o n s y s t e m . I t i s u s u a l l y d e n o t e d b y f i . H e n c e , t h e c o m m o n l o g a r i t h m o f a n y n u m b e r i s e q u a l t o t h e N a p i e r i a n l o g a r i t h m o f t h e s a m e n u m b e r m u l t i p l i e d b y t h e m o d u l u s o f t h e c o m m o n s y s t e m , . 4 3 4 2 9 4 4 8 . M u l t i p l y i n g ( 4 ) o f A r t . 1 3 0 b y / a , w e o b t a i n a s e r i e s b y w h i c h c o m m o n l o g a r i t h m s m a y b e c o m p u t e d ; t h u s , h > g i o ( n + l ) = l o g 1 0 n + 2 / x — L — + 1 + 1 1 V S + -C 1 ) | _ 2 n + l 3 ( 2 n + l ) , i 5 ( 2 n + l ) s C o m m o n L o g a r i t h m s . l o g , „ 2 = m l o g e 2 = . 4 3 4 2 9 4 4 8 x . 6 9 3 1 4 7 1 8 = . 3 0 1 0 3 0 0 . l o g 1 0 3 = / x l o g „ 3 = . 4 3 4 2 9 4 4 8 x 1 . 0 9 8 6 1 2 2 8 = . 4 7 7 1 2 1 3 . l o g 1 0 4 = 2 l o g 1 0 2 = . 6 0 2 0 6 0 0 . l o g 1 0 5 = ^ l o g e 5 = . 4 3 4 2 9 4 4 8 x 1 . 6 0 9 4 3 7 9 0 = . 6 9 8 9 7 0 0 . A n d s o o n . 1 3 2 . I f 6 b e t h e C i r c u l a r M e a s u r e o f a n A c u t e A n g l e , s i n 6 , 6 , a n d t a n 6 a r e i n A s c e n d i n g O r d e r o f M a g n i t u d e . W i t h c e n t r e O , a n d a n y r a d i u s , d e s c r i b e a n a r c B A B ' . B i s e c t t h e a n g l e B O B ' b y O A ; j o i n B B ' , a n d d r a w t h e t a n g e n t s B T , B ' T . L e t A O B = A O B ' = 6 . T h e n B B ' < a r c B A B ' < B T + B ' T ( G e o m . , A r t . 2 4 6 ) . - . B C < a r c B A < B T . L I M I T I N G V A L U E S O F S I N 6 . 2 0 9 - B C B A B T " O B O B O B ' . - . s i n 0 < 0 < t a n 0 . . 1 3 3 . T h e L i m i t o f ^ 5 ^ , w h e n 6 i s I n d e f i n i t e l y D i m i n i s h e d , i s U n i t y . W e h a v e s i n 6 < 6 < t a n 6 ( A r t . 1 3 2 ) . - . l < - ^ - < s e c 6 . s i n 6 N o w a s 6 i s d i m i n i s h e d i n d e f i n i t e l y , s e c 6 a p p r o a c h e s t h e l i m i t u n i t y ; t h e n w h e n 6 = 0 , w e h a v e s e c 6 = 1 . Q . : t h e l i m i t o f , w h i c h l i e s b e t w e e n s e c O a n d u n i t y , . . s i n 6 i s u n i t y . . - . a l s o S l n -a p p r o a c h e s t h e l i m i t u n i t y . 6 A s t J ^ l = m i x s e c a t h e l i m i t o f ^ S i . w h e n 0 i s i n -d e f i n i t e l y d i m i n i s h e d , i s a l s o u n i t y . T h i s i s o f t e n s t a t e d b r i e f l y t h u s : ™i = l , a n d ^ = l , w h e n 0 = O . 0 6 N o t e . — F r o m t h i s i t f o l l o w s t h a t t h e s i n e s a n d t h e t a n g e n t s o f v e r y s m a i l a n g l e s a r e p r o p o r t i o n a l t o t h e a n g l e s t h e m s e l v e s . 1 3 4 . I f 0 i s t h e C i r c u l a r M e a s u r e o f a n A c u t e A n g l e , s i n 9 l i e s b e t w e e n 6 a n d 6 ; a n d c o s 6 l i e s b e t w e e n 1 a n d 4 2 2 1 6 ( 1 ) W e h a v e t a n - > -( A r t . 1 3 2 ) 2 2 . 6 ^ 6 6 . : s i n - > - c o s - -2 2 2 2 1 0 P L A N E T R I G O N O M E T R Y . 2 s i n - c o s j > 6 c o s 2 j ~ 2 2 2 s i n 6 > ( l - s i n 2 f ) ( A r t . 1 3 2 ) s i n 0 < 0 a n d > 0 — - . 4 ( 2 ) c o s 0 = l - 2 s i n 2 ^ -A t > l - 2 ( . - . c o s 0 > 1 -- . 2 A l s o , . s i n | > | _ V | Y b y ( 1 ) . c o s 0 < 1 - 2 -2 3 2 _ f l 2 6 6 2 1 6 5 1 2 ' . - . c o s 6 > 1 --a n d < 1 --+ — -2 2 1 6 N o t e . — I t m a y b e p r o v e d t h a t s i n 0 > 6 , a s f o l l o w s : 6 W e h a v e 3 s i n - -s i n 9 = 4 s i n 3 | ( A r t . 5 0 ) ( 1 ) . - . 3 s i n £ -s i n | = 4 s i n 3 £ ( b y p u t t i n g ? f o r 8 ) ( 2 ) 3 s i n £ -s i n — = 4 s i n 3 £ ( n ) 3 » 3 " - 1 3 " M u l t i p l y ( 1 ) , ( 2 ) , - - - ( n ) b y 1 , 3 , — 3 n _ 1 , r e s p e c t i v e l y , a n d a d d t h e m , 8 " s i n £ -s i n 9 = i ( s i n 3 - + 3 s i n 3 £ + . . . 3 " - 1 s i n 3 £ V 8 " \ 8 8 » 3 V S I N E A N D C O S I N E O F 1 0 " A N D O F 1 , . 2 1 1 s R —3 " / 0 3 0 3 9 3 \ 8 i n 9 < 4 ( 3 3 + 3 i + - i ^ + T ) ( A r t . 1 3 2 ) 3 " < ± » ( i + X + . . . - L - ) . s i n —3 " I f n = o o , t h e n — g — = 1 ( A r t . 1 3 3 ) 3 " 4 „ , A 1 1 \ « 1 6 3 § i S ( 1 + 3 i + - ^ = 2 ) = i r - 1 — = 6 -3 ! . - . « - s i n f l < - , a n d . - . « i n f l > 9 — — . 6 6 T h i s m a k e s t h e l i m i t s f o r s i n 9 c l o s e r t h a n i n ( 1 ) o f t h i s A r t . 1 3 5 . T o c a l c u l a t e t h e S i n e a n d C o s i n e o f 1 0 " a n d o f V . ( 1 ) L e t 6 b e t h e c i r c u l a r m e a s u r e o f 1 0 " . T h e n 1 0 3 . 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 -6 : 1 8 0 x 6 0 x 6 0 6 4 8 0 0 o r 6 = . 0 0 0 0 4 8 4 8 1 3 6 8 1 1 0 - - - , c o r r e c t t o 1 5 d e c i m a l p l a c e s . . - . - = . 0 0 0 0 0 0 0 0 0 0 0 0 0 3 2 — , " " " " 4 . - . = . 0 0 0 0 4 8 4 8 1 3 6 8 0 7 8 - , " " « 4 a t H e n c e t h e t w o q u a n t i t i e s 6 a n d 6 a g r e e t o 1 2 d e c i -4 m a l p l a c e s ; a n d s i n c e s i n 6 < 6 a n d > 6 — -( A r t . 1 3 4 ) , 4 . - . s i n 1 0 " = . 0 0 0 0 4 8 4 8 1 3 6 8 , t o 1 2 d e c i m a l p l a c e s . W e h a v e c o s 1 0 " = V I - s i n 2 1 0 " = 1 -£ s i n 2 1 0 " = . 9 9 9 9 9 9 9 9 8 8 2 4 8 — , t o 1 3 d e c i m a l p l a c e s . O r w e m a y u s e t h e r e s u l t s e s t a b l i s h e d i n ( 2 ) o f A r t . 1 3 4 , a n d o b t a i n t h e s a m e v a l u e . 2 1 2 P L A N E T R I G O N O M E T R Y . ( 2 ) L e t 6 b e t h e c i r c u l a r m e a s u r e o f 1 ' . T h e n f l = „ Q g r > = . 0 0 0 2 9 0 8 8 8 2 0 8 6 6 5 , t o 1 5 d e c i m a l p l a c e s . 1 8 0 x 6 0 . - . ^ = . 0 0 0 0 0 0 0 0 0 0 0 6 t o 1 2 " " 4 . - . 0 - ? = . 0 0 0 2 9 0 8 8 8 2 0 t o l l « " 4 H e n c e 6 a n d 6 — — d i f f e r o n l y i n t h e t w e l f t h d e c i m a l . . - . s i n 1 ' = . 0 0 0 2 9 0 8 8 8 2 0 t o 1 1 d e c i m a l p l a c e s . c o s 1 ' = V I -s i n 2 1 ' = . 9 9 9 9 9 9 9 5 7 6 9 2 0 2 5 t o 1 5 d e c i m a l p l a c e s , O t h e r w i s e t h u s : 1 -^ = . 9 9 9 9 9 9 9 5 7 6 9 2 0 2 5 0 2 9 t o 1 8 d e c i m a l p l a c e s . 2 r a n d — = . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 t o 1 7 d e c i m a l p l a c e s . 1 6 r B u t c o s V > 1 -f a n d < 1 -^ + ~ . . . ( A r t . 1 3 4 ) 2 2 1 6 . - . c o s 1 ' = . 9 9 9 9 9 9 9 5 7 6 9 2 0 2 5 t o 1 5 d e c i m a l p l a c e s , a s b e f o r e . C o r . 1 . T h e s i n e o f 1 0 " e q u a l s t h e c i r c u l a r m e a s u r e o f 1 0 " , t o 1 2 d e c i m a l p l a c e s ; a n d t h e s i n e o f V e q u a l s t h e c i r c u l a r m e a s u r e o f V t o 1 1 d e c i m a l p l a c e s . C o r . 2 . I f n d e n o t e a n y n u m b e r o f s e c o n d s l e s s t h a n 6 0 , w e s h a l l h a v e a p p r o x i m a t e l y s i n n " = n s i n 1 " , f o r t h e s i n e o f n " = t h e c i r c u l a r m e a s u r e o f n " , a p p r o x i m a t e l y , = n t i m e s t h e c i r c u l a r m e a s u r e o f 1 " . o c i r c u l a r m e a s u r e o f n " . . . . . C o r . 3 . n = — — — , a p p r o x i m a t e l y ; t h a t s i n 1 " i s , t h e n u m b e r o f s e c o n d s i n a n y s m a l l a n g l e i s f o u n d T A B L E S O F N A T U R A L S I N E S A N D C O S I N E S . 2 1 3 a p p r o x i m a t e l y b y d i v i d i n g t h e c i r c u l a r m e a s u r e o f t h a t a n g l e b y t h e s i n e o f o n e s e c o n d . 1 3 6 . T o c o n s t r u c t a T a b l e o f N a t u r a l S i n e s a n d C o s i n e s a t I n t e r v a l s o f 1 ' . W e h a v e , b y A r t . 4 5 , • s i n ( a ; + y ) = 2 s i n a ; c o s y — s i n ( a ; — y ) , c o s ( a ; + y ) = 2 c o s x c o s y — c o s ( a ; — y ) . S u p p o s e t h e a n g l e s t o i n c r e a s e b y 1 ' ; p u t t i n g y = V , w e h a v e , s i n ( a ; + 1 ' ) = 2 s i n x c o s 1 ' — s i n ( a ; — 1 ' ) . . . . ( I ) c o s ( a ; + 1 ' ) = 2 c o s x c o s 1 ' — c o s ( a ; — 1 ' ) . . . . ( 2 ) P u t t i n g x = V , 2 ' , 3 ' , 4 ' , e t c . , i n ( 1 ) a n d ( 2 ) , w e g e t f o r t h e s i n e s s i n 2 ' = 2 s i n 1 ' c o s 1 ' -s i n 0 ' = . 0 0 0 5 8 1 7 7 6 4 , s i n 3 ' = 2 s i n 2 ' c o s 1 ' -s i n 1 ' = . 0 0 0 8 7 2 6 6 4 6 , s i n 4 ' = 2 s i n 3 ' c o s 1 ' -s i n 2 ' = . 0 0 1 1 6 3 5 5 2 6 ; a n d f o r t h e c o s i n e s c o s 2 ' = 2 c o s 1 ' c o s 1 ' -c o s 0 ' = . 9 9 9 9 9 9 8 3 0 8 , c o s 3 ' = 2 c o s 2 ' c o s 1 ' -c o s 1 ' = . 9 9 9 9 9 9 6 1 9 3 , c o s 4 ' = 2 c o s 3 ' c o s 1 ' -c o s 2 ' = . 9 9 9 9 9 9 3 2 2 3 . W e c a n p r o c e e d i n t h i s m a n n e r u n t i l w e f i n d t h e v a l u e s o f t h e s i n e s a n d c o s i n e s o f a l l a n g l e s a t i n t e r v a l s o f 1 ' f r o m 0 ° t o 3 0 ° . 1 3 7 . A n o t h e r M e t h o d . L e t a d e n o t e a n y a n g l e . T h e n , i n t h e i d e n t i t y , s i n ( n + 1 ) « = 2 s i n n a c o s a — s i n ( n — l ) a , p u t 2 ( 1 — c o s a ) = & , a n d w e g e t s i n ( n + l ) « — s i n n r c = s i n n a — s i r \ ( n — l ) n — k s i n n a . ( 1 ) T h i s m e t h o d l s d u e t o T h o m a s S i m p s o n , a n E n g l i s h g e o m e t r i c i a n . 2 1 4 P L A N E T R I G O N O M E T R Y . T h i s f o r m u l a e n a b l e s u s t o c o n s t r u c t a t a b l e o f s i n e s o f a n g l e s w h o s e c o m m o n d i f f e r e n c e i s a . T h u s , s u p p o s e a = 1 0 " , a n d l e t n = 1 , 2 , 3 , 4 , e t c . T h e n s i n 2 0 " -s i n 1 0 " = s i n 1 0 " -f t s i n 1 0 " , s i n 3 0 " -s i n 2 0 " = s i n 2 0 " -s i n 1 0 " -f t s i n 2 0 " , s i n 4 0 " -s i n 3 0 " = s i n 3 0 " -s i n 2 0 " -f t s i n 3 0 " , e t c . T h e s e e q u a t i o n s g i v e i n s u c c e s s i o n s i n 2 0 " , s i n 3 0 " , e t c . I t w i l l b e s e e n t h a t t h e m o s t l a b o r i o u s p a r t o f t h i s w o r k i s t h e m u l t i p l i c a t i o n o f f t b y t h e s i n e s o f 1 0 " , 2 0 " , e t c . , a s t h e y a r e s u c c e s s i v e l y f o u n d . B u t f r o m t h e v a l u e o f c o s 1 0 " , w e h a v e f t = 2 ( 1 -c o s 1 0 " ) = . 0 0 0 0 0 0 0 0 2 3 5 0 4 , t h e s m a l l n e s s o f w h i c h f a c i l i t a t e s t h e p r o c e s s . I n t h e s a m e m a n n e r a t a b l e o f c o s i n e s c a n b e c o n s t r u c t e d b y m e a n s o f t h e f o r m u l a , c o s ( n + l ) a — c o s n a = c o s n a — c o s ( n — l ) a — f t c o s n e t , w h i c h i s o b t a i n e d f r o m t h e i d e n t i t y , c o s ( n + 1 ) « = 2 c o s n a c o s a — c o s ( n — l ) a , b y p u t t i n g 2 ( 1 — c o s a ) = f t , a s b e f o r e . 1 3 8 . T h e S i n e s a n d C o s i n e s f r o m 3 0 ° t o 6 0 ° . — I t i s n o t n e c e s s a r y t o c a l c u l a t e i n t h i s w a y t h e s i n e s a n d c o s i n e s o f a n g l e s b e y o n d 3 0 ° , a s w e c a n o b t a i n t h e i r v a l u e s f o r a n g l e s f r o m 3 0 ° t o 6 0 ° m o r e e a s i l y b y m e a n s o f t h e f o r m u l a e ( A r t . 4 5 ) : s i n ( 3 0 ° + a ) = c o s « -s i n ( 3 0 ° -a ) , c o s ( 3 0 ° + « ) = c o s ( 3 0 ° — « ) — s i n a , b y g i v i n g a a l l v a l u e s u p t o 3 0 ° . T h u s , s i n 3 0 ° 1 ' = c o s 1 ' -s i n 2 9 ° 5 9 ' , c o s 3 0 ° 1 ' = c o s 2 9 ° 5 9 ' -s i n 1 ' , a n d s o o n . T A B L E S O F T A N G E N T S A N D S E C A N T S . 2 1 5 1 3 9 . S i n e s o f A n g l e s g r e a t e r t h a n 4 5 ° . — W h e n t h e s i n e s o f a n g l e s u p t o 4 5 ° h a v e b e e n c a l c u l a t e d , t h o s e o f a n g l e s b e t w e e n 4 5 ° a n d 9 0 ° m a y b e d e d u c e d b y t h e f o r m u l a s i n ( 4 5 ° + a ) — s i n ( 4 5 ° -a ) = V 2 s i n a . ( A r t . 4 5 ) A l s o , w h e n t h e s i n e s o f a n g l e s u p t o 6 0 ° h a v e b e e n f o u n d , t h e r e m a i n d e r u p t o 9 0 ° c a n b e f o u n d s t i l l m o r e e a s i l y f r o m t h e f o r m u l a s i n ( 6 0 ° + a ) -s i n ( 6 0 ° - « ) = s i n a . H a v i n g c o m p l e t e d a t a b l e o f s i n e s , t h e c o s i n e s a r e k n o w n , S 1 U c 6 . / a a q \ c o s a = s i n ( 9 0 — a ) . O t h e r w i s e t h u s : W h e n t h e s i n e s a n d c o s i n e s o f t h e a n g l e s u p t o 4 5 ° h a v e b e e n o b t a i n e d , t h o s e o f a n g l e s b e t w e e n 4 5 ° a n d 9 0 ° a r e o b t a i n e d f r o m t h e f a c t t h a t t h e s i n e o f a n a n g l e i s e q u a l t o t h e c o s i n e o f i t s c o m p l e m e n t , s o t h a t i t i s n o t n e c e s s a r y t o p r o c e e d i n t h e c a l c u l a t i o n b e y o n d 4 5 ° . N o t e . — A m o r e m o d e r n m e t h o d o f c a l c u l a t i n g t h e s i n e s a n d c o s i n e s o f a n g l e s i s t o u s e s e r i e B ( 3 ) a n d ( 4 ) o f A r t . 1 5 6 . 1 4 0 . T a b l e s o f T a n g e n t s a n d S e c a n t s . — T o f o r m a t a b l e o f t a n g e n t s , w e f i n d t h e t a n g e n t s o f a n g l e s u p t o 4 5 ° , f r o m t h e t a b l e s o f s i n e s a n d c o s i n e s , b y m e a n s o f t h e f o r m u l a c o s a T h e n t h e t a n g e n t s o f a n g l e s f r o m 4 5 ° t o 9 0 ° m a y b e o b t a i n e d b y m e a n s o f t h e i d e n t i t y t a n ( 4 5 ° + a ) = t a n ( 4 5 ° -a ) + 2 t a n 2 a . W h e n t h e t a n g e n t s h a v e b e e n f o u n d , t h e c o t a n g e n t s a r e k n o w n , s i n c e t h e c o t a n g e n t o f a n y a n g l e i s e q u a l t o t h e t a n g e n t o f i t s c o m p l e m e n t . A t a b l e o f c o s e c a n t s m a y b e o b t a i n e d b y c a l c u l a t i n g t h e r e c i p r o c a l s o f t h e s i n e s ; o r t h e y m a y b e o b t a i n e d m o r e C a l l e d C a g n o l i ' s f o r m u l a . 2 1 6 P L A N E T R I G O N O M E T R Y . e a s i l y f r o m t h e t a b l e s o f t h e t a n g e n t s b y m e a n s o f t h e f o r m u l a c o s e c a = t a n - + c o t « . T h e s e c a n t s a r e t h e n k n o w n , s i n c e t h e s e c a n t o f a n y a n g l e i s e q u a l t o t h e c o s e c a n t o f i t s c o m p l e m e n t . 1 4 1 . F o r m u l a e o f V e r i f i c a t i o n . — F o r m u l a e u s e d t o t e s t t h e a c c u r a c y o f t h e c a l c u l a t e d s i n e s o r c o s i n e s o f a n g l e s a r e c a l l e d F o r m u l a s o f V e r i f i c a t i o n . I t i s n e c e s s a r y t o h a v e m e t h o d s o f v e r i f y i n g f r o m t i m e t o t i m e t h e c o r r e c t n e s s o f t h e v a l u e s o f t h e s i n e s a n d c o s i n e s o f a n g l e s c a l c u l a t e d b y t h e p r e c e d i n g m e t h o d , s i n c e a n y e r r o r m a d e i n o b t a i n i n g t h e v a l u e o f o n e o f t h e f u n c t i o n s w o u l d b e r e p e a t e d t o t h e e n d o f t h e w o r k . F o r t h i s p u r p o s e w e m a y c o m p a r e t h e v a l u e o f t h e s i n e o f a n y a n g l e o b t a i n e d b y t h e p r e c e d i n g m e t h o d w i t h i t s v a l u e o b t a i n e d i n d e p e n d e n t l y . T h u s , f o r e x a m p l e , w e k n o w t h a t s i n 1 8 ° = - — ( A r t . 5 7 ) ; h e n c e t h e s i n e o f 1 8 ° m a y b e c a l c u l a t e d t o a n y d e g r e e o f a p p r o x i m a t i o n , a n d b y c o m p a r i s o n w i t h t h e v a l u e o b t a i n e d i n t h e t a b l e s , w e c a n j u d g e h o w f a r w e c a n r e l y u p o n t h e t a b l e s . S i m i l a r l y , w e m a y c o m p a r e o u r r e s u l t s f o r t h e a n g l e s 2 2 £ ° , 3 0 ° , 3 6 ° , 4 5 ° , e t c . , c a l c u l a t e d b y t h e p r e c e d i n g m e t h o d w i t h t h e s i n e s a n d c o s i n e s o f t h e s a m e a n g l e s a s o b t a i n e d i n A r t s . 2 6 , 2 7 , 5 6 , 5 7 , 5 8 , e t c . T h e r e a r e , h o w e v e r , c e r t a i n w e l l - k n o w n f o r m u l a e o f v e r i f i c a t i o n w h i c h c a n b e u s e d t o v e r i f y a n y p a r t o f t h e c a l c u l a t e d t a b l e s ; t h e s e a r e E u l e i J s F o r m u l a ! : s i n ( 3 6 ° + A ) -s i n ( 3 6 ° -A ) + s i n ( 7 2 ° -A ) -s i n ( 7 2 ° + A ) = s i n A . c o s ( 3 6 ° + A ) + c o s ( 3 6 ° -A ) -c o s ( 7 2 ° + A ) -c o s ( 7 2 ° -A ) = c o s A . L O G A R I T H M I C T R I G O N O M E T R I C F U N C T I O N S . 2 1 7 L e g e n d r e , s F o r m u l a : s i n ( 5 4 ° + A ) + s i n ( 5 4 ° -A ) -s i n ( 1 8 ° + A ) — s i n ( 1 8 ° — A ) = c o s A . T h e v e r i f i c a t i o n c o n s i s t s i n g i v i n g t o A a n y v a l u e , a n d t a k i n g f r o m t h e t a b l e s t h e s i n e s a n d c o s i n e s o f t h e a n g l e s i n v o l v e d : t h e s e v a l u e s m u s t s a t i s f y t h e a b o v e e q u a t i o n s . T o p r o v e E u l e r ' s F o r m u l a s : s i n ( 3 6 ° + A ) -s i n ( 3 6 ° -A ) = 2 c o s 3 6 ° s i n A . ( A r t . 4 5 ) = V 5 + l a i n A ( A r t . 5 8 ) t s i n ( 7 2 ° + A ) -s i n ( 7 2 ° -A ) = 2 c o s 7 2 ° s i n A . ( A r t . 4 5 ) = ^ ~ 1 s i n A . ( A r t . 5 7 ) S u b t r a c t i n g t h e l a t t e r f r o m t h e f o r m e r , w e g e t s i n A . S i m i l a r l y , E u l e r , s s e c o n d f o r m u l a m a y b e p r o v e d . B y s u b s t i t u t i n g 9 0 ° — A f o r A i n t h i s f o r m u l a w e o b t a i n L e g e n d r e , s F o r m u l a . 1 4 2 . T a b l e s o f L o g a r i t h m i c T r i g o n o m e t r i c F u n c t i o n s . — T o s a v e t h e t r o u b l e o f r e f e r r i n g t w i c e t o t a b l e s — f i r s t t o t h e t a b l e o f n a t u r a l f u n c t i o n s f o r t h e v a l u e o f t h e f u n c t i o n , a n d t h e n t o a t a b l e o f l o g a r i t h m s f o r t h e l o g a r i t h m o f t h a t f u n c t i o n — i t i s c o n v e n i e n t t o c a l c u l a t e t h e l o g a r i t h m s o f t r i g o n o m e t r i c f u n c t i o n s , a n d a r r a n g e t h e m i n t a b l e s , c a l l e d t a b l e s o f l o g a r i t h m i c s i n e s , c o s i n e s , e t c . W h e n t a b l e s o f n a t u r a l s i n e s a n d c o s i n e s h a v e b e e n c o n s t r u c t e d , t a b l e s o f l o g a r i t h m i c s i n e s a n d c o s i n e s m a y b e m a d e b y m e a n s o f t a b l e s o f o r d i n a r y l o g a r i t h m s , w h i c h w i l l g i v e t h e l o g a r i t h m o f t h e c a l c u l a t e d n u m e r i c a l v a l u e o f t h e s i n e o r c o s i n e o f a n y a n g l e ; a d d i n g 1 0 t o t h e l o g a r i t h m s o f o u n d w e h a v e t h e c o r r e s p o n d i n g t a b u l a r l o g a r i t h m . T h e l o g a r i t h m i c t a n g e n t s m a y b e f o u n d b y t h e r e l a t i o n l o g t a n A = 1 0 + l o g s i n A — l o g c o s A ; a n d t h u s a t a b l e o f l o g a r i t h m i c t a n g e n t s m a y b e c o n s t r u c t e d . 2 1 8 P L A N E T R I G O N O M E T R Y . P R O P O R T I O N A L P A R T S . 1 4 3 . T h e P r i n c i p l e o f P r o p o r t i o n a l P a r t s . — I t i s o f t e n n e c e s s a r y t o f i n d f r o m a t a b l e o f l o g a r i t h m s , t h e l o g a r i t h m o f a n u m b e r c o n t a i n i n g m o r e d i g i t s t h a n a r e g i v e n i n t h e t a b l e . I n o r d e r t o d o t h i s , w e a s s u m e d , i n C h a p t e r I V . , t h e p r i n c i p l e o f p r o p o r t i o n a l p a r t s , w h i c h i s a s f o l l o w s : T h e d i f f e r e n c e s b e t w e e n t h r e e n u m b e r s a r e p r o p o r t i o n a l t o t h e c o r r e s p o n d i n g d i f f e r e n c e s b e t w e e n t h e i r l o g a r i t h m s , p r o v i d e d t h e d i f f e r e n c e s b e t w e e n t h e n u m b e r s a r e s m a l l c o m p a r e d w i t h t h e n u m b e r s . B y m e a n s o f t h i s p r i n c i p l e , w e a r e e n a b l e d t o u s e t a b l e s o f a m o r e m o d e r a t e s i z e t h a n w o u l d o t h e r w i s e b e n e c e s s a r y . W e s h a l l n o w i n v e s t i g a t e h o w f a r , a n d w i t h w h a t e x c e p t i o n s , t h e p r i n c i p l e o r r u l e o f p r o p o r t i o n a l i n c r e a s e i s t r u e . 1 4 4 . T o p r o v e t h e R u l e f o r t h e T a b l e o f C o m m o n L o g a r i t h m s . W e h a v e • l o g ( n + d ) -l o g n = l o g ^ ± ^ = l o g f l + n \ = n ( - - f + & - - ) . ( A r t . 1 3 0 ) \ n I n 1 3 n s ) w h e r e / i = . 4 3 4 2 9 4 4 8 a q u a n t i t y < \ . N o w l e t n b e a n i n t e g e r n o t < 1 0 0 0 0 , a n d d n o t > 1 ; t h e n -i s n o t g r e a t e r t h a n . 0 0 0 1 . n . - . i s n o t > > ; ( . 0 0 0 1 ) 2 , i . e . , n o t > . 0 0 0 0 0 0 0 0 2 5 ; a n d i s m u c h l e s s t h a n t h i s . 3 n 3 . - . l o g ( n + d ) — l o g n = / - , c o r r e c t a t l e a s t a s f a r a s s e v e n d e c i m a l p l a c e s . n t ) R U L E O F P R O P O R T I O N A L P A R T S . 2 1 9 H e n c e i f t h e n u m b e r b e c h a n g e d f r o m n t o n + d , t h e ' c o r r e s p o n d i n g c h a n g e i n t h e l o g a r i t h m i s a p p r o x i m a t e l y i x d n T h e r e f o r e , t h e c h a n g e o f t h e l o g a r i t h m i s a p p r o x i m a t e l y p r o p o r t i o n a l t o t h e c h a n g e o f t h e n u m b e r . 1 4 5 . T o p r o v e t h e R u l e f o r t h e T a b l e o f N a t u r a l S i n e s . s i n ( 6 + f t ) — s i n 6 = s i n h c o s 6 — s i n 6 ( 1 — c o s h ) = s i n h c o s 6 ( l — t a n 6 t a n | ^ . ( A r t . 5 1 ) I f h i s t h e c i r c u l a r m e a s u r e o f a v e r y s m a l l a n g l e , s i n h = h n e a r l y , a n d t a n ^ = -n e a r l y . 2 2 . : s i n ( 6 + h ) — s i n 6 = h c o s 6 ^ 1 — t a n 6 t a n | ^ f t 2 = h c o s 6 s i n 6 . 2 I f f t i s t h e c i r c u l a r m e a s u r e o f a n a n g l e n o t > 1 ' , t h e n h i s n o t > . 0 0 0 3 ( A r t . 1 3 5 ) . . - . -i s n o t > . 0 0 0 0 0 0 0 5 ; a n d s i n 6 i s n o t > 1 . . - . s i n ( 0 + 7 i ) — s i n 6 = h c o s 6 , a s f a r a s s e v e n d e c i m a l p l a c e s , w h i c h p r o v e s t h e p r o p o s i t i o n . S i m i l a r l y , s i n ( 6 — h ) — s i n 6 = — h c o s 6 , a p p r o x i m a t e l y . 1 4 6 . T o p r o v e t h e R u l e f o r a T a b l e o f N a t u r a l C o s i n e s . c o s ( 6 — h ) — c o s 6 = s i n h s i n 6 — c o s 6 ( 1 — c o s h ) s i n h s i n 6 f l — c o t 6 t a n I f f t i s t h e c i r c u l a r m e a s u r e o f a v e r y s m a l l a n g l e , s i n h = h a r l y , a n d t a n - = -n e a r l y . 2 2 . : c o s ( 6 — h ) — c o s 6 = h s i n 6 ^ 1 — c o t 6 t a n | j = 7 i s i u 6 — — c o s 0 . 2 2 2 0 P L A N E T R I G O N O M E T U Y . W e m a y p r o v e , a s i n A r t . 1 4 5 , t h a t -c o s 6 i s n o t > . 0 0 0 0 0 0 0 5 . 2 . : c o s ( 6 — h ) — c o s 6 = 7 i s i n 6 , a s f a r a s s e v e n d e c i m a l p l a c e s , w h i c h p r o v e s t h e p r o p o s i t i o n . S i m i l a r l y , c o s ( 6 + h ) — c o s 6 = — h s i n 6 , a p p r o x i m a t e l y . 1 4 7 . T o p r o v e t h e R u l e f o r a T a b l e o f N a t u r a l T a n g e n t s . t a n ( f l + 7 Q - t a n f l = s i n ( 0 + ) _ g i n j = s i n A c o s ( 6 + h ) c o s 0 c o s ( 0 + h ) c o s 0 t a n h c o s 2 0 ( 1 — t a n 6 t a n f t ) I f 7 i i s t h e c i r c u l a r m e a s u r e o f a v e r y s m a l l a n g l e , t a n 7 t = h n e a r l y . . - . t a n ( f l + 7 i ) - t a n f l = T i s e c 2 f l , v ' l - f t t a n 0 = h s e c 2 6 + h 2 s i n 6 s e c 3 0 . . : t a n ( 6 + f t ) — t a n 0 = h s e c 2 f l , a p p r o x i m a t e l y , u n l e s s s i n 6 s e c 3 0 i s l a r g e , w h i c h p r o v e s t h e p r o p o s i t i o n . S i m i l a r l y , c o t ( 6 — h ) — c o t 6 = h c o s e c 2 6 , a p p r o x i m a t e l y . S c h . 1 . I f h i s t h e c i r c u l a r m e a s u r e o f a n a n g l e n o t > V , t h e n h i s n o t > . 0 0 0 3 . H e n c e t h e g r e a t e s t v a l u e o f f t 2 s i n O s e c 3 6 i s n o t > . 0 0 0 0 0 0 0 9 s i n 6 s e c 3 6 . T h e r e f o r e , w h e n 6 > ~ , 4 w e a r e l i a b l e t o a n e r r o r i n t h e s e v e n t h p l a c e o f d e c i m a l s . H e n c e t h e r u l e i s n o t t r u e f o r t a b l e s o f t a n g e n t s c a l c u l a t e d f o r e v e r y m i n u t e , w h e n t h e a n g l e i s b e t w e e n 4 5 ° a n d 9 0 ° . S c h . 2 . S i n c e t h e c o t a n g e n t o f a n a n g l e i s e q u a l t o t h e t a n g e n t o f i t s c o m p l e m e n t , i t f o l l o w s i m m e d i a t e l y t h a t t h e r u l e m u s t n o t b e u s e d f o r a t a b l e o f c o t a n g e n t s , c a l c u l a t e d f o r e v e r y m i n u t e , w h e n t h e a n g l e l i e s b e t w e e n 0 ° a n d 4 5 ° . R U L E O F P R O P O R T I O N A L P A R T S . 2 2 1 1 4 8 . T o p r o v e t h e R u l e f o r a T a b l e o f L o g a r i t h m i c S i n e s . s i n ( 6 + h ) -s i n 6 = h c o s 6 --s i n 6 . . ( A r t . 1 4 o } . - . s i ° ( f l + f t ) = 1 + f t c 0 t 0 _ f t 2 . s i n 0 2 . - . l o g s i n ( 0 + h ) — l o g s i n 0 ^ / x l o g ^ l + f c c o t f l - ! ^ = f i ^ h c o t 0 -| 2 c o t 0 - | y + . . . J ( A r t . 1 3 0 ) = / i f t c o t 6 - ^ ( 1 + c o t 2 0 ) + -= / u . A c o t 0 — c o s e c 2 0 + - - . I f f t i s t h e c i r c u l a r m e a s u r e o f a n a n g l e n o t > 1 0 " , t h e n h i s n o t > . 0 0 0 0 5 , a n d t h e r e f o r e , u n l e s s c o t 6 i s s m a l l o r c o s e c 2 6 l a r g e , w e h a v e l o g s i n ( 6 + h ) — l o g s i n 6 = f t h c o t 0 , a s f a r a s s e v e n d e c i m a l p l a c e s , w h i c h p r o v e s t h e r u l e t o b e g e n e r a l l y t r u e . S c h . 1 . W h e n 6 i s s m a l l , c o s e c 6 i s l a r g e . I f t h e l o g s i n e s a r e c a l c u l a t e d t o e v e r y 1 0 " , t h e n h i s n o t > . 0 0 0 0 5 , a n d / x i s n o t > . 5 . i i 2 m . a . ^ 6 c o s e c 2 6 . : I f i h 2 c o s e c 2 6 i s n o t > -1 0 1 0 I n o r d e r t h a t t h i s e r r o r m a y n o t a f f e c t t h e s e v e n t h d e c i m a l p l a c e , 6 c o s e c 2 0 m u s t n o t b e > 1 0 3 , t h a t i s , 6 m u s t n o t b e l e s s t h a n a b o u t 5 ° . W h e n 6 i s s m a l l , c o t 6 i s l a r g e . H e n c e , w h e n t h e a n g l e s 2 2 2 P L A N E T R I G O N O M E T R Y . a r e s m a l l , t h e d i f f e r e n c e s o f c o n s e c u t i v e l o g s i n e s a r e i r r e g u -. I q g , a n d t h e y a r e n o t i n s e n s i b l e . T h e r e f o r e t h e r u l e d o e s n o t a p p l y t o t h e l o g s i n e w h e n t h e a n g l e i s l e s s t h a n 5 ° . S c h . 2 . W h e n 6 i s n e a r l y a r i g h t a n g l e , c o t 6 i s s m a l l , a n d c o s e c 6 a p p r o a c h e s u n i t y . H e n c e , w h e n t h e a n g l e s a r e n e a r l y r i g h t a n g l e s , t h e d i f f e r e n c e s o f c o n s e c u t i v e l o g s i n e s a r e i r r e g u l a r a n d n e a r l y i n s e n s i b l e . 1 4 9 . T o p r o v e t h e R u l e f o r a T a b l e o f L o g a r i t h m i c C o s i n e s . h 2 c o s ( 0 — h ) — c o s 0 = h s i n 6 — — c o s 0 . ( A r t . 1 4 6 ) c o s ( f l — K ) 1 + h t a n 6 -^ . c o s 6 . : l o g c o s ( 6 — h ) — l o g c o s 6 = / i l o g ^ l + A t a n 0 - ^ = u | f c t a n 0 - ^ - - / A t a n 0 - -1 2 2 ^ 52 = / u 7 i t a n 0 - ^ s e c 2 0 + - - -I n t h i s c a s e t h e d i f f e r e n c e s w i l l b e i r r e g u l a r a n d l a r g e w h e n 6 i s n e a r l y a r i g h t a n g l e , a n d i r r e g u l a r a n d i n s e n s i b l e w h e n 6 i s n e a r l y z e r o . T h i s i s a l s o c l e a r b e c a u s e t h e s i n e o f a n a n g l e i s t h e c o s i n e o f i t s c o m p l e m e n t . 1 5 0 . T o p r o v e t h e R u l e f o r a T a b l e o f L o g a r i t h m i c T a n g e n t s . t a n ( 0 + h ) -t a n 6 = h s e c 2 6 + h 2 s i n 6 s e c 3 6 . ( A r t . 1 4 7 ) R U L E O F P R O P O R T I O N A L P A R T S . 2 2 3 . - . l o g t a n ( 6 + h ) — l o g t a n 6 | _ s i n 0 c o s 0 2 ^ s i n 0 c o s 0 J J = + u t f ( s e c O V -s i n 6 c o s 0 ^ V 2 s i n 2 0 c o s 2 0 / . - . l o g t a n ( 6 + f t ) — l o g t a n 0 _ f i h _ n t j j o o s 2 ^ ~ ~ s i n 0 c o s 0 s i n 2 2 6 1 5 1 . C a s e s w h e r e t h e P r i n c i p l e o f P r o p o r t i o n a l P a r t s i s I n a p p l i c a b l e . I t a p p e a r s f r o m t h e l a s t s i x A r t i c l e s t h a t i f h i s s m a l l e n o u g h , t h e d i f f e r e n c e s a r e p r o p o r t i o n a l t o f t , f o r v a l u e s o f 6 w h i c h a r e n e i t h e r v e r y s m a l l n o r n e a r l y e q u a l t o a r i g h t a n g l e . T h e f o l l o w i n g e x c e p t i o n a l c a s e s a r i s e : ( 1 ) T h e d i f f e r e n c e s i n ( 6 + h ) — s i n 6 i s i n s e n s i b l e w h e n 6 i s n e a r l y 9 0 ° , f o r i n t h a t c a s e h c o s 6 i s v e r y s m a l l ; i t i s t h e n a l s o i r r e g u l a r , f o r £ f t 2 s i n 0 m a y b e c o m e c o m p a r a b l e w i t h h c o s 6 . ( 2 ) T h e d i f f e r e n c e c o s ( 6 + h ) — c o s 6 i s b o t h i n s e n s i b l e a n d i r r e g u l a r w h e n 6 i s s m a l l . ( 3 ) T h e d i f f e r e n c e t a n ( 6 + h ) — t a n 6 i s i r r e g u l a r w h e n 0 i s n e a r l y 9 0 ° , f o r A 2 s i n 0 s e c 3 6 m a y t h e n b e c o m e c o m p a r a b l e w i t h h s e c 2 6 ; i t i s n e v e r i n s e n s i b l e , s i n c e s e c 0 i s n o t < 1 . ( 4 ) T h e d i f f e r e n c e l o g s i n ( 6 + h ) — l o g s i n 6 i s i r r e g u l a r w h e n 6 i s s m a l l , a n d b o t h i r r e g u l a r a n d i n s e n s i b l e w h e n 0 i s n e a r l y 9 0 ° . ( 5 ) T h e d i f f e r e n c e l o g c o s ( 0 + f t ) — l o g c o s 0 i s i n s e n s i b l e a n d i r r e g u l a r w h e n 0 i s s m a l l , a n d i r r e g u l a r w h e n 0 i s n e a r l y 9 0 ° . ( 6 ) T h e d i f f e r e n c e l o g t a n ( 0 + f t ) — l o g t a n 0 i s i r r e g u l a r w h e n 0 i s e i t h e r s m a Z i o r n e a r l y 9 0 ° . 2 2 4 P L A N E T R I G O N O M E T R Y . A d i f f e r e n c e w h i c h i s i n s e n s i b l e i s a l s o i r r e g u l a r ; b u t t h e c o n v e r s e d o e s n o t h o l d . W h e n t h e d i f f e r e n c e s f o r a f u n c t i o n a r e i n s e n s i b l e t o t h e n u m b e r o f d e c i m a l p l a c e s o f t h e t a b l e s , t h e t a b l e s w i l l g i v e t h e f u n c t i o n s w h e n t h e a n g l e i s k n o w n , b u t w e c a n n o t u s e t h e t a b l e s t o f i n d a n y i n t e r m e d i a t e a n g l e b y m e a n s o f t h i s f u n c t i o n ; t h u s , w e c a n n o t d e t e r m i n e 6 f r o m t h e v a l u e l o g c o s 6 , f o r s m a l l a n g l e s , o r f r o m t h e v a l u e l o g s i n 6 , f o r a n g l e s n e a r l y 9 0 ° . W h e n t h e d i f f e r e n c e s f o r a f u n c t i o n a r e i r r e g u l a r w i t h o u t b e i n g i n s e n s i b l e , t h e a p p r o x i m a t e m e t h o d o f p r o p o r t i o n a l p a r t s i s n o t s u f f i c i e n t f o r t h e d e t e r m i n a t i o n o f t h e a n g l e b y m e a n s o f t h e f u n c t i o n , n o r t h e f u n c t i o n b y m e a n s o f t h e a n g l e ; t h u s , t h e a p p r o x i m a t i o n i s i n a d m i s s i b l e f o r l o g s i n 6 , w h e n 6 i s s m a l l , f o r l o g c o s 6 , w h e n 6 i s n e a r l y 9 0 ° , a n d f o r l o g t a n 6 i n e i t h e r c a s e . ( C o m p a r e A r t . 8 1 . ) I n t h e s e c a s e s o f i r r e g u l a r i t y w i t h o u t i n s e n s i b i l i t y , t h e f o l l o w i n g t h r e e m e a n s m a y b e u s e d t o e f f e c t t h e p u r p o s e o f f i n d i n g t h e a n g l e c o r r e s p o n d i n g t o a g i v e n v a l u e o f t h e f u n c t i o n , o r o f t h e f u n c t i o n c o r r e s p o n d i n g t o a g i v e n a n g l e . 1 5 2 . T h r e e M e t h o d s t o r e p l a c e t h e R u l e o f P r o p o r t i o n a l P a r t s . ( 1 ) T h e s i m p l e s t p l a n i s t o h a v e t a b l e s o f l o g s i n e s a n d l o g t a n g e n t s , f o r e a c h s e c o n d , f o r t h e f i r s t f e w d e g r e e s o f t h e q u a d r a n t , a n d o f l o g c o s i n e s a n d l o g c o t a n g e n t s , f o r e a c h s e c o n d , f o r t h e f e w d e g r e e s n e a r 9 0 ° . S u c h t a b l e s a r e g e n e r a l l y g i v e n i n t r i g o n o m e t r i c t a b l e s o f s e v e n p l a c e s ; w e c a n t h e n u s e t h e p r i n c i p l e o f p r o p o r t i o n a l p a r t s f o r a l l a n g l e s w h i c h a r e n o t e x t r e m e l y n e a r 0 ° o r 9 0 ° . ( 2 ) D e l a m b r e , s M e t h o d . I n t h i s m e t h o d a t a b l e i s c o n s t r u c t e d w h i c h g i v e s t h e v a l u e o f l o g + l o g s i n 1 " f o r e v e r y s e c o n d f o r t h e f i r s t f e w d e g r e e s o f t h e q u a d r a n t . T h i s a r t i c l e h a s b e e n t a k e n s u b s t a n t i a l l y f r o m H o b s o n s T r i g o n o m e t r y . M E T H O D S T O R E P L A C E T H E R U L E . 2 2 5 L e t 6 b e t h e c i r c u l a r m e a s u r e o f n s e c o n d s . T h e n , w h e n 6 i s s m a l l , w e h a v e 6 = n s i n 1 " , a p p r o x i m a t e l y . , s i n 6 , s i n n " , - h i i . t » . : l o g — — = l o g — :— — = l o g s i n n " — l o g n — l o g s i n 1 " . . - . l o g s i n n " = l o g n + ^ l o g s n - + l o g s i n 1 " ^ -H e n c e , i f t h e a n g l e i s k n o w n , t h e t a b l e g i v e s t h e v a l u e o f t h e e x p r e s s i o n i n p a r e n t h e s i s , a n d l o g n c a n b e f o u n d f r o m t h e o r d i n a r y t a b l e o f t h e l o g s o f n u m b e r s ; t h u s l o g s i n n " c a n b e f o u n d . I f l o g s i n n " i s g i v e n , w e c a n f i n d a p p r o x i m a t e l y t h e v a l u e o f n , a n d t h e n f r o m t h e t a b l e w e h a v e t h e v a l u e o f t h e e x p r e s s i o n i n p a r e n t h e s i s ; t h u s w e c a n f i n d l o g n , a n d t h e n n f r o m a n o r d i n a r y t a b l e o f l o g s o f n u m b e r s . R e m . W h e n 6 i s s m a l l ( l e s s t h a n 5 ° ) , 5 1 5 - ^ = l — — , a p p r o x i m a t e l y . . ( A r t . 1 3 4 , N o t e ) 6 6 H e n c e a s m a l l e r r o r i n 6 w i l l n o t p r o d u c e a s e n s i b l e e r r o r i n t h e r e s u l t , s i n c e l o g w i l l v a r y m u c h l e s s r a p i d l y t h a n 6 . ( 3 ) M a s k e l y n e , s M e t h o d . T h e p r i n c i p l e o f t h i s m e t h o d i s t h e s a m e a s t h a t o f D e l a m b r e , s . I f 6 i s a s m a l l a n g l e , w e h a v e s i n 6 = 6 — — , a p p r o x i m a t e l y , 6 a n d c o s 0 = 1 - ^ , « . . ( A r t . 1 3 4 ) = ( c o s 0 ) , " . - . l o g s i n 6 = l o g 6 + ^ l o g c o s 6 , a p p r o x i m a t e l y . 2 2 6 P L A N E T R I G O N O M E T R Y . W h e n 6 i s a s m a l l a n g l e , t h e d i f f e r e n c e s o f l o g c o s 6 a r e i n s e n s i b l e ( A r t . 1 4 9 ) ; h e n c e , i f 6 b e g i v e n , w e c a n f i n d l o g 6 a c c u r a t e l y f r o m t h e t a b l e o f n a t u r a l l o g a r i t h m s , a n d a l s o a n a p p r o x i m a t e v a l u e o f l o g c o s 6 ; t h e f o r m u l a t h e n g i v e s l o g s i n 6 a t o n c e . I f l o g s i n 6 b e g i v e n , w e m u s t f i r s t f i n d a n a p p r o x i m a t e v a l u e o f 6 f r o m t h e t a b l e , a n d u s e t h a t f o r f i n d i n g l o g c o s 6 , a p p r o x i m a t e l y ; 6 . i s t h e n o b t a i n e d f r o m t h e f o r m u l a . E X A M P L E S . 1 . P r o v e l + 2 + | + | + -= 2 e . 2 . P r o v e l o g - ^ - f — ? — + — § — + . . . Y 3 . P r o v e - e = 1 + L ± j + 1 + 2 + 3 + -2 | 2 T [ 3 [ 4 ^ - d 1 2 , 4 , 6 , 4 . P r o v e - = ,— . e 1 3 ( 5 | 7 5 . P r o v e t a n 0 + 1 t a n 3 0 + £ t a n 5 0 + - - -/ c o s ^ 0 . = 4 l o g y c o s 6 . P r o v e l o g c l l = 2 . 3 9 7 8 9 5 2 7 b y ( 4 ) o f A r t . 1 3 0 . 7 . P r o v e l o g . 1 3 = 2 . 5 6 4 9 4 9 3 5 " » 8 . P r o v e l o g e 1 7 = 2 . 8 3 3 2 1 3 3 4 - , " a 9 . P r o v e l o g e 1 9 = 2 . 9 4 4 4 3 9 4 . u 1 0 . F i n d , b y m e a n s o f t h e t a b l e o f c o m m o n l o g a r i t h m s a n d t h e m o d u l u s , t h e N a p i e r i a n l o g a r i t h m s o f 1 3 2 5 . 0 7 , 5 2 . 9 3 8 1 , a n d . 0 8 5 6 2 3 . A n s . 7 . 1 8 9 2 3 , 3 . 9 6 9 1 3 , -2 . 4 5 7 8 . E X A M P L E S . 2 2 7 1 1 . P r o v e t h a t t h e l i m i t o f m s i n — i s 6 , w h e n m = o o . m 1 2 " " " m t a n — i s 6 , " m = » . m 1 3 " « " — s i n — i s t i t 2 , " n = o o . 2 n 1 4 . « " " t t t 2 t a n _ i s i r r 2 , " n = a o . n 1 5 « « « v e r s a g . g g ? < ( e = 0 v e r s 6 0 I f 1 6 . " " " ^ c o s - ^ i s l , " n = o o . 1 7 . " i " ^ s i n - ^ i s l , " n = o o . 1 8 . " " " ^ c o s . J ^ s l , " n = o o . . 8 " / s i n " v 1 9 . " " " i s 1 , " 1 1 = o o . 0 1 2 0 . " " " f c o s - j i s e 2 , " n = o o . / O \ n 3 2 1 . " " " ( c o s - ) i s z e r o , w h e n n = 0 0 . V n J 2 2 . I f O i s t h e c i r c u l a r m e a s u r e o f a n a c u t e a n g l e , p r o v e ( 1 ) c o s 0 < l - ^ + ^ , a n d ( 2 ) t a n f l > 0 + f . 2 3 . G i v e n ? H L = : p r o v e t h a t 0 = 4 ° 2 4 ' , n e a r l y . 6 1 0 1 4 F ' J 2 4 . G i v e n _ ^ l 6 ^ p r 0 v e t h a t O = 3 ° , n e a r l y . 6 2 1 6 6 r ' J 2 2 8 P L A N E T R I G O N O M E T R Y . 2 5 . G i v e n s i n < f > = n s i n 6 , t a n < j > = 2 t a n 0 : f i n d t h e l i m i t i n g v a l u e s o f n t h a t t h e s e e q u a t i o n s m a y c o e x i s t . A r t s , n m u s t l i e b e t w e e n 1 a n d 2 , o r b e t w e e n — 1 a n d — 2 . 2 6 . F i n d t h e l i m i t o f ( c o s a x ) ™" " " , w h e n x = 0 . A n s . e 2 M . 2 7 . F r o m a t a b l e o f n a t u r a l t a n g e n t s o f s e v e n d e c i m a l p l a c e s , s h o w t h a t w h e n a n a n g l e i s n e a r 6 0 ° i t m a y b e d e t e r m i n e d w i t h i n a b o u t o f a s e c o n d . 2 8 . W h e n a n a n g l e i s v e r y n e a r 6 4 ° 3 6 ' , s h o w t h a t t h e a n g l e c a n b e d e t e r m i n e d f r o m i t s l o g s i n e w i t h i n a b o u t ^ o f a s e c o n d ; h a v i n g g i v e n ( l o g e 1 0 ) t a n 6 4 ° 3 6 ' = 4 . 8 4 9 2 , a n d t h e t a b l e s r e a d i n g t o s e v e n d e c i m a l p l a c e s . B E M O I V R E ' S T H E O R E M . 2 2 9 C H A P T E R I X . D E M O I V R E , S T H E O R E M . — A P P L I C A T I O N S . 1 5 3 . D e M o i v r e ' s T h e o r e m . — F o r a n y v a l u e o f n , p o s i t i v e o r n e g a t i v e , i n t e g r a l o r f r a c t i o n a l . ( c o s 0 + V ^ l s i n 6 ) n = c o s n 0 + V ^ 1 s i n n f ? . . ( 1 ) I . W h e n n i s a p o s i t i v e i n t e g e r . W e h a v e t h e p r o d u c t ( c o s a + V — 1 s i n « ) ( c o s / 8 + V — 1 s i n / ? ) = ( c o s a c o s ^ — s i n a s i n / 3 ) + V — 1 ( c o s « s i n / J + s i n a c o s / J ) = c o s ( « + / 3 ) + V ^ T s i n ( a + / 8 ) . S i m i l a r l y , t h e p r o d u c t [ c o s ( a + / ? ) + V — 1 s i n ( « + / 3 ) ] [ c o s y + V — 1 s i n y ] = c o s ( a + 0 + y ) + V ^ T s i n ( a + / J + y ) . P r o c e e d i n g i n t h i s w a y w e f i n d t h a t t h e p r o d u c t o f a n y n u m b e r n o f f a c t o r s , e a c h o f t h e f o r m c o s a + V — 1 s i n « = c o s ( « + / 3 + y + - . - n t e r m s ) + V — 1 s i n ( « + / 3 + y H n t e r m s ) . S u p p o s e n o w t h a t « = f 3 = y = e t c . = 6 , t h e n w e h a v e ( c o s 6 + V — 1 s i n 6 ) n — c o s n O + V — 1 s i n n 0 , w h i c h p r o v e s t h e t h e o r e m w h e n n i s a p o s i t i v e i n t e g e r . F r o m t h e n a m e o f t h e F r e n c h g e o m e t e r w h o d i s c o v e r e d i t . 2 3 0 P L A N E T R I G O N O M E T R Y . I I . W h e n n i s a n e g a t i v e i n t e g e r . L e t n = — m ; t h e n m i s a p o s i t i v e i n t e g e r . T h e n ( c o s 6 + y / ^ T s i n 6 ) " = ( c o s 6 + V ^ T E s i n 6 ) ~ ™ — : ( f e y L ) ( c o s 0 + V — 1 s i n 0 ) " c o s m 6 + V — 1 s i n m f l 1 c o s m g — V — 1 s i n m 0 c o s m 0 + V — 1 s i n m 6 c o s m 0 — V — 1 s i n m 6 c o s m 0 — V — 1 s i n m 6 a / — - r . a = -= c o s m 0 — V — 1 s i n m a c o s 2 m 6 + s i n 2 m 6 = c o s ( — m 6 ) + V — 1 s i n ( — m 6 ) . . : ( c o s 6 + V ^ T s i n 0 ) " = c o s n 0 + V ^ l s i n n 6 , w h i c h p r o v e s t h e t h e o r e m w h e n n i s a n e g a t i v e i n t e g e r . I I I . W h e n n i s a f r a c t i o n , p o s i t i v e o r n e g a t i v e . P L e t n = - , w h e r e p a n d q a r e i n t e g e r s . T h e n ( c o s t f + V ^ s i n t f ^ c o s p t f + V ^ s i n p f l ( b y I . a n d I I . ) . ( p p \ f f c o s - d + V — 1 s i n - d ) = c o s p f l + V — 1 s i n p 0 . . - . ( c o s f l + s i n 6 ) p = ( c o s ^ 6 + V ^ T s i n ^ t f . - . ( c o s 0 + V ^ s i n 0 ) « = c o s | 0 + V ^ s i n ^ ; r t h a t i s , o n e o f t h e v a l u e s o f ( c o s 6 + V — 1 s i n 6 ) < i s c o s — + V — 1 s i n — -I n l i k e m a n n e r , ( c o s 6 — V — 1 s i n 6 ) n = c o s n 6 — V — 1 s i n n 6 . T h u s , D e M o i v r e , s T h e o r e m i s c o m p l e t e l y e s t a b l i s h e d . . I t s h o w s t h a t t o r a i s e t h e b i n o m i a l c o s 6 + V — 1 s i n 6 t o D E M O I V R E ' S T H E O R E M . 2 3 1 a n y p o w e r , w e h a v e o n l y t o m u l t i p l y t h e a r c 6 b y t h e e x p o n e n t o f t h e p o w e r . T h i s t h e o r e m i s a f u n d a m e n t a l o n e i n A n a l y t i c M a t h e m a t i c s . 1 5 4 . T o f i n d A l l t h e V a l u e s o f ( c o s f l + V - l s i n 6 y . — W h e n n i s a n i n t e g e r , t h e e x p r e s s i o n ( c o s 0 - f V — l s i n 0 ) n P c a n h a v e o n l y o n e v a l u e . B u t i f n i s a f r a c t i o n = - , t h e e x p r e s s i o n b e c o m e s ( c o s 6 + V ^ T s i n O y = V ( c o s 6 + V ^ 1 s i n O y , w h i c h h a s q d i f f e r e n t v a l u e s , f r o m t h e p r i n c i p l e o f A l g e b r a ( A r t . 2 3 5 ) . I n I I I . o f A r t . 1 5 3 , w e f o u n d o n e o f t h e v a l u e s V o f ( c o s 6 + V — l s i n 0 ) ? ; w e s h a l l n o w f i n d a n e x p r e s s i o n . z w h i c h w i l l g i v e a l l t h e q v a l u e s o f ( c o s 6 + V — 1 s i n 6 ) q . N o w b o t h c o s 6 a n d s i n 6 r e m a i n u n c h a n g e d w h e n 6 i s i n c r e a s e d b y a n y m u l t i p l e o f 2 - i r ; t h a t i s , t h e e x p r e s s i o n c o s 0 + V — l s i n 0 i s u n a l t e r e d i f f o r 6 w e p u t ( 6 + 2 r i r ) , w h e r e r i s a n i n t e g e r ( A r t . 3 6 ) . . - . ( c o s 6 + s i n 0 ) « = [ c o s ( 0 + 2 n r ) + V ^ l s i n ( 6 + 2 r i r ) ] « = c o s — - + v — l s i n — — ( A r t . 1 5 3 ) ( 1 ) T h e s e c o n d m e m b e r o f ( 1 ) h a s q d i f f e r e n t v a l u e s , a n d n o m o r e ; t h e s e q v a l u e s a r e f o u n d b y p u t t i n g r = 0 , 1 , 2 , - - - q — 1 , s u c c e s s i v e l y , b y w h i c h w e o b t a i n t h e f o l l o w i n g s e r i e s o f a n g l e s . p ( 0 + 2 r i r ) p O W h e n r = 0 , c o s = c o s — -" r = l , " = c o s ^ v ' — u r = 2 , « = c o s ^ 0 + 4 ^ e t c . e t c . ' / 2 3 2 P L A N E T R I G O N O M E T R Y . W h e n r = q - l , c o s ^ + ^ -= c o s ^ [ ^ + 2 , ( g -1 ) ] q q = c o s q A l l t h e s e q v a l u e s a r e d i f f e r e n t . p ( 6 + 2 r ) p ( 6 + 2 q ) W h e n r = q , c o s = c o s -= c o s ( — + 2 p i r 1 = p 6 c o s — , q > t h e s a m e v a l u e a s w h e n r = 0 . w . , 1 P ( g + 2 r r ) [ f l + ( 2 g + 2 ) , r ] W h e n r = g + 1 , c o s — = c o s = c O S t - l — ! _ - l , q t h e s a m e a s w h e n r = 1 , e t c . , f r o m w h i c h i t a p p e a r s t h a t t h e r e a r e q a n d o n l y q d i f f e r e n t v a l u e s o f c o s - ^ — — ^ > n " ) s i n c e t h e s a m e v a l u e s a f t e r w a r d s r e c u r i n t h e s a m e o r d e r . S i m i l a r l y f o r s i n T h e r e f o r e t h e e x p r e s s i o n p ( 6 + 2 r i r ) . p ( 6 + 2 r i r ) c o s \ - V — 1 s i n q q p g i v e s a l l t h e q v a l u e s o f ( c o s 6 + V — 1 s i n 6 ) « a n d n o m o r e . A . n d t h i s a g r e e s w i t h t h e T h e o r y o f E q u a t i o n s t h a t t h e r e m u s t b e q v a l u e s o f x , a n d n o m o r e , w h i c h s a t i s f y t h e e q u a t i o n a f = c , w h e r e c i s e i t h e r r e a l o r o f t h e f o r m o + i V — 1 . A P P L I C A T I O N S O F D E M O I V R E ' S T H E O R E M . 2 3 3 A P P L I C A T I O N S O F D E M O I V R E ' S T H E O R E M . 1 5 5 . T o d e v e l o p c o s n 6 a n d s i n n O i n P o w e r s o f s i n 6 a n d c o s 6 . W e s h a l l g e n e r a l l y i n t h i s c h a p t e r w r i t e i f o r V — 1 i n a c c o r d a n c e w i t h t h e u s u a l n o t a t i o n . B y D e M o i v r e , s T h e o r e m ( A r t . 1 5 3 ) w e h a v e c o s n 6 + i s i n n O = ( c o s 6 + i s i n 6 ) " ( 1 ) L e t n b e a p o s i t i v e i n t e g e r . E x p a n d t h e s e c o n d m e m b e r o f ( 1 ) b y t h e b i n o m i a l t h e o r e m , r e m e m b e r i n g t h a t l 2 = — 1 , = — ? , a n d t h a t i = + 1 ( A l g e b r a , A r t . 2 1 9 ) . E q u a t e t h e r e a l a n d i m a g i n a r y p a r t s o f t h e t w o m e m b e r s . T h u s , c o s n 6 = c o s " 6 -n i n - 1 ) c o s " - 2 6 s i n 2 6 n ( n _ l ) ( n -2 ) ( W -3 ) < f l ^ _ ( 2 ) [ 4 w s i n n 0 = n c o s " - 0 s i n 6 -n ( M -1 ) ( n -2 ) c 0 8 . - 3 g s i n 8 ^ [ 3 n ( n - l ) ( n - 2 ) ( n - 8 ) ( n - 4 ) c o s „ _ 5 ^ s i n ^ _ e t c . ( 3 ) [ 6 T h e l a s t t e r m s i n t h e s e r i e s f o r c o s a n d f o r s i n n 0 w i l l b e d i f f e r e n t a c c o r d i n g a s n i s e v e n o r o d d . T h e l a s t t e r m i n t h e e x p a n s i o n o f ( c o s 6 + i s i n g ) " i s i ' " s i n " 0 ; a n d t h e l a s t t e r m b u t o n e i s n f " 1 c o s 0 s i n ~ l 0 . T h e r e f o r e : W h e n n i s e v e n , t h e l a s t t e r m o f c o s n 0 i s i " s i n " 0 o r ( — 1 ) 2 s i n " 6 , a n d t h e l a s t t e r m o f s i n n O i s ? u ' " - 2 c o s 6 s i n " - 1 6 n - 2 o r n ( — 1 ) 2 c o s 6 s i n " - 1 f t W h e n n i s o d d , t h e l a s t t e r m o f c o s n 6 i s n i " _ 1 c o s 6 s i n n ~ l 6 n - l o r n ( — 1 ) 2 c o s 0 s i n " - 1 6 , a n d t h e l a s t t e r m o f s i n ? i 6 i s n — 1 f l - l s i n " 0 o r ( — 1 ) 2 s i n " 0 . 2 3 4 P L A N E T R I G O N O M E T R Y . E X A M P L E S . P r o v e t h e f o l l o w i n g s t a t e m e n t s : 1 . s i n 4 0 = 4 c o s 3 0 s i n 0 — 4 c o s 0 s i n 3 f t 2 . c o s t 6 = c o s 4 0 — 6 c o s 2 0 s i n 2 0 + s i n 4 f t 1 5 6 . T o d e v e l o p s i n 6 a n d c o s 0 i n S e r i e s o f P o w e r s o f f t P u t n 6 = a i n ( 2 ) a n d ( 3 ) o f A r t . 1 5 5 ; a n d l e t n b e i n c r e a s e d w i t h o u t l i m i t w h i l e « r e m a i n s u n c h a n g e d . T h e n s i n c e 0 = - , 0 m u s t d i m i n i s h w i t h o u t l i m i t . T h e r e f o r e t h e i i a b o v e f o r m u l a e m a y b e w r i t t e n c o s a = c o s » 0 -" ( " - ) e o 3 » - f l f e g y I " ( « -g ) ( " -2 g ) ( " ~ 3 f > ) c 0 E „ - 4 ^ g i n 0 J ( 1 ) a n d s i n « = a c o s " - 1 0 s i n f t -" ( « ~ ) _ | " ~ 2 ) c o s - » ^ 8 ^ 8 + . . . . . ( 2 ) I f n = o o , t h e n 0 = 0 , a n d t h e l i m i t o f c o s 6 a n d i t s p o w e r s i s 1 ; a l s o t h e l i m i t o f ^ s m ^ \ a n ( j i t s p o w e r s i s 1 . H e n c e ( 1 ) a n d ( 2 ) b e c o m e ^ 6 ' c o s « = l - -+ - - -+ - -( 3 ) 2 | 4 [ 6 v ' s i n « = « — ^ + ^ ( 4 ) [ 3 [ 5 _ S c h . I n t h e s e r i e s f o r s i n a a n d c o s a , j u s t f o u n d , a i s < f t e c i r c u l a r m e a s u r e o f t h e a n g l e c o n s i d e r e d . C O N V E R G E N C E O F T H E S E R I E S . 2 3 5 C o r . 1 . I f a b e a n a n g l e s o s m a l l t h a t a 2 a n d h i g h e r p o w e r s o f « m a y b e n e g l e c t e d w h e n c o m p a r e d w i t h u n i t y , ( 3 ) b e c o m e s c o s a = 1 , a n d ( 4 ) , s i n a = a . I f a 2 , « 3 b e r e t a i n e d , b u t h i g h e r p o w e r s o f a b e n e g l e c t e d , ( 3 ) a n d ( 4 ) g i v e ( f t s i n « = a — — ; c o s a = 1 — — ( C o m p a r e A r t . 1 3 4 ) C o r . 2 . B y d i v i d i n g ( 3 ) b y ( 4 ) , w e o b t a i n , « 3 , 2 « 5 , 1 7 a 7 , . , K N t a n a = « + - + _ + 3 - 3 ^ + e t c . " " ( 5 ) 1 5 7 . C o n v e r g e n c e o f t h e S e r i e s . — T h e s e r i e s ( 3 ) a n d ( 4 ) o f A r t . 1 5 6 m a y b e p r o v e d t o b e c o n v e r g e n t , a s f o l l o w s : T h e n u m e r i c a l v a l u e o f t h e r a t i o o f t h e s u c c e s s i v e p a i r s o f c o n s e c u t i v e t e r m s i n t h e s e r i e s f o r s i n a a r e a 2 a 2 a 2 a 2 , , , , , e t c . 2 - 3 4 -5 G - 7 8 - 9 H e n c e t h e r a t i o o f t h e ( n + l ) t h t e r m t o t h e n t h t e r m i s a 2 ; a n d w h a t e v e r b e t h e v a l u e o f a , w e c a n t a k e 2 n ( 2 n + l ) n s o l a r g e t h a t f o r s u c h v a l u e o f n a n d a l l g r e a t e r v a l u e s , t h i s f r a c t i o n c a n b e m a d e l e s s t h a n a n y a s s i g n a b l e q u a n t i t y ; h e n c e t h e s e r i e s i s c o n v e r g e n t . S i m i l a r l y , i t m a y b e s h o w n t h a t t h e s e r i e s f o r c o s a i s a l w a y s c o n v e r g e n t . 1 5 8 . E x p a n s i o n o f c o s " 0 i n T e r m s o f C o s i n e s o f M u l t i p l e s o f 6 , w h e n n i s a P o s i t i v e I n t e g e r . L e t x = c o s 6 + i s i n 6 ; t h e n - = = c o s 6 — i s i n 6 . x c o s 6 + i s i n 6 . : x + - = 2 c o s 6 ; a n d x — - = 2 i s i n 6 ( 1 ) 2 3 6 P L A N E T R I G O N O M E T R Y . A l s o x n = ( c o s 6 + i s m 6 ) n = e o s n 6 + i s m n 6 ( A r t . 1 5 3 ) ( 2 ) a n d — = ( c o s 6 — i s i n 6 ) " = c o s n O — i s i n n 6 . . . ( 3 ) " . - . 2 c o s n O = x n + — , a n d 2 i s i n ? i 0 = — i . ( 4 ) • x n x n H e n c e ( 2 c o s 0 ) " = ( a ; + a r 1 ) " , b y ( 1 ) , = a s " + n x n - 2 + -( w ~ ^ a ; " - 4 + e t c . + M a ; - ' " ~ 2 ' + a r " = 2 c o s n f ? + n 2 c o s ( n -2 ) 0 + n ( n ~ 1 ^ 2 c o s ( n -4 ) 0 + e t c . [ f . - . 2 n _ 1 c o s " 0 = c o s n 6 + n c o s ( n — 2 ) 0 + c o s ( n - 4 ) 0 + e t c . ( 5 ) N o t e . — I n t h e e x p a n s i o n o f ( x + x ~ l ) n t h e r e a r e n + 1 t e r m s ; t h u s w h e n n i s e v e n t h e r e i s a m i d d l e t e r m , t h e ^ n + l ) t h , w h i c h i s i n d e p e n d e n t o f 9 , a n d w h i c h i s n ( . n - l ) . . . ( - + l ) n ( n - l ) . . . ( n - j n + l ) v 2 > 5 5 ' ' " i i » H e n c e w h e n n i s e v e n t h e l a s t t e r m i n t h e e x p a n s i o n o f 2 n _ 1 c o s n 0 i s « ( n - l ) . . . ( | + l ) 5 5 W h e n n i s © d d t h e l a s t t e r m i n t h e e x p a n s i o n o f 2 n — 1 c o s n 0 ~ i s n ( n r l ) . . J ( n ± _ 3 ) c o s » . l i ( » - i y 1 5 9 . E x p a n s i o n o f s i n " 0 i n T e r m s o f C o s i n e s o f M u l t i p l e s o f 6 , w h e n n i s a n E v e n P o s i t i v e I n t e g e r . ( 2 f s i n 6 ) n = ( x -^ J ' b y ( 1 ) o f A r t . 1 5 8 = 3 f -n x n ' 1 + n ( n - 1 ) x n ~ + . . . ( 2 n ( n - l ) ( » _ 4 , _ - ( n _ 2 > , - » \ 2 r , m V T ^ J l [ 8 § T V A j o ) f ) ] 0 i « l g ( s + M ) f - . . ( T _ M ) M ^ ) l _ ) + z l - - g ) t - " ) u i s l g ) l _ M ) M + f x \ ) T - " ) t | . \ T / ) 8 + n ) f - ( T - n ) n _ ■ ) ! ~ ) + i — « f t - " 5 5 N z l \ " B _ . . ' v t ~ , - u X J { i - ^ + \ i r i - J 1 „ _ a ; ( j _ U ) _ x n - + -( ^ - u ) - / _ z l ^ ) T -m ) n + 2 - " 9 ! " a ; ~ ~ 8 S T W j o ) l ) £ < \ J ^ n i s ? s ) e r g t e u j p p o s i m n a q A j o s a j d r n n j i j o s a m g j o s r a i a x u i 6 „ n i s j o u o i s i r B d x g ; 0 9 1 n H z . + . . . _ ( t + ! ) . . . ) T - n ) » i ) T - ) z l f ( f -n ) s o o ( ] - _ l t ) M + 0 ( g -n ) s o o n -6 ) n s o o = 0 „ m s , ) t - ) i - 2 u T + r J -) T - n ) n ! ) ! - ) - + - . -+ ( ^ ) ^ + ( ^ M t + ) -2 3 8 P L A N E T R I G O N O M E T R Y . W h e n c e d i v i d i n g b y 2 i , w e h a v e n - l 2 » - I ( - l ) 2 s i n " 0 = s i n n 0 -n s i n ( n -2 ) 6 » + n ( n ~ 1 ) s i n ( » -4 ) 0 + — ( - l ) ^ ( » - l ) - | ( » + 3 ) s i n f t H ( » - l ) E X A M P L E S . P r o v e t h a t 1 . 1 2 8 c o s 8 0 = c o s 8 0 + 8 c o s 6 0 + 2 8 c o s 4 0 + 5 6 c o s 2 0 + 3 5 . 2 . 6 4 c o s i 0 = c o s 7 0 + 7 c o s 5 0 + 2 1 c o s 3 0 + 3 5 c o s 0 . 1 6 1 . E x p o n e n t i a l V a l u e s o f S i n e a n d C o s i n e . S i n c e e = 1 + x + ^ + g + £ + . . . . . . ( A r t . 1 2 9 ) [ 2 [ 3 [ 4 . . . e = 1 - ^ + @ - e t c - + \ e - [ 3 + ^ - e t c j = c o s 0 + i s i n 0 ( A r t . 1 5 6 ) a n d e - « = 1 -£ + £ -e t c . -i ( O -£ + ? -e t c . " ) [ 2 [ 4 \ ( 3 [ 6 ; = c o s 0 — i s i n 0 . . - . 2 c o s 0 = e M + e - i 9 , a n d 2 i s i n 0 = e i e — e ~ i e . . ( 1 ) . - . c o s 0 = - — , a n d s i n 0 = — — . . . ( 2 ) 2 2 i w h i c h a r e c a l l e d t h e e x p o n e n t i a l v a l u e s o f t h e c o s i n e a n d s i n e . C o r . F r o m t h e s e e x p o n e n t i a l v a l u e s w e m a y d e d u c e s i m i l a r v a l u e s f o r t h e o t h e r t r i g o n o m e t r i c f u n c t i o n s . T h u s , t a n 0 = — ( 3 ) i - ( e ' O + e - ' 8 ) v ' C a l l e d a l s o E u l e r ' s e q u a t i o n s , a f t e r E u l e r , t b e l r d i s c o v e r e r . G R E G O R Y ' S S E R I E S . 2 3 9 S c h . T h e s e r e s u l t s m a y b e a p p l i e d t o p r o v e a n y g e n e r a l f o r m u l a i n e l e m e n t a r y T r i g o n o m e t r y , a n d a r e o f " g r e a t i m p o r t a n c e i n t h e H i g h e r M a t h e m a t i c s . E X A M P L E S . i t i s i n 2 6 . a 1 . P r o v e - = t a n 0 . l + c o s 2 0 w v 2 i s i n 2 0 e 2 » _ g - 2 » » W e h a v e - — - = — b y ( 1 ) 2 + 2 c o s 2 0 2 + e 2 ' » + e - 2 W -' g U g — t t- = £ t a n 0 b y ( 3 ) . . - . e t c . e , e + e P r o v e t h e f o l l o w i n g , b y t h e e x p o n e n t i a l v a l u e s o f t h e s i n e a n d c o s i n e . 2 . c o s 2 « = c o s 2 « — s i n 2 a . 3 . s i n 0 = — s i n ( — 0 ) . 4 . c o s 3 0 = 4 c o s 3 0 - 3 c o s 0 . R e n t . — I f w e o m i t t h e t f r o m t h e e x p o n e n t i a l v a l u e s o f t h e s i n e , c o s i n e , a n d t a n g e n t o f 9 , t h e r e s u l t s a r e c a l l e d r e s p e c t i v e l y t h e h y p e r b o l i c s i n e , c o s i n e , a n d t a n g e n t o f 9 , a n d a r e w r i t t e n s i n h 9 , c o s h 9 , a n d t a n h 9 , r e s p e c t i v e l y . T h u s w e h a v e s i n h 9 — — i s i n i 9 , c o s h 6 = c o s i 9 , t a n h 9 = — i t a n i 9 . H y p e r b o l i c f u n c t i o n s a r e s o c a l l e d , b e c a u s e t h e y h n v e g e o m e t r i c r e l a t i o n s w i t h t h e e q u i l a t e r a l h y p e r b o l a a n a l o g o u s t o t h o s e b e t w e e n t h e c i r c u l a r f u n c t i o n s a n d t h e c i r c l e . A c o n s i d e r a t i o n o f h y p e r b o l i c f u n c t i o n s i s c l e a r l y b e y o n d t h e l i m i t s o f t h i s t r e a t i s e . F o r a n e x c e l l e n t d i s c u s s i o n o f s u c h f u n c t i o n s , t h e s t u d e n t i s r e f e r r e d t o s u c h w o r k s a s C a s e y ' s T r i g o n o m e t r y , H o b s o n ' s T r i g o n o m e t r y , L o c k ' s H i g h e r T r i g o n o m e t r y , e t c . 1 6 2 . G r e g o r y ' s S e r i e s . — T o e x p a n d 6 i n p o w e r s o f t a n 6 i e r e 6 l i e s b e t w e e n — -a n d 4 - ~ -2 2 B y ( 3 ) o f A r t . 1 6 1 , w e h a v e e i 9 Q - i O i t a n 6 — . , - i 8 e t 9 + e " l - i t a n 0 2 e ~ i d 2 4 0 P L A N E T R I G O N O M E T R Y . . : l o g e 2 i » = l o g ( l + i t a n 6 ) — l o g ( l -i t a n 6 ) . . : 2 i 6 = 2 £ ( t a n 6 -£ t a n 3 g + \ t a n 5 0 -e t c . ) ( A r t . 1 3 0 ) . - . 0 = t a n 0 - | t a n 3 0 + £ t a n 5 0 - e t c ( 1 ) w h i c h i s G r e g o r y , s S e r i e s . T h i s s e r i e s i s c o n v e r g e n t i f t a n 6 = o r < 1 , i . e . , i f 0 l i e s b e t w e e n — - a n d - , o r b e t w e e n f i r a n d 4 i r . 4 4 a S c A . T h i s s e r i e s m a y a l s o b e o b t a i n e d b y r e v e r t i n g ( 5 ) i n C o r . 2 , A r t . 1 5 6 . C o r . 1 . I f t a n 6 = x , w e h a v e f r o m ( 1 ) t a n - j a ; = x - ~ + ~ -e t c ( 2 ) C o r . 2 . I f 0 = ^ j w e h a v e f r o m ( 1 ) | = l - | + W + e t c ( 3 ) a s e r i e s w h i c h i s v e r y s l o w l y c o n v e r g e n t , s o t h a t a l a r g e n u m b e r o f t e r m s w o u l d h a v e t o b e t a k e n t o c a l c u l a t e i t t o a c l o s e a p p r o x i m a t i o n . W e s h a l l t h e r e f o r e s h o w h o w s e r i e s , w h i c h a r e m o r e r a p i d l y c o n v e r g e n t , m a y b e o b t a i n e d f r o m G r e g o r y , s s e r i e s . 1 6 3 . E u l e r ' s S e r i e s . t a n - 1 - + t a n - 1 - = -. . . ( b y E x . 2 , A r t . 6 0 ) 2 3 4 P u t 6 = t a n - 1 . : t a n 0 » = ^ , w h i c h i n ( 1 ) o f A r t . 1 6 2 2 2 g i v e s t a n - 1 - ^ - — + — - ^ -+ e t c ( 1 ) 2 2 3 - 2 3 5 - 2 5 7 - 2 i v P u t 6 = t a n 1 - -. - . t a n 6 = 1 , a n d ( 1 ) b e c o m e s t a n - 1 ! ^ -- ^ - + — — + e t c ( 2 ) 3 3 3 - 3 3 5 - 3 ' 7 - 3 i M A C H I N ' S S E R I E S . 2 4 1 A d d i n g ( 1 ) a n d ( 2 ) w e h a v e 4 ^ 2 3 - 2 3 5 - 2 3 J \ 3 3 . 3 3 5 . 3 5 J K ' a s e r i e s w h i c h c o n v e r g e s m u c h m o r e r a p i d l y t h a n ( 3 ) o f A r t . 1 6 2 . 1 6 4 . M a c h i n ' s S e r i e s . S i n c e 2 t a n ~ 1 - = t a n - 1 - i — ( b y E x . 3 , A r t . 6 0 ) = t a n - 1 — , 5 l - ^ 1 2 . - . 4 t a n - 1 - = 2 t a n - 1 ^ = t a n - 1 = t a n - 1 — . 5 1 2 1 - t ¥ t 1 1 9 A l s o , t a n - 1 — -t a n - 1 1 = t a n - 1 T T £ = 1 = t a n - 1 — -1 1 9 1 + t H 2 3 9 . , 4 t a n - | - ^ = t a n - ^ . , . ^ = 4 t a n - 1 | - t a n - 1 ^ . 4 \ 5 3 - 5 3 5 - 5 5 V 2 3 9 3 - ( 2 3 9 ) 3 5 ( 2 3 9 ) 5 J I n t h i s w a y i t i s f o u n d t h a t t t = 3 . 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 - . C o r . S i n c e t a n - 1 — + t a n - 1 — = t a n - 1 A + A 9 9 ^ 2 3 9 1 - ^ X j b = t a n - 1 — , 7 0 , . - . - = 4 t a n - 1 - -t a n - 1 — + t a n - 1 — -4 5 7 0 9 9 N o t e . — T h e s e r i e s f o r t a n - 1 - - a n d t a n - 1 — a r e m u c h m o r e c o n v e n i e n t f o r p u r -7 0 9 9 p o s e s o f n u m e r i c a l c a l c u l a t i o n t h a n t h e s e r i e s f o r t a n - 1 2 3 9 E x a m p l e . — F i n d t h e n u m e r i c a l v a l u e o f i r t o 6 f i g u r e s b y M a c h i n , s s e r i e s , 2 4 2 P L A N E T R I G O N O M E T R Y . 1 6 5 . G i v e n s i n 6 = x s i n ( 6 + « ) ; e x p a n d 6 i n a S e r i e s o f A s c e n d i n g P o w e r s o f x . W e h a v e e i e — e ~ i 9 = x [ e w + ' a — e - ' 9 - ' " ] . . ( A r t . 1 6 1 ) . - . e 2 W — 1 = x [ e 2 i 9 - e « — e - a ] . . . e 2 « = l - « - • 1 — x e i a . - . 2 £ 0 = l o g ( l — a ; e - i a ) — l o g ( l — a ; e i a ) = a ; ( e » ' 1 — e - » ) + ^ ( e 2 ' " — e ~ 2 i a ) + ^ ( e 3 i a — e - 3 i a ) . - -2 3 ( A r t . 1 3 0 ) . - . e = x s i n a + ^ s i n 2 « + ^ s i n 3 « + - -( A r t . 1 6 1 ) ( 1 ) 2 3 E x a m p l e . I f a = i r — 2 0 , t h e n g = 1 . . - . ( 1 ) b e c o m e s 0 = s i n 2 0 - | s i n 4 0 + 1 s i n 6 0 - £ s i n 8 0 + - . 1 6 6 . G i v e n t a n x = n t a n 0 ; e x p a n d x i n P o w e r s o f n . n i x p — i x p i B p - i Q - . — = = n - . . . . . ( A r t . 1 6 1 ) e « + e - K e u + e - u e 2 i x _ i _ e 2 i » — 1 e 2 « + 1 — e 2 W + l -c ^ = ( e 2 i , , + l ) + t t ( C 2 " - l ) ( e 2 i » + l ) - ? i ( e 2 i 9 - l ) _ ( l + n ) e " » 4 - l — n ~ ( 1 — n ) e 2 1 » + l + n e 2 ' « + m / l — n \ — ( w h e r e m = m e 2 , 9 + l V 1 + n J \ 1 + m e ™ J . - . 2 i x = 2 i 0 + l o g ( l + m e -2 i e ) — l o g ( l + m e ™) = 2 1 " 0 -m ( e ™-e - 2 i « ) + ^ ( e w -e ~ 4 i 9 ) . . - . = 0 - m s i n 2 0 + ^ - 2 s i n 4 0 . . ( A r t . 1 6 1 ) R E S O L U T I O N I N T O F A C T O R S . 2 4 3 R E S O L U T I O N O F E X P R E S S I O N S I N T O F A C T O R S . 1 6 7 . R e s o l v e x n — 1 i n t o F a c t o r s . S i n c e c o s 2 r i r ± V — 1 s i n 2 r r = 1 , w h e r e r i s a n y i n t e g e r , a n d x n = 1 , . - . x " = c o s 2 « r ± V — 1 s i n 2 r i r . i . - . k = ( c o s 2 r i t ± V ^ l s i n 2 r i r ) " = c o s — ± V ^ l s i n — . . ( A r t . 1 5 3 ) ( 1 ) n n ( 1 ) W f t e n n i s e v e n . I f r = 0 , w e o b t a i n f r o m ( 1 ) a r e a l r o o t 1 ; i f r = ^ , w e o b t a i n a r e a l r o o t — 1 , a n d t h e t w o c o r -r e s p o n d i n g f a c t o r s a r e x — 1 a n d x + 1 . I f w e p u t r = l , 2 , 3 . . . | - l , i n s u c c e s s i o n i n ( 1 ) , w e o b t a i n n — 2 a d d i t i o n a l r o o t s , s i n c e e a c h v a l u e o f r g i v e s t w o r o o t s . T h e p r o d u c t o f t h e t w o f a c t o r s , w h i c h a r e [ x — c o s V — 1 s i n \ n n J a n d f s - c o s ^ + y ^ T l s i n ^ V n n J = [ x — c o s ) + s i n 2 n J n = a ; 2 - 2 a ; c o s — + 1 ( 2 ) w h i c h i s a r e a l q u a d r a t i c f a c t o r . . - . x " - l = ( x 2 - l ) ^ a r - 2 e o s ~ + 1 ^ -2 a ; c o s ^ + l j -. . . ^ a r - 2 a ; c o s ^ ^ + l ^ a ; 2 - 2 a ; c o s ' ^ i r + l ) - . . ( 3 ) T h i s e x p r e s s i o n g i v e s t h e n n t h r o o t s o f u n i t y . 2 4 4 P L A N E T R I G O N O M E T R Y . ( 2 ) W h e n n i s o d d . T h e o n l y r e a l r o o t i s 1 , f o u n d b y p u t t i n g r = 0 i n ( 1 ) ; t h e o t h e r n — 1 r o o t s a r e f o u n d b y n — 1 p u t t i n g r = 1 , 2 , 3 , — i n ( 1 ) o r ( 2 ) i n s u c c e s s i o n . L i . : a f - l = ( x - l ) ( a ? - 2 » c o s ^ + l Y a ; 2 - 2 a ; c o s ^ + 1 ) -. - - ^ i t 2 — 2 x c o s ^ - ^ i r + l ^ x 2 — 2 a ; c o s M ^ ( 4 ) 1 6 8 . R e s o l v e x " + l i n t o F a c t o r s . S i n c e c o s ( 2 r + 1 ) t t ± V ^ T s i n ( 2 r + 1 ) j r = — 1 , w h e r e r i s a n y i n t e g e r , a n d x " = — 1 , . - . x " = c o s ( 2 r + 1 ) t t ± s i n ( 2 r + l ) i r . i . - . x = [ c o s ( 2 r + 1 ) t t ± V -1 s i n ( 2 r + l ) i r ] n 2 r + l ,— 7 . 2 r + l = c O S J r ± V — 1 S i n i r . . . ( 1 ) n n w h i c h i s a r o o t o f t h e e q u a t i o n x n = — 1 ; i . e . , — 1 i s a r o o t . ( 1 ) T T A e n n . i s e v e n . T h e r e i s n o r e a l r o o t ; t h e n r o o t s a r e a l l i m a g i n a r y , a n d a r e f o u n d b y p u t t i n g r = 0 , l , 2 , . . . | - l , s u c c e s s i v e l y , i n ( 1 ) . T h e p r o d u c t o f t h e t w o f a c t o r s , ( x _ c o s T ± 1 W - V 3 T s i n , \ \ n n I a n d / ' x - c o s ^ i i i r + V ^ l s i n ^ l + i ^ = ! r ? - 2 a ; c o s 2 - ^ ± ^ , r + l ( 2 ) w h i c h i s a r e a l q u a d r a t i c f a c t o r . R E S O L U T I O N I N T O F A C T O R S . 2 4 5 . - . x n + l = ( x > -2 x c o s £ + l j ( x -2 a r c o s ~ + l j - -. . . ^ - 2 o ; c o s ^ - ^ w + l Y ! e ! - 2 o o s ^ ^ a - + l V . ( 3 ) ( 2 ) W h e n n i s o d d . T h e o n l y r e a l r o o t i s — 1 ; t h e o t h e r n — 1 r o o t s a r e f o u n d b y p u t t i n g r = 0 , 1 , 2 , - - - M ~ -i n ( 1 ) , i n s u c c e s s i o n . . - . x + l = ( a ; + l ) ^ e ! - 2 a ; c o s j [ + 1 ^ - 2 a ? c o s ^ + l V -- - - ^ - 2 - 2 a ; c o s ^ i i r + l ^ : r 2 - 2 a ; c o s J . ^ ^ i r + l ^ ( 4 ) E X A M P L E S . 1 . F i n d t h e r o o t s o f t h e e q u a t i o n X s — 1 = 0 . A n s . 1 , c o s £ ( 2 « r ) + i s i n i ( 2 r i r ) , w h e r e r = 1 , 2 , 3 , 4 . 2 . F i n d t h e q u a d r a t i c f a c t o r s o f X s — 1 . 4 n s . ( a -1 ) ( a r -V 2 x + 1 ) ( x 2 + 1 ) ( a ? + y / 2 x + 1 ) . 3 . F i n d t h e r o o t s o f t h e e q u a t i o n x + 1 = 0 , a n d w r i t e d o w n t h e q u a d r a t i c f a c t o r s o f x + 1 . A n s . ± J - ± V ^ T — : ( a r -a ; V 2 + 1 ) ( z 2 + x V 2 + 1 ) . V 2 V 2 1 6 9 . R e s o l v e a r " — 2 a ; " c o s 6 + 1 i n t o F a c t o r s . L e t a r " — 2 a f c o s 0 + 1 = 0 . . - . x 2 " -2 x n c o s 0 + c o s 2 6 = — s i n 2 0 . . - . a ; " — c o s 0 = ± V — 1 s i n 6 = ± i s i n 6 . . : x = ( c o s 6 ± i s i n = c o s 2 r " + 6 ± i s i n 2 r " + ° ( 1 ) n n s i n c e c o s 0 i s u n a l t e r e d i f f o r 6 w e p u t 0 + 2 t v . I f w e p u t r = 0 , 1 , 2 , - - m — 1 , s u c c e s s i v e l y i n ( 1 ) , w e f i n d 2 n d i f f e r e n t r o o t s , s i n c e e a c h v a l u e o f r g i v e s t w o r o o t s . 2 4 6 P L A N E T M G O N O M E T B Y . T h e p r o d u c t o f t h e t w o f a c t o r s i n ( 1 ) / 2 m + 6 . . 2 n r + 6 \ = [ x — c o s — i s i n — ^ n n ) X [ x — c o s x -+ 1 s i n X _ ^ n n J = x 2 - 2 x c o s 2 r ' + 6 + 1 ( 2 ) n . - . a ? " — 2 x " c o s 6 + 1 = ^ 8 -2 c o s ^ + l j ( a ? - 2 x c o s 2 i r + 0 + l ^ . . . - f X s - 2 x c o s -J + l j -/ „ ( 2 n - 2 ) i r + O A . . . f ^ - 2 a ; c o s - ^ ^ + ( 3 ) C o r . C h a n g e x i n t o -i n ( 3 ) a n d c l e a r o f f r a c t i o n s , a n d a w e g e t x 2 n — 2 a " x n c o s 9 + a 2 n = ^ x — 2 a x c o s - + a 2 ^ - - -. - . f x 2 - 2 a x c o s 2 j l ~ - + a ^ ( 2 -2 a x c o s + a 2 ^ - -- . - t o n f a c t o r s ( 4 ) E X A M P L E S . F i n d t h e q u a d r a t i c f a c t o r s o f t h e f o l l o w i n g : 1 . x 8 - 2 < k 4 c o s 6 0 ° + 1 = 0 . A n s . ( x 2 -2 x c o s 1 5 ° + 1 ) ( a ; 2 -2 x c o s 1 0 5 ° + 1 ) X ( x 2 -2 x c o s 1 9 5 ° + 1 ) ( x 2 -2 x c o s 2 8 5 ° + 1 ) = 0 . 2 . z 1 0 - 2 ^ c o s 1 0 ° + 1 = 0 . A n s . ( x 2 -2 x c o s 2 ° + 1 ) ( x 2 -2 x c o s 7 4 ° + 1 ) X ( X 2 -2 x c o s 1 4 6 ° + l ) ( x -2 c o s 2 1 8 ° + 1 ) { x 2 - 2 x c o s 2 y u 3 + 1 ) = 0 . D E M O I V R E ' S P R O P E R T Y O F T H E C I R C L E . 2 4 7 1 7 0 . D e M o i v r e ' s P r o p e r t y o f t h e C i r c l e . — L e t O b e t h e c e n t r e o f a c i r c l e , P a n y p o i n t i n i t s p l a n e . D i v i d e t h e c i r c u m f e r e n c e i n t o n e q u a l p a r t s B C , C D , D E , - - , b e g i n n i n g a t a n y p o i n t B ; a n d j o i n O a n d P w i t h t h e p o i n t s o f d i v i s i o n B , C , D , L e t P O B = 0 ; t h e n w i l l O B 2 " - 2 O B " - O P " c o s n O + O P 2 " = P B 2 - P C 2 . P D 2 -- - t o n t e r m s . F o r , p u t O B = a , O P = x , a n d 6 = - : t h e n 7 1 P B 2 = O P 2 + O B 2 -2 O P - O B c o s 6 = x 2 + a 2 — 2 a x c o s -n P C 2 = O P 2 + O C 2 -2 O P - O C c o s " + 2 , r n = x 2 + a 2 — 2 a x c o s " + 2 " . a n d s o o n . . n M u l t i p l y i n g ( 1 ) , ( 2 ) , ( 3 ) , - -t o g e t h e r , w e h a v e P B 2 - P C 2 - P D 2 — t o n t e r m s , « + 2 i r = ( a ? — 2 a x c o s a + a P j ^ a ? — 2 x ^ a ? — 2 a x c o s a x c o s — h a 2 ( 1 ) ( 2 ) = x 2 " — 2 a ' a f c o s a + a 2 " . [ b y ( 4 ) o f A r t . 1 6 9 ] = O T 5 2 » - 2 O P n - O B " c o s n 0 + O B 2 n . . . ( 3 ) w h i c h p r o v e s t h e p r o p o s i t i o n . 2 4 8 P L A N E T R I G O N O M E T R Y . 1 7 1 . C o t e ' s P r o p e r t i e s o f t h e C i r c l e . — T h e s e a r e p a r t i c u l a r c a s e s o f D e M o i v r e , s p r o p e r t y o f t h e c i r c l e . ( 1 ) L e t O P , p r o d u c e d i f n e c e s s a r y , m e e t t h e c i r c l e a t A , a n d l e t A B = B C = C D , e t e . , = — ; n t h e n n 6 i s a m u l t i p l e o f 2 i r . H e n c e w e h a v e f r o m ( 3 ) o f A r t . 1 7 0 , a f t e r t a k i n g t h e s q u a r e r o o t o f b o t h m e m b e r s , O B " - O F = P B - P C - P D -t o n f a c t o r s . . . I . ( 2 ) L e t t h e a r c s A B , B C , - . . b e b i s e c t e d i n t h e p o i n t s a , b , - - -; t h e n w e h a v e , b y ( 1 ) , O B " " _ O P = P a - P B - F b - P C - P c - - -t o 2 n f a c t o r s . H e n c e , b y d i v i s i o n , O B " + O P " = P a . P 6 - P c - . -t o n f a c t o r s . . . I I . C o r . I f t h e a r c s A B , B C , - - -b e t r i s e c t e d i n t h e p o i n t s a l t a 2 , 6 j , 6 2 , t h e n w e h a v e O B + O B " - O P " + O P 2 " = P ^ - P a 2 - P 6 j - P 6 2 - - - t o 2 n f a c t o r s . 1 7 2 . R e s o l v e s i n 6 i n t o F a c t o r s . ( 1 ) P u t » = 1 ; t h e n w e g e t f r o m ( 3 ) o f A r t . 1 6 9 2 ( l - c o s . ) = 2 » ( l - c o s ^ ( l - ' c 0 s ^ ) ( l - c o s 4 - ^ ) . ( 1 ) . . . ( l _ c 0 S ( 2 ^ L ± f ) . . . . P u t 6 = 2 n < j > i n ( 1 ) , a n d l e t 2 n a = i r . 1 — c o s 6 = 1 — c o s 2 n < £ = 2 s i n 2 n < £ ; t h e n e x t r a c t i n g t h e s q u a r e r o o t , w e h a v e s i n r u j > = 2 " _ 1 s i n < f > . s i n ( < £ + 2 a ) s i n ( < £ + 4 r e ) x - - - x s i n 0 + 2 ? i « - 2 « ) ( 2 ) R E S O L U T I O N I N T O F A C T O R S . 2 4 9 B u t s i n ( < £ + 2 i 1 a — 2 « ) = s i n ( < f > + t t — 2 a ) = s i n ( 2 « — < £ ) , s i n ( < £ + 2 n a — 4 « ) = s i n ( 4 « — < £ ) , a n d s o o n . H e n c e , w h e n n i s o d t i , m u l t i p l y i n g t o g e t h e r t h e s e c o n d f a c t o r a n d t h e l a s t , t h e t h i r d a n d t h e l a s t b u t o n e , a n d s o o n , w e h a v e s i n n < j > = 2 " _ 1 s i n < f > s i n ( 2 a + < f > ) s i n ( 2 a — < j > ) s i n ( 4 « - f c £ ) s i n ( 4 a — < £ ) . - - x s i n [ ( n — 1 ) « + < £ ] s i n [ ( n — l ) « — < £ ] . B u t s i n ( 2 a + < f > ) s i n ( 2 « — < j > ) = s i n a 2 « — s i n 2 < £ , a n d s o o n . . - . s i n M < £ = 2 n ~ 1 s i n $ ( s i n 2 2 « — s i n 2 < £ ) ( s i n 2 4 « — s i n 2 < £ ) x - - -- - - x [ s i n 2 ( n - l ) « - s i n 2 < £ ] . . . . ( 3 ) D i v i d e b o t h m e m b e r s o f ( 3 ) b y s i n < £ , a n d t h e n d i m i n i s h < j > i n d e f i n i t e l y . S i n c e t h e l i m i t o f s i n n < j > - f - s i n < j > i s n , w e g e t n = 2 " ~ 1 s m 2 2 « s i n 2 4 « s i n 2 6 « x - - - X s i n 2 ( n — 1 ) « ( 4 ) D i v i d e ( 3 ) b y ( 4 ) ; t h u s s i n ^ = n s i n / l - ^ - Y l - 4 ^ ) x . . . ( 5 ) \ s m 2 2 u j \ s i n M o / P u t n < j > = 6 , a n d l e t n b e i n c r e a s e d w h i l e < f > i s d i m i n i s h e d w i t h o u t l i m i t , 6 r e m a i n i n g u n c h a n g e d ; t h e n s i n c e 2 n a = t t , t h e l i m i t o f s i n 2 t f 2 S i n " ! ( j i f & I i D ^ = — 7 ^ = ^ x 7 ? y x ~ 7 ^ = ^ ( A r t 1 3 3 ) s i n ' --s i n " 1 n j n a n d t h e l i m i t o f n s i n d > = t h a t o f n s i n - = 9 ; a n d s o o n . n H e n c e ( 5 ) b e c o m e s N o t e . — T h e s a m e r e s u l t w i l l b e o b t a i n e d i f w e s u p p o s e n e v e n . 2 5 0 P L A N E T R I G O N O M E T R Y . R e m . — W h e n 0 > O a n d < i r , s i n 9 i s + , a n d e v e r y f a c t o r i n t h e s e c o n d m e m b e r o f ( 6 ) i s p o s i t i v e ; w h e n 9 > n - a n d < 2 i r , s i n 8 i s — , a n d o n l y t h e s e c o n d f a c t o r i s n e g a t i v e ; w h e u # > 2 t t a n d < 3 t t , b o t h m e m b e r s a r e p o s i t i v e , s i n c e o n l y t h e s e c o n d a n d t h i r d f a c t o r s a r e n e g a t i v e ; a n d s o o n . H e n c e t h e + s i g n w a s t a k e n i n e x t r a c t i n g t h e s q u a r e r o o t o f ( 1 ) . C o r . L e t 6 = t h e n s i n - = 1 , a n d -2 , 2 , i T c o m e s 2 \ 2 2 J \ 2 1 - 2 2 J \ 1 ^ 3 3 - 5 5 - 7 _ 2 ~ 2 2 ' 4 2 ' 6 2 = _ 2 ^ _ _ 4 _ J > ! -w h i c h i s T F i i M i s , s e x p r e s s i o n f o r i r . 1 7 3 . R e s o l v e c o s 6 i n t o F a c t o r s . — I n ( 2 ) o f A r t . 1 7 2 , c h a n g e i n t o < f > + a , t h e n n < £ b e c o m e s n < £ + n u , i . e . , n < j > + — . H e n c e ( 2 ) b e c o m e s c o s n < £ = 2 " _ 1 s i n ( < £ + « ) s i n ( o J 4 - 3 « ) s i n ( o i + 5 « ) X - s i n [ > + ( 2 n - l ) a ] ( 1 ) B u t s i n ( < £ + 2 n « — « ) = s i n ( < £ + i r — a ) = s i n ( a — r / > ) , s i n ( < £ + 2 n « — 3 « ) = s i n ( 3 « — < £ ) , a n d s o o n . H e n c e w h e n n i s e v e n w e h a v e f r o m ( 1 ) c o s n r / > = 2 " ~ 1 s i n ( « + < j > ) s i n ( a — < j > ) s i n ( 3 « + < £ ) s i n ( 3 a — < f > ) X - - - X s i n [ ( n — l ) a + r / > ] s i n [ ( n — 1 ) « — < £ ] = 2 " ~ 1 ( s i n 2 a — s i n 2 < £ ) ( s i n 2 3 « — s i n 2 < £ ) X - - - X [ s i n 2 ( n — l ) a — s i n 2 < £ ] ( 2 ) T h e r e f o r e , p u t t i n g n < j > = 6 , a s i n A r t . 1 7 2 , w e o b t a i n c o s 6 = N o t e . — F o r a n a l t e r n a t i v e p r o o f o f t h e p r o p o s i t i o n s o f A r t s . 1 7 2 a n d 1 7 3 , s e e L o c k ' s H i g h e r T r i g o n o m e t r y , p p . 9 2 - 9 5 . : ~ . H e n c e ( 6 ) b e -2 v ' _ _ L A 3 2 - 2 7 S U M M A T I O N O F S E R I E S . 2 5 1 E X A M P L E S . 1 . I f a = — , p r o v e t h a t i n s i n « s i n 5 a s i n 9 a - - . s i n ( 4 n — 3 ) « = 2 _ n + i . 2 . S h o w t h a t 1 6 c o s 0 c o s ( 7 2 ° -6 > ) c o s ( 7 2 ° + 6 ) c o s ( 1 4 4 ° - 0 ) c o s ( 1 4 4 ° + 0 ) = c o s 5 f t S U M M A T I O N O F T R I G O N O M E T R I C S E R I E S . 1 7 4 . S u m t h e S e r i e s s i n a + s i n ( a + 0 ) + s i n ( « + 2 / 3 ) H h s i n [ « + ( n — 1 ) 0 ] . W e h a v e 2 s i n a s i n £ / 3 = c o s ( a c o s f a + @ \ ( A r t . 4 5 ) 2 s i n ( « + / 3 ) s i n £ / 3 = c o s j j + ^ — c o s ( « + f / J ) , 2 s i n ( « + 2 / 3 ) s i n ^ / S = c o s ( a + f / 3 ) -c o s ( a + f / 3 ) , e t c . - e t c . 2 s i n [ a + ( n — l ) / 3 ] s i n | / 3 2 n -= c o s « H ; ( 8 -. c o s v . T h e r e f o r e , i f S „ d e n o t e t h e s u m o f n t e r m s , w e h a v e , b y a d d i t i o n , 2 S „ s i n £ / 3 = c o s ( « - ^ / 3 ) - c o s « + - ? l ^ / 3 j = 2 s i n u + - ~ — P s i n ^ n / S . . ( A r t . 4 5 ) 2 s i n s « = -i 1 > — 1 , « H / 2 ' s i n ^ n / 3 a i n £ / 8 2 5 2 P L A N E T R I G O N O M E T R Y . 1 7 5 . S u m t h e S e r i e s c o s a + c o s ( a + f 3 ) + c o s ( « + 2 / 3 ) H h c o s [ a + ( n — 1 ) / 3 ] . W e h a v e 2 c o s « s i n £ / 3 = s i n ( « + — s i n ( « — £ / ? ) , 2 c o s ( r e + / 3 ) s i n = s i n ( a + f / 3 ) — s i n ( « + e t c . = e t c . 2 c o s [ a + ( n — 1 ) 0 ] s i n = s i n « H - — / 3 — s m « H — / 8 -D e n o t i n g t h e s u m o f n t e r m s b y S „ , a n d a d d i n g , w e g e t 2 S „ s i n I P = s i n ^ « + ^ ™/ 3 - s i n + s i n | n / 3 c o s | . - -S n = -s i n £ / 3 H e m . — T h e s u m o f t h e s e r i e s i n t h i s a r t i c l e m a y b e d e d u c e d f r o m t h a t i n A r t . 1 7 4 b y p u t t i n g a + ~ f o r a -T n e s u m s o f t h e s e t w o s e r i e s a r e o f t e n u s e f u l ; a n d t h e s t u d e n t i s a d v i s e d t o c o m m i t t h e m t o m e m o r y . 2 i T C o r . I f w e p u t / ? = — , t h e n s i n $ n / 3 = s i n t t = 0 . H e n c e w e h a v e f r o m A r t s . 1 7 4 a n d 1 7 5 s i n a + s i n + ^ + s i n ( a + + - - - s i n j ^ t + 2 ( n ~ ^ j = 0 . / « + 2 t t \ , / , 4 t t \ , r , 2 ( n - l ) 1 r . c o s r e + c o s f — ' ) + c o s ( a - \ H c o s a - \ — 5 i v = 0 . \ n J . \ n J | _ n J N o t e . — T h e s e t w o r e s u l t s a r e v e r y i m p o r t a n t , a n d t h e s t u d e n t s h o u l d c a r e f u l l y n o t i c e t h e m . 1 7 6 . S u m t h e S e r i e s s i n ™a + s i n ™( a - | - £ ) + s i n ™( a + 2 / ? ) H h s i n ™[ « + ( n — 1 ) 0 ] . T h i s m a y b e d o n e b y t h e a i d o f A r t . 1 5 9 o r A r t . 1 6 0 . S e e T h o m p s o n ' s D y n a m o - E l e c t r i c M a c h i n e r y . 3 d e d . ( p p . 3 4 5 , 3 4 6 . S U M M A T I O N O F S E R I E S . 2 5 3 T h u s , i f m i s e v e n , w e h a v e f r o m A r t . 1 5 9 ™ 2 ™ 1 s i n ™« = ( — l ) ^ [ c o s m a — m c o s ( m — 2 ) a + - - - ] . ( 1 ) 2 — 1 s i n " ( a + = ( - I f [ c o s m ( « + / ? ) - m c o s ( m - 2 ) ( « + 0 ) + - - - ] ( 2 ) a n d s o o n ; a n d t h e r e q u i r e d s u m m a y b e o b t a i n e d f r o m t h e k n o w n s u m o f t h e s e r i e s [ c o s m a + c o s m ( a + / ? ) + c o s m ( a + 2 / 8 ) + - - - ] a n d { c o s ( m — 2 ) a - f c o s [ ( m — 2 ) ( « + / ? ) ] + c o s [ ( m — 2 ) ( « + 2 0 ) ] + - - - } , e t c . , W e m a y f i n d t h e s u m o f t h e s e r i e s c o s ™a + c o s ™( « + / ? ) + c o s ™( a + 2 / ? ) + e t c . t o n t e r m s i n a s i m i l a r m a n n e r b y t h e a i d o f A r t . 1 5 8 . E X A M P L E S . 1 . S u m t o n t e r m s t h e s e r i e s s i n 2 a + s i n 2 ( a + / 3 ) + s i n 2 ( a + 2 0 ) H W e h a v e 2 s i n 2 a = — ( c o s 2 a — 1 ) b y ( 1 ) , 2 s i n 2 ( a + / ? ) = -[ c o s 2 ( a + / ? ) -1 ] b y ( 2 ) , 2 s i n 2 ( a + 2 / ? ) = -[ c o s 2 ( a + 2 0 ) -1 ] , a n d s o o n . H e n c e 2 S „ = n — [ c o s 2 « + c o s 2 ( « + / 3 ) + c o s 2 ( « + 2 / 3 ) + . . - ] = n c o s [ 2 « + ( n - ! ) / ? ] s i n m ) s i n p ' v ' g _ n c o s [ 2 r e + ( n — l ) / 8 ] s i n n / 3 " n _ 2 2 s i n / J 2 5 4 P L A N E T R I G O N O M E T R Y . 2 . S u m t o n t e r m s t h e s e r i e s c o s 3 a + c o s 3 2 a + c o s 3 3 a + 2 c o s [ 3 « + | ( ? i — 1 ) 3 « ] s i n f n « 8 s i n | a 6 c o s 1 ~ -a 2 . n a s i n -" r 8 s i n | « " ~ 2 1 7 7 . S u m t h e S e r i e s s i n a — s i n ( « + / J ) + s i n ( « + 2 / 3 ) — - - - t o n t e r m s . ( 1 ) C h a n g e f } i n t o / 3 + i r , a n d ( 1 ) b e c o m e s s i n a + s i n ( r e + i r + / ? ) + s i n ( « + 2 i r + 2 / ? ) H . . ( 2 ) T h e r e f o r e w e h a v e f r o m A r t . 1 7 4 ( n - l ) Q r + , 8 ) - | s i n n ( + p ) a + ) ~ l • w ( " s i n — S l n 2 S i m i l a r l y , c o s a — c o s ( « + / ? ) + c o s ( « + 2 / 3 ) t o n t e r m s r , ( n -1 ) ( i r + p y c o s a + — — — ( 3 ) s i n — i l — t L L s i n — 2 1 7 8 . S u m t h e S e r i e s c o s e c 6 + c o s e c 2 6 + c o s e c 4 0 + - . -t o n t e r m s . a W e h a v e c o s e c 6 = c o t - — c o t 6 , 2 c o s e c 2 6 = c o t 6 — c o t 2 0 , e t c . = e t c . c o s e c 2 " ~ 1 0 = c o t 2 " - 2 0 -c o t 2 2 » - 1 0 . T h e r e f o r e , b y a d d i t i o n , a s i n A r t . 1 7 4 , S „ = c o t £ 0 - c o t 2 2 " - 1 0 . ( 4 ) S U M M A T I O N O F S E R I E S . 2 5 5 N o t e . — T h e a r t i f i c e e m p l o y e d i n t h i s A r t . , o f r e s o l v i n g e a c h t e r m i n t o t h e d i f f e r e n c e o f t w o o t h e r s , i s e x t e n s i v e l y u s e d i n t h e s u m m a t i o n o f s e r i e s . P r a c t i c e a l o n e w i l l g i v e t h e s t u d e n t r e a d i n e s s i n e f f e c t i n g s u c h t r a n s f o r m a t i o n s . I f h e c a n n o t d i s c o v e r t h e m o d e o f r e s o l u t i o n i n a n y e x a m p l e , h e w i l l o f t e n e a s i l y r e c o g n i z e i t w h e n h e s e e s t h e r e s u l t o f s u m m a t i o n . T h e s t u d e n t , h o w e v e r , i s a d v i s e d t o r e s o r t t o t h i s m e t h o d o f s o l u t i o n o n l y a s a l a s t r e s o u r c e . 1 7 9 . S u m t h e S e r i e s 6 6 t a n 6 - f £ t a n - + J t a n - + - - -t o n t e r m s . W e h a v e t a n 6 = c o t 6 — 2 c o t 2 6 , 6 6 | t a n - = \ c o t ; — c o t 6 , 2 2 i t a n | = i c o t | - ^ c o t | e t c = e t c . J L t a n J L = J -c o t J L - J L c o t J - . 2 » - i 2 " _ 1 2 " - 1 2 " - 1 2 " ~ 2 2 " - 2 . - . S n = - i - c o t — - 2 c o t 2 0 . 2 " ~ 1 2 " _ I 1 8 0 . S u m t h e S e r i e s s i n « + a ; s i n ( « + 0 ) + a r s i n ( « + 2 / 8 ) + - - - a ; " - 1 s i n [ a + ( n - l ) / 3 ] . D e n o t e t h e s u m b y S „ , a n d s u b s t i t u t e f o r t h e s i n e s t h e i r e x p o n e n t i a l v a l u e s ( A r t . 1 6 1 ) . T h u s , 2 i S „ = ( e i a — e - « ) + x ( e ' ( a + ' J > — e - » ( a + « ) + x 2 ( e , ' ( a + 2 « — e - » ( " + 2 P ) ) - f a ; " - 1 [ e ' ( a + " / 3 - « _ e - i ( a + " P - M ] l _ a ; e » 3 l — x e - ' P [ A l g . ( 3 ) A r t . 1 6 3 ] _ e ' ° — e - f a — a ; [ e ' ( " - P ) — e - ' ^ - P ) ] — g " [ e ' ( a + " P ) — e - > ( ° + " P ) ] ~ 1 — x ( e ^ + e ~ i P ) + x 2 + [ e » ( » + n | 3 - ) 3 ) _ e - i ( a + n p - p ) ] 1 — a ; ( e i 3 + e - i ^ ) + a ^ 2 5 6 P L A N E T R I G O N O M E T R Y . . . . s „ = s i n « — a ; s i n ( « — ; 8 ) — a ; " s i n ( « + n / 3 ) + a ' " + 1 s i n [ " + ( n — ^ 1 — 2 x c o s / 2 + ^ C o r . I f z < 1 , a n d n b e i n d e f i n i t e l y i n c r e a s e d , g _ s i n a — x s i n ( a — ( 3 ) ^ 1 — 2 a ; c o s / ? + a r ' S W L S i m i l a r l y , c o s a + a ; c o s ( « + / ? ) + a r c o s ( a + 2 ) 3 ) + - . - t o n t e r m s = c o s a — a ; e o s ( « — / 3 ) — a ; " c o s ( « + n / 8 ) + a ; n + 1 c o s [ a + ( n — 1 ) / ? ] ^ l — 2 x c o s ( 3 + X 2 W e m a y o b t a i n ( 3 ) f r o m ( 1 ) b y c h a n g i n g a t o a + ~ A l s o S m = c o s « -x c o s ( « - / ? ) . . . . ( 4 ) 1 - 2 0 ; c 0 8 ) 8 + a ? 1 8 1 . S u m t h e I n f i n i t e S e r i e s a ; s i n ( « + £ ) + 5 : s i n ( « + 2 / 8 ) + ^ s i n ( « + 3 / 8 ) + I f I f a n d c o s ( « + ) 8 ) + ^ c o s ( a + 2 ) 8 ) + ^ c o s ( « + 3 ) 8 ) + I f i f L e t S d e n o t e t h e f o r m e r s e r i e s , a n d C t h e l a t t e r . T h e n C + t S = x ( » < . a + v + - & < . a + 2 & + - e t ' ( « + 3 « + - - -[ 2 [ 3 = e i a ( e ^ _ i ) . . [ b y ( 3 ) o f A r t . 1 2 9 ] _ g i a ( e x ( c o s 0 + i s i n / 3 ) _ J ) , , . ( A r t 1 6 1 ) q x c o s ) 3 g t C a + x s i n 0 ) e a _ g x c o s p j - c o s ( « + x s i n / 8 ) 4 - 1 s i n ( a + a ; s i n / 3 ) ] — ( c o s « + £ s i n a ) . . . . ( A r t . 1 6 1 ) E X A M P L E S . 2 5 7 E q u a t i n g r e a l a n d i m a g i n a r y p a r t s , w e h a v e C = e x c o ^ c o s ( a + x s i n / 3 ) — c o s a , S = e a x o s ' 3 s i n ( « + x s i n / ? ) — s i n « . E X A M P L E S . P r o v e t h e f o l l o w i n g s t a t e m e n t s : 1 . T h e t w o v a l u e s o f ( c o s 4 6 + V — 1 s i n 4 0 ) % a r e ± ( c o s 2 0 + V ^ T s i n 2 0 ) . . . ( A r t . 1 5 4 ) 2 . T h e t h r e e v a l u e s o f ( c o s 0 + V — 1 s i n 0 ) » a r e 0 , / — r . 6 2 i r + 0 , / — r -f c o s - + V — 1 s i n - , c o s — ^ ( - V — 1 s i n -3 3 3 3 s i n „ 4 i r + 0 , / z r . 4 t t + 0 c o s . r - V — I F " ' 3 . T h e t h r e e v a l u e s o f ( — 1 ) a r e - 1 , ( A r t . 1 5 4 ) 4 . T h e s i x v a l u e s o f ( — 1 ) ^ a r e c o n t a i n e d i n c o s < 2 r + 1 ) , r ± V ^ l s i n + w h e r e r = 0 , 1 , o r 2 . 6 6 5 . T h e t h r e e v a l u e s o f ( 1 + V — 1 ) a r e c o n t a i n e d i n 2 r c o s ^ + V 3 1 s i n - l , w h e r e 6 = i v , o r V " -L 3 3 J 4 6 . T h e t h r e e v a l u e s o f ( 3 + 4 V — 1 ) a r e c o n t a i n e d i n V o c o s — ' .— + v — 1 s m — - — , w h e r e r = 0 , 1 , o r 2 . o o 7 . c o s 6 6 = c o s 6 — 1 5 c o s 4 0 s i n 2 0 + 1 5 c o s 2 6 s i n 4 0 — s i n 6 0 . 8 . s i n 9 6 = 9 c o s 8 6 s i n 0 -8 4 c o s 6 6 s i n 3 0 + 1 2 6 c o s 4 6 s i n 5 0 -3 6 c o s 2 0 s i n 7 0 + s i n 9 0 . 2 5 8 P L A N E T R I G O N O M E T R Y . 9 . t a n n 6 1 1 t a n 6 n ( n - ! ) ( » -2 ) t a n 8 g + , t a n 2 0 + n ( n — l ) ( ? i — 2 ) ( w - 3 ) L i t a n 4 0 s i n 6 2 1 6 5 1 0 . G i v e n = -: s h o w t h a t 6 i s n e a r l y t h e c i r c u l a r m e a s u r e o f 3 ° . P r o v e t h e f o l l o w i n g : 1 1 . -6 4 s i n i 0 = s i n 7 6 — 7 s i n 5 0 + 2 1 s i n 3 0 — 3 5 s i n f t 1 2 . -2 9 s i n 1 0 0 = c o s 1 0 0 -1 0 c o s 8 0 + 4 5 c o s 6 0 -1 2 0 c o s 4 0 + 2 1 0 c o s 2 0 -1 2 6 . 1 3 . 2 6 ( c o s 8 0 + s i n 8 0 ) = c o s 8 0 + 2 8 c o s 4 0 + 3 5 . 1 4 . c o s « 0 + s i n 6 0 = £ ( 5 + 3 c o s 4 0 ) . 1 5 . E x p a n d ( s i n 0 ) 4 " + z i n t e r m s o f c o s i n e s o f m u l t i p l e s o f 0 . 1 6 . E x p a n d ( s i n 0 ) 4 n + 1 i n t e r m s o f s i n e s o f m u l t i p l e s o f 0 . 1 7 . E x p a n d ( c o s 0 ) 2 " i n t e r m s o f c o s i n e s o f m u l t i p l e s o f 0 . U s e t h e e x p o n e n t i a l v a l u e s o f t h e s i n e a n d c o s i n e t o p r o v e t h e f o l l o w i n g : 1 9 . I f l o g ( a ; + y V — 1 ) = a + f t V — 1 , p r o v e t h a t 3 ? + i f = e 2 , a n d y = x t a n f t . 2 0 . I f s i n ( « + f t V — 1 ) = x + ? / V — 1 , p r o v e t h a t a ; 2 c o s e c 2 « — y 2 s e c 2 « = 1 . 1 8 . s i n 0 = c o t —2 1 — c o s 0 + ( a ; _ V - l V i - a r ) " . E X A M P L E S . 2 5 9 I T 2 2 . ( V ^ y ^ e ' . 2 3 . e » ( c o s 6 + V ^ T s i n 0 ) = e < v / c o s | + V ^ T s i n 2 4 . T h e c o e f f i c i e n t s o f x n i n t h e e x p a n s i o n , ( 1 ) o f e ° c o s b x , a n d ( 2 ) o f e " 1 s i n ¿ > a ¡ , i n p o w e r s o f x , a r e j — — ¿ - c o s n o a n d ^ — ^ — s i n n 0 . [ n \ n 2 5 . T h e c o e f f i c i e n t o f x n i n t h e e x p a n s i o n o f e c o s x i n f t t . 2 M i T p o w e r s o f a ; i s — c o s F [ n 4 2 6 . I f t h e s i d e s o f a r i g h t t r i a n g l e a r e 4 9 a n d 5 1 , t h e n t h e a n g l e s o p p o s i t e t h e m a r e 4 3 ° 5 1 ' 1 5 " a n d 4 6 " 8 ' 4 5 " n e a r l y . 2 7 . I f a a n d 6 b e t h e s i d e s o f a t r i a n g l e , A a n d B t h e o p p o s i t e a n g l e s , t h e n w i l l l o g b — l o g a = c o s 2 A — c o s 2 B + ^ ( c o s 4 A — c o s 4 B ) + ^ ( c o s 6 A — c o s 6 B ) + 2 8 . I f A + i B = l o g ( m + i n ) , t h e n t a n B = — , a n d 2 A = l o g ( n 2 + m z ) . m 2 9 . c o s ( 0 + i < t > ) = c o s A ^ - 6 ) + s i n f l ^ - " ' ' ~ e ' ^ . 3 0 . s i n ( 0 + < ' < £ ) = s i n 0 ( ~ + -i c o s 0 ^ ~ ~ C j -3 1 . 2 c o s ( « - f i f } ) = c o s « ( e 9 + e ~ P ) — i s i n « ( e £ — e ~ P ) . 3 2 . ( o + i 6 ) ( « + « _ r ° e " - ^ r [ c o s ( / 3 l o g r + a r ) + i s i n ( / J l o g r + o r ) ] , w h e r e a + ¿ 6 = ? - ( c o s r + i s i n r ) . 3 3 . l o g ( a + ¿ 6 ) = I l o g ( a 2 + 6 2 ) + 1 t a n - 1 2 6 0 P L A N E T R I G O N O M E T R Y . 3 4 . [ s i n ( « — 0 ) + e ± i a s i n 0 ] " = s i n " ~ , « [ s i n ( « — n O ) + e ± i a s i n n 0 ] . 3 5 . - = - 1 _ + - L - + - ! _ + . . . . 8 1 - 3 5 - 7 9 - 1 1 3 6 . W r i t e d o w n t h e q u a d r a t i c f a c t o r s o f x m — 1 . . 4 n s . ( a ; — 1 ) [ a ; 2 — 2 a ; c o s y 1 ^ ( 2 m ) + 1 ] , s i x f a c t o r s , p u t t i n g r — l } 2 , 3 , 4 , 5 , 6 . 3 7 . S o l v e t h e e q u a t i o n a ^ — 1 = 0 . A n s . ( a ; 2 -1 ) ( a ? -x + l ) ( x + x + 1 ) = 0 . 3 8 . G i v e t h e g e n e r a l q u a d r a t i c f a c t o r o f a r " — a 2 0 . A n s . x 2 — 2 a x c o s ^ ( r i r ) + a 2 . 3 9 . F i n d a l l t h e v a l u e s o f V l . A n s . c o s £ ( j i r ) + i s i n £ ( r i r ) , r h a v i n g e a c h i n t e g r a l v a l u e f r o m 0 t o 1 1 . 4 0 . W r i t e d o w n t h e q u a d r a t i c f a c t o r s o f x + 1 . A n s . ( x i - ^ 3 x + l ) ( x ' + l ) ( x > + V 3 x ' + l ) . 4 1 . W r i t e d o w n t h e g e n e r a l q u a d r a t i c f a c t o r o f a ; 2 0 + 1 . A n s . a r - 2 a ; c o s ( l + 2 r ) 9 ° + l . 4 2 . F i n d t h e f a c t o r s o f x w + 1 = 0 . A n s . ( a ; + l ) [ x 2 — 2 a ; c o s T l j + 2 r i r ) + 1 ] , s e v e n f a c t o r s i n a l l . 4 3 . F i n d a g e n e r a l e x p r e s s i o n f o r a l l t h e v a l u e s o f V — T . A n s . c o s — ! 1 - i s i n — . , w h e r e r m a y h a v e n n a n y i n t e g r a l v a l u e . 4 4 . S o l v e a ; , ! - 2 a " c o s $ i r + l = 0 . A n s . a ? — 2 x c o s £ ( 3 r v + + 1 = 0 , s i x q u a d r a t i c s . E X A M P L E S . 2 6 1 4 5 . S o l v e x 1 0 + V 3 x 5 + 1 = 0 . A n s . x 2 + 2 x c o s ( r x 7 2 ° + 6 ° ) + 1 = 0 , f i v e q u a d r a t i c s . 4 6 . W r i t e d o w n t h e q u a d r a t i c f a c t o r s o f x 2 " — 2 x n y " c o s a + y 2 n . A n s . a ? — 2 x y c o s - — ^ r 7 r + y 1 , n f a c t o r s . n P r o v e t h e f o l l o w i n g : 4 7 . t a n t a n ^ + ^ t a n + . . t a n ^ + ^ — ! : ^ = ( — l ) 2 , w h e r e n i s e v e n . [ U s e ( 2 ) o f A r t . 1 7 2 . ] 4 8 . s i n 5 6 -c o s 5 6 = 1 6 c o s ( 0 -2 7 ° ) c o s ( 6 + 9 ° ) x X s i n ( 6 + 2 7 ° ) s i n ( 6 -9 ° ) ( c o s 6 -s i n 0 ) . 5 1 1 + 1 + 1 4 - 1 + = ^ 3 4 ' 6 " 5 2 . 1 + 1 + 1 + 1 + . . . = ! [ ! . I 2 3 2 5 7 8 5 3 ^ = 3 g g 1 4 4 3 2 4 5 7 6 ° ' ' 3 5 ' 1 4 3 ' 3 2 3 ' 5 7 5 " " 5 4 f f _ 2 - 2 - 4 - 4 - 6 - 6 - 8 - 8 — 2 1 - 3 - 3 - 5 . 5 - 7 - 7 - 9 - " 5 5 / q _ 4 - 3 6 - 1 0 0 - 1 9 6 - 3 2 4 -3 - 3 5 -9 9 - 1 9 5 -3 2 3 - - - ' 5 6 . ^ 3 ^ 8 . 8 0 . 2 2 4 - 4 4 0 - , 9 - 8 1 - 2 2 5 - 4 4 1 — 2 6 2 P L A N E T R I G O N O M E T R Y . 5 7 . c o s x + t a n V . s i n x = 2 V - » A + » A S i r + y A 5 t - & / 5 8 . c o s a ; — c o t i s i n a ; = 5 9 . B y a i d o f t h e f o r m u l a c o s f l - 8 l n 2 g a n d A r t . 1 7 2 , 2 s i n 0 , d e d u c e t h e v a l u e f o r c o s 6 o b t a i n e d i n A r t . 1 7 3 . 6 0 . B y e x p a n d i n g b o t h s i d e s o f E x . 5 7 i n p o w e r s o f x a n d e q u a t i n g t h e c o e f f i c i e n t s o f x , p r o v e t h a t t a n 2 / = — 2 ? — + _ 2 2 _ + _ 2 -? _ + . . . 2 n — y i r + y 3 i r — y 3 v + y 5 n — y 5 n + y 6 1 . P r o v e i n l i k e m a n n e r f r o m E x . 5 8 t h a t c o a = 2 ? _ + ^ + 2 y 2 i r — y 2 n + y i i r - y i v + y 6 2 . P r o v e = 1 -1 + 1 -1 + 1 -1 + 1 -. . . 3 V 3 2 4 5 7 8 1 0 C O T > -1 1 . 1 1 , 1 1 , 1 6 3 . P r o v e r = = 1 2 V 3 5 7 1 1 1 3 1 7 1 9 6 4 . P r o v e t h a t — i — = s i n y I L . + ^ _ + ^ 1 ? _ + . . . y v — y 2 - r r — y i r + y 2 i r + y 3 - n — y 4 t t -? / 3 i r + y S u m t h e f o l l o w i n g s e r i e s t o n t e r m s : -n + 1 . n « s i n — 1— a s i n — 2 2 6 5 . s i n « + s i n 2 « + s i n 3 a H | - s i n n a = . a s i n -2 E X A M P L E S . 2 6 3 n + 1 . n a c o s — .— a s i n — 2 2 6 6 . c o s a + c o s 2 « + e o s 3 < H | - c o s ? j a = s i n ^2 6 7 . s i n « + s i n 3 a + s i n 5 « + - - - = S 1 B ! ^ . s i n a 6 8 . c o s « + c o s 3 « + c o s 5 « + . . - = ^ 2 - ^ -2 s i n a 6 9 . s i n 2 « + s i n 2 2 « + s i n 2 3 a + . . . = " s i n « - s i n n « c o s ( n + l ) « 2 s i n « 7 0 . c o s 2 « + c o s 2 2 f t + c o s 2 3 a + . - - = » s j n j < + c o s ( n + 1 ) a s i n n « 2 s i n « 7 1 . s i n 3 « + s i n 3 ( « + + s i n 3 ( « + 2 / 3 ) H 3 s i n ( a + s i n 5 0 j s i n « + ^ S / s j s i — 4 s i n ^ / 3 4 s i n f / 2 7 2 . s i n 3 a + s i n 3 2 c t + s i n 3 3 « + - --n « -/ n + l \ . 3 n « . 3 ( n + l ) « s i n — s i n [ — — ) a s i n s i n — >— '— — 3 2 I 2 J 2 2 4 . « . . 3 a s i n -4 s i n — . 2 2 7 3 . s i n a s i n 2 a + s i n 2 « s i n 3 « + s i n 3 a s i n 4 a + - - . _ n s i n a c o s a — s i n n « c o s ( n + 2 ) a 2 s i n a 7 4 . t a n « + 2 t a n 2 « + 2 2 t a n 2 2 a H = c o t « — 2 n c o t 2 " « . 7 5 . ( t a n a + c o t a ) + ( t a n 2 c t + c o t 2 « ) + ( t a n 2 2 a + c o t 2 2 « ) - f - - - = 2 c o t a — 2 c o t 2 " a . 7 6 . s e c a s e c 2 a + s e c 2 a s e c 3 a + . - -= c o s e c a [ t a n ( n + 1 ) a — t a n « ] . 2 6 4 P L A N E T R I G O N O M E T R Y . 7 7 . c o s e c a c o s e c 2 a + c o s e c 2 a c o s e c 3 a + - - -= c o s e c « [ c o t « — c o t ( n + 1 ) a ] . 7 8 s i n 2 0 s i n 4 0 s e c ( 2 n + l ) 0 — s e c 0 c o s 0 c o s 3 0 c o s 3 0 c o s 5 0 2 s i n 0 7 9 . c o s 4 a + c o s 4 ( a + / 3 ) + c o s 4 ( a + 2 / 3 ) H = f ? i + c o s r 2 « + ( n — l ) / 3 ] s i n n / ? c o s [ 4 « + ( » — 1 ) 2 ) 8 ] s i n 2 n / 3 2 S Ü 1 0 8 s i n 2 / 3 o n . n s i n 0 + s i n 3 0 + s i n 5 0 + - - - t o n t e r m s 8 0 . t a n n i = — ! — — — — c o s 0 + c o s 3 0 + c o s 5 0 + - - - t o n t e r m s 8 1 . c o s 0 c o s ( 0 + a ) + c o s ( 0 + a ) c o s ( 0 + 2 « ) + c o s ( 0 + 2 « ) c o s ( 0 + 3 « ) H n , c o s ( 2 0 + n a ) s i n ? i « = - c o s a - \ ' -2 2 s i n a g 2 s i n 0 — s i n 2 0 + s i n 3 0 t o ? i t e r m s _ ^ a n n + l / ^ ^ c o s 0 — c o s 2 0 + c o s 3 0 t o n t e r m s 2 8 3 . s i n ( p + l ) 0 c o s 0 + s i n ( p + 2 ) 0 c o s 2 0 + - - - n s i n p 6 s i n ( p + 1 + n ) 0 s i n n 6 ~ 2 2 s i n 0 8 4 . s i n 3 0 s i n 0 + s i n 6 0 s i n 2 0 + s i n 1 2 0 s i n 4 0 H = £ ( c o s 2 0 - c o s 2 " + 1 0 ) . 8 5 . s i n 0 ^ s i n | J + 2 s i n | ^ s i n 0 + 4 s i n | ^ s i n | J = 2 " - 2 s i n - A -1 s i n 2 0 . 0 0 0 0 0 0 8 6 . t a n s e c 0 + t a n - s e c - + t a n - s e c . A — = t a n 0 — t a n — -2 4 2 8 4 2 " 8 7 . c o t 0 c o s e c 0 + 2 c o t 2 0 c o s e c 2 0 + 2 2 c o t 2 2 0 c o s e c 2 2 0 + - . -1 2 » - 1 ~ - 2 0 s i n ! 2 — » 0 ' 2 s i n ¿ -2 + , E X A M P L E S . 2 6 5 8 8 . 1 h -- _ f t „ -o f t 1 o f t v f t 1 „ : s i n 0 s i n 2 0 s i n 2 0 s i n 3 0 s i n 3 0 s i n 4 0 1 ( c o t 3 6 - c o t 4 0 ) . s i n 0 v y « Q - . _ L . s i n 0 c o s 2 0 c o s 2 0 s i n 3 0 s i n 3 0 c o s 4 0 = c o s e c ^ 0 + | ) t a n ( n + l ) ^ + | ^ - t a n ^ + | ^ -9 0 . t a n - 1 \ - t a n - , h t a n - 1 l + l + l ^ ^ 1 + 2 + 2 2 1 + 3 + 3 2 1 + - - . = t a n 4 n + 1 9 1 . t a n - 1 x + t a n - 1 \ - t a n - 1 : + . 1 + 1 - 2 - x 2 ^ 1 + 2 - 3 - x 2 = t a n - 1 n x . 9 2 . s i n a s i n 3 a + s i n - s i n h s i n - s i n — — + . 2 2 2 2 I f a . \ - I c o s c o s 4 a ] -2 ^ 2 " - 2 y 9 4 . - s e c 0 + — s e c 0 s e c 2 0 + i s e c 0 s e c 2 0 s e c 2 2 0 + 2 2 2 2 9 3 . 1 H + - - -c o s 0 + c o s 3 0 c o s 0 + c o s 5 0 c o s 0 + c o s 7 0 = £ c o s e c 0 [ t a n ( n + 1 ) 0 — t a n 0 ] . s e c 2 0 s e c 2 2 0 + - - -= s i n 0 ( c o t 0 - c o t 2 " 0 ) . 9 5 . ^ l o g t a n 2 0 + ~ l o g t a n 2 2 0 + ~ l o g t a n 2 3 0 + . - . A A Z = l o g 2 s i n 2 0 --l o g 2 s i n 2 n + 1 6 . S u m t h e f o l l o w i n g s e r i e s t o i n f i n i t y : n a a , c O S 0 o a i c o s 2 0 o a , c O S 2 0 . , 9 6 . c o s 0 H c o s 2 0 H — , c o s 3 0 H c o s 4 0 + - -1 [ 2 [ 3 = e c ° ' e c o s ( 0 + s i n 0 c o s 0 ) . 2 6 6 P L A N E T R I G O N O M E T R Y . 9 7 . s i n 6 _ n M + « B M _ . . . = e - « - s i n ( s i n [ 2 [ 3 9 8 . 1 -+ . . . = | c o s ( c o s 0 ) ( e ° " > » + « - • ) . 9 9 . 2 c o s 0 + f c o s 2 0 + f c o s 3 0 + £ c o s 4 0 + - - -c o s 0 1 — c o s 0 - l o g ( l — c o s 0 ) . 1 A A -/ j / , , s i n 2 0 c o s 2 0 , s i n 3 0 c o s 3 0 , 1 0 0 . s i n 0 c o s 0 - \ — 1 h , [ 2 [ 3 1 0 1 . c o s 0 + S — c o s 2 0 + ^ I B l ^ c o s 3 0 + 1 [ 2 _ e c o s 2 » s i n ( s i n ^ c o s ) + . . . e a i a e ° o a B c o s ( 6 + s m 2 6 ) . 1 0 2 . s i n 0 + — s i n 2 0 + s u — s i n 3 0 + 1 1 2 _ e » i n S c o s 9 s i n ^ + s i n 2 6 ) . 1 0 3 . c o s 0 - £ c o s 2 0 + | c o s 3 0 = l o g ^ 2 c o s | ^ -1 0 4 . c o s 2 0 + £ c o s 6 0 + £ c o s l 0 0 + - - - = £ l o g c o t - -1 0 5 . x s i n t f _ ^ s i n 2 0 ^ s i n 3 t f _ . . . = c o t - , / c o s e c t f f c f l \ 2 3 V ■ / 1 0 6 . x c o s 0 -- c o s 2 0 + - c o s 3 0 - - c o s 4 0 + - - -2 3 4 = l o g ( l + 2 x c o s 0 + x J . -/ > s i n 0 . O f l s i n 2 0 . -o / 1 s i n 3 0 1 0 7 . s i n 0 s i n 2 0 h s i n 3 0 — 1 o ' - -1 2 3 = c o t - 1 ( l + c o t 2 0 + c o t 0 ) . 1 0 8 . i + i + I + i + . . . = i r . I 4 2 3 4 4 4 9 0 1 0 9 . I + 1 + 1 + A + . . . = z . I 4 ^ 3 4 5 4 7 4 9 6 P A R T I I . S P H E R I C A L T R I G O N O M E T R Y . C H A P T E R X . F O R M U L A R E L A T I V E T O S P H E R I C A L T R I A N G L E S . 1 8 2 . S p h e r i c a l T r i g o n o m e t r y h a s f o r i t s o b j e c t t h e s o l u t i o n o f s p h e r i c a l t r i a n g l e s . A s p h e r i c a l t r i a n g l e i s t h e f i g u r e f o r m e d b y j o i n i n g a n y t h r e e p o i n t s o n t h e s u r f a c e o f a s p h e r e b y a r c s o f g r e a t c i r c l e s . T h e t h r e e p o i n t s a r e c a l l e d t h e v e r t i c e s o f t h e t r i a n g l e ; t h e t h r e e a r c s a r e c a l l e d t h e s i d e s o f t h e t r i a n g l e . A n y t w o p o i n t s o n t h e s u r f a c e o f a s p h e r e c a n b e j o i n e d b y t w o d i s t i n c t a r c s , w h i c h t o g e t h e r m a k e u p a g r e a t c i r c l e p a s s i n g t h r o u g h t h e p o i n t s . H e n c e , w h e n t h e p o i n t s a r e n o t d i a m e t r i c a l l y o p p o s i t e , t h e s e a r c s a r e u n e q u a l , o n e o f t h e m b e i n g l e s s , t h e o t h e r g r e a t e r , t h a n 1 8 0 ? . I t i s n o t n e c e s s a r y t o c o n s i d e r t r i a n g l e s i n w h i c h a s i d e i s g r e a t e r t h a n 1 8 0 ° , s i n c e w e m a y a l w a y s r e p l a c e s u c h a s i d e b y t h e r e m a i n i n g a r c o f t h e g r e a t c i r c l e t o w h i c h i t b e l o n g s . 1 8 3 . G e o m e t r i c P r i n c i p l e s . — I t i s s h o w n i n g e o m e t r y ( A r t . 7 0 2 ) , t h a t i f t h e v e r t e x o f a t r i e d r a l a n g l e i s m a d e t h e c e n t r e o f a s p h e r e , t h e n t h e p l a n e s w h i c h f o r m t h e t r i e d r a l a n g l e w i l l c u t t h e s u r f a c e o f t h e s p h e r e i n t h r e e a r c s o f g r e a t c i r c l e s , f o r m i n g a s p h e r i c a l t r i a n g l e . T h u s , l e t O b e t h e v e r t e x o f a t r i e d r a l a n g l e , a n d A O B , B O C , C O A i t s f a c e - a n g l e s . W e m a y c o n s t r u c t a s p h e r e w i t h i t s c e n t r e a t O , a n d w i t h a n y r a d i u s O A . L e t A B , 2 6 7 2 6 8 S P H E R I C A L T R I G O N O M E T R Y . B C , C A b e t h e a r c s o f g r e a t c i r c l e s i n w h i c h t h e p l a n e s o f t h e f a c e - a n g l e s A O B , B O C , C O A c u t t h e s u r f a c e o f t h i s s p h e r e ; t h e n A B C i s a s p h e r i c a l t r i a n g l e , a n d t h e a r c s A B , B C , C A a r e i t s s i d e s . N o w i t i s s h o w n i n g e o m e t r y t h a t t h e t h r e e f a c e - a n g l e s A O B , B O C , C O A a r e m e a s u r e d b y t h e s i d e s A B , B C , C A , r e s p e c t i v e l y , o f t h e s p h e r i c a l t r i a n g l e , a n d t h a t t h e d i e d r a l a n g l e s O A , O B , O C a r e e q u a l t o t h e a n g l e s A , B , C , r e s p e c t i v e l y , o f t h e s p h e r i c a l t r i a n g l e A B C , a n d a l s o t h a t a d i e d r a l a n g l e i s m e a s u r e d b y i t s p l a n e a n g l e . T h e r e i s t h e n a c o r r e s p o n d e n c e b e t w e e n t h e t r i e d r a l a n g l e O - A B C a n d t h e s p h e r i c a l t r i a n g l e A B C : t h e s i x p a r t s o f t h e t r i e d r a l a n g l e a r e r e p r e s e n t e d b y t h e c o r r e s p o n d i n g s i x p a r t s o f t h e s p h e r i c a l t r i a n g l e , a n d a l l t h e r e l a t i o n s a m o n g t h e p a r t s o f t h e f o r m e r a r e t h e s a m e a s t h e r e l a t i o n s a m o n g t h e c o r r e s p o n d i n g p a r t s o f t h e l a t t e r . 1 8 4 . F u n d a m e n t a l D e f i n i t i o n s a n d P r o p e r t i e s . — T h e f o l l o w i n g d e f i n i t i o n s a n d p r o p e r t i e s a r e f r o m G e o m e t r y , B o o k V I I I . : I n e v e r y s p h e r i c a l t r i a n g l e E a c h s i d e i s l e s s t h a n t h e s u m o f t h e o t h e r t w o . T h e s u m o f t h e t h r e e s i d e s l i e s b e t w e e n 0 ° a n d 3 6 0 ° . T h e s u m o f t h e t h r e e a n g l e s l i e s b e t w e e n 1 8 0 ° a n d 5 4 0 ° . E a c h a n g l e i s g r e a t e r t h a n t h e d i f f e r e n c e b e t w e e n 1 8 0 ° a n d t h e s u m o f t h e o t h e r t w o . I f t w o s i d e s a r e e q u a l , t h e a n g l e s o p p o s i t e t h e m a r e e q u a l ; a n d c o n v e r s e l y . I f t w o s i d e s a r e u n e q u a l , t h e g r e a t e r s i d e l i e s o p p o s i t e t h e g r e a t e r a n g l e ; a n d c o n v e r s e l y . T h e p e r p e n d i c u l a r f r o m t h e v e r t e x t o t h e b a s e o f a n i s o s c e l e s t r i a n g l e b i s e c t s b o t h t h e v e r t i c a l a n g l e a n d t h e b a s e . D E F I N I T I O N S A N D P R O P E R T I E S . 2 6 9 T h e a x i s o f a c i r c l e i s t h e d i a m e t e r o f t h e s p h e r e p e r p e n d i c u l a r t o t h e p l a n e o f t h e c i r c l e . T h e p o l e s o f a c i r c l e a r e t h e t w o p o i n t s i n w h i c h i t s a x i s m e e t s t h e s u r f a c e o f t h e s p h e r e . O n e s p h e r i c a l t r i a n g l e i s c a l l e d t h e p o l a r t r i a n g l e o f a s e c o n d s p h e r i c a l t r i a n g l e w h e n t h e s i d e s o f t h e f i r s t t r i a n g l e h a v e t h e i r p o l e s a t t h e v e r t i c e s o f t h e s e c o n d . I f t h e f i r s t o f t w o s p h e r i c a l t r i a n g l e s i s t h e p o l a r t r i a n g l e o f t h e s e c o n d , t h e n t h e s e c o n d i s t h e p o l a r t r i a n g l e o f t h e f i r s t . T w o s u c h t r i a n g l e s a r e s a i d t o b e p o l a r w i t h r e s p e c t t o e a c h o t h e r . T h u s : , I f A ' B ' C i s t h e p o l a r t r i a n g l e o f A B C , t h e n A B C i s t h e p o l a r t r i a n g l e o f A ' B ' C . I n t w o p o l a r t r i a n g l e s , e a c h a n g l e o f o n e i s m e a s u r e d b y t h e s u p p l e m e n t o f t h e c o r r e s p o n d i n g s i d e o f t h e o t h e r . T h u s : A = 1 8 0 ° -a ' , B = 1 8 0 ° -6 ' , C = 1 8 0 ° -c ' , a = 1 8 0 ° - A ' , 6 = 1 8 0 ° - B ' , c = 1 8 0 ° - C . T h i s r e s u l t i s o f g r e a t i m p o r t a n c e ; f o r i f a n y g e n e r a l e q u a t i o n b e e s t a b l i s h e d b e t w e e n t h e s i d e s a n d a n g l e s o f a s p h e r i c a l t r i a n g l e , i t h o l d s o f c o u r s e f o r t h e p o l a r t r i a n g l e a l s o . H e n c e , b y m e a n s o f t h e a b o v e f o r m u l a e a n y t h e o r e m o f a s p h e r i c a l t r i a n g l e m a y b e a t o n c e t r a n s f o r m e d i n t o a n o t h e r t h e o r e m b y s u b s t i t u t i n g f o r e a c h s i d e a n d a n g l e r e s p e c t i v e l y t h e s u p p l e m e n t s o f i t s o p p o s i t e a n g l e a n d s i d e . I f a s p h e r i c a l t r i a n g l e h a s o n e r i g h t a n g l e , i t i s c a l l e d a r i g h t t r i a n g l e ; i f . i t h a s t w o r i g h t a n g l e s , i t i s c a l l e d a b i -r e c t a n g u l a r t r i a n g l e ; a n d i f i t h a s t h r e e r i g h t a n g l e s , i t i s c a l l e d a t r i - r e c t a n g u l a r t r i a n g l e . I f i t h a s o n e s i d e e q u a l t o a q u a d r a n t , i t i s c a l l e d a q u a d r a n t a l t r i a n g l e ; a n d i f i t h a s t w o s i d e s e q u a l t o a q u a d r a n t , i t i s c a l l e d a b i - q u a d r a n t a l t r i a n g l e . 2 7 0 S P H E R I C A L T R I G O N O M E T R Y . N o t e . — I t i s s h o w n i n g e o m e t r y t h a t a s p h e r i c a l t r i a n g l e m a y , i n g e n e r a l , b e c o n s t r u c t e d w h e n a n y t h r e e o f i t s s i x p a r t s a r e g i v e n ( n o t e x c e p t i n g t h e c a s e i n w h i c h t h e g i v e n p a r t s a r e t h e t h r e e a n g l e s ) . I n s p h e r i c a l t r i g o n o m e t r y w e i n v e s t i g a t e t h e m e t h o d s b y w h i c h t h e u n k n o w n p a r t s o f a s p h e r i c a l t r i a n g l e m a y b e c o m p u t e d f r o m t h e a b o v e d a t a . E X A M P L E S . 1 . I n t h e s p h e r i c a l t r i a n g l e w h o s e a n g l e s a r e A , B , C , p r o v e B + C - A < , r ( 1 ) C + A - B < ^ ( 2 ) A + B -C < i t ( 3 ) 2 . I f C i s a r i g h t a n g l e , p r o v e A + B < f i r ( 1 ) , a n d A -B < | ( 2 ) . 3 . T h e a n g l e s o f a t r i a n g l e a r e A , 4 5 ° , a n d 1 2 0 ° ; f i n d t h e m a x i m u m v a l u e o f A . A n s . A < 1 0 5 ° . 4 . T h e a n g l e s o f a t r i a n g l e a r e A , 3 0 ° , a n d 1 5 0 ° ; f i n d t h e m a x i m u m v a l u e o f A . A n s . A < 6 0 ° . 5 . T h e a n g l e s o f a t r i a n g l e a r e A , 2 0 ° , a n d 1 1 0 ° ; f i n d t h e m a x i m u m v a l u e o f A . A n s . A < 9 0 ° . 6 . A n y s i d e o f a t r i a n g l e i s g r e a t e r t h a n t h e d i f f e r e n c e b e t w e e n t h e o t h e r t w o . R I G H T S P H E R I C A L T R I A N G L E S . 1 8 5 . F o r m u l a e f o r R i g h t T r i a n g l e s s p h e r i c a l t r i a n g l e i n w h i c h C i s a r i g h t a n g l e , a n d l e t O b e t h e c e n t r e o f t h e s p h e r e ; t h e n w i l l O A , O B , O C b e r a d i i : l e t a , b , c d e n o t e t h e s i d e s o f t h e t r i a n g l e 0 < ^ o p p o s i t e t h e a n g l e s A , B , C , r e s p e c t i v e l y ; t h e n a , b , a n d c a r e t h e m e a s u r e s o f t h e a n g l e s B O C , C O A , a n d A O B . R I G H T S P H E R I C A L T R I A N G L E S . 2 7 1 F r o m a n y p o i n t D i n O A d r a w D E _ L t o b Q , a n d f r o m E d r a w E F _ L t o O B ^ S n d j o i n D F . T h e n r > f i i s J L t o E F ( G e o m . A r t . 5 3 7 ) . H e n c e ( G e o m . A r t . 5 0 7 ) , t R T D F i s J . t o O ^ f . - . ^ D F E = Z ^ < J 2 . ( A r t . 1 8 3 ) , T O F O F O E . , . -w b c o s a ' c o s ^ s -. ( 1 ) N o w — — = — : t h a t i s , O D O E O D , , c O S 8 = D E D E D F . . s i n b = s i n B s i n c . . ( 2 ) = — -: t h a t i s , I n t e r c h a n g i n g a ' s a n d 6 , s , O D D F O D , s i n a = s i n A s i n c . . ( 3 ) E F E F D F . . — = - — : t h a t i s , t a n a = c o s B t a n c . . ( 4 ) I n t e r c h a n g i n g a , s a n d b , s , O F D F O F , t a n 6 = c o s A t a n c . . ( 5 ) D E D E E F . . . . t a n 6 t a n B s i n a . . ( 6 ) = -; t h a t i s , I n t e r c h a n g i n g a ' s a n d 6 ' s , O E E F O E , a = t a n A s i n h . . ( 7 ) M u l t i p l y ( 6 ) a n d ( 7 ) t o g e t h e r , a n d w e g e t t a n A t a n B = = ^ , b y ( 1 ) c o s a c o s b c o s c . - . c o s c = c o t A c o t B ( § ) M u l t i p l y c r o s s w i s e ( 3 ) a n d ( 4 ) , a n d w e g e t s i n a c o s B t a n c = t a n a s i n A s i n c . - r , s i n A c o s c . , , „ , . - . c o s B = = s i n A c o s b , b y ( 1 ) . . . ( 9 ) c o s « > J \ J v > I n t e r c h a n g i n g a , s a n d 6 , s , c o s A = s i n B c o s a ( 1 0 ) S c h . B y t h e s e t e n f o r m u l a e , e v e r y c a s e o f r i g h t t r i a n g l e s c a n b e s o l v e d ; f o r e v e r y o n e o f t h e s e t e n f o r m u l a e i s a d i s t i n c t c o m b i n a t i o n , i n v o l v i n g t h r e e o u t o f t h e f i v e q u a n t i t i e s , a , b , c , A , B , a n d t h e r e c a n b e b u t t e n c o m b i n a t i o n s i n a l l . H e n c e , a n y t w o o f t h e f i v e q u a n t i t i e s b e i n g g i v e n a n d a t h i r d r e q u i r e d , t h a t t h i r d q u a n t i t y m a y b e d e t e r m i n e d b y s o m e o n e o f t h e a b o v e t e n f o r m u l a ? . 2 7 2 S P H E R I C A L T R I G O N O M E T R Y . 1 8 6 . N a p i e r ' s R u l e s . — T h e t e n p r e c e d i n g f o r m u l a e , w h i c h m a y b e f o u n d d i f f i c u l t t o r e m e m b e r , h a v e b e e n i n c l u d e d u n d e r t w o s i m p l e r u l e s , c a l l e d a f t e r t h e i r i n v e n t o r , N a p i e r ' s R u l e s o f t h e C i r c u l a r P a r t s . L e t A B C b e a r i g h t s p h e r i c a l t r i a n g l e . O m i t t h e r i g h t a n g l e C . T h e n t h e t w o s i d e s a a n d b , w h i c h i n c l u d e t h e r i g h t a n g l e , t h e c o m p l e m e n t o f t h e h y p o t e n u s e c , a n d t h e c o m p l e m e n t s o f t h e o b l i q u e a n g l e s A a n d B , a r e c a l l e d t h e c i r c u l a r p a r t s o f t h e t r i a n g l e . T h u s , t h e r e a r e f i v e c i r c u l a r p a r t s , a r r a n g e d i n t h e f i g u r e i n t h e f o l l o w i n g o r d e r : a , b , c o . A , c o . c , c o . B . A n y o n e o f t h e s e f i v e p a r t s m a y b e s e l e c t e d a n d c a l l e d t h e m i d d l e p a r t ; t h e n t h e t w o p a r t s n e x t t o i t a r e c a l l e d a d j a c e n t p a r t s , a n d t h e r e m a i n i n g t w o p a r t s a r e c a l l e d o p p o s i t e p a r t s . T h u s , i f c o . A i s s e l e c t e d a s t h e m i d d l e p a r t , t h e n b a n d c o . c a r e t h e a d j a c e n t p a r t s , a n d a a n d c o . B a r e t h e o p p o s i t e p a r t s . T h e n N a p i e r , s K u l e s a r e : . ( 1 ) T h e s i n e o f t h e m i d d l e p a r t e q u a l s t h e p r o d u c t o f t h e t a n g e n t s o f t h e a d j a c e n t p a r t s . ( 2 ) T J i e s i n e o f t h e m i d d l e p a r t e q u a l s t h e p r o d u c t o f t h e c o s i n e s o f t h e o p p o s i t e p a r t s . N o t e 1 . — I t w i l l a s s i s t t h e s t u d e n t i n r e m e m b e r i n g t h e s e r u l e s t o n o t i c e t h e o c c u r r e n c e o f t h e v o w e l i i n s i n e a n d m i d d l e , o f t h e v o w e l a i n t a n g e n t a n d a d j a c e n t , a n d o f t h e v o w e l o i n c o s i n e a n d o p p o s i t e . N a p i e r ' s R u l e s m a y h e m a d e e v i d e n t b y t a k i n g i n d e t a i l e a c h o f t h e f i v e p a r t s a s m i d d l e p a r t , a n d c o m p a r i n g t h e e q u a t i o n s t h u s f o u n d w i t h t h e f o r m u l a ) o f A r t . 1 8 s . T h u s , l e t c o . c b e t h e m i d d l e p a r t . T h e r u l e s g i v e s i n ( c o . c ) = t a n ( c o . A ) t a n ( c o . B ) ; . . c o s c = c o t A c o t B ( 8 ) s i n ( c o . c ) = c o s a c o s 6 ; . - . c o s c ~ c o s a c o s b ( 1 ) c o . B t h e m i d d l e p a r t . s i n ( c o . B ) = t a n a t a n ( c o . c ) ; . ' . c o s B = t a n a c o t c ( 4 ) s i n ( c o . B ) = c o s 6 c o s ( c o . A ) ; . . c o s B = c o s b s i n A ( 9 ) W h i l e s o m e f i n d t h e s e r u l e s t o b e u s e f u l a i d s t o t h e m e m o r y , o t h e r s q u e s t i o n t h e i r u t i l i t y . T H E S P E C I E S O F T H E P A R T S . 2 7 3 a t h e m i d d l e p a r t . s i n a = t a n b t a n ( c o . B ) ; . - . s i n a = t a n 6 c o t B ( 6 ) s i n a = c o s ( c o . A ) c o s ( c o . c ) ; , \ s i n a = s i n A s i n c ( 3 ) b t h e m i d d l e p a r t . s i n 6 = t a n a U n ( c o . A ) ; / . s i n 6 = t a n a c o t A ( 7 ) s i n 6 = c o s ( c o . c ) c o s ( c o . B ) ; . . s i n b = s i n e s i n B ( 2 ) c o . A t h e m i d d l e p a r t . s i n ( c o . A ) = t a n b t a n ( c o . c ) ; . \ c o s A = t a n b c o t c ( 5 ) s i n f c o . A ) = c o s a c o s ( c o . B ) ; . . . c o s A = c o s a s i n B ( 1 0 ) N o t e 2 . — I n a p p l y i n g t h e s e r u l e s i t i s n o t n e c e s s a r y t o u s e t h e n o t a t i o n c o . c , c o . A , c o . B , s i n c e w e m a y w r i t e a t o n c e c o s c f o r s i n ( c o . c ) , e t c . 1 8 7 . T h e S p e c i e s o f t h e P a r t s . — I f t w o p a r t s o f a s p h e r i c a l t r i a n g l e a r e e i t h e r b o t h l e s s t h a n 9 0 ° o r b o t h g r e a t e r t h a n 9 0 ° , t h e y a r e s a i d t o b e o f t h e s a m e s p e c i e s . B u t i f o n e p a r t i s l e s s t h a n 9 0 ° a n d t h e o t h e r p a r t i s g r e a t e r t h a n 9 0 ° , t h e y a r e o f d i f f e r e n t s p e c i e s . I n o r d e r t o d e t e r m i n e w h e t h e r t h e r e q u i r e d p a r t s a r e l e s s o r g r e a t e r t h a n 9 0 ° , i t w i l l b e n e c e s s a r y c a r e f u l l y t o o b s e r v e t h e i r a l g e b r a i c s i g n s . I f t h e r e q u i r e d p a r t i s d e t e r m i n e d b y m e a n s o f i t s c o s i n e , t a n g e n t , o r c o t a n g e n t , t h e a l g e b r a i c s i g n o f t h e r e s u l t w i l l s h o w w h e t h e r i t i s l e s s o r g r e a t e r t h a n 9 0 ° . B u t w h e n a r e q u i r e d p a r t i s f o u n d i n t e r m s o f i t s s i n e , i t w i l l b e a m b i g u o u s , s i n c e t h e s i n e s a r e p o s i t i v e i n b o t h t h e f i r s t a n d s e c o n d q u a d r a n t s . T h i s a m b i g u i t y , h o w e v e r , m a y g e n e r a l l y b e r e m o v e d b y e i t h e r o f t h e f o l l o w i n g p r i n c i p l e s : ( 1 ) I n a r i g h t s p h e r i c a l t r i a n g l e , e i t h e r o f t h e s i d e s c o n t a i n i n g t h e r i g h t t r i a n g l e i s o f t h e s a m e s p e c i e s a s t h e o p p o s i t e a n g l e . ( 2 ) T h e t h r e e s i d e s o f a r i g h t s p h e r i c a l t r i a n g l e ( o m i t t i n g b i - r e c t a n g u l a r o r t r i - r e c t a n g u l a r t r i a n g l e s ) a r e e i t h e r a l l a c u t e , o r e l s e o n e i s a c u t e a n d t h e o t h e r t w o o b t u s e . T h e f i r s t f o l l o w s f r o m t h e e q u a t i o n c o s A = c o s a s i n B , 2 7 4 S P H E R I C A L T R I G O N O M E T R Y . i n w h i c h , s i n c e s i n B i s a l w a y s p o s i t i v e ( B < 1 8 0 ° ) , c o s A a n d c o s a m u s t h a v e t h e s a m e s i g n ; i . e . , A a n d a m u s t b e e i t h e r b o t h < o r b o t h > 9 0 ° . T h e s e c o n d f o l l o w s f r o m t h e e q u a t i o n c o s c = c o s a c o s b . 1 8 8 . A m b i g u o u s S o l u t i o n . — W h e n t h e g i v e n p a r t s o f a r i g h t t r i a n g l e a r e a s i d e a n d i t s o p p o s i t e a n g l e , t h e t r i a n g l e c a n n o t b e d e t e r m i n e d . F o r t w o r i g h t s p h e r i c a l t r i a n g l e s A B C , A ' B C , r i g h t a n g l e d a t C , m a y a l w a y s b e f o u n d , h a v i n g t h e a n g l e s A ^ a n d A ' e q u a l , a n d B C , t h e A < C ^ / ^ ~ ^ > A s i d e o p p o s i t e t h e s e a n g l e s , ^ . — " " ^ t h e s a m e i n b o t h t r i a n g l e s , b u t t h e r e m a i n i n g s i d e s , A B , A C , a n d t h e r e m a i n i n g a n g l e A B C o f t h e o n e t r i a n g l e a r e t h e s u p p l e m e n t s o f t h e r e m a i n i n g s i d e s A ' B , A ' C , a n d t h e r e m a i n i n g a n g l e A ' B C o f t h e o t h e r t r i a n g l e . I t i s t h e r e f o r e a m b i g u o u s w h e t h e r A B C o r A ' B C b e t h e t r i a n g l e r e q u i r e d . T h i s a m b i g u i t y w i l l a l s o b e f o u n d t o e x i s t , i f i t b e a t t e m p t e d t o d e t e r m i n e t h e t r i a n g l e b y t h e e q u a t i o n s i n b = t a n a c o t A , s i n c e i t c a n n o t b e d e t e r m i n e d f r o m t h i s e q u a t i o n w h e t h e r t h e s i d e A C i s t o b e t a k e n o r i t s s u p p l e m e n t A ' C . 1 8 9 . Q u a d r a n t a l T r i a n g l e s . — T h e p o l a r t r i a n g l e o f a r i g h t t r i a n g l e h a s o n e s i d e a q u a d r a n t , a n d i s t h e r e f o r e a q t u x d r a n t a l t r i a n g l e ( A r t . 1 8 4 ) . T h e f o r m u l a e f o r q u a d -r a n t a l t r i a n g l e s m a y b e o b t a i n e d b y a p p l y i n g t h e t e n f o r m u l a e o f A r t . 1 8 5 t o t h e p o l a r t r i a n g l e . T h e y a r e a s f o l l o w s , c b e i n g t h e q u a d r a n t a l s i d e : c o s C = — c o s A c o s B s i n B = s i n b s i n C . ( 1 ) ( 2 ) E X A M P L E S . 2 7 5 s i n A = s i n a s i n C ( 3 ) c o s b = — t a n A c o t C ( 4 ) c o s a = — t a n B c o t C ( 5 ) s i n A = t a n B c o t b ( 6 ) s i n B = t a n A c o t a ( 7 ) c o s C = — c o t a c o t b ( 8 ) c o s b = c o s B s i n a ( 9 ) c o s a = c o s A s i n b ( 1 0 ) E X A M P L E S . I n t h e r i g h t t r i a n g l e A B C i n w h i c h t h e a n g l e C i s t h e r i g h t a n g l e , p r o v e t h e f o l l o w i n g r e l a t i o n s : 1 . s i n 2 a + s i n 2 b — s i n 2 c = s i n 2 a s i n 2 6 . 2 . c o s 2 A s i n 2 c = s i n 2 c — s i n 2 a . 3 . s i n 2 A c o s 2 c = s i n 2 A — s i n a . 4 . s i n 2 A c o s 2 6 s i n 2 c = s i n 2 c — s i n 2 6 . 5 . 2 c o s c = c o s ( a + b ) + c o s ( a — b ) . 6 . t a n £ ( c + a ) t a n | ( c — a ) = t a n 2 £ i > . . s i n r - = s u r -c o s ' - + c o s -s i n - -2 2 2 2 2 8 . s i n ( c — 6 ) = t a n 2 — s i n ( c + 6 ) . 2 i 9 . I f b = c — p r o v e c o s a = c o s A . 1 0 . I f a = b = c , p r o v e s e c A = 1 + s e c a . 1 1 . I f c < 9 0 ° , s h o w t h a t a a n d 6 a r e o f t h e s a m e s p e c i e s . 1 2 . I f c > 9 0 ° , a a n d b a r e o f d i f f e r e n t s p e c i e s . 1 3 . A s i d e a n d t h e h y p o t e n u s e a r e o f t h e s a m e o r o p p o s i t e s p e c i e s , a c c o r d i n g a s t h e i n c l u d e d a n g l e < , o r > - . 2 7 6 S P H E R I C A L T R I G O N O M E T R Y . O B L I Q U E S P H E R I C A L T R I A N G L E S . 1 9 0 . L a w o f S i n e s . — I n a n y s p h e r i c a l t r i a n g l e t h e s i n e s o f t h e s i d e s a r e p r o p o r t i o n a l t o t h e s i n e s o f t h e o p p o s i t e a n g l e s . L e t A B C b e a s p h e r i c a l t r i a n g l e , O t h e c e n t r e o f t h e s p h e r e ; a n d l e t a , b , c d e n o t e t h e s i d e s o f t h e t r i a n g l e o p p o s i t e t h e a n g l e s A , B , C , r e s p e c t i v e l y . T h e n a , b , a n d c a r e t h e m e a s u r e s o f t h e a n g l e s B O C , C O A , a n d A O B . F r o m a n y p o i n t D i n O A d r a w D G ± t o t h e p l a n e B O C , a n d f r o m G d r a w G E , G F _ L t o O B , O C . J o i n D E , D F , a n d G O . T h e n D G i s _ L t o G E , G F , a n d G O ( G e o m . A r t . 4 8 7 ) . H e n c e , D E i s _ L t o O B , a n d D F _ L t o O C ( G e o m . A r t . 5 0 7 ) . . - . Z D E G = f Z B , a n d Z D F G = Z C . . ( A r t . 1 8 3 ) I n t h e r i g h t p l a n e t r i a n g l e s D G E , D G F / O D E , O D F , D G = D E s i n B = O D s i n D O E s i n B = O D s i n c s i n B , D G = D F s i n C = O D s i n D O F s i n C = O D s i n 6 s i n C . . - . s i n c s i n B = s i n b s i n C ; o r s i n 6 : s i n c : : s i n B : s i n C . S i m i l a r l y , i t m a y b e s h o w n t h a t s i n a : s i n c : : s i n A : s i n C . s i n a _ s i n b _ s i n c s i n A s i n B s i n C N o t e . — T h e c o m m o n v a l u e o f t h e s e t h r e e r a t i o s i s c a l l e d t h e m o d u l u s o f t h e s p h e r i c a l t r i a n g l e . S c h . I n t h e f i g u r e , B , C , b , c a r e e a c h l e s s t h a n a r i g h t a n g l e ; b u t i t w i l l b e f o u n d o n e x a m i n a t i o n t h a t t h e p r o o f w i l l h o l d w h e n t h e f i g u r e i s m o d i f i e d t o m e e t a n y c a s e w h i c h c a n o c c u r . F o r e x a m p l e , i f B a l o n e i s g r e a t e r t h a n L A W O F C O S I N E S . 2 7 7 c I V ^ — / c l \ \ w A / / / a B 9 0 ° , t h e p o i n t G w i l l f a l l o u t s i d e o f O B i n s t e a d o f b e t w e e n O B a n d O C . T h e n D E G w i l l b e t h e s u p p l e m e n t o f B , a n d t h u s w e s h a l l s t i l l h a v e s i n D E G = s i n B . 1 9 1 . L a w o f C o s i n e s . — I n a n y s p h e r i c a l t r i a n g l e , t h e c o s i n e o f e a c h s i d e i s e q u a l t o t h e p r o d u c t o f t h e c o s i n e s o f t h e o t h e r t w o s i d e s , p l u s t h e p r o d u c t o f t h e s i n e s o f t h o s e s i d e s i n t o t h e c o s i n e o f t h e i r i n c l u d e d a n g l e . L e t A B C b e a s p h e r i c a l t r i a n g l e , O t h e c e n t r e o f t h e s p h e r e , a n d a , b , c t h e s i d e s o f t h e t r i a n g l e o p p o s i t e t h e a n g l e s A , B , C , r e s p e c t i v e l y . T h e n a = Z B O C , b = Z C O A , c = Z A O B . F r o m a n y p o i n t D i n O A d r a w , i n t h e p l a n e s A O B , A O C , r e s p e c t i v e l y , t h e l i n e s D E , D F J _ t o O A . T h e n Z E D F = Z A ( A r t . 1 8 3 ) J o i n E F ; t h e n i n t h e p l a n e t r i a n g l e s E O F , E D F , w e h a v e E F 2 = O E 2 + O F 2 - 2 O E . O F c o s E O F . . ( 1 ) E F 2 = D E 2 + D F 2 - 2 D E . D F c o s E D F . . ( 2 ) a l s o i n t h e r i g h t t r i a n g l e s E O D , F O D , w e h a v e O E 2 -D E 2 = O D 2 , a n d O F 2 - D F 2 = O D 2 . ( 3 ) S u b t r a c t i n g ( 2 ) f r o m ( 1 ) , a n d r e d u c i n g b y ( 3 ) , a n d t r a n s p o s i n g , w e g e t 2 O E . O F c o s E O F = 2 O D 2 + 2 D E - D F c o s E D F . . - . c O S E O r = — — -- — — c o s h t D h , O F O E T O F O E , o r c o s a = c o s 6 c o s c + s i n b s i n c c o s A ( 4 ) 2 7 8 S P H E R I C A L T R I G O N O M E T R Y . B y t r e a t i n g t h e o t h e r e d g e s i n o r d e r i n t h e s a m e w a y , o r b y a d v a n c i n g l e t t e r s ( s e e N o t e , A r t . 9 0 ) w e g e t c o s b = c o s c c o s a + s i n c s i n a c o s B . . ( u ) c o s c = c o s a c o s b + s i n a s i n b c o s C . . ( 6 ) S c h . F o r m u l a ( 4 ) h a s b e e n p r o v e d o n l y f o r t h e c a s e i n w h i c h t h e s i d e s b a n d c a r e l e s s t h a n q u a d r a n t s ; b u t i t m a y b e s h o w n t o b e t r u e w h e n t h e s e s i d e s a r e n o t l e s s t h a n q u a d r a n t s , a s f o l l o w s : C n ' ( 1 ) S u p p o s e c i s g r e a t e r t h a n 9 0 ° . P r o d u c e B A , B C B ^ ^ 6 t o m e e t i n B ' , a n d p u t A B ' = c ' , C B ' = a ' . T h e n , f r o m t h e t r i a n g l e A B ' C , w e h a v e b y ( 4 ) c o s a ' = c o s b c o s c ' + s i n b s i n c ' c o s B ' A C , o r c o s ( i r — a ) = c o s b c o s ( i r — c ) + s i n b s i n ( i r — c ) c o s ( i r — A ) . . - . c o s a = c o s b c o s c + s i n b s i n c c o s A . ( 2 ) S u p p o s e b o t h b a n d c t o B _ b e g r e a t e r t h a n 9 0 ° . P r o d u c e A B , A C t o m e e t i n A ' , a n d p u t A ' B = c ' , A ' C = b , . b ~ C T h e n , f r o m t h e t r i a n g l e A ' B C , w e h a v e b y ( 4 ) c o s a = c o s b , c o s c ' + s i n b , s i n c ' c o s A ' ; b u t V = i r — b , c ' = i r — c , A ' = A . . - . c o s a = c o s b c o s c + s i n 6 s i n c c o s A . T h e t r i a n g l e A B ' C i s c a l l e d t h e c o l u n a r t r i a n g l e o f A B C . 1 9 2 . R e l a t i o n b e t w e e n a S i d e a n d t h e T h r e e A n g l e s . — / > ( a n y s p h e r i c a l t r i a n g l e A B C , c o s A = — c o s B c o s C + s i n B s i n C c o s a . R E L A T I O N B E T W E E N S I D E A N D A N G L E S . 2 7 9 L e t A ' B ' C b e t h e p o l a r t r i a n g l e o f A B C , a n d d e n o t e i t s a n g l e s a n d s i d e s b y A ' , B ' , C , a , V , c ' ; t h e n w e h a v e b y ( 4 ) o f A r t . 1 9 1 c o s a ' = c o s b , c o s c ' + s i n b , s i n c ' c o s A ' ; b u t a , = i r -A , V = t t -B , c ' = i r -C , e t c . . ( A r t . 1 8 4 ) H e n c e , s u b s t i t u t i n g , w e g e t c o s A = — c o s B c o s C + s i n B s i n C c o s a . . . . ( 1 ) S i m i l a r l y , c o s B = — c o s C c o s A + s i n C s i n A c o s 6 . . . . ( 2 ) c o s C = — c o s A c o s B + s i n A s i n B c o s c . . . . ( 3 ) R e m . — T h i s p r o c e s s i s c a l l e d " a p p l y i n g t h e f o r m u l a t o t h e p o l a r t r i a n g l e . " B y m e a n s o f t h e p o l a r t r i a n g l e , a n y f o r m u l a o f a s p h e r i c a l t r i a n g l e m a y b e i m m e d i a t e l y t r a n s f o r m e d i n t o a n o t h e r , i n w h i c h a n g l e s t a k e t h e p l a c e o f s i d e s , a n d s i d e s o f a n g l e s . 1 9 3 . T o s h o w t h a t i n a s p h e r i c a l t r i a n g l e A B C , c o t a s i n b = c o t A s i n C + c o s C c o s b . M u l t i p l y ( 6 ) o f A r t . 1 9 1 b y c o s b , a n d s u b s t i t u t e t h e r e s u l t i n ( 4 ) o f A r t . 1 9 1 , a n d w e g e t c o s a = c o s a c o s 2 b + s i n a s i n b c o s 6 c o s C + s i n b s i n c c o s A . T r a n s p o s e c o s a c o s 2 b , a n d d i v i d e b y s i n a s i n b ; t h u s , , -, , „ , s i n c c o s A c o t a s i n o = c o s b c o s C H s i n a = c o s b c o s C + c o t A s i n C . ( b y A r t . 1 9 0 ) B y i n t e r c h a n g i n g t h e l e t t e r s , w e o b t a i n l i v e o t h e r f o r m u l a e l i k e t h e p r e c e d i n g o n e . T h e s i x f o r m u l a e a r e a s f o l l o w s : c o t a s i n 6 = c o t A s i n C + c o s C c o s b . . . . ( 1 ) c o t a s i n c = c o t A s i n B + c o s B c o s c c o t b s i n a = c o t B s i n C + c o s C c o s a c o t b s i n c = c o t B s i n A + c o s A c o s c c o t c s i n a = c o t C s i n B + c o s B c o s a c o t c s i n b = c o t C s i n A + c o s A c o s b ( 2 ) ( 3 ) ( 4 ) ( 5 ) ( 6 ) 2 8 0 S P H E R I C A L T R I G O N O M E T R Y . E X A M P L E S . 1 . I f a , b , c b e t h e s i d e s o f a s p h e r i c a l t r i a n g l e , a , , b ' , c 1 t h e s i d e s o f i t s p o l a r t r i a n g l e , p r o v e s i n o : s i n b : s i n c = s i n a , : s i n b , : s i n c ' . 2 . I f t h e b i s e c t o r A D o f t h e a n g l e A o f a s p h e r i c a l t r i a n g l e d i v i d e t h e s i d e B C i n t o t h e s e g m e n t s C D = b ' , B D = c ' , p r o v e s i n 6 : s i n c = s i n 6 ' : s i n c ' . 3 . I f D b e a n y p o i n t o f t h e s i d e B C , p r o v e t h a t c o t A B s i n D A C + c o t A C s i n D A B = c o t A D s i n B A C . c o t A B C s i n D C + c o t A C B s i n B D = c o t A D B s i n B C . 4 . I f a , / J , y b e t h e p e r p e n d i c u l a r s o f a t r i a n g l e , p r o v e t h a t s i n a s i n « = s i n 6 s i n / 3 = s i n c s i n y . 5 . I n E x . 4 p r o v e t h a t s i n a c o s a = V c o s 2 6 + c o s 2 c — 2 c o s a c o s 6 c o s c . 1 9 4 . U s e f u l F o r m u l a e . — S e v e r a l o t h e r g r o u p s o f u s e f u l f o r m u l a e a r e e a s i l y o b t a i n e d f r o m t h o s e o f A r t . 1 9 1 ; t h e f o l l o w i n g a r e l e f t a s e x e r c i s e s f o r t h e s t u d e n t : s i n a c o s B = c o s b s i n c — s i n b c o s c c o s A -( 1 ) s i n a c o s C = s i n b c o s c — c o s b s i n c c o s A . ( 2 ) s i n b c o s A = c o s a s i n c — s i n a c o s c c o s B -( 3 ) s i n b c o s C = s i n a c o s c — c o s a s i n c c o s B . ( 4 ) s i n c c o s A = c o s a s i n 6 — s i n a c o s 6 c o s C -( 5 ) s i n c c o s B = s i n a c o s b — c o s a s i n b c o s C -( 6 ) F O R M U L A F O R T H E H A L F A N G L E S . 2 8 1 A p p l y i n g t h e s e s i x f o r m u l a e t o t h e p o l a r t r i a n g l e , w e o b t a i n t h e f o l l o w i n g s i x : s i n A c o s b = c o s B s i n C + s i n B c o s C c o s a s i n A c o s c = s i n B c o s C + c o s B s i n C c o s a s i n B c o s a = c o s A s i n C + s i n A c o s C c o s b s i n B c o s c = s i n A c o s C + c o s A s i n C c o s 6 s i n C c o s a = c o s A s i n B - + - s i n A c o s B c o s c s i n C c o s b = s i n A c o s B + c o s A s i n B c o s c ( 7 ) ( 8 ) ( 9 ) ( 1 0 ) ( 1 1 ) ( 1 2 ) 1 9 5 . F o r m u l a e f o r t h e H a l f A n g l e s . — T o e x p r e s s t h e s i n e , c o s i n e , a n d t a n g e n t o f h a l f a n a n g l e o f a s p h e r i c a l t r i a n g l e i n t e r m s o f t h e s i d e s . I . B y ( 4 ) o f A r t . 1 9 1 w e h a v e a c o s a — c o s 6 c o s e H n . . A / A . i r , \ c o s A = = 1 — 2 s i n 2 — ( A r t . 4 9 ) s i n b s i n c 2 c . 2 A c o s a — c o s b c o s c . ' . £ i S i n — - = X ; ; . 2 s i n b s i n c _ c o s ( b — c ) — c o s a s i n b s i n c . - . z j n A = s i n I ( a + ° - c ) s i n £ ( a - f e + c ) ^ 4 g ) 2 s i n b s i n c L e t 2 s = a + b + c ; s o t h a t s i s h a l f t h e s u m o f t h e s i d e s o f t h e t r i a n g l e ; t h e n a + b — c = 2 ( s — c ) , a n d a — b + c = 2 ( s — b ) . . : s i n 2 A = s i n ( s ~ 6 ) s i n ( s ~ c ) -2 s i n b s i n c 2 8 2 S P H E R I C A L T R I G O N O M E T R Y . A d v a n c i n g l e t t e r s , s i n M _ / s i n ( s - c ) s i n ( s - a ) 2 \ s i n c s i n a ( 2 ) s i n —2 _ / s i n ( 8 -a ) s i n ( s -6 ) . V s i n a s i n 6 ■ --\ ) I I . 2 c o s 2 ^ = 1 + c o s A ( A r t . 4 9 ) 2 _ ^ _ j _ c o s a — c o s b c o s c s i n f t s i n e _ c o s a — c o s ( 6 + c ) s i n b s i n e . - . c o s 2 A = s i n + 6 + c ) s i n H 6 + c ~ f f ) 2 s i n 6 s i n e s i n s s m i n ( s — a ) s i n b s i n e . - . c o s A = J s i i l s s i 1 l _ ( « - ^ 0 . . . f 4 ) 2 V s h i f t s i n e W A d v a n c i n g l e t t e r s , / s i n . q s i n ^ . e — 7 i \ . . . . ( 5 ) c o s g = - v E a i n ^ - 6 l 2 \ s i n c s i n a o o s g = J 8 i n s s i n ( s - « - - ) ( 6 ) 2 \ s i n a s i n 6 ' I I I . B y d i v i s i o n , w e o b t a i n t a n ± = / « n ( « - f t ) 8 i » ( « - c ) 2 V s i n s s i n ( s — a ) ( 7 ) t a n g _ / a i n ( « - c ) s i n ( s - q ) 2 \ s i n s s i n ( « — 6 ) v ' t a n 2 / s m ^ f f l ) s i n _ ( s - 6 ) 2 V s i n s s i n ( s — c ) v ' F O R M U L A F O R T H E H A L F A N G L E S . 2 8 3 S c h . T h e p o s i t i v e s i g n m u s t b e g i v e n t o t h e r a d i c a l s i n e a c h c a s e i n t h i s a r t i c l e , b e c a u s e | - A , J - B , £ C a r e e a c h l e s s t h a n 9 0 ° . C o r . 1 . t a n ^ t a n ? = ^ - ^ S l ( 1 0 ) 2 2 s i n s . ' , B , C s i n ( s — a ) t a n — t a n = — ' -( 1 1 ) 2 2 s i n s v y . C . A s i n ( s — b ) / 1 0 N t a n — t a n — = - — ' -( 1 2 ) 2 2 s i n s v ' A A C o r . 2 . S i n c e s i n A = 2 s i n — c o s — , 2 2 . - . s i n A = 2 V s i n s s i " ( - s — " ) s m ( g -b ) s i " ( -< Q n 3 \ s i n b s i n e = . l n . ( 1 4 ) S i l l 0 S i l l c w h e r e n 2 = s i n s s i n ( . s — a ) s i n ( s — b ) s i n ( s — c ) . E X A M P L E S . - , r , -o . 1 — c o s 2 a — o o s L 7 > — e o s 2 e + 2 c o s a c o s 6 c o s c 1 . P r o v e s i n 2 A = — — ^ . s u r i s u r e — 4 n 2 s i n 2 b s i n 2 c , w h e r e 4 h 2 = 1 — c o s 2 a — c o s 2 b — c o s 2 c + 2 c o s a c o s 6 c o s c . C C 2 . P r o v e c o s e = c o s ( a + ¿ > ) s i n 2 — h c o s ( a — 6 ) c o s 2 . -2 2 o - , , . A . B . C s i n ( s — a ) s i n ( s ^ i ) s i n ( s — c ) ó . P r o v e s m — s m - s i n = — 5 L 5 ' v L . 2 2 2 s m a s m b s m c , - , , c o s A + c o s B s i n ( a + b ) 4 . P r o v e — = \ — — ' — 1 — c o s C s i n e k i > - ¡ c o s A + c o s B . , , \ -A 5 . P r o v e 5 s i n ( a — o ) s i n e = 0 . 1 — c o s C - , , c o s A — c o s B s i n ( a ~ 6 ) 6 . P r o v e = - — — -1 4 - c o s C s i n e 2 8 4 S P H E R I C A L T R I G O N O M E T R Y . 1 9 6 . F o r m u l a e f o r t h e H a l f S i d e s . — T o e x p r e s s t h e s i n e , c o s i n e , a n d t a n g e n t o f h a l f a s i d e o f a s p h e r i c a l t r i a n g l e i n t e r m s o f t h e a n g l e s . B y ( 1 ) o f A r t . 1 9 2 , w e h a v e c o s A + c o s B c o s C - , n . . a , A . A n s c o s a = - t -= 1 — 2 s m 2 . ( A r t . 4 9 ) s i n B s m C 2 -2 s i n 2 a = = c o s - A - + c o s ( B + C ) 2 s i n B s i n C -z i n a = c o 8 j ( A + B + C ) c o s i ( B + C - A ) , A r t ^ " 2 s i n B s i n G \ -) L e t 2 S = A + B + C ; t h e n B + C -A = 2 ( S -A ) . P r o c e e d i n g i n t h e s a m e w a y a s i n A r t . 1 9 5 , w e f i n d t h e f o l l o w i n g e x p r e s s i o n s f o r t h e s i d e s , i n t e r m s o f t h e t h r e e a n g l e s : a _ I c o s S c o s ( S — A ) 2 ~ \ s i n B s i n C ( ' s i n -s i n ^ = A / - c o s S c o s ( S - B ) . . . . ( 2 ) 2 \ s i n C s i n A v ' . c s i n -2 V _ c o s S c o s ( S — C ) . o \ ~ ~ s i n A s i n B ( ' a _ I c o s ( S -B ) c o s ( S -C ) . . . 2 \ s i n B s i n C ( V c o s ( S — C ) c o s ( 8 - A ) , _ . s i n C s i n A ( ' V c o s ( S — A ) c o s ( S — B ) , „ s . s i n A s i n B ' t a n a = J - -c o s S c o s ( S - A ) -( 7 ) 2 \ c o s ( S -B ) c o s ( S -C ) b c o s -2 c i — : 2 F O R M U L A E F O R T H E H A L F S I D E S . 2 8 5 t a i ^ = J c o s S c o s ( S - B ) 2 \ c o s f S - C " ) c o s f S - A ^ V ; t a n £ = J c o s S c o s ( S - C ) ( g 2 \ c o s f S -A U o s r S -B 1 ) v ; 1 . T h e s e f o r m u l a e m a y a l s o b e o b t a i n e d i m m e d i a t e l y f r o m t h o s e o f A r t . 1 9 5 b y m e a n s o f t h e p o l a r t r i a n g l e . S c h . 2 . T h e p o s i t i v e s i g n m u s t b e g i v e n 1 t o t h e a b o v e r a d i c a l s , b e c a u s e - . - , - , a r e e a c h l e s s t h a n 9 0 ° . , 2 2 2 S c h . 3 . T h e s e v a l u e s o f t h e s i n e s , c o s i n e s , a n d t a n g e n t s o f t h e h a l f s i d e s a r e a l w a y s r e a l . F o r S i s > 9 0 ° a n d < 2 7 0 ° ( A r t . 1 8 4 ) , s o t h a t c o s S i s a l w a y s n e g a t i v e . A l s o , i n t h e p o l a r t r i a n g l e , a n y s i d e i s l e s s t h a n t h e s u m o f t h e o t h e r t w o ( A r t . 1 8 4 ) . . - . i T — A < T T — B + i T — C . . . . b + C - A < t t . . - . c o s ( S — A ) i s p o s i t i v e . S i m i l a r l y , c o s ( S — B ) a n d c o s ( S — C ) a r e p o s i t i v e . C o r . S i n c e s i n a = 2 s i n -c o s 2 2 s i n a = 2 ^ ~ c o s S c o s ( s ~ A ) c o s ( S -B ) c o s ( S - C ) ( 1 0 . s i n B s i n C 2 N s i n B s i n C w h e r e N = V -c o s S c o s ( S -A ) c o s ( S -B ) c o s ( S -C ) . 2 8 6 S P H E R I C A L T R I G O N O M E T R Y . E X A M P L E S . 1 . P r o v e c o s C = — c o s ( A + B ) c o s 2 - — c o s ( A — B ) s i n 2 - -2 2 n - o . a . b . c — N c o s S 2 . P r o v e s i n -s i n - s m - = . 2 s i n A s i n B s i n C w h e r e N « = V — c o s S c o s ( S — A ) c o s ( S — B ) c o s ( S — C ) . 1 9 7 . N a p i e r ' s A n a l o g i e s . L e t m = = ^ 5 ( A r t . 1 9 0 ) ( 1 ) s i n a s i n b s i n A + s i n B , . , , . = -7 . ( A l g e b r a ) ( 2 ) s i n a + s m o s i n A — s i n B / o x o r -t - t ----( 3 ) s i n a — s i n o c o s A + c o s B c o s C = s i n B s i n C c o s a ( A r t . 1 9 2 ) = m s i n C s i n 6 c o s a , b y ( 1 ) ( 4 ) a n d c o s B + c o s C c o s A = s i n C s i n A c o s 6 = m s i n C s i n a c o s b . . ( 5 ) . - . ( c o s A + c o s B ) ( 1 + c o s C ) = m s i n C s i n ( a + 6 ) , ( 6 ) f r o m ( 4 ) a n d ( 5 ) D i v i d i n g ( 2 ) b y ( 6 ) , s i n A + s i n B _ s i n a + s i n 6 1 + c o s C c o s A + c o s B s i n ( a + 6 ) s i n C . - . t a n \ ( A + B ) = c 0 S H « - 6 ) c o t ° ( 7 ) V ' c o s ( a + 6 ) 2 ? v ' ( A r t s . 4 5 , 4 6 , a n d 4 9 ) S i m i l a r l y , t a n \ ( A -B ) = S l n \ ( a ~ b ) c o t -. . ( 8 ) s i n \ ( a + b ) 2 D E L A M B R E ' S A N A L O G I E S . 2 8 7 W r i t i n g n — A f o r a , e t c . , b y A r t . 1 8 4 , w e o b t a i n f r o m ( 7 ) a n d ( 8 ) » ( « + » ) - ; ? t , ( i i S ° | ----w c o s £ ( A + B ) 2 t a n H a - 6 ) = s i D ( A - B ) t a n £ . . . . ( 1 0 ) v y s i n £ ( A + B ) 2 v ' # c f t . T h e f o r m u l a e ( 7 ) , ( 8 ) , ( 9 ) , ( 1 0 ) a r e k n o w n a s N a p i e r , s A n a l o g i e s , a f t e r t h e i r d i s c o v e r e r . T h e l a s t t w o m a y b e p r o v e d w i t h o u t t h e p o l a r t r i a n g l e b y s t a r t i n g w i t h t h e f o r m u l a e o f A r t . 1 9 1 . C o r . I n a n y s p h e r i c a l t r i a n g l e w h o s e p a r t s a r e p o s i t i v e , a n d l e s s t h a n 1 8 0 ° , t h e h a l f - s u m o f a n y t i v o s i d e s a n d t h e h a l f -s u m o f t h e i r o p p o s i t e a n g l e s a r e o f t h e s a m e s p e c i e s . n F o r , s i n c e c o s £ ( a — b ) a n d c o t — a r e n e c e s s a r i l y p o s i t i v e , t h e r e f o r e b y ( 7 ) t a n £ ( A + B ) a n d c o s J ( a + 6 ) a r e b o t h p o s i t i v e o r b o t h n e g a t i v e . . - . $ ( A + B ) a n d £ ( a + b ) a r e b o t h > o r b o t h < o r b o t h = 9 0 ° . 1 9 8 . D e l a m b r e ' s ( o r G a u s s ' s ) A n a l o g i e s . s i n £ ( A + B ) -A B , A . B = s i n — c o s — \ - c o s — s i n — 2 2 2 2 V s i n ( s — b ) s i n ( s — c ) / s i n s s i n ( s — b ) s i n b s i n c \ s i n c s i n a + / s i n g s i n ( . s - a ) / s i n ( s - e ) s i n ( s - a ) 1 9 5 ^ V s i n b s i n c \ s i n c s i n a _ s i n ( s — b ) + s i n ( s — a ) / s i n s s i n ( s — c ) s i n c V s i n a s i n 6 = c 0 S H « - 6 ) c o s C ( A r t s 4 5 a n d 1 9 5 ) c 2 c o s -2 2 8 8 S P H E R I C A L T R I G O N O M E T R Y . c C . - . s i n £ ( A + B ) c o s - = c o s £ ( a — b ) c o s -. . . ( 1 ) 2 2 S i m i l a r l y , w e o b t a i n t h e f o l l o w i n g t h r e e e q u a t i o n s : s i n £ ( A -B ) s i n £ = s i n c o s -. . . ( 2 ) 2 2 r C c o s £ ( A + B ) c o s j = c o s \ ( a + b ) s i n -. . . ( 3 ) 2 2 c C c o s % ( A — B ) s i n - = s i n £ ( « + b ) s i n — . . . . ( 4 ) 2 2 / S e / i , . 1 . W h e n t h e s i d e s a n d a n g l e s a r e a l l l e s s t h a n 1 8 0 ° , b o t h m e m b e r s o f t h e s e e q u a t i o n s a r e p o s i t i v e . S c h . 2 . N a p i e r , s a n a l o g i e s m a y b e o b t a i n e d f r o m D e -l a m b r e , s b y d i v i s i o n . N o t e . — D e l a m b r e ' s a n a l o g i e s w e r e d i s c o v e r e d b y h i m i n 1 8 0 7 , a n d p u b l i s h e d i n t h e C o n n a i s s a n c e d e s T e m p s f o r 1 8 0 9 , p . 4 4 3 . T h e y w e r e s u b s e q u e n t l y d i s c o v e r e d i n d e p e n d e n t l y b y G a u s s , a n d p u b l i s h e d b y h i m , a n d a r e s o m e t i m e s i m p r o p e r l y c a l l e d G a u s s ' s e q u a t i o n s . B o t h s y s t e m s m a y b e p r o v e d g e o m e t r i c a l l y . T h e g e o m e t r i c p r o o f i s t h e o n e o r i g i n a l l y g i v e n b y D e l a m b r e . I t w a s r e d i s c o v e r e d b y P r o f e s s o r C r o f t o n i n 1 8 6 9 , a n d p u b l i s h e d i n t h e P r o c e e d i n g s o f t h e L o n d o n M a t h e m a t i c a l S o c i e t y , V o l . I I I . [ C a s e y ' s T r i g o n o m e t r y , p . 4 1 ] . E X A M P L E S . I n t h e r i g h t t r i a n g l e A B C , i n w h i c h C i s t h e r i g h t a n g l e , p r o v e t h e f o l l o w i n g r e l a t i o n s i n E x s . 1 - 4 5 : 1 . s i n 2 a c o s 2 6 = s i n ( c + b ) s i n ( c — 6 ) . 2 . t a n 2 a : t a n 2 6 = s i n 2 c — s i n 2 b : s i n 2 c — s i n 2 « . 3 . c o s 2 a c o s 2 B = s i n 2 A — s i n 2 a . 4 . c o s 2 A - f c o s 2 c = c o s 2 A c o s 2 c + c o s 2 a . 5 . s i n 2 A — c o s 2 B = s i n 2 a s i n 2 B . 6 . I f o n e o f t h e s i d e s o f a r i g h t t r i a n g l e b e e q u a l t o t h e o p p o s i t e a n g l e , t h e r e m a i n i n g p a r t s a r e e a c h e q u a l t o 9 0 ° . E X A M P L E S . 2 8 9 7 . I f t h e a n g l e A o f a r i g h t t r i a n g l e b e a c u t e , s h o w t h a t t h e d i f f e r e n c e o f t h e s i d e s w h i c h c o n t a i n i t i s l e s s t h a n 9 0 ° . 8 . P r o v e t a n g = s i n < -a ) -2 s i n s 9 . P r o v e ( 1 ) 2 M = s i n a s i n 6 ; ( 2 ) 2 N = s i n a s i n B . 1 0 . P r o v e s i n 2 a s i n 2 6 = s i n 2 a . + s i n 2 6 — s i n 2 c . 1 1 . P r o v e t a n 2 A = s i " ( c - 6 ) . 2 s i n ( c + b ) 1 2 . P r o v e 2 s i n 2 - = s i n 2 £ ( a + b ) + s i n 2 £ ( a -6 ) . 2 1 3 . I n a s p h e r i c a l t r i a n g l e , i f c = 9 0 ° , p r o v e t h a t t a n a t a n 6 4 - s e c C = 0 . 1 4 . I n a s p h e r i c a l t r i a n g l e , i f c = 9 0 ° , p r o v e t h a t s i n 2 p = c o t 6 c o t < j > , w h e r e p i s t h e p e r p e n d i c u l a r o n c , a n d 0 a n d a r e t h e s e g m e n t s o f t h e v e r t i c a l a n g l e . 1 5 . S h o w t h a t t h e r a t i o o f t h e c o s i n e s o f t h e s e g m e n t s o f t h e b a s e m a d e b y t h e p e r p e n d i c u l a r f r o m t h e v e r t e x i s e q u a l t o t h e r a t i o o f t h e c o s i n e s o f t h e s i d e s . 1 6 . I f B b e t h e b i s e c t o r o f t h e h y p o t e n u s e , s h o w t h a t s i n 2 a + s i n 2 6 s i n 2 B = ' 4 c o s 2 -2 1 7 . P r o v e t a n S = c o t - c o t - . 2 2 1 8 . C o n s t r u c t a t r i a n g l e , b e i n g g i v e n t h e h y p o t e n u s e a n d ( 1 ) t h e s u m o f t h e b a s e a n g l e s , a n d ( 2 ) t h e d i f f e r e n c e o f t h e b a s e a n g l e s . 1 9 . G i v e n t h e h y p o t e n u s e a n d t h e s u m o r d i f f e r e n c e o f t h e s i d e s : c o n s t r u c t t h e t r i a n g l e . 2 9 0 S P H E R I C A L T R I G O N O M E T R Y . 2 0 . G i v e n t h e s u m o f t h e s i d e s a a n d b , a n d t h e s u m o f t h e b a s e a n g l e s : s o l v e t h e t r i a n g l e . 0 1 0 1 , 1 , . A v s i n c + s i n a + V s i n c — s i n a 2 1 . S h o w t h a t s i n — = — — — . 2 2 V s i n c 0 0 1 » / s i n ( c — b 2 2 . s i n | A = ^ \ , -» 2 c o s 0 s m 2 3 . c o s £ A = J ^ + i ) . V 2 c o s b s i n e ' 2 c o s b s i n c 2 4 . s i n ( a + 6 ) t a n $ ( A + B ) = s i n ( a — 6 ) c o t £ ( A — B ) . c o s a + c o s 6 2 5 . s i n ( A + B ) = 2 6 . s i n ( A - B ) = 1 + c o s a c o s 6 c o s b — c o s a 2 7 . c o s ( A + B ) = -2 8 . c o s ( A - B ) = 1 — c o s a c o s b s i n a s i n 6 1 + c o s a c o s 6 s i n a s i n 6 1 — c o s a c o s 6 2 9 . s i n 2 c - = s i n 2 -c o s 2 - + c o s 2 -s i n 2 - . 2 2 2 2 2 3 0 . s i n ( c — 6 ) = s i n ( c + 6 ) t a n 2 — -A . B 3 1 . s i n ( a — b ) = s i n a t a n s i n 6 t a n — V ' 2 2 3 2 . s i n ( c — a ) = c o s a s i n b t a n 5 . V ' 2 3 3 . I f A B C i s a s p h e r i c a l t r i a n g l e , r i g h t - a n g l e d a t C , a n d c o s A = c o s 2 a , s h o w t h a t i f A b e n o t a r i g h t a n g l e , 6 - ) - c = ^ i r o r Q t t , a c c o r d i n g a s b a n d c a r e b o t h < o r b o t h > i t E X A M P L E S . 2 9 1 3 4 . I f a , f t b e t h e a r c s d r a w n f r o m t h e r i g h t a n g l e r e s p e c t i v e l y p e r p e n d i c u l a r t o a n d b i s e c t i n g t h e h y p o t e n u s e c , s h o w t h a t s i n 2 ^ ( 1 + s i n 2 « ) = s i n 2 / ? . 2 3 5 . I n a t r i a n g l e , i f C b e a r i g h t a n g l e a n d D t h e m i d d l e p o i n t o f A B , s h o w t h a t 4 c o s 2 -s i n 2 C D = s i n 2 a + s i n 2 b . 2 I n a r i g h t t r i a n g l e , i f p b e t h e l e n g t h o f t h e a r c d r a w n f r o m t h e r i g h t a n g l e C p e r p e n d i c u l a r t o t h e h y p o t e n u s e A B , p r o v e : 3 6 . c o t 2 / > = c o t 2 a + c o t 2 b . 3 7 . c o s 2 / } = c o s 2 A + c o s 2 B . 3 8 . t a n 2 a = T t a n a ' t a n c . 3 9 . t a n 2 b = ± t a n b ' t a n c . 4 0 . t a n 2 o : t a n 2 b = t a n a , : t a n b ' . 4 1 . s i n 2 / > = s i n a 1 s i n b , . 4 2 . s i n / > s i n c = s i n a s i n b . 4 3 . t a n a t a n b = t a n c s i n p . 4 4 . t a n 2 a + t a n 2 6 = t a n 2 c c o s 2 / > . 4 5 . c o t A : c o t B = s i n a , : s i n b ' . I n t h e o b l i q u e t r i a n g l e A B C , p r o v e t h e f o l l o w i n g : 4 6 . I f t h e d i f f e r e n c e b e t w e e n a n y t w o a n g l e s o f a t r i a n g l e i s 9 0 ° , t h e r e m a i n i n g a n g l e i s l e s s t h a n 9 0 ° . 4 7 . I f a t r i a n g l e i s e q u i l a t e r a l o r i s o s c e l e s , i t s p o l a r t r i a n g l e i s e q u i l a t e r a l o r i s o s c e l e s . 4 8 . I f t h e s i d e s o f a t r i a n g l e a r e e a c h ^ , f i n d t h e s i d e s o f t h e p o l a r t r i a n g l e . 2 9 2 S P H E R I C A L T R I G O N O M E T R Y . 4 9 . I f i n a t r i a n g l e t h e s i d e a = 9 0 ° , s h o w t h a t c o s A + c o s B c o s C = 0 . 5 0 . I f 6 a n d f f a r e t h e a n g l e s w h i c h t h e i n t e r n a l a n d e x t e r n a l b i s e c t o r s o f t h e v e r t i c a l a n g l e o f a t r i a n g l e m a k e w i t h t h e b a s e , s h o w t h a t . c o s A ~ c o s B , c o s A + c o s B c o s 6 = , a n d c o s 6 ' 2 c o s -2 s i n ^ 2 2 5 1 . G i v e n t h e b a s e c a n d c o s — = — c o s C : f i n d t h e l o c u s o f t h e v e r t e x . c o s 5 2 . P r o v e 4 N 2 = 1 -c o s 2 A -c o s 2 B -c o s 2 C — 2 c o s A c o s B c o s C . 5 3 . I f p , q , r b e t h e p e r p e n d i c u l a r s f r o m t h e v e r t i c e s o n t h e o p p o s i t e s i d e s , s h o w t h a t ( 1 ) s i n a s h i p = s i n b s i n q = s i n c s i n r = 2 n . ( 2 ) s i n A s i n p = s i n B s i n < 7 = s i n C s i n r = 2 N . 5 4 . P r o v e 8 n 3 = s i n 2 a s i n 2 b s i n 2 c s i n A s i n B s i n C . 5 5 P r o v e s 1 n 2 ^ + s i n 2 B + s i n 2 C _ 1 + c o s A c o s B c o s C s i n 2 a + s i n 2 6 + s i n 2 c 1 — c o s a c o s b c o s c 5 6 . I f I b e t h e l e n g t h o f t h e a r c j o i n i n g t h e m i d d l e p o i n t o f t h e b a s e t o t h e v e r t e x , f i n d a n e x p r e s s i o n f o r i t s l e n g t l i i n t e r m s o f t h e s i d e s . . , c o s a + c o s b A n s . c o s Z = o c 2 c o s -2 5 7 . I f C D , C D ' a r e t h e i n t e r n a l a n d e x t e r n a l b i s e c t o r s o f t h e a n g l e C o f a t r i a n g l e , p r o v e t h a t . n T . c o t a + c o t 6 , . c o t a ~ c o t 6 c o t C D = — , a n d c o t C D ' = 2 c o s — 2 s i n — 2 2 E X A M P L E S . 2 9 3 5 8 . S h o w t h a t t h e a n g l e s 6 a n d m a d e b y t h e b i s e c t o r s o f t h e a n g l e C i n E x . 5 5 w i t h t h e o p p o s i t e s i d e c , a r e t h u s g i v e n : , . c o t a — c o t b . „ _ c o t 6 = — s i n C D , 2 s i n ^2 . . . c o t a + c o t b . c o t 6 ' = y < — s i n C D -2 c o s -5 9 . S h o w t h a t t h e a r c i n t e r c e p t e d o n t h e b a s e b y t h e b i s e c t o r s i n E x . 5 5 i s t h u s g i v e n : , T v r . , s i n 2 A — s i n 2 B c o t D D ' = 2 s i n A s i n B s i n C 6 0 . P r o v e t h a t c o s 2 6 — c o s 2 c c o s 2 c — c o s c o s b c o t B — c o s c c o t C c o s c c o t C — c o s a c o t A c o s 2 a — c o s 2 b c o s a c o t A — c o s b c o s B 6 1 . I f s a n d s , a r e t h e s e g m e n t s o f t h e b a s e m a d e b y t h e p e r p e n d i c u l a r f r o m t h e v e r t e x , a n d m a n d m , t h o s e m a d e b y t h e b i s e c t o r o f t h e v e r t i c a l a n g l e , s h o w t h a t . s — s ' m — i t ' . „ a — b t a n t a n = t a n ' . -2 2 2 6 2 . P r o v e s i n b s i n c + c o s b c o s c c o s A = s i n B s i n C — c o s B c o s C c o s a . 6 3 . S h o w t h a t t h e a r c I j o i n i n g t h e m i d d l e p o i n t s o f t h e t w o s i d e s a a n d 6 o f a t r i a n g l e i s t h u s g i v e n : . 1 + c o s a + c o s b + c o s e c o s I = . a b 4 c o s c o s -2 9 4 S P H E R I C A L T R I G O N O M E T R Y . 6 4 . I f t h e s i d e c o f a t r i a n g l e b e 9 0 ° , a n d 8 t h e a r c d r a w n a t r i g h t a n g l e s t o i t f r o m t h e o p p o s i t e v e r t e x , s h o w t h a t c o t 2 8 = c o t 2 A + c o t 2 B . 6 5 . P r o v e t h a t t h e a n g l e b e t w e e n t h e p e r p e n d i c u l a r f r o m t h e v e r t e x o n t h e b a s e a n d t h e b i s e c t o r o f t h e v e r t i c a l a n g l e i s t h u s g i v e n : t a n = t a n ( A -B ) . r c o s £ ( a + f c ) 7 V ' 6 6 . I n a n i s o s c e l e s t r i a n g l e , i f e a c h o f t h e b a s e a n g l e s b e d o u b l e t h e v e r t i c a l a n g l e , p r o v e t h a t a ( , a c o s a c o s - = c o s f c + -2 \ 2 6 7 . I f a s i d e c o f a t r i a n g l e b e 9 0 ° , s h o w t h a t ( 1 ) c o t a c o t b + c o s C = 0 . ( 2 ) c o s S c o s ( S -C ) + c o s ( S -A ) c o s ( S -B ) = 0 . 6 8 . I n a n y t r i a n g l e p r o v e c o s a — c o s 6 s i n ( A — B ) _ q 1 — c o s c s i n C 6 9 . t a n $ ( A + B ) : t a n £ ( A -B ) = t a n £ ( a + b ) : t a n £ ( a — 6 ) . 7 0 . t a n £ ( A + a ) : t a n £ ( A — a ) = t a n % ( B + 6 ) : t a n \ ( B -6 ) . 7 1 . I f t h e b i s e c t o r o f t h e e x t e r i o r a n g l e , f o r m e d b y p r o d u c i n g B A t h r o u g h A , m e e t t h e b a s e B C i n D ' , a n d i f B D = c " , C D ' = b " , p r o v e s i n b : s i n c = s i n b " : s i n c " . 7 2 . I f D b e a n y p o i n t i n t h e s i d e B C o f a t r i a n g l e , p r o v e s i n B D _ s i n B A D s i n C s i n C D s i n C A D s i n B E X A M P L E S . 2 9 5 7 3 . I f A = a , s h o w t h a t B a n d b a r e e i t h e r e q u a l o r s u p p l e m e n t a l , a s a l s o C a n d c . 7 4 . I f A = B + C , a n d D b e t h e m i d d l e p o i n t o f a , s h o w t h a t a = 2 A D . 7 5 . ' W h e n d o e s t h e p o l a r t r i a n g l e c o i n c i d e w i t h t h e p r i m i t i v e t r i a n g l e ? 7 6 . I f D b e t h e m i d d l e p o i n t o f c , s h o w t h a t c o s a + c o s b = 2 c o s c c o s C D . 2 7 7 . I n a n e q u i l a t e r a l t r i a n g l e s h o w t h a t ( 1 ) 2 c o s -s i n — = 1 . w 2 2 ( 2 ) t a n 2 ^ + 2 c o s A = l . 7 8 . I f b + c = i T , s h o w t h a t s i n 2 B + s i n 2 C = 0 . 7 9 . S h o w t h a t s i n b s i n c + c o s 6 c o s c c o s A = s i n B s i n C — c o s B c o s C c o s a . 8 0 . I f D b e a n y p o i n t i n t h e s i d e B C o f a t r i a n g l e , s h o w t h a t c o s A D s i n a = c o s c s i n D C + c o s b s i n B D . 8 1 . P r o v e c o s 2 - = c o s 2 £ ( a + 6 ) s i n 2 ~ + c o s 2 £ ( a — 6 ) c o s 2 ^ -2 ^ 2 8 2 . " s i n 2 5 = s i n 2 ^ ( a + ^ s i n 2 ^ + s i n 2 £ ( a - 6 ) c o s 2 p . 2 2 2 8 3 . " s i n s s i n ( s — a ) s i n ( s — b ) s i n ( s — c ) = £ ( 1 — c o s 2 a — c o s 2 b — c o s 2 c + 2 c o s a c o s 6 c o s c ) . 8 4 . I f A D b e t h e b i s e c t o r o f t h e a n g l e A , p r o v e t h a t ( 1 ) c o s B + c o s C = 2 s i n — s i n A D B c o s A D . ( 2 ) c o s C — c o s B = 2 c o s ( c o s A D B . 2 2 9 6 S P H E R I C A L T R I G O N O M E T R Y . 8 5 . P r o v e c o s a s i n b = s i n a o o s b c o s C + c o s A s i n c . 8 6 . " s i n C c o s a = c o s A s i n B + s i n A c o s B c o s C . 8 7 . I n a t r i a n g l e i f A = - , B = - , C = ^ , s h o w t h a t 5 3 2 a + b + c = - -2 0 0 „ . . „ . . 1 + c o s a — c o s 6 — c o s e 8 8 . P r o v e s i n ( S — A ) = ; v ' . a . b . c 4 c o s - s i n - s m -2 2 2 8 9 . I f 8 b e t h e l e n g t h o f t h e a r c f r o m t h e v e r t e x o f a n i s o s c e l e s t r i a n g l e , d i v i d i n g t h e b a s e i n t o s e g m e n t s a a n d / ? , p r o v e t h a t t a n - t a n ^ = t a n a — - t a n a — ^ -2 2 2 2 9 0 . I f 6 = c , s h o w t h a t . a A s m -c o s — 2 2 s i n 6 -t - , a n d s i n B = . A a s i n — c o s -2 2 9 1 . I f A B , A C b e p r o d u c e d t o B ' , C , s o t h a t B B ' , C C s h a l l b e t h e s e m i - s u p p l e m e n t s o f A B , A C r e s p e c t i v e l y , p r o v e t h a t t h e a r c B ' C w i l l s u b t e n d a n a n g l e a t t h e c e n t r e o f t h e s p h e r e e q u a l t o t h e a n g l e b e t w e e n t h e c h o r d s o f A B , A C . P R E L I M I N A R Y O B S E R V A T I O N S . 2 9 7 C H A P T E R X I . S O L U T I O N O P S P H E R I C A L T R I A N G L E S . 1 9 9 . P r e l i m i n a r y O b s e r v a t i o n s . — I n e v e r y s p h e r i c a l t r i a n g l e t h e r e a r e s i x e l e m e n t s , t h e t h r e e s i d e s a n d t h e t h r e e a n g l e s , b e s i d e s t h e r a d i u s o f t h e s p h e r e , w h i c h i s s u p p o s e d c o n s t a n t . T h e s o l u t i o n o f s p h e r i c a l t r i a n g l e s i s t h e p r o c e s s b y w h i c h , w h e n t h e v a l u e s o f a n y t h r e e e l e m e n t s a r e g i v e n , w e c a l c u l a t e t h e v a l u e s o f t h e r e m a i n i n g t h r e e ( A r t . 1 8 4 , N o t e ) . I n m a k i n g t h e c a l c u l a t i o n s , a t t e n t i o n m u s t b e p a i d t o t h e a l g e b r a i c s i g n s o f t h e f u n c t i o n s . W h e n a n g l e s g r e a t e r t h a n 9 0 ° o c c u r i n c a l c u l a t i o n , w e r e p l a c e t h e m b y t h e i r s u p p l e m e n t s ; a n d i f t h e f u n c t i o n s o f s u c h a n g l e s b e e i t h e r c o s i n e , t a n g e n t , c o t a n g e n t , o r s e c a n t , w e t a k e a c c o u n t o f t h e c h a n g e o f s i g n . I t i s n e c e s s a r y t o a v o i d t h e c a l c u l a t i o n o f v e r y s m a l l a n g l e s b y t h e i r c o s i n e s , o r o f a n g l e s n e a r 9 0 ° b y t h e i r s i n e s , f o r t h e i r t a b u l a r d i f f e r e n c e s v a r y t o o s l o w l y ( A r t . 8 1 ) . I t i s b e t t e r t o d e t e r m i n e s u c h a n g l e s , f o r e x a m p l e , b y m e a n s o f t h e i r t a n g e n t s . W e s h a l l b e g i n w i t h t h e r i g h t t r i a n g l e ; h e r e t w o e l e m e n t s , i n a d d i t i o n t o t h e r i g h t a n g l e , w i l l b e s u p p o s e d k n o w n . S O L U T I O N O F R I G H T S P H E R I C A L T R I A N G L E S . 2 0 0 . T h e S o l u t i o n o f R i g h t S p h e r i c a l T r i a n g l e s p r e s e n t s S i x C a s e s , w h i c h m a y b e s o l v e d b y t h e f o r m u l a e o f A r t . 1 8 5 . I f t h e f o r m u l a r e q u i r e d f o r a n y c a s e b e n o t r e m e m b e r e d , i t i s a l w a y s e a s y t o f i n d i t b y N a p i e r , s R u l e s ( A r t . 2 9 8 S P H E R I C A L T R I G O N O M E T R Y . 1 8 G ) . I n a p p l y i n g t h e s e r u l e s , w e m u s t c h o o s e t h e m i d d l e p a r t a s f o l l o w s : W h e n t h e t h r e e p a r t s c o n s i d e r e d a r e a l l a d j a c e n t , t h e o n e b e t w e e n i s , o f c o u r s e , t h e m i d d l e p a r t . W h e n o n l y t w o a r e a d j a c e n t , t h e o t h e r o n e i s t h e m i d d l e p a r t . L e t A B C b e a s p h e r i c a l t r i a n g l e , r i g h t - a n g l e d a t C , a n d l e t a , b , c d e n o t e t h e s i d e s o p p o s i t e t h e a n g l e s A , B , C , r e s p e c t i v e l y . W e s h a l l a s s u m e t h a t t h e p a r t s a r e a l l p o s i t i v e a n d l e s s t h a n 1 8 0 ° ( A r t . 1 8 2 ) . 2 0 1 . C a s e I . — G i v e n t h e h y p o t e n u s e c a n d a n a n g l e A ; t o f i n d a , b , B . B y ( . ' 5 ) , ( 5 ) , a n d ( 8 ) o f A r t . 1 8 5 , o r b y N a p i e r , s R u l e s , w e h a v e s i n a = s i n c s i n A , t a n b = t a n c c o s A , c o t B = c o s c t a n A . S i n c e a i s f o u n d b y i t s s i n e , i t w o u l d b e a m b i g u o u s , b u t t h e a m b i g u i t y i s r e m o v e d b e c a u s e a a n d A a r e o f t h e s a m e s p e c i e s [ A r t . 1 8 7 , ( 1 ) ] . B a n d b a r e d e t e r m i n e d i m m e d i a t e l y w i t h o u t a m b i g u i t y . I f o b e v e r y n e a r 9 0 ° , w e c o m m e n c e b y c a l c u l a t i n g t h e v a l u e s o f 6 a n d B , a n d t h e n d e t e r m i n e a b y e i t h e r o f t h e f o r m u l a e t a n a = s i n b t a n A , t a n a = t a n c c o s B . C h e c k . — A s a f i n a l s t e p , i n o r d e r t o g u a r d a g a i n s t n u m e r i c a l e r r o r s , i t i s o f t e n e x p e d i e n t t o c h e c k t h e l o g a r i t h m i c w o r k , w h i c h m a y b e d o n e i n e v e r y c a s e w i t h o u t t h e n e c e s s i t y o f n e w l o g a r i t h m s . T o c h e c k t h e w o r k , w e m a k e u p a f o r m u l a b e t w e e n t h e t h r e e r e q u i r e d p a r t s , a n d s e e w h e t h e r C A S E I I . 2 9 9 i t i s s a t i s f i e d b y t h e r e s u l t s . I n t h e p r e s e n t c a s e , w h e n t h e t h r e e p a r t s a , b , B h a v e b e e n f o u n d , t h e c J i e c k f o r m u l a i s s i n a = t a n 6 c o t B . . . . [ ( 6 ) o f A r t . 1 8 5 ] E x . 1 . G i v e n c = 8 1 ° 2 9 ' 3 2 " , A = 3 2 ° 1 8 ' 1 7 " ; f i n d a , b , B . S o l u t i o n . l o g s i n c l o g s i n A l o g s i n a . : a : 9 . 9 9 5 1 9 4 5 : 9 . 7 2 7 8 8 4 3 : 9 . 7 2 3 0 7 8 8 : 3 1 ° 5 4 ' 2 5 " . l o g c o s e = 9 . 1 7 0 0 9 6 0 l o g t a n A = 9 . 8 0 0 9 1 5 7 l o g c o t B = 8 . 9 7 1 0 1 1 7 . - . B = 8 4 ° 3 9 ' 2 1 " . 3 3 . l o g t a n c = 1 0 . 8 2 5 0 9 8 2 l o g c o s A = 9 . 9 2 6 9 6 8 7 l o g t a n b = 1 0 . 7 5 2 0 6 6 9 . - . b = 7 9 ° 5 1 ' 4 8 " . 6 5 . C h e c k . l o g t a n b = 1 0 . 7 5 2 0 6 6 9 l o g c o t B = 8 . 9 7 1 0 1 1 7 l o g s i n a = 9 . 7 2 3 0 7 8 6 ' E x . 2 . G i v e n c = 1 1 0 ° 4 6 ' 2 0 " , A = 8 0 ° 1 0 ' 3 0 " ; f i n d a , 6 , B . A n s . a = 6 7 ° 5 ' 5 2 " . 7 , b = 1 5 5 ° 4 6 ' 4 2 " . 7 , B = 1 5 3 ° 5 8 ' 2 4 " . 5 . 2 0 2 . C a s e I I . — G i v e n t h e h y p o t e n u s e c a n d a s i d e a ; t o f i n d b , A , B . B y ( 1 ) , ( 3 ) , ( 4 ) o f A r t . 1 8 5 , o r b y N a p i e r , s E u l e s , w e h a v e , c o s c -. s i n a r , t a n a c o s o = , s i n A = — — , c o s B = c o s a s i n c t a n c T h e c h e c k f o r m u l a i n v o l v e s b , A , B ; t h e r e f o r e , f r o m ( 9 ) o f A r t . 1 8 5 w e h a v e c o s B = s i n A c o s b . I n t h i s c a s e t h e r e i s a n a p p a r e n t a m b i g u i t y i n t h e v a l u e o f A , b u t t h i s i s r e m o v e d b y c o n s i d e r i n g t h a t A a n d a a r e a l w a y s o f t h e s a m e s p e c i e s ( A r t . 1 8 7 ) . 3 0 0 S P H E R I C A L T R I G O N O M E T R Y . E x . 1 . G i v e n c = 1 4 0 ° , a = 2 0 ° ; f i n d b , A , B . S o l u t i o n . l o g c o s e = 9 . 8 8 4 2 5 4 0 -c o l o g c o s a = 0 . 0 2 7 0 1 4 2 l o g c o s 6 = 9 . 9 1 1 2 6 8 2 -. - . b = 1 4 4 ° 3 6 ' 2 8 " . 4 . l o g t a n a = 9 . 5 6 1 0 6 5 9 c o l o g t a n c = 0 . 0 7 6 1 8 6 5 -l o g c o s B = 9 . 6 3 7 2 5 2 4 -. - . B = 1 1 5 ° 4 2 ' 2 3 " . 8 . l o g s i n a = 9 . 5 3 4 0 5 1 7 c o l o g s i n e = 0 . 1 9 1 9 3 2 5 l o g s i n A = 9 . 7 2 5 9 8 4 2 . - . A = 3 2 ° 8 ' 4 8 " . l . C h e c k . l o g s i n A = 9 . 7 2 5 9 8 4 2 l o g c o s b = 9 . 9 1 1 2 6 8 2 l o g c o s B = 9 . 6 3 7 2 5 2 4 E x . 2 . G i v e n c = 7 2 ° 3 0 ' , o = 4 5 ° 1 5 ' ; f i n d b , A , B . A n s . b = 6 4 ° 4 2 ' 5 2 " , A = 4 8 ° V 4 4 " . 5 , B = 7 1 ° 2 7 ' 1 5 " . 2 0 3 . C a s e I I I . — G i v e n a s i d e a a n d t h e a d j a c e n t a n g l e B ; t o f i n d A , b , c . B y ( 1 0 ) , ( 6 ) , ( 4 ) o f A r t . 1 8 5 , w e h a v e c o s A = c o s a s i n B , t a n 6 = s i n a t a n B , c o t c = c o t a c o s B . C h e c k f o r m u l a , c o s A = t a n b c o t c . I n t h i s c a s e t h e r e i s e v i d e n t l y n o a m b i g u i t y . E x . 1 . G i v e n a = 3 1 ° 2 0 ' 4 5 " , B = 5 5 ° 3 0 ' 3 0 " ; f i n d A , b , c . S o l u t i o n . l o g c o s a = 9 . 9 3 1 4 7 9 7 l o g s i n B = 9 . 9 1 6 0 3 7 1 l o g c o s A = 9 . 8 4 7 5 1 6 8 . - . A = 4 5 ° 1 5 ' 3 0 " . 6 . l o g c o t a = 0 . 2 1 5 3 0 7 3 l o g c o s B = 9 . 7 5 3 0 3 6 1 l o g c o t c = 9 . 9 6 8 3 4 3 4 . - . c = 4 7 ° 5 ' 1 1 " . l o g s i n a = 9 . 7 1 6 1 7 2 4 l o g t a n B = 0 . 1 6 3 0 0 1 0 l o g t a n b = 9 . 8 7 9 1 7 3 4 . - . 6 = 3 7 ° 7 ' 5 0 " . C h e c k . l o g t a n b = 9 . 8 7 9 1 7 3 4 l o g c o t e = 9 . 9 6 8 3 4 3 4 l o g c o s A = 9 . 8 4 7 5 1 6 8 C A S E I V . 3 0 1 E x . 2 . G i v e n a = 1 1 2 ° 0 ' 0 " , B = 1 5 2 ° 2 3 ' 1 " . 3 ; f i n d A , b , c . A n s . A = 1 0 0 ° , b = 1 5 4 ° V 2 6 " . 5 , c = 7 0 ° 1 8 ' 1 0 " . 2 . 2 0 4 . C a s e I V . — G i v e n a s i d e a a n d t h e o p p o s i t e a n g l e A ; t o f i n d b , c , B . B J ( 7 ) > ( 3 ) , ( 1 0 ) o f A r t . 1 8 5 , w e h a v e . -, . . . s i n a . - p . c o s A s i n o = t a n a c o t A , s i n c = , s i n B = s i n A c o s a C h e c k f o r m u l a , s i n b = s i n c s i n B . I n t h i s c a s e t h e r e i s a n a m b i g u i t y , a s t h e p a r t s a r e d e t e r m i n e d b y t h e i r s i n e s , a n d t w o v a l u e s f o r e a c h a r e i n g e n e r a l a d m i s s i b l e . B u t f o r e a c h v a l u e o f b t h e r e w i l l , i n g e n e r a l , b e o n l y o n e v a l u e f o r c , s i n c e c a n d 6 a r e c o n n e c t e d b y t h e r e l a t i o n c o s c = c o s a c o s 6 ( A r t . 1 8 5 ) ; a n d a t t h e s a m e t i m e o n l y o n e a d m i s s i b l e v a l u e f o r B , s i n c e c o s c = c o t A c o t B . H e n c e t h e r e w i l l b e , i n g e n e r a l , o n l y t w o t r i a n g l e s h a v i n g t h e g i v e n p a r t s , e x c e p t w h e n t h e s i d e a i s a q u a d r a n t a n d t h e a n g l e A i s a l s o 9 0 ° , i n w h i c h c a s e t h e s o l u t i o n b e c o m e s i n d e t e r m i n a t e . I t i s a l s o e a s i l y s e e n f r o m a f i g u r e t h a t t h e a m b i g u i t y m u s t o c c u r i n g e n e r a l ( A r t . 1 8 8 ) . W h e n a = A , t h e f o r m u l a e , a n d a l s o t h e f i g u r e , s h o w t h a t b , c , a n d B a r e e a c h 9 0 ° . E x . 1 . G i v e n a = 4 6 ° 4 5 ' , A = 5 9 ° 1 2 ' ; f i n d b , c , B . S o l u t i o n . l o g t a n a l o g c o t A l o g s i n b 0 . 0 2 6 5 4 6 1 9 . 7 7 5 3 3 3 4 9 . 8 0 1 8 7 9 5 l o g s i n a l o g s i n A l o g s i n c 9 . 8 6 2 3 5 2 6 9 . 9 3 3 9 7 2 9 9 . 9 2 8 3 7 9 7 b = 3 9 ° 1 9 ' 2 3 " . 5 , o r 1 4 0 ° 4 0 ' 3 6 " . 5 . c : = 5 7 ° 5 9 ' 2 9 " , o r 1 2 2 ° 0 ' 3 1 " . 3 0 2 S P H E R I C A L T R I G O N O M E T R Y . l o g c o s A = 9 . 7 0 9 3 0 0 3 l o g c o s « = 9 . 8 3 5 8 0 0 6 l o g s i n e = 9 . 9 2 8 3 7 9 7 l o g s i n B = 9 . S 7 3 4 9 9 7 C h e c k . l o g s i n B = 9 . 8 7 3 4 9 9 7 . - . B = 4 8 ° 2 1 ' 2 8 " , l o g s i n 6 = 9 . 8 0 1 8 7 9 4 o r 1 3 1 ° 3 8 ' 3 2 " . | E x . 2 . G i v e n a = 1 1 2 ° , A = 1 0 0 ° ; f i n d b , c , B . n s . 6 = 1 5 4 ° 7 ' 2 6 " . 5 , c = 7 0 ° 1 8 ' 1 0 " . 2 , B = 1 5 2 ° 2 3 ' 1 " . 3 , o r 2 5 ° 5 2 ' 3 3 " . 5 , o r 1 0 9 ° 4 1 ' 4 9 " . 8 , o r 2 7 ° 3 6 ' 5 8 " . 7 . 2 0 5 . C a s e V . — G i v e n t h e t w o s i d e s a a n d b ; t o f i n d A , B y ( 7 ) , ( 6 ) , ( 1 ) o f A r t . 1 8 5 , w e h a v e c o t A = c o t a s i n b , c o t B = c o t b s i n a , c o s c = c o s a c o s b . I n t h i s c a s e t h e r e i s n o a m b i g u i t y . E x . 1 . G i v e n a = 5 4 ° 1 6 ' , b = 3 3 ° 1 2 ' ; f i n d A , B , c . A n s . A = 6 8 ° 2 9 ' 5 3 " , B = 3 8 ° 5 2 ' 2 6 " , c = 6 0 ° 4 4 ' 4 6 " . E x . 2 . G i v e n a = 5 6 ° 3 4 ' , 6 = 2 7 ° 1 8 ' ; f i n d A , B , c . A n s . A = 7 3 ° 9 ' 1 3 " , B = 3 1 ° 4 4 ' 9 " , c = 6 0 ° 4 1 ' 9 " . 2 0 6 . C a s e V I . — G i v e n t h e t w o a n g l e s A a n d B ; t o f i n d a , C h e c k f o r m u l a , c o s c = c o s a c o s b . T h e r e i s n o a m b i g u i t y i n t h i s c a s e . E x . 1 . G i v e n A = 7 4 ° 1 5 ' , B = 3 2 ° 1 0 ' ; f i n d a , b , c . A n s . a = 5 9 ° 2 0 ' 4 4 " , b = 2 8 ° 2 4 ' 5 4 " , c = 6 3 ° 2 1 ' 2 4 " . 5 . E x . 2 . G i v e n A = 9 1 ° 1 1 ' , B = 1 1 1 ° 1 1 ' ; f i n d a , b , c . A n s . a = 9 1 ° 1 6 ' 8 " , b = 1 1 1 ° 1 1 ' 1 6 " . c = 8 9 ° 3 2 ' 2 9 " . B , c . C h e c k f o r m u l a , c o s c = c o t A c o t B . c o s B s i n A ' ' , c o s c = c o t A c o t B . E X A M P L E S . 3 0 3 2 0 7 . Q u a d r a n t a l a n d I s o s c e l e s T r i a n g l e s . — S i n c e t h e p o l a r t r i a n g l e o f a q u a d r a n t a l t r i a n g l e i s a r i g h t t r i a n g l e ( A r t . 1 8 4 ) , w e h a v e o n l y t o s o l v e t h e p o l a r t r i a n g l e b y t h e f o r m u l a e o f A r t . 1 8 5 , a n d t a k e t h e s u p p l e m e n t s o f t h e p a r t s t h u s f o u n d f o r t h e r e q u i r e d p a r t s o f t h e g i v e n t r i a n g l e ; o r w e c a n s o l v e t h e q u a d r a n t a l t r i a n g l e i m m e d i a t e l y b y t h e f o r m u l a e o f A r t . 1 8 9 . A b i q u a d r a n t a l t r i a n g l e i s i n d e t e r m i n a t e u n l e s s e i t h e r t h e b a s e o r t h e v e r t i c a l a n g l e b e g i v e n . A n i s o s c e l e s t r i a n g l e i s e a s i l y s o l v e d b y d i v i d i n g i t i n t o t w o e q u a l r i g h t t r i a n g l e s b y d r a w i n g a n a r c f r o m t h e v e r t e x t o t h e m i d d l e o f t h e b a s e . T h e s o l u t i o n o f t r i a n g l e s i n w h i c h a + b = i t , o r A + B = i t , c a n b e m a d e t o d e p e n d o n t h e s o l u t i o n o f r i g h t t r i a n g l e s . T h u s ( s e e t h e s e c o n d f i g u r e o f A r t . 1 9 1 ) t h e t r i a n g l e B ' A C h a s t h e t w o e q u a l s i d e s , a ' a n d b , g i v e n , o r t h e t w o e q u a l a n g l e s , A a n d B ' , g i v e n , a c c o r d i n g a s a + 6 = i r o r A + B = i r i n t h e t r i a n g l e A B C . ' E X A M P L E S . S o l v e t h e f o l l o w i n g r i g h t t r i a n g l e s : 1 . G i v e n c = 3 2 ° 3 4 ' , a = 4 4 ° 4 4 ' ; f i n d a = 2 2 ° 1 5 ' 4 3 " , b = 2 4 ° 2 4 ' 1 9 " , B = 5 0 ° 8 ' 2 1 " . 2 . G i v e n c = 6 9 ° 2 5 ' l l " , A = 5 4 ° 5 4 ' 4 2 " ; f i n d o = 5 0 ° 0 ' 0 " , 6 = 5 6 ° 5 0 ' 4 9 " , B = 6 3 ° 2 5 ' 4 " . 3 . G i v e n c = 5 5 ° 9 ' 3 2 " , a = 2 2 ° 1 5 ' 7 " ; f i n d 6 = 5 1 ° 5 3 ' , A = 2 7 ° 2 8 ' 3 7 ' . 5 , B = 7 3 ° 2 7 ' l l " . l 4 . G i v e n c = 1 2 7 ° 1 2 ' , a = 1 4 1 ° l l ' ; f i n d 6 = 3 9 ° 6 ' 2 5 " , A = 1 2 8 ° 5 ' 5 4 " , B = 5 2 ° 2 1 ' 4 9 " . 5 . G i v e n a = 1 1 8 ° 5 4 ' , B = 1 2 ° 1 9 ' ; f i n d A = 9 5 ° 5 5 ' 2 " , 6 = 1 0 ° 4 9 ' 1 7 " , c = 1 1 8 ° 2 0 ' 2 0 " . Q u a d r a n t a l t r i a n g l e s a r e g e n e r a l l y a v o i d e d i n p r a c t i c e , b u t w h e n u n a v o i d a b l e , t h e y a r e r e a d i l y s o l v e d b y e i t h e r o f t h e s e m e t h o d s . 3 0 4 S P H E R I C A L T R I G O N O M E T R Y . 6 . G i v e n a = 2 9 ° 4 6 ' 8 " , B = 1 3 7 ° 2 4 ' 2 1 " j f i n d A = 5 4 ° 1 ' 1 6 " , 6 = 1 5 5 ° 2 7 ' 5 4 " , c = 1 4 2 ° 9 ' 1 3 " . 7 . G i v e n a = 7 7 ° 2 1 ' 5 0 " , A = 8 3 ° 5 6 ' 4 0 " ; f i n d 6 = 2 8 ° 1 4 ' 3 1 " . l , c = 7 8 ° 5 3 ' 2 0 " , B = 2 8 ° 4 9 ' 5 7 " . 4 , o r b ' = 1 5 1 ° 4 5 ' 2 8 " . 9 , c ' = 1 0 1 ° 6 ' 4 0 " , B ' = 1 5 1 ° 1 0 ' 2 " . 6 . 8 . G i v e n a = 6 8 ° , A = 8 0 ° ; f i n d 6 = 2 5 ° 5 2 ' 3 3 " . 5 , c = 7 0 ° 1 8 ' 1 0 " . 2 , B = 2 7 ° 3 6 ' 5 8 " . 7 , o r 6 ' = 1 5 4 ° 7 ' 2 6 " . 5 , c ' = 1 0 9 ° 4 1 ' 4 9 " . 8 , B ' = 1 5 2 ° 2 3 ' 1 " . 3 . 9 . G i v e n a = 1 4 4 ° 2 7 ' 3 " , 6 = 3 2 ° 8 ' 5 6 " ; f i n d A = 1 2 6 ° 4 0 ' 2 4 " , B = 4 7 ° 1 3 ' 4 3 " , = 1 3 3 ° 3 2 ' 2 6 " . 1 0 . G i v e n a = 3 6 ° 2 7 ' , 6 = 4 3 ° 3 2 ' 3 1 " ; c f i n d A = 4 6 ° 5 9 ' 4 3 " . 3 , B = 5 7 ° 5 9 ' 1 9 " . 2 , = 5 4 ° 2 0 ' . 1 1 . G i v e n A 6 3 ° 1 5 ' 1 2 " , B = 1 3 5 ° 3 3 ' 3 9 " ; c t i n t ! a — 4 9 ° 5 9 ' 5 6 " , b = 1 4 3 ° 5 ' 1 2 " , = 1 2 0 ° 5 5 ' 3 4 " . 1 2 . G i v e n A = 6 7 ° 5 4 ' 4 7 " , . B = 9 9 ° 5 7 ' 3 5 " ; c f i n d a = 6 7 ° 3 3 ' 2 7 " , 6 = 1 0 0 ° 4 5 ' , c = 9 4 ° 5 ' . 1 3 . S o l v e t h e q u a d r a n t a l t r i a n g l e i n w h i c h c = 9 0 ° , A = 4 2 ° l ' , B = 1 2 1 ° 2 0 ' . A n s . C = 6 7 ° 1 6 ' 2 2 " , b = 1 1 2 ° 1 0 ' 2 0 " , a = 4 6 ° 3 1 ' 3 0 " . 1 4 . S o l v e t h e q u a d r a n t a l t r i a n g l e i n w h i c h a = 1 7 4 ° 1 2 ' 4 9 " . l , 6 = 9 4 ° 8 ' 2 0 " , c = 9 0 ° . A n s . A = 1 7 5 ° 5 7 ' 1 0 " , B = 1 3 5 ° 4 2 ' 5 5 " , C = 1 3 5 ° 3 4 ' 8 " . S O L U T I O N O F O B L I Q U E S P H E R I C A L T R I A N G L E S . 2 0 8 . T h e S o l u t i o n o f O b l i q u e S p h e r i c a l T r i a n g l e s p r e s e n t s S i x C a s e s ; a s f o l l o w s : I . G i v e n t w o s i d e s a n d t h e i n c l u d e d a n g l e , a , b , C . I I . G i v e n t w o a n g l e s a n d t h e i n c l u d e d s i d e , A , B , c . I I I . G i v e n t w o s i d e s a n d a n a n g l e o p p o s i t e o n e o f t h e m , a , b , A . C A S E I . 3 0 5 I V . G i v e n t w o a n g l e s a n d a s i d e o p p o s i t e o n e o f t h e m , A , B , a . V . G i v e n t h e t h r e e s i d e s , a , b , c . V I . G i v e n t h e t h r e e a n g l e s , A , B , C . T h e s e s i x c a s e s a r e i m m e d i a t e l y r e s o l v e d i n t o t h r e e p a i r s o f c a s e s b y t h e a i d o f t h e p o l a r t r i a n g l e ( A r t . 1 8 4 ) . F o r w h e n t w o s i d e s a n d t h e i n c l u d e d a n g l e a r e g i v e n , a n d t h e r e m a i n i n g p a r t s a r e r e q u i r e d , t h e a p p l i c a t i o n o f t h e d a t a t o t h e p o l a r t r i a n g l e t r a n s f o r m s t h e p r o b l e m i n t o t h e s u p p l e m e n t a l p r o b l e m : g i v e n t w o a n g l e s a n d t h e i n c l u d e d s i d e , t o f i n d t h e r e m a i n i n g p a r t s . S i m i l a r l y , c a s e s I I I . a n d I V . a r e s u p p l e m e n t a l , a l s o V . a n d V I . T h e p a r t s a r e a l l p o s i t i v e a n d l e s s t h a n 1 8 0 ° ( A r t . 1 8 2 ) . T h e a t t e n t i o n o f t h e s t u d e n t i s c a l l e d t o A r t . 1 9 9 . 2 0 9 . C a s e I . — G i v e n t w o s i d e s , a , b , a n d t h e i n c l u d e d a n g l e C ; t o f i n d A , B , c . B y N a p i e r , s A n a l o g i e s , ( 7 ) a n d ( 8 ) o f A r t . 1 9 7 , t a n $ ( A + B ) = < ( « - 6 ) o o t & 7 V ' c o s \ ( a + b ) 2 t a n $ ( A -B ) = s i " c o t g . T V ; s i n £ ( « + 6 ) 2 T h e s e d e t e r m i n e £ ( A + B ) a n d £ ( A — B ) , a n d t h e r e f o r e A a n d B ; t h e n c c a n b e f o u n d b y A r t . 1 9 0 , o r b y o n e o f G a u s s , s e q u a t i o n s ( A r t . 1 9 8 ) . S i n c e c i s f o u n d f r o m i t s s i n e i n A r t . 1 9 0 , i t m a y b e u n c e r t a i n w h i c h o f t w o v a l u e s i s t o b e g i v e n t o i t : i f w e d e t e r m i n e c f r o m o n e o f G a u s s , s e q u a t i o n s , i t i s f r e e f r o m a m b i g u i t y . W e m a y t h e r e f o r e f i n d c f r o m ( 3 ) o f A r t . 1 9 8 . T h u s c C c o s \ ( A + B ) c o s - = c o s % ( a + b ) s i n - -2 3 0 6 S P H E R I C A L T R I G O N O M E T R Y . C h e c k , t a n £ ( a + b ) c o s £ ( A + B ) = c o s £ ( A — B ) t a n - -T h e r e i s n o a m b i g u i t y i n t h i s c a s e . E x . 1 . G i v e n a = 4 3 ° 1 8 ' , 6 = 1 9 ° 2 4 ' , C = 7 4 ° 2 2 ' ; f i n d A , B , c . S o l u t i o n . \ ( a + b ) = 3 1 ° 2 1 ' , £ ( a -6 ) = 1 1 ° 5 7 ' , £ C = 3 7 ° 1 1 ' . l o g c o s I ( a -b ) = 9 . 9 9 0 4 8 4 8 l o g s e c \ ( a + 6 ) = 0 . 0 6 8 5 3 9 5 l o g c o t ^ = 1 0 . 1 1 9 9 9 6 9 8 2 l o g t a n £ ( A + B ) = 1 0 . 1 7 9 0 2 1 2 . - . £ ( A + B ) = 5 6 ° 2 9 ' 1 7 " £ ( A - B ) = 2 7 ° 4 1 ' 0 " . 5 . - . A = 8 4 ° 1 0 ' 1 7 " . 5 B = 2 8 ° 4 8 ' 1 6 " . 5 . c = 4 1 ° 3 5 ' 4 8 " . 5 . l o g s i n £ ( a - 6 ) = 9 . 3 1 6 0 9 2 1 l o g c o s e c £ ( a + 6 ) = 0 . 2 8 3 7 7 5 7 l o g c o t ^ = 1 0 . 1 1 9 9 9 6 9 l o g t a n £ ( A - B ) = 9 . 7 1 9 8 6 4 7 . - . £ ( A - B ) = 2 7 ° 4 1 ' 0 " . 5 . l o g c o s \ ( a + b ) = 9 . 9 3 1 4 6 0 5 l o g s e c £ ( A + B ) = 0 . 2 5 7 9 7 3 7 l o g s i n 9 . 7 8 1 3 0 1 0 l o g c o s 1 = 9 . 9 7 0 7 3 5 2 = 2 0 ° 4 7 ' 5 4 " . 2 5 . O t h e r w i s e t h u s : L e t f a l l t h e p e r p e n d i c u l a r B D , d i v i d i n g t h e t r i a n g l e A B C i n t o t w o r i g h t t r i a n g l e s , B D A , B D C . D e n o t e A D b y m , t h e a n g l e A B D b y < f > , a n d A -B D b y p . T h e n b y N a p i e r ' s R u l e s , w e h a v e c o s C = t a n ( b — m ) c o t a ; s i n ( b — m ) = c o t C t a n p ; s i n m = c o t A t a n p . . : t a n ( 6 — m ) = t a n a c o s C ( 1 ) a n d t a n A s i n m = t a n C s i n ( b — m ) ( 2 ) C A S E I I . 3 0 7 F r o m ( 1 ) m i s d e t e r m i n e d , a n d f r o m ( 2 ) A i s d e t e r m i n e d . I n a s i m i l a r m a n n e r B m a y b e f o u n d . • A l s o , f r o m t h e s a m e t r i a n g l e s , w e h a v e b y N a p i e r , s R u l e s c o s a = c o s ( b — m ) c o & p ; c o s c = c o s m c o s p . . : c o s c c o s ( b — m ) = c o s m c o s a , f r o m w h i c h c i s f o u n d . N o t e . — T h i s m e t h o d h a s t h e a d v a n t a g e t h a t , i n u s i n g i t , n o t h i n g n e e d b e r e m e m b e r e d e x c e p t N a p i e r ' s R u l e s . I f o n l y t h e s i d e c i s w a n t e d , i t m a y b e f o u n d f r o m ( 4 ) o f A r t . 1 9 1 , w i t h o u t p r e v i o u s l y d e t e r m i n i n g A a n d B . T h i s f o r m u l a m a y b e a d a p t e d t o l o g a r i t h m s b y t h e u s e o f a s u b s i d i a r y a n g l e ( A r t . 9 0 ) . E x . 2 . G i v e n 6 = 1 2 0 ° 3 0 ' 3 0 " , c = 7 0 ° 2 0 ' 2 0 " , A = 5 0 ° 1 0 ' 1 0 " ; f i n d B , C , o . A n s . B = 1 3 5 ° 5 ' 2 8 " . 8 , C = 5 0 ° 3 0 ' 8 " . 4 , a = 6 9 ° 3 4 ' 5 6 " . 2 1 0 . C a s e I I . — G i v e n t w o a n g l e s , A , B , a n d t h e i n c l u d e d s i d e c ; t o f i n d a , b , C . B y N a p i e r , s A n a l o g i e s ( 9 ) a n d ( 1 0 ) o f A r t . 1 9 7 , t a n £ ( q + 6 ) = c 0 S H A - B ) c 7 V ' c o s £ ( A + B ) 2 t a n I ( a - b ) = ^ ( A - B ) t a n « ^ v ; s i n £ ( A + B ) 2 f r o m w h i c h a a n d b a r e f o u n d . T h e r e m a i n i n g p a r t C m a y b e f o u n d b y ( 2 ) o f A r t . 1 9 8 . s i n \ ( a — b ) c o s — = s i n £ ( A — B ) s i n 2 2 f t C h e c k , c o s % ( a — b ) c o t — = c o s £ ( a + b ) t a n \ ( A + B ) . 2 T h e r e i s n o a m b i g u i t y i n t h i s c a s e . 3 0 8 S P H E R I C A L T R I G O N O M E T R Y . E x . 1 . G i v e n A = 6 8 ° 4 0 ' , B = 5 6 ° 2 0 ' , c = 8 4 ° 3 0 ' ; f i n d a , b , C . S o l u t i o n . \ ( A + B ) = 6 2 ° 3 0 ' , \ ( A -B ) = 6 ° 1 0 ' , £ = 4 2 ° 1 5 ' . l o g c o s £ ( A - B ) = 9 . 9 9 7 4 7 9 7 l o g s e c £ ( A + B ) = 0 . 3 3 5 5 9 4 4 l o g t a n - = 9 . 9 5 8 2 4 6 5 6 2 l o g t a n \ ( a + b ) = 1 0 . 2 9 1 3 2 0 6 . - . £ ( < x + 6 ) = 6 2 ° 5 5 ' 9 " \ ( a - b ) = 6 ° 1 6 ' 3 9 " a = 6 9 ° 1 1 ' 4 8 " b = 5 6 ° 3 8 ' 3 0 " . C = 9 7 ° 1 9 ' 3 " . 5 . l o g s i n ^ ( A -B ) = 9 . 0 3 1 0 8 9 0 l o g c o s e c | ( A + B ) = 0 . 0 5 2 0 7 1 1 l o g t a n £ = 9 . 9 5 8 2 4 6 5 l o g t a n \ ( a -b ) = 9 . 0 4 1 4 0 6 6 . - . ± ( a - b ) = < 5 ° l & 3 9 " . l o g s i n \ ( A - B ) = 9 . 0 3 1 0 8 9 0 l o g c o s e c £ ( a -b ) = 0 . 9 6 1 2 0 5 0 l o g s i n £ = 9 . 8 2 7 6 0 6 3 l o g c o s ~ = 9 . 8 1 9 9 0 0 3 . § = 4 8 ° 3 9 ' 3 1 f " . O t h e r w i s e t h u s : L e t f a l l t h e p e r p e n d i c u l a r B D ( s e e l a s t f i g u r e ) . D e n o t e , a s b e f o r e , A D b y m , t h e a n g l e A B D b y < f > , a n d B D b y p . T h e n b y N a p i e r , s R u l e s , w e h a v e c o s c = c o t < f > c o t A ; c o s < j > = c o t c t a n p ; c o s ( B — < £ ) = c o t a t a n p . . - . c o t = t a n A c o s c ( 1 ) t a n a c o s ( B — < £ ) = c o s < f > t a n c ( 2 ) F r o m ( 1 ) < f > i s d e t e r m i n e d , a n d f r o m ( 2 ) a i s f o u n d . S i m i l a r l y 6 m a y b e f o u n d . A l s o , f r o m t h e s a m e t r i a n g l e s , w e h a v e c o s C s i n < f > = c o s A s i n ( B — < f > ) , f r o m w h i c h C i s f o u n d . C A S E 1 1 1 . 3 0 9 E x . 2 . G i v e n A = 1 3 5 ° 5 ' 2 8 " . 0 , C = 5 0 ° 3 0 ' 8 " . 6 , b = 6 9 ° 3 4 ' 5 6 " . 2 ; f i n d a , c , B . A n s . a = 1 2 0 ° 3 0 ' 3 0 " , c = 7 0 ° 2 0 ' 2 0 " , B = 5 0 ° 1 0 ' 1 0 " . 2 1 1 . C a s e I I I . — G i v e n t w o s i d e s , a , b , a n d t h e a n g l e A o p p o s i t e o n e o f t h e m ; t o f i n d B , C , c . T h e a n g l e B i s f o u n d f r o m t h e f o r m u l a , s i n B = 5 H ^ s i n A ( A r t . 1 9 0 ) ( 1 ) s i n s T h e n C a n d c a r e f o u n d f r o m N a p i e r , s A n a l o g i e s , C = s m i ( a - 6 ) A _ 2 s m \ ( a + b ) ^ . > ^ > c = s j n j i A ± B ) . . . ( 3 ) 2 s i n i ( A - B ) ¥ V ; ' C h e c k , s i n A _ s i n B _ s i n C s i n a s i n b s i n c S i n c e B i s f o u n d f r o m i t s s i n e i n ( 1 ) , i t w i l l h a v e t w o v a l u e s , i f s i n A s i n b < s i n a , a n d t h e t r i a n g l e , i n g e n e r a l , w i l l a d m i t o f t w o s o l u t i o n s . W h e n s i n A s i n 6 > s i n a , t h e r e w i l l b e n o s o l u t i o n , f o r t h e n s i n B > 1 . I n o r d e r t h a t e i t h e r o f t h e s e v a l u e s f o u n d f o r B m a y b e a d m i s s i b l e , i t i s n e c e s s a r y a n d s u f f i c i e n t t h a t , w h e n s u b s t i t u t e d i n ( 2 ) a n d ( 3 ) , t h e y g i v e p o s i t i v e v a l u e s f o r C c t a n — a n d t a n - , o r w h i c h i s t h e s a m e t h i n g , t h a t A — B a n d 2 2 a — b h a v e t h e s a m e s i g n . H e n c e w e h a v e t h e f o l l o w i n g R u l e . — I f b o t h v a l u e s o f B o b t a i n e d f r o m ( 1 ) b e s u c h a s t h a t A — B a n d a — b h a v e l i k e s i g n s , t h e r e a r e t i v o c o m p l e t e s o l u t i o n s . I f o n l y o n e o f t h e v a l u e s o f B s a t i s f i e s t h i s c o n d i t i o n , t h e r e i s o n l y o n e t r i a n g l e t h a t s a t i s f i e s t h e p r o b l e m , s i n c e 3 1 0 S P H E R I C A L T R I G O N O M E T R Y . i n t h i s c a s e C , o r c > 1 8 0 ° . I f n e i t h e r o f t h e v a l u e s o f B m a k e s A — B a n d a — b o f t h e s a m e s i g n s , t h e p r o b l e m i s i m p o s s i b l e . T h i s c a s e i s k n o w n a s t h e a m b i g u o u s c a s e , a n d i s l i k e t h e a n a l o g o u s a m b i g u i t y i n P l a n e T r i g o n o m e t r y ( A r t . 1 1 6 ) , t h o u g h i t i s s o m e w h a t m o r e c o m p l e x . F o r a c o m p l e t e d i s c u s s i o n o f t h e A m b i g u o u s C a s e , t h e s t u d e n t i s r e f e r r e d t o T o d h u n t e r , s S p h e r i c a l T r i g o n o m e t r y , p p . 5 3 - 5 8 ; M c C o l -l e n d a n d P r e s t o n , s S p h e r i c a l T r i g o n o m e t r y , p p . 1 3 7 - 1 4 3 ; S e r r e t , s T r i g o n o m e t r y , p p . 1 9 1 - 1 9 5 , e t c . E x . 1 . G i v e n a = 4 2 ° 4 5 ' , 6 = 4 7 ° 1 5 ' , A = 5 6 ° 3 0 ' ; f i n d B , C , a S o l u t i o n . l o g s i n b = 9 . 8 6 5 8 8 6 8 c o l o g s i n a = 0 . 1 6 8 2 5 7 7 l o g s i n A = 9 . 9 2 1 1 0 6 6 l o g s i n B = 9 . 9 5 5 2 5 1 1 . - . B = 6 4 ° 2 6 ' 4 " , B ' = 1 1 5 ° 3 3 ' 5 6 " . i ( a + 6 ) = 4 5 ° 0 ' 0 " . i ( a - 6 ) = -2 ° 1 5 ' 0 " . £ ( A + B ) = 6 0 ° 2 8 ' 2 " . i ( A - B ) = -3 ° 5 8 ' 2 " . £ ( A + B ' ) = 8 6 ° 1 ' 5 8 " . i ( A - B ' ) = - 2 9 ° 3 1 ' 5 8 " . S i n c e b o t h v a l u e s o f B a r e s u c h t h a t A — B , A — B ' , a n d a — b , a r e a l l n e g a t i v e , t h e r e a r e t w o s o l u t i o n s , b y t h e a b o v e R u l e . ( 1 ) W h e n B = 6 4 ° 2 6 ' 4 " . l o g s i n | ( a - 6 ) = 8 . 5 9 3 9 4 8 3 -c o l o g s i n | ( a + 6 ) = 0 . 1 5 0 5 1 5 0 l o g c o t | ( A -B ) = 1 . 1 5 8 9 4 1 3 -l o g t a n 5 = 9 . 9 0 3 4 0 4 6 L i . : 5 = 3 8 ° 4 0 ' 4 8 " . _ . - . C = 7 7 ° 2 1 ' 3 6 " . l o g s i n \ ( A + B ) = 9 . 9 3 9 5 5 6 0 c o l o g s i n ! ( A - B ) = 1 . 1 5 9 9 8 3 2 -l o g t a n \ ( a -b ) = 8 . 5 9 4 2 8 3 2 -l o g t a n | = 9 . 6 9 3 8 2 2 4 " = 2 6 ° 1 7 ' 4 0 " . 2 c = 5 3 ° 3 5 ' 2 0 " . C A S E I I I . 3 1 1 ( 2 ) W h e n W l o g s i n $ ( a -6 ) = 8 . 5 9 3 9 4 8 3 -c o l o g s i n $ ( a + b ) = 0 . 1 5 0 5 1 5 0 l o g c o t \ ( A - B ' ) = 0 . 2 4 6 7 7 8 4 -l o g t a n - - = 8 . 9 9 1 2 4 1 7 . - . - = 5 ° 3 5 ' 5 0 1 " . 2 4 . - . c = i r i r 4 0 £ " . l o g s i n | ( A + B ' ) = 9 . 9 9 8 9 5 8 1 c o l o g s i n $ ( A - B ' ) = 0 . 3 0 7 2 2 2 3 -l o g t a n £ ( « -b ) = 8 . 5 9 4 2 8 3 2 -l o g t a n | ' = 8 . 9 0 0 4 6 3 6 . : | ' = 4 ° 8 2 ' 4 7 i " . . - . c ' = 9 ° 5 ' 3 4 £ " . ^ n s . B = 6 4 ° 2 6 ' 4 " , C = 7 7 ° 2 1 ' 3 6 " , c = 5 3 ° 3 5 ' 2 0 " ; B ' = 1 1 5 ° 3 5 ' 5 6 " , C = 1 1 ° 1 1 ' 4 0 £ " , c ' = 9 ° 5 ' 3 4 £ " . O t h e r w i s e t h u s : L e t f a l l t h e p e r p e n d i c u l a r C D ; d e n o t e A D b y m , t h e a n g l e A C D b y < f > , a n d C D b y p . T h e n w e h a v e ( 1 ) c o t < f > = c o s b t a n A c o s A = t a n t o c o t b ; . - . t a n m = c o s A t a n b c o s b = c o t A c o t f j > . A g a i n , c o s a = c o s ( c — t o ) c o s p ; c o s 6 = c o s t o c o s p . . - . c o s ( c — m ) = c o s a c o s t o - 4 - 6 . . . A l s o , c o s ( C — < f > ) = c o t a t a n p ; c o s < £ = c o t b t a n p . . - . c o s ( C — $ ) = c o t a t a n b c o s . . . s i n 6 L a s t l y , s i n B : s i n A s m a ( 2 ) ( 3 ) ( 4 ) ( 5 ) T h e r e q u i r e d p a r t s a r e g i v e n b y ( 1 ) , ( 2 ) , ( 3 ) , ( 4 ) , ( 5 ) . E x . 2 . G i v e n a = 7 3 ° 4 9 ' 3 8 " , 6 = 1 2 0 ° 5 3 ' 3 5 " , A = 8 8 ° 5 2 ' 4 2 " : f i n d B , C , c . A n s . B = 1 1 6 ° 4 4 ' 4 8 " , C = 1 1 6 ° 4 4 ' 4 8 " , c = 1 2 0 ° 5 5 ' 3 5 " . 3 1 2 S P H E R I C A L T R I G O N O M E T R Y . 2 1 2 . C a s e I V . — G i v e n t w o a n g l e s , A , B , a n d t h e s i d e a o p p o s i t e o n e o f t h e m ; t o f i n d b , c , C . T h i s c a s e r e d u c e s , b y a i d o f t h e p o l a r t r i a n g l e , t o t h e p r e c e d i n g c a s e , a n d g i v e s r i s e t o t h e s a m e a m b i g u i t i e s . H e n c e t h e s a m e r e m a r k s m a d e i n A r t . 2 1 1 a p p l y i n t h i s c a s e a l s o , a n d t h e d i r e c t s o l u t i o n m a y b e o b t a i n e d i n t h e s a m e w a y a s i n C a s e I I I . T h u s , T h e s i d e 6 i s f o u n d f r o m t h e f o r m u l a . , s i n B -s i n o — — — - s i n a s i n A T h e n c a n d C a r e f o u n d f r o m N a p i e r , s A n a l o g i e s . c = c o s £ ( A + B ) H } 2 c o s ^ ( A - B ) ^ K T ' . C c o s - A - ( a — b ) , , , . , t i n t a n — = ' c o t + ( A + B ) . . 2 c o s £ ( < + 6 ) ( 1 ) ( 2 ) ( 3 ) C h e c k , s i n A s i n B s i n C s i n o s i n b s i n e E x . 1 . G i v e n A = 6 6 ° 7 ' 2 0 " , B = 5 2 ° 5 0 ' 2 0 " , a = 5 9 ° 2 8 ' 2 7 " ; f i n d 6 , c , C . S o l u t i o n . B y ( 1 ) w e f i n d 6 = 4 8 ° 3 9 ' 1 6 " , o r 1 3 1 ° 2 0 ' 4 4 " . £ ( A + B ) = 5 9 ° 2 8 ' 5 0 " . £ ( A - B ) = 6 ° 3 8 ' 3 0 " . l o g c o s $ ( A + B ) = 9 . 7 0 5 7 1 9 0 c o l o g c o s | ( A -B ) = 0 . 0 0 2 9 2 4 4 l o g t a n $ ( a + b ) = 1 0 . 1 3 9 7 6 4 3 l o g t a n - = 9 . 8 4 8 4 0 7 7 - = 3 5 ° l l ' 5 0 f " . £ ( a + 6 ) = 5 4 ° 3 ' 5 1 £ " . £ ( a - 6 ) = 5 ° 2 4 ' 3 5 i " . l o g c o s $ ( a -b ) = 9 . 9 9 8 0 6 1 2 c o l o g c o s £ ( « + 6 ) = 0 . 2 3 1 4 5 3 0 l o g c o t £ ( A + B ) = 9 . 7 7 0 4 8 5 4 l o g t a n ? = 9 . 9 9 9 9 9 9 6 c = 4 5 ° . T h e s e c o n d v a l u e o f b i s i n a d m i s s i b l e ( s e e R u l e o f A r t . 2 1 1 ) , a n d t h e r e f o r e t h e r e i s o n l y o n e s o l u t i o n . A n s . b = 4 8 ° 3 9 ' 1 6 " , c = 7 0 ° 2 3 ' 4 H " , C = 9 0 ° . C A S E V . 3 1 3 E x . 2 . G i v e n A = 1 1 0 ° 1 0 ' , B = 1 3 3 ° 1 8 ' , a = 1 4 7 ° 5 ' 3 2 " ; f i n d b , c , C . A n a . b = 1 5 5 ° 5 ' 1 8 " , c = 3 3 ° 1 ' 4 5 " , C = 7 0 ° 2 0 ' 5 0 " . 2 1 3 . C a s e V . — G i v e n t h e t h r e e s i d e s , a , b , c ; t o f i n d t h e a n g l e s . T h e a n g l e s m a y b e c o m p u t e d b y a n y o f t h e f o r m u l a e o f A r t . 1 9 5 ; b u t s i n c e a n a n g l e n e a r 9 0 ° c a n n o t b e a c c u r a t e l y d e t e r m i n e d b y i t s s i n e , n o r o n e n e a r 0 ° b y i t s c o s i n e ( A r t . 1 5 1 ) , n e i t h e r o f t h e f i r s t s i x f o r m u l a e c a n b e u s e d w i t h a d v a n t a g e i n a l l c a s e s . T h e f o r m u l a e f o r t h e t a n g e n t s h o w e v e r a r e a c c u r a t e i n a l l p a r t s o f t h e q u a d r a n t , a n d a r e t h e r e f o r e t o b e p r e f e r r e d f o r t h e s o l u t i o n o f a t r i a n g l e i n w h i c h a l l t h r e e s i d e s o r a l l t h r e e a n g l e s a r e g i v e n . B y ( 7 ) o f A r t . 1 9 5 w e h a v e t a n - = J s i " ( s -b ) s i n ( s -c ) 2 \ s i n s s i n ( s — a ) _ 1 / s i n ( s — a ) s i n ( s — b ) s i n ( g — c ) s i n ( s — a ) y l s i n s S i n c e t h e p a r t u n d e r t h e r a d i c a l i s a s y m m e t r i c f u n c t i o n o f t h e s i d e s , i t i s i n t h e f o r m u l a e f o r d e t e r m i n i n g a l l t h r e e a n g l e s A , B , C , a n d w h e n o n c e c a l c u l a t e d , i t m a y b e u t i l i z e d i n t h e c a l c u l a t i o n o f e a c h a n g l e . F o r c o n v e n i e n c e i n c o m p u t a t i o n , d e n o t e t h i s t e r m b y t a n r . T h e n t a n r = J s i n ( s ~ a ) s i n p -b ) s i n ( s ~ < 0 ; A l s i n s a n d ( 7 ) , ( 8 ) , ( 9 ) o f A r t . 1 9 5 b e c o m e . A t a n r t a n ~ = ( 1 ) 2 s m ( s — a ) . B t a n r / O N t a n — = ( 2 ) 2 s i n ( s - 6 ) v ' t a n ° = . t f n r ( 3 ) 2 s i n ( s — c ) 3 1 4 S P H E R I C A L T R I G O N O M E T R Y . C h e c k , s i n A s i n B s i n C s i n a s i n b s i n e E x . 1 . G i v e n a = 4 6 ° 2 4 ' , b = 6 7 ° 1 4 ' , c = 8 1 ° 1 2 ' ; f i n d A , B , C . S o l u t i o n . a = 4 6 ° 2 4 ' l o g s i n ( s -a ) = 9 . 8 9 0 6 0 4 9 6 = 6 7 ° 1 4 ' l o g s i n ( s -b ) = 9 . 7 0 1 3 6 8 1 c — 8 1 ° 1 2 ' l o g s i n ( s -c ) = 9 . 4 4 6 0 2 5 1 2 s = 1 9 4 ° 5 0 ' c o l o g s i n s = 0 . 0 0 3 6 4 8 7 s = 9 7 ° 2 5 ' , l o g t a n 2 r = 9 . 0 4 1 6 4 6 8 s — a = 5 1 ° 1 ' , l o g t a n r = 9 . 5 2 0 8 2 3 4 . s - 6 = 3 0 ° 1 1 ' , s — c = 1 6 ° 1 3 ' . t a n r = 9 . 5 2 0 8 2 3 4 s i n ( s - a ) = 9 . 8 9 0 6 0 4 9 t a n : = 9 . 6 3 0 2 1 8 5 - = 2 3 ° 6 ' 4 5 " . 2 A = 4 6 ° 1 3 ' 3 0 " . t a n r = 9 . 5 2 0 8 2 3 4 s i n ( s - b ) = 9 . 7 0 1 3 6 8 1 t a n 1 5 : 9 . 8 1 9 4 5 5 3 = 3 3 ° 2 5 ' 1 0 " . B 2 : B = 6 6 ° 5 0 ' 2 0 " . t a n r = 9 . 5 2 0 8 2 3 4 s i n ( s - c ) = 9 . 4 4 6 0 2 5 1 t a n ^ = 1 0 . 0 7 4 7 9 8 3 . - . £ = 4 9 ° 5 4 ' 3 5 " . C = 9 9 ° 4 9 ' 1 0 " . E x . 2 . G i v e n a = 1 0 0 ° , b = 3 7 ° 1 8 ' , c = 6 2 ° 4 6 ' ; f i n d A , B , C . A n s . A = 1 7 6 ° 1 5 ' 4 6 " . 5 6 , B = 2 ° 1 7 ' 5 5 " . 0 8 , C = 3 ° 2 2 ' 2 5 " . 4 6 . 2 1 4 . C a s e V I . — G i v e n t h e t h r e e a n g l e s , A , B , C ; t o f i n d t h e s i d e s . A s i n A r t . 2 1 3 , t h e f o r m u l a e f o r t h e t a n g e n t s a r e t o b e p r e f e r r e d . P u t t i n g t a n B = J - c o s 8 6 \ c o s ( S - A ) c o s ( S - B ) c o s ( S - C ) w e h a v e , f r o m ( 7 ) , ( 8 ) , ( 9 ) o f A r t . 1 9 6 , t a n - = t a n E c o s ( S — A ) , C A S E V I . 3 1 5 t a n - = t a n R c o s ( S — B ) , 2 t a n - = t a n R c o s ( S — C ) , 2 b y w h i c h t h e t h r e e s i d e s m a y b e f o u n d . s i n A s i n B s i n C C h e c k , s m a s i n b s i n c E x . 1 . G i v e n A = 6 8 ° 3 0 ' , B = 7 4 ° 2 0 ' , C = 8 3 ° 1 0 ' ; f i n d a , b , c . S o l u t i o n . A = 6 8 ° 3 0 ' B = 7 4 ° 2 0 ' C = 8 3 ° 1 0 ' 2 S = 2 2 6 ° 0 ' l o g ( -c o s S ) = 9 . 5 9 1 8 7 8 0 l o g c o s ( S -A ) = 9 . 8 5 3 2 4 2 1 l o g c o s ( S -B ) = 9 . 8 9 2 5 3 0 5 l o g c o s ( S -C ) = 9 . 9 3 8 2 5 7 6 l o g t a n - ' R = 9 . 9 0 7 8 4 1 8 l o g t a n R = 9 . 9 5 3 9 2 0 9 S = 1 1 3 ° 0 ' S - A = 4 4 ° 3 0 ' S - B = 3 8 ° 4 0 ' S -C = 2 9 ° 5 0 ' l o g t a n - = 9 . 8 0 7 1 6 3 0 l o g t a n - = 9 . 8 4 6 4 5 7 4 l o g t a n - = 9 . 8 9 2 1 7 8 5 2 a = 6 5 ° 2 1 ' 2 2 ^ " b = 7 Q ° 9 ' 9 \ " , c = 7 5 ° 5 5 ' 9 " . C h e c k , s m a s i n b s i n c s i n A s i n B s i n C E x . 2 . G i v e n A = 5 9 ° 5 5 ' 1 0 " , B = 8 5 ° 3 6 ' 5 0 ° , C = 5 9 ° 5 5 ' 1 0 " ; f i n d a , b , c . A n s . a = 1 2 9 ° 1 1 ' 4 0 " , 6 = 6 3 ° 1 5 ' 1 2 " , c = 1 2 9 ° 1 1 ' 4 0 " . T h e n e c e s s a r y a d d i t i o n s m a y b e c o n v e n i e n t l y p e r f o r m e d b y w r i t i n g t o l t o B o . a s l i p o f p a p e r , a n d h o l d i n g i t s u c c e s s i v e l y o v e r l o g c o s ( S - A ) , l o g c o s ( S - B ) , a n d l o g c o s ( S — C ) . 3 1 6 S P H E R I C A L T R I G O N O M E T M Y . E X A M P L E S . S o l v e t h e f o l l o w i n g r i g h t t r i a n g l e s : 1 . G i v e n c = 8 4 ° 2 0 ' , A = 3 5 ° 2 5 ' ; f i n d 0 = 3 5 ° 1 3 ' 4 " , b = 8 3 ° 3 ' 2 9 " , B = 8 5 ° 5 9 ' 1 " . 2 . G i v e n c — 6 7 ° 5 4 ' , A = 4 3 ° 2 8 ' ; f i n d a = 3 9 ° 3 5 ' 5 1 " , b = 6 0 ° 4 6 ' 2 5 1 " , B = 7 0 ° 2 2 ' 2 1 " . 3 . G i v e n C — 2 2 ° 1 8 ' 3 0 " , A = 4 7 ° 3 9 ' 3 6 " ; f i n d a = 1 6 ° 1 7 ' 4 1 " , b = 1 5 ° 2 6 ' 5 3 " , B = 4 4 ° 3 3 ' 5 3 " . 4 . 4 . G i v e n c = 1 4 5 ° , A = 2 3 ° 2 8 ' ; f i n d a = 1 3 ° 1 2 ' 1 2 " , 6 = 1 4 7 ° 1 7 ' 1 5 " , B = 1 0 9 ° 3 4 ' 3 3 " . 5 . G i v e n c = 9 8 ° 6 ' 4 3 " , A = 1 3 8 ° 2 7 ' 1 8 " ; f i n d a = 1 3 7 ° 6 ' , b = 7 7 ° 5 1 ' , B = 8 0 ° 5 5 ' 2 7 " . 6 . G i v e n c = 4 6 ° 4 0 ' 1 2 " , A = 3 7 ° 4 6 ' 9 " ; f i n d a = 2 6 ° 2 7 ' 2 3 " . 8 , 6 = 3 9 ° 5 7 ' 4 1 " . 4 , B = 6 2 ° 0 ' 4 " . 7 . G i v e n c = 7 6 ° 4 2 ' , a = 4 7 ° 1 8 ' ; f i n d 6 = 7 0 ° 1 0 ' 1 3 " , A = 4 9 ° 2 ' 2 4 " . 5 , 1 5 = 7 5 ° 9 ' 2 4 " . 7 5 . 8 . G i v e n c = 9 1 ° 1 8 ' , a = 7 2 ° 2 7 ' ; f i n d b = 9 4 ° 1 8 ' 5 3 " . 8 , A = 7 2 ° 2 9 ' 4 8 " , B = 9 4 ° 6 ' 5 3 " . 3 . 9 . G i v e n c = 8 6 ° 5 1 ' , 6 = 1 8 ° 1 ' 5 0 " ; f i n d a = 8 6 ° 4 1 ' 1 4 " , A = 8 8 ° 5 8 ' 2 5 " , B = 1 8 ° 3 ' 3 2 " . 1 0 . G i v e n c = 2 3 ° 4 9 ' 5 1 " , a = 1 4 ° 1 6 ' 3 5 " ; f i n d 6 = 1 9 ° 1 7 ' , A = 3 7 ° 3 6 ' 4 9 " . 3 , B = 5 4 ° 4 9 ' 2 3 " . 3 . 1 1 . G i v e n c = 9 7 ° 1 3 ' 4 " , a = 1 3 2 ° 1 4 ' 1 2 " ; f i n d 7 9 ° 1 3 ' 3 8 " . 2 , A = 1 3 1 ° 4 3 ' 5 0 " , B = 8 1 ° 5 8 ' 5 3 " . 3 . 1 2 . G i v e n c = 3 7 ° 4 0 ' 2 0 " , a — 3 7 ° 4 0 ' 1 2 " ; f i n d 6 = 0 ° 2 6 ' 3 7 " . 2 , A = 8 9 ° 2 5 ' 3 7 " , B = 0 ° 4 3 ' 3 3 " . E X A M P L E S . 3 1 7 1 3 . G i v e n a = 8 2 ° 6 ' , B = 4 3 ° 2 8 ' ; f i n d A = 8 4 ° 3 4 ' 2 8 " , b = 4 3 ° 1 1 ' 3 8 " , c = 8 4 ° 1 4 ' 5 7 " . 1 4 . G i v e n a = 4 2 ° 3 0 ' 3 0 " , B = 5 3 ° 1 0 ' 3 0 " ; f i n d A = 5 3 ° 5 0 ' 1 2 " , b = 4 2 ° 3 ' 4 7 " , c = 5 6 ° 4 9 ' 8 " . 1 5 . G i v e n a = 2 0 ° 2 0 ' 2 0 " , B = 3 8 ° 1 0 ' 1 0 " ; f i n d A = 5 4 0 3 5 ' 1 6 " . 7 , b = 1 5 ° 1 6 ' 5 0 " . 4 , c = 2 5 ° 1 4 ' 3 8 " . 2 . 1 6 . G i v e n a = 9 2 ° 4 7 ' 3 2 " ; B = 5 0 ° 2 ' 1 " ; f i n d A = 9 2 ° 8 ' 2 3 " , 6 = 5 0 ° , c = 9 1 ° 4 7 ' 4 0 " . 1 7 . G i v e n 6 = 5 4 ° 3 0 ' , A = 3 5 ° 3 0 ' ; f i n d B = 7 0 ° 1 7 ' 3 5 " , a = 3 0 ° 8 ' 3 9 " . 2 , C — 5 9 ° 5 1 ' 2 0 " . 1 8 . G i v e n 6 = 1 5 5 ° 4 6 ' 4 2 " . 7 , A = 8 0 ° 1 0 ' 3 0 " ; f i n d B = 1 5 3 ° 5 8 ' 2 4 " . 5 , a = 6 7 ° 6 ' 5 2 " . 6 , c = 1 1 0 ° 4 6 ' 2 0 " . 1 9 . G i v e n a = 3 5 ° 4 4 ' , A = 3 7 ° 2 8 ' ; f i n d b = 6 9 ° 5 0 ' 2 4 " , c = 7 3 ° 4 5 ' 1 5 " , B = 7 7 ° 5 4 ' , o r b , = 1 1 0 ° 9 ' 3 6 " , c ' = 1 0 6 ° 1 4 ' 4 5 " , B = 1 0 2 ° 6 ' . 2 0 . G i v e n a = 1 2 9 ° 3 3 ' , A = 1 0 4 ° 5 9 ' ; f i n d 6 = 1 8 ° 5 4 ' 3 8 " , c = 1 2 7 ° 2 ' 2 7 " , B = 2 3 ° 5 7 ' 1 9 " , o r 6 ' = 1 6 1 ° 5 ' 2 2 " , c ' = 5 2 ° 5 7 ' 3 3 " , B ' = 1 5 6 ° 2 ' 4 1 " . 2 1 . G i v e n a = 2 1 ° 3 9 ' , A = 4 2 ° 1 0 ' 1 0 " ; f i n d 6 = 2 5 ° 5 9 ' 2 7 " . 8 , c = 3 3 ° 2 0 ' 1 3 " . 4 , B = 5 2 ° 2 3 ' 2 " . 8 , o r b ' = 1 5 4 ° 0 ' 3 2 " . 2 , c ' = 1 4 6 ° 3 9 ' 4 6 " . 6 , B ' = 1 2 7 ° 3 6 ' 5 7 " . 2 . 2 2 . G i v e n a = 4 2 ° 1 8 ' 4 5 " , A = 4 6 ° 1 5 ' 2 5 " ; f i n d 6 = 6 0 ° 3 6 ' 1 0 " , c = 6 8 ° 4 2 ' 5 9 " , B = 6 9 ° 1 3 ' 4 7 " , o r 6 ' = 1 1 9 ° 2 3 ' 5 0 " , c ' = 1 1 1 ° 1 7 ' 1 " , B ' = 1 1 0 ° 4 6 ' 1 3 " . 2 3 . G i v e n 6 = 1 6 0 ° , B = 1 5 0 ° ; f i n d a = 3 9 ° 4 ' 5 0 " . 7 , c = 1 3 6 ° 5 0 ' 2 3 " . 3 , A = 6 7 ° 9 ' 4 2 " . 7 , o r a ' = 1 4 0 ° 5 5 ' 9 " . 3 , c ' = 4 3 ° 9 ' 3 6 " . 7 , A ' = 1 1 2 ° 5 0 ' 1 7 " . 3 . 3 1 8 S P H E R I C A L T R I G O N O M E T R Y . 2 4 . G i v e n a — 2 5 ° 1 8 ' 4 5 " , A = 1 5 ° 5 8 ' 1 5 " . A n s . I m p o s s i b l e ; w h y ? 2 5 . G i v e n a = 3 2 ° 9 ' 1 7 " , 6 = . 3 2 ° 4 1 ' ; f i n d A = 4 9 ° 2 0 ' 1 7 " , B = 5 0 ° 1 9 ' 1 6 " , c = 4 4 ° 3 3 ' 1 7 " . 2 6 . G i v e n a = 5 5 ° 1 8 ' , b = 3 9 ° 2 7 ' ; f i n d A = 6 6 ° 1 5 ' 6 " , B = 4 5 ° 1 ' 3 1 " , c = 6 3 ° 5 5 ' 2 1 " . 2 7 . G i v e n a = 5 6 ° 2 0 ' , b = 7 8 ° 4 0 ' ; f i n d A = 5 6 ° 5 1 ' 7 " , B = 8 0 ° 3 1 ' 4 8 " , c = 8 3 ° 4 4 ' 4 4 ^ " . 2 8 . G i v e n a = 8 6 ° 4 0 ' , b = 3 2 ° 4 0 ' ; f i n d A = 8 8 ° 1 1 ' 5 7 " . 8 , B = 3 2 ° 4 2 ' 3 7 " . 8 , c = 8 7 ° 1 1 ' 3 9 " . 8 . 2 9 . G i v e n a = 3 7 ° 4 8 ' 1 2 " , b = 5 9 ° 4 4 ' 1 6 " ; f i n d A = 4 1 ° 5 5 ' 4 5 " , B = 7 0 ° 1 9 ' 1 5 " , c = 6 6 ° 3 2 ' 6 " . 3 0 . G i v e n a = 1 1 6 ° , 6 = 1 6 ° ; f i n d A = 9 7 ° 3 9 ' 2 4 " . 4 , B = 1 7 ° 4 1 ' 3 9 " . 9 , c = 1 1 4 ° 5 5 ' 2 0 " . 4 . 3 1 . G i v e n A = 5 2 ° 2 6 ' , B = 4 9 ° 1 5 ' ; f i n d a = 3 6 ° 2 4 ' 3 4 " . 5 , 6 = 3 4 ° 3 3 ' 4 0 " , c = 4 8 ° 2 9 ' 2 0 " . 3 2 . G i v e n A = 6 4 ° 1 5 ' , B = 4 8 ° 2 4 ' ; f i n d a = 5 4 ° 2 8 ' 5 3 " , b = 4 2 ° 3 0 ' 4 7 " , c = 6 4 ° 3 8 ' 3 8 " . 3 3 . G i v e n A = 5 4 ° 1 ' 1 5 " , B = 1 3 7 ° 2 4 , 2 1 " ; f i n d a = 2 9 ° 4 6 ' 8 " , b = 1 5 5 ° 2 7 ' 5 5 " , c = 1 4 2 ° 9 ' 1 2 " . 3 4 . G i v e n A = 4 6 ° 5 9 ' 4 2 " , B = 5 7 ° 5 9 ' 1 7 " ; f i n d a = 3 6 ° 2 7 ' , b = 4 3 ° 3 2 ' 3 7 " , c = 5 4 ° 2 0 ' 3 " . 3 5 . G i v e n A = 5 5 ° 3 2 ' 4 5 " , B = 1 0 1 ° 4 7 ' 5 6 " ; f i n d a = 5 4 ° 4 1 ' 3 5 " , b = 1 0 4 ° 2 1 ' 2 8 " , c = 9 8 ° 1 4 ' 2 4 " . 3 6 . G i v e n A = 6 0 ° 2 7 ' 2 4 " . 3 , B = 5 7 ° 1 6 ' 2 0 " . 2 ; f i n d a = 5 4 ° 3 2 ' 3 2 " . l , 6 = 5 1 ° 4 3 ' 3 6 " . l , c = 6 8 ° 5 6 ' 2 8 " . 9 . E X A M P L E S . 3 1 9 S o l v e t h e f o l l o w i n g q u a d r a n t a l t r i a n g l e s : G i v e n B = 7 4 ° 4 5 ' , f i n d b = 8 5 ° 1 7 ' 1 5 " . 5 , G i v e n A = 1 1 0 ° 4 7 ' 5 0 " , f i n d a = 1 0 4 ° 5 3 ' 0 " . 8 , a = 1 8 ° 1 2 ' , A = 1 7 ° 3 4 ' 2 " , B = 1 3 5 ° 3 5 ' 3 4 ' ' . 5 , 6 = 1 3 3 ° 3 9 ' 4 7 " . 7 , : 9 0 ° ; = 1 0 4 ° 3 1 ' 1 3 " . : 9 0 ° ; : 1 0 4 ° 4 1 ' 3 7 " . 2 . S o l v e t h e f o l l o w i n g o b l i q u e t r i a n g l e s : 3 9 . G i v e n a = 7 3 ° 5 8 ' , b f i n d A = 1 1 6 ° 8 ' 2 8 " , B 4 0 . G i v e n a = 9 6 ° 2 4 ' 3 0 " , b f i n d A = 9 7 ° 5 3 ' 0 £ " , B 4 1 . G i v e n a = 7 6 ° 2 4 ' 4 0 " , b f i n d A = 6 3 C 4 8 ' 3 5 V , B 4 2 . G i v e n a = 8 6 ° 1 8 ' 4 0 " , b f i n d A = 6 4 ° 4 8 ' 5 3 | " , B 4 3 . G i v e n a = 8 8 ° 2 4 ' , 6 f i n d A = 6 5 ° 1 3 ' 3 £ " , B 4 4 . G i v e n a = 6 8 ° 2 0 ' 2 5 " , b f i n d A = 5 6 ° 1 6 ' 1 5 " , B 4 5 . G i v e n a = 8 8 ° 1 2 ' 2 0 " , 6 f i n d A = 6 3 ° 1 5 ' 1 2 " , B 4 6 . G i v e n a = 3 2 ° 2 3 ' 5 7 " , b f i n d A = 6 0 ° 5 3 ' 2 " , B 4 7 . G i v e n 6 = 9 9 ° 4 0 ' 4 8 " , c f i n d B = 9 5 ° 3 8 ' 4 " , C 4 8 . G i v e n A = 3 1 ° 3 4 ' 2 6 " , B f i n d a = 4 0 ° 1 ' 5 ^ " , b 3 8 ° 4 5 ' , C = 4 6 ° 3 3 ' 3 9 " ; 3 5 ° 4 6 ' 3 9 " , c = 5 1 ° I ' l l " . 6 8 ° 2 7 ' 2 6 " , C = 8 4 ° 4 6 ' 4 0 " ; 6 7 ° 5 9 ' 3 9 £ " , c = 8 7 ° 3 1 ' 3 7 " . 5 8 ° 1 8 ' 3 6 " , C = 1 1 6 ° 3 0 ' 2 8 " ; 5 1 ° 4 6 ' 1 2 V , c = 1 0 4 ° 1 3 ' 2 7 " . 4 5 ° 3 6 ' 2 0 " , C = 1 2 0 ° 4 6 ' 3 0 " ; 4 0 ° 2 3 ' 1 5 f " , c = 1 0 8 ° 3 9 ' 1 1 1 " 5 6 ° 4 8 ' , C = 1 2 8 ° 1 6 ' ; 4 9 ° 2 7 ' 5 1 " , c = 1 2 0 ° 1 0 ' 5 2 " . 5 2 ° 1 8 ' 1 5 " , C = = 1 1 7 ° 1 2 ' 2 0 " ; 4 5 ° 4 ' 4 1 " , c = 9 6 ° 2 0 ' 4 4 " . = 1 2 4 ° 7 ' 1 7 " , = 1 3 2 ° 1 7 ' 5 9 " , = 3 2 ° 2 3 ' 5 7 " , = 6 0 ° 5 3 ' 2 " , = 1 0 0 ° 4 9 ' 3 0 " , = 9 7 ° 2 6 ' 2 9 " . l , = 3 0 ° 2 8 ' 1 2 " , = 3 8 ° 3 1 ' 3 V , C = c -C = C : A = a -C z : 5 0 ° 2 ' 1 " ; : 5 9 ° 4 ' 2 5 " . = 6 6 ° 4 9 ' 1 7 " ; = 3 4 ° 1 9 ' 1 1 " . : 6 5 ° 3 3 ' 1 0 " ; = 6 4 ° 2 3 ' 1 5 " . l . : 7 0 ° 2 ' 3 " ; = 1 3 0 ° 3 ' 5 0 " . 8 2 0 S P H E R I C A L T R I G O N O M E T R Y . 4 9 . 5 0 . 5 1 . 5 2 . 5 3 . 5 4 . 5 5 . 5 6 . 5 7 . 5 8 . 5 9 . 6 0 . G i v e n A : f i n d a : G i v e n A : f i n d a -G i v e n A : f i n d a -G i v e n A : f i n d a : G i v e n A : f i n d a - -G i v e n A : f i n d a : G i v e n A : f i n d a : G i v e n A : f i n d a : G i v e n O : f i n d B : o r B ' : G i v e n a -f i n d B : o r B ' : G i v e n a -f i n d B : o r B ' : G i v e n a : f i n d B : o r B ' : = 1 3 0 ° 5 ' 2 2 " . 4 , = 8 4 ° 1 4 ' 2 9 " , = 9 6 ° 4 6 ' 3 0 " , = 1 0 2 ° 2 1 ' 4 2 " , = 8 4 ° 3 0 ' 2 0 " , = 9 4 ° 3 4 ' 5 2 V , = 1 0 7 ° 4 7 ' 7 " , = 7 0 ° 2 0 ' 5 0 " , = 1 2 8 ° 4 1 ' 4 9 " , = 1 2 5 ° 4 4 ' 4 4 " , = 1 2 9 ° 5 8 ' 3 0 " , = 8 5 ° 5 9 ' , = 9 5 ° 3 8 ' 4 " , = 9 9 ° 4 0 ' 4 8 " , = 7 0 ° , 5 7 ° 5 6 ' 5 3 " , = 6 2 ° 1 5 ' 2 4 " , = 6 2 ° 2 4 ' 2 4 " . 8 , = 1 1 7 ° 3 5 ' 3 5 " . 2 , = 5 2 ° 4 5 ' 2 0 " , = 5 9 ° 2 4 ' 1 5 f " , = 1 2 0 ° 3 5 ' 4 4 | " , = 4 8 ° 4 5 ' 4 0 " , = 5 5 ° 3 9 ' 5 7 " , = 1 2 4 ° 2 0 ' 3 " , = 4 6 ° 2 0 ' 4 5 " , = 5 4 ° 6 ' 1 9 " , = 1 2 5 ° 5 3 ' 4 1 " , B = 3 2 ° 2 6 ' 6 " . 4 1 , b = 4 4 ° 1 3 ' 4 5 " , 1 5 = 8 4 ° 3 0 ' 2 0 " , b = 7 8 ° 1 7 ' 2 " , B = 7 6 ° 2 0 ' 4 0 " , b = 7 6 ° 4 0 ' 4 8 1 " , B = 3 8 ° 5 8 ' 2 7 " , b = 3 8 ° 2 7 ' 5 9 " , B = 1 0 7 ° 3 3 ' 2 0 " , 6 = 8 2 ° 4 7 ' 3 5 " , B = 3 4 ° 2 9 ' 3 0 " , b = 4 7 ° 2 9 ' 2 0 " , C = 9 7 ° 2 6 ' 2 9 " , c = 1 0 0 ° 4 9 ' 3 0 " , B = 1 3 1 ° 1 8 ' , b = 1 3 7 ° 2 0 ' 3 3 " , 6 = 1 0 3 ° 1 8 ' 4 7 " , C = 1 5 5 ° 4 3 ' l l " . 3 ; C ' = 5 9 ° 6 ' 1 0 " . 6 , b = 7 1 ° 1 2 ' 4 0 " , C = 1 1 5 ° 3 9 ' 5 5 £ " , C ' = 2 6 ° 5 9 ' 5 5 " . 2 , b = 6 7 ° 1 2 ' 2 0 " , C = 1 1 6 ° 3 4 ' 1 8 " , C ' = 2 4 ° 3 2 ' 1 5 " , b = 6 5 ° 1 8 ' 1 5 " , C = 1 1 6 ° 5 5 ' 2 6 " , C ' = 2 4 ° 1 2 ' 5 3 " . 3 , c = c = c = C = c = C = c = G = c = C = 6 = B = c = C = A c -c ' = A = c - . c ' = A = C = c ' = A = c = = 5 1 ° 6 ' 1 1 " . 0 ; = 3 6 ° 4 5 ' 2 6 " . = 1 2 6 ° 4 6 ' ; = 1 2 5 ° 2 8 ' 1 3 | " . = 1 3 0 ° 4 6 ' ; = 1 3 0 ° 5 1 ' 3 3 - i " . = 5 1 ° 4 1 ' 1 4 " ; = 5 2 ° 2 9 ' 4 5 " . = 1 2 4 ° 1 2 ' 3 1 " ; = 1 2 7 ° 2 2 ' 7 " = 5 0 ° 6 ' 2 0 " ; = 3 6 ° 6 ' 5 0 " . = 6 4 ° 2 3 ' 1 5 " ; = 6 5 ° 3 3 ' 1 0 " . = 1 1 6 ° ; = 9 4 ° 4 8 ' 1 2 " . = 5 3 ° 4 2 ' 3 8 " ; = 1 5 3 ° 9 ' 3 5 i " , > " . 4 6 ° 2 2 ' 1 0 " ; 9 7 ° 3 3 ' 1 8 " . S , 2 9 ° 5 7 ' 1 0 " . 5 . 4 2 ° 2 0 ' 3 0 " ; 9 3 ° 8 ' 9 " . 6 , 2 7 ° 3 7 ' 2 0 " . 4 0 ° 1 0 ' 3 0 " ; 9 0 ° 3 1 ' 4 6 " , 2 7 ° 2 3 ' 1 4 " . E X A M P L E S . 3 2 1 G i v e n o = 1 6 0 ° 5 7 ' 5 " , f i n d B = 1 2 0 ° 4 7 ' 4 4 " , o r B ' = 5 9 ° 1 2 ' 1 6 " , G i v e n a = 5 0 ° 4 5 ' 2 0 " , f i n d B = 5 7 ° 3 4 ' o l " . 4 , o r B ' = 1 2 2 ° 2 5 ' 8 " . 6 , G i v e n a = 4 0 ° 5 ' 2 5 " . 6 , f i n d B = 4 2 ° 3 7 ' 1 7 " . o , o r B ' = 1 3 7 ° 2 2 ' 4 2 " . 5 , G i v e n o = 9 9 ° 4 0 ' 4 8 " , f i n d B = 6 5 ° 3 3 ' 1 0 " , ( N o a m b i g u i t y ; w h y ? ) G i v e n A = 7 9 ° 3 0 ' 4 5 " , f i n d 6 = 3 6 ° 5 ' 3 4 £ " , ( X o a m b i g u i t y ; w h y ? ) G i v e n A = 7 3 ° 1 1 ' 1 8 " , f i n d 6 = 4 1 ° 5 2 ' 3 4 f " , ( O n l y o n e s o l u t i o n ; w h y ? ) G i v e n A = 4 6 ° 3 0 ' 4 0 " , f i n d 6 = 3 3 ° 1 8 ' 4 7 f , ( O n l y o n e s o l u t i o n ; w h y ? ) G i v e n A = 6 1 ° 2 9 ' 3 0 " , f i n d 6 = 1 5 ° 3 0 ' 3 0 " . 5 , ( O n l y o n e s o l u t i o n ; w h y ? ) G i v e n A = 3 6 6 2 0 ' 2 0 " , f i n d b = 5 5 ° 2 5 ' 2 | " , o r b ' = 1 2 4 ° 3 4 ' 5 7 $ " , 6 = C = C , = 6 = C = C ' = b = C = C ' = 6 = C = B = c = B = c -B = c = B = c = B = c = c ' = = 1 3 4 ° 1 5 ' 5 4 " , : 9 7 ° 4 2 ' 5 5 " , : 2 9 ° 9 ' 9 " , = 6 9 ° 1 2 ' 4 0 " , = 1 1 5 ° 5 7 ' 5 0 " . 6 , = 2 5 ° 4 4 ' 3 1 " . 6 , : 1 4 4 ° 2 2 ' 4 2 " ; : 5 5 ° 4 2 ' 8 " , : 2 3 ° 5 7 ' 2 9 " . A = 4 4 ° 2 2 ' 1 0 " ; c = 7 : 1 1 8 ° 2 2 ' 7 " . 3 , : 1 6 0 ° l ' 2 4 " . 4 , j c - -: 5 0 ° 1 8 ' 5 5 " . 2 , J c ' = : 6 4 ° 2 3 ' 1 5 " , A : : 9 7 ° 2 6 ' 2 9 " , = 4 6 ° 1 5 ' 1 5 " , = 5 0 ° 2 4 ' 5 7 " , = 6 1 ° 1 8 ' 1 2 " , = 4 1 ° 3 5 ' 4 " , = 3 6 " 2 0 ' 2 0 " , = 6 0 ° 3 2 ' 6 " , = 2 4 ° 3 0 ' 3 0 " , = 3 9 ° 3 3 ' 5 2 " , = 4 6 ° 3 0 ' 4 0 " , = 8 1 ° 2 7 ' 2 6 V , = 1 6 2 ° 3 4 ' 2 7 " , 9 5 ° 1 8 ' J ^ " . 4 , 2 8 ° 4 5 ' 5 " . 2 . : 2 9 ° 4 2 ' 3 3 " . 8 ; : 1 5 3 ° 3 8 ' 4 2 " . 4 , : 9 0 ° 5 ' 4 1 " . 0 . : 9 5 ° 3 8 ' 4 " ; : 1 0 0 ° 4 9 ' 3 0 " . a = 5 3 ° 1 8 ' 2 0 " ; C = 7 0 ° 5 5 ' 3 5 " . a = 4 6 ° 4 5 ' 3 0 " ; C = 6 0 ° 4 2 ' 4 6 " . 5 . a = 4 2 ° 1 5 ' 2 0 " ; C = 1 1 0 ° 3 ' 1 4 " . 6 . a = 3 4 ° 3 0 " ; C = 9 8 ' 4 8 ' 5 8 " . 5 . a = 4 2 ° 1 5 ' 2 0 " ; C = 1 1 9 ° 2 2 ' 2 7 £ " , C ' = 1 6 4 ° 4 1 ' 5 5 " . 3 2 2 S P H E R I C A L T R I G O N O M E T R Y . 7 0 . G i v e n A -f i n d b = o r 6 ' = 5 2 ° 5 0 ' 2 0 " , 8 1 ° 1 5 ' 1 5 " , 9 8 ° 4 4 ' 4 5 " , I I I I I I 6 6 ° V 2 0 " , 1 1 0 ° 1 0 ' 5 0 f , 1 3 8 ° 4 5 ' 2 6 " , c c = C = C ' = 5 9 ° 2 8 ' 2 7 " ; 1 1 9 ° 4 3 ' 4 8 " , 1 4 2 ° 2 4 ' 5 9 " . 7 1 . G i v e n A = f i n d a = 1 1 5 ° 3 6 ' 4 5 " , 1 1 4 ° 2 6 ' 5 0 " , B = c = 8 0 ° 1 9 ' 1 2 " , 8 2 ° 3 3 ' 3 1 " , b = C = 8 4 ° 2 1 ' 5 6 " ; 7 9 ° 1 0 ' 3 0 " . 7 2 . G i v e n A = f i n d a = o r a ' = ( i l ° 3 7 ' 5 2 " . 7 , 4 2 ° 3 7 ' 1 7 " . 5 , 1 3 7 ° 2 2 ' 4 2 " . 5 , B = c ' = 1 3 9 ° 5 4 ' 3 4 " . 4 , 1 2 9 ° 4 1 ' 4 " . 8 , 1 9 ° 5 8 ' 3 5 " . 6 , b = C = C ' = 1 5 0 ° 1 7 ' 2 0 " . 2 ; 8 9 ° 5 4 ' 1 9 " . 0 , 2 6 ° 2 1 ' 1 7 " . 6 . 7 3 . G i v e n A = 7 0 ° , A r t s . I m p o s s i b l e ; w h y ? B = 1 2 0 ° , 6 = 8 0 ° . 7 4 . G i v e n o = f i n d A = 1 0 8 ° 1 4 ' , 1 2 3 ° 5 3 ' 4 7 " , 6 = B = 7 5 ° 2 9 ' , 5 7 ° 4 6 ' 5 6 " , c = C = 5 6 ° 3 7 ' ; 4 6 ° 5 1 ' 5 1 " . 5 . 7 5 . G i v e n a = f i n d A = 5 7 ° 1 7 ' , 2 1 ° 1 ' 2 " , 6 = B = 2 0 ° 3 9 ' , 8 ° 3 8 ' 4 6 " , C = 7 6 ° 2 2 ' ; 1 5 5 ° 3 1 ' 3 6 " . 5 . 7 6 . G i v e n a = f i n d A = 6 8 ° 4 5 ' , 9 4 ° 5 2 ' 4 0 " , 6 = B = 5 3 ° 1 5 ' , 5 8 ° 5 ' 1 0 " , c = C = 4 6 ° 3 0 ' ; 5 0 ° 5 0 ' 5 2 £ " . 7 7 . G i v e n a = f i n d A = 6 3 ° 5 4 ' , 8 6 ° 3 0 ' 4 0 " , 6 = B = 4 7 ° 1 8 ' , 5 4 ° 4 6 ' 1 4 " , c = C = 5 3 ° 2 6 ' ; 6 3 ° 1 2 ' 5 5 V -7 8 . G i v e n a = f i n d A = 7 0 ° 1 4 ' 2 0 " , 1 1 0 ° 5 1 ' 1 6 " , 6 = B = 4 9 ° 2 4 ' 1 0 " , 4 8 ° 5 6 ' 4 " , c = C = 3 8 ° 4 6 ' 1 0 " ; 3 8 ° 2 6 ' 4 8 " . 7 9 . G i v e n a — f i n d A = 1 2 4 ° 1 2 ' 3 1 " , 1 2 7 ° 2 2 ' 7 " , 6 = B = 5 4 ° 1 8 ' 1 6 " , 5 1 ° 1 8 ' 1 1 " , c = C = 9 7 ° 1 2 ' 2 5 " ; 7 2 ° 2 6 ' 4 0 " . 8 0 . G i v e n a = f i n d A = 5 0 ° 1 2 ' 4 " , 5 9 ° 4 ' 2 5 " , 6 = B = 1 1 6 ° 4 4 ' 4 8 " , 9 4 ° 2 3 ' 1 0 " , c = C = 1 2 9 ° 1 1 ' 4 2 " ; 1 2 0 ° 4 ' 5 0 " . 8 1 . G i v e n a = f i n d A = 1 0 0 ° , 6 = B = 5 0 ° , c = C = 6 0 ° ; 1 3 8 ° 1 5 ' 4 5 " . 4 , 3 1 ° 1 1 ' 1 4 " . 0 , 3 5 ° 4 9 ' 5 8 " . 2 . 8 2 . G i v e n A = f i n d a = 8 6 ° 2 0 ' , 8 7 ° 2 0 ' 2 8 " , B = b = 7 6 ° 3 0 ' , 7 6 ° 4 4 ' 2 | " , C = c = 9 4 ° 4 0 ' ; 9 3 ° 5 5 ' 3 1 " . E X A M P L E S . 3 2 3 8 3 . G i v e n A = 9 6 ° 4 5 ' , B = 1 0 8 ° 3 0 ' , C = 1 1 6 ° 1 5 ' ; f i n d a = 8 8 ° 2 7 ' 4 9 " , b = 1 0 7 ° 1 9 ' 5 2 " , c = 1 1 5 ° 2 8 ' 1 3 i " . 8 4 . G i v e n A = 7 8 ° 3 0 ' , B = 1 1 8 ° 4 0 ' , C = 9 3 ° 2 0 ' ; f i n d « = 7 4 ° 5 7 ' 4 6 " , b = 1 2 0 ° 8 ' 4 9 " , c = 1 0 0 ° 1 8 ' l l f " . 8 5 . G i v e n A = 5 7 ° 5 0 ' , B = 9 8 ° 2 0 ' , C = 6 3 ° 4 0 ' ; f i n d a = 5 8 ° 8 ' 1 9 " , b = 8 3 ° 5 ' 3 6 " , c = 6 4 ° 3 ' 2 0 " . 8 6 . G i v e n A = 1 2 9 ° 5 ' 2 8 " , B = 1 4 2 ° 1 2 ' 4 2 " , C = 1 0 5 ° 8 ' 1 0 " ; f i n d a = 1 3 5 ° 4 9 ' 2 0 " , b = 1 4 4 ° 3 7 ' 1 5 " , c = 6 0 ° 4 ' 5 4 " . 8 7 . G i v e n A = 1 3 8 ° 1 5 ' 5 0 " , B = 3 1 ° 1 1 ' 1 0 " , C = 3 5 ° 5 0 ' ; f i n d a = 1 0 0 ° 0 ' 8 " . 4 , 6 = 4 9 ° 5 9 ' 5 6 " . 4 , c = 6 0 ° 0 ' 1 1 " . 2 . 8 8 . G i v e n A = 1 0 2 ° 1 4 ' 1 2 " , B = 5 4 ° 3 2 ' 2 4 " , C = 8 9 ° 5 ' 4 6 " ; f i n d a — 1 0 4 ° 2 5 ' 8 " , b = 5 3 ° 4 9 ' 2 5 " , c = 9 7 ° 4 4 ' 1 8 " . 8 9 . G i v e n A = 2 0 ° 9 ' 5 6 " , B = 5 5 ° 5 2 ' 3 2 " , C = 1 1 4 ° 2 0 ' 1 4 " ; f i n d « = 2 0 ° 1 6 ' 3 8 " , 6 = 5 6 ° 1 9 ' 4 1 " , c = 6 6 ° 2 0 ' 4 3 " . 9 0 . I f a , 6 , c a r e e a c h < s h o w t h a t t h e g r e a t e r a n g l e m a y e x c e e d - -9 1 . I f a a l o n e > ^ i r , s h o w t h a t A m u s t e x c e e d - . 2 9 2 . I f a a n d 6 a r e e a c h a n d c < f r ^ i r , p r o v e t h a t : ( 1 ) T h e g r e a t e s t a n g l e A m u s t b e ( 2 ) B m a y b e > % i r ; ( 3 ) C m a y o r m a y n o t b e < \ t t . 9 3 . I f c o s a , c o s b , c o s c a r e a l l n e g a t i v e , p r o v e t h a t c o s A , c o s B , c o s C a r e a l l n e c e s s a r i l y n e g a t i v e . 9 4 . I n a s p h e r i c a l t r i a n g l e , o f t h e f i v e p r o d u c t s , c o s a c o s A , c o s b c o s B , c o s c c o s C , c o s a c o s b c o s c , — c o s A c o s B c o s C , s h o w t h a t o n e i s n e g a t i v e , t h e o t h e r f o u r b e i n g p o s i t i v e . 3 2 4 S P H E R I C A L T R I G O N O M E T R Y . C H A P T E R X I I . T H E I N - O I R O L E S A N D E X - O I K O L E S . — A R E A S . 2 1 5 . T h e I n - C i r c l e ( I n s c r i b e d C i r c l e ) . — T o f i n d t h e a n g u l a r r a d i u s o f t h e i n - c i r c l e o f a t r i a n g l e . L e t A B C b e t h e t r i a n g l e ; b i s e c t t h e a n g l e s A a n d B b y t h e a r c s A O , B O ; f r o m O d r a w O D , O E , O F p e r p e n d i c u l a r t o t h e 1 s i d e s . T h e n i t m a y b e s h o w n t h a t / 1 \ 0 i s t h e i n - c e n t r e , a n d t h a t t h e p e r -/ M p e n d i c u l a r s O D , O E , O F a r e e a c h A ^ \ L - ^ P e q u a l t o t h e r e q u i r e d a n g u l a r r a d i u s . / \ / / ^ ^ v ^ A L e t 2 s = t h e s u m o f t h e s i d e s o f / / ^ ^ S ^ ^ t h e t r i a n g l e A B C . T h e r i g h t t r i a n g l e s — — - ^ ^ E O A E , O A F a r e e q u a l . A . - . A F = A E . S i m i l a r l y , B D = B F , a n d C D = C E . . - . B C + A F A C + B F = s . . . . A F s — B C = s — a . N o w t a n O F t a n O A F s i n A F ( A r t . 1 8 6 ) o r , d e n o t i n g t h e r a d i u s O F b y r , w e h a v e A t a n r = t a n — s i n ( s — a ) . . . . ( 1 ) n ( A r t . 1 9 5 ) ( 2 ) s i n s T H E E S C R I B E D C I R C L E S . 3 2 5 A l s o , s i n ( s — a ) = s i n £ ( 6 + c ) - | J = s i n £ ( 6 + c ) c o s ^ a — c o s £ ( 6 + c ) s i n j a s i n ^ - a c o s \ a . A s i n —2 [ c o s | ( B - C ) - c o s £ ( B + C ) ] ( A r t . 1 9 8 ) _ s i n a s i n ^ B s i n ^ C s i n £ A w h i c h i n ( 1 ) g i v e s . B . C s i n — s i n — t a n r = r - r — s i n a ( 3 ) c o s _ J - A N . . ( A r t . 1 9 6 ) ( 4 ) 2 c o s £ A c o s £ B c o s £ C a n e q u a t i o n w h i c h i s e q u i v a l e n t t o t h e f o l l o w i n g : c o t r = - i - [ c o s S + c o s ( S - A ) + c o s ( S - B ) + c o s ( S - C ) ] ( 5 ) 2 1 6 . T h e E x - C i r c l e s . — T o f i n d t h e a n g u l a r r a d i i o f t h e e x - c i r c l e s o f a t r i a n g l e . A c i r c l e w h i c h t o u c h e s o n e s i d e o f a t r i a n g l e a n d t h e o t h e r t w o s i d e s p r o d u c e d , i s c a l l e d a n e s c r i b e d c i r c l e , o r e x - c i r c l e , o f t h e t r i a n g l e . I t i s c l e a r t h a t t h e t h r e e e x - c i r c l e s o f a n y t r i a n g l e a r e t h e i n - c i r c l e s o f i t s c o l u n a r t r i a n g l e s ( A r t . 1 9 1 , S c h . ) . S i n c e t h e c i r c l e e s c r i b e d t o t h e s i d e a o f t h e t r i a n g l e A B C i s t h e i n - c i r c l e o f t h e c o l u n a r t r i a n g l e A ' B C , t h e p a r t s o f w h i c h a r e a , i t — b , v — c , A , i r — B , i r — C , t h e p r o b l e m b e c o m e s i d e n t i c a l w i t h t h a t o f A r t . 2 1 5 ; a n d w e o b t a i n B t h e v a l u e f o r t h e i n - r a d i u s o f t h e c o l u n a r t r i a n g l e A ' B C , b y s u b s t i t u t i n g f o r 6 , c , B , C , t h e i r s u p p l e m e n t s i n t h e f i v e e q u a t i o n s o f t h a t a r t i c l e . 3 2 6 S P H E R I C A L T R I G O N O M E T R Y . H e n c e , d e n o t i n g t h e r a d i u s b y r a , w e g e t t a n r „ = t a n £ A s i n s ( 1 ) = -( 2 ) s i n ( s — a ) c o s A B c o s A C -/ c > \ = » . . . . ( 4 ) 2 c o s £ A s i n | B s i n A , C v ' c o t r . = ^ - [ - c o s S - c o s ( S - A ) + c o s ( S - B ) + c o s ( S - C ) ] ( 5 ) T h e s e f o r m u l a e m a y a l s o b e f o u n d i n d e p e n d e n t l y b y m e t h o d s s i m i l a r t o t h o s e e m p l o y e d i n A r t . 2 1 5 , f o r t h e i n - c i r c l e , a s t h e s t u d e n t m a y s h o w . S c h . S i m i l a r l y , a n o t h e r t r i a n g l e m a y b e f o r m e d b y p r o d u c i n g B C , B A t o m e e t a g a i n , a n d a n o t h e r b y p r o d u c i n g C A , C B t o m e e t a g a i n . T h e c o l u n a r t r i a n g l e s o n t h e s i d e s 6 a n d c h a v e e a c h t w o p a r t s , b a n d B , c a n d C , e q u a l t o p a r t s o f t h e p r i m i t i v e t r i a n g l e , w h i l e t h e i r r e m a i n i n g p a r t s a r e t h e s u p p l e m e n t s i n t h e f o r m e r c a s e o f a , c , A , C , a n d i n t h e l a t t e r , o f a , b , A , B . T h e v a l u e s f o r t h e r a d i i r b a n d r c a r e t h e r e f o r e f o u n d i n t h e s a m e w a y a s t h e a b o v e v a l u e s f o r r a ; o r t h e y m a y b e o b t a i n e d f r o m t h e v a l u e s o f r „ b y a d v a n c i n g t h e l e t t e r s . T h u s , t a n r b = t a n i B s i n s = , e t c . , s i n ( s — 6 ) a n d t a n r c = t a n A C s i n s = , e t c . s i n ( s — c ) 2 1 7 . T h e C i r e u m c i r c l e . — T o f i n d t h e a n g u l a r r a d i u s o f t h e c i r e u m c i r c l e o f a t r i a n g l e . T h e s m a l l c i r c l e p a s s i n g t h r o u g h t h e v e r t i c e s o f a s p h e r i c a l t r i a n g l e i s c a l l e d t h e c i r c u m s c r i b i n g c i r c l e , o r c i r e u m c i r c l e , o f t h e t r i a n g l e . T H E C I R C U M C I R C L E . 3 2 7 L e t A B C b e t h e t r i a n g l e ; b i s e c t t h e s i d e s C B , C A a t D , E , a n d l e t O b e t h e i n t e r s e c t i o n o f p e r p e n d i c u l a r s t o C B , C A , a t D , E ; t h e n O i s t h e c i r c u m -c e n t r e . F o r , j o i n O A , O B , O C ; t h e n ( A r t . 1 8 6 ) c o s O B = c o s B D c o s O D , c o s O C = c o s D C c o s O D . . - . O B = O C . S i m i l a r l y , O C = O A . N o w t h e a n g l e O A B = O B A , O B C = O C B , O C A = O A C . . - . O C B + A = | ( A + B + C ) = S . . - . O C B = S - A . L e t O C = R ; t h e n , i n t h e t r i a n g l e O D C , w e h a v e c o s O C D = t a n C D c o t C O = t a n | a c o t R . ( A r t . 1 8 6 ) t a n + a t a n R : o r t a n R = c o s ( S -A ) c o s S . . . . ( 1 ) ( A r t . 1 9 6 ) ( 2 ) A l s o c o s ( S - A ) = c o s | [ ( B + C ) -A ] = c o s | ( B + C ) c o s | A + s i n £ ( B + C ) s i n i A = ? H L i A ^ i i A [ c o s ^ ( 6 + c ) + c o s | ( 6 - c ) ] ( A r t . 1 9 8 ) c O S G t s i n A c o s £ b c o s \ c , c o s ^ a w h i c h i n ( 1 ) g i v e s t a n R = s i n A c o s \ b c o s ^ c 2 s i n ^ q s i n \ b s i n ^ c n . . . . ( 3 ) ( A r t . 1 9 5 ) ( 4 ) 3 2 8 S P H E R I C A L T R I G O N O M E T R Y . w h i c h m a y b e r e d u c e d t o t h e f o l l o w i n g : t a n R = J - [ s i n ( s — a ) + s i n ( s — b ) + s i n ( s — c ) — s i n s ] ( 5 ) 2 1 8 . C i r c u m e i r c l e s o f C o l u n a r T r i a n g l e s . — T o f i n d t h e a n g u l a r r a d i i o f t h e c i r c u m e i r c l e s o f t h e t h r e e c o l u n a r t r i a n g l e s . L e t R „ R 3 b e t h e a n g u l a r r a d i i o f t h e c i r c u m e i r c l e s o f t h e c o l u n a r t r i a n g l e s o n t h e s i d e s a , b , c , r e s p e c t i v e l y . T h e n , s i n c e R , i s t h e c i r c u m r a d i u s o f t h e t r i a n g l e A ' B G w h o s e p a r t s a r e a , i r — b , i r — c , A , i r — B , i r — C , w e h a v e , f r o m A r t . 2 1 7 , t a n R i = -( ! ) c o s S t a n R ^ c 0 8 < t ~ A ) ( 2 ) t a n R 1 = ( 3 ) s i n A s i n £ 6 s i n ^ c t a n R _ ^ s i n ^ a c o s \ b c o s $ c n ( 4 ) t a n R j = - i - [ s i n s — s i n ( s — a ) + s i n ( s — 6 ) + s i n ( s — c ) ] ( 5 ) S i m i l a r l y , t a n R , t g £ i ^ = c o s ( S ~ B ) = e t c . , 2 c o s S N a n d t a n R 3 = -= c o s ( S ~ ° ) = e t c . 3 c o s S N E X A M P L E S . P r o v e t h e f o l l o w i n g : 1 . c o s s + c o s ( s — a ) + c o s ( s — b ) + c o s ( s — c ) = 4 c o s £ a c o s $ b c o s £ c . 2 . c o s ( s — 6 ) - f c o s ( s — c ) — c o s ( s — a ) — c o s s = 4 c o s i a s i n £ 6 s i n c . P R O B L E M . 3 2 9 3 . t a n n = c 0 S £ C c ° S A s i n 5 = : 5 c o s £ B 2 c o s | B s i n £ C s i n £ A . , c o s A A c o s A B . N 4 . t a n r c = — — s i n c = c o s £ C -2 c o s £ C s i n £ A s i n £ B 5 . c o t r : c o t r j : c o t r 2 : c o t r 3 = s i n s : s i n ( s — a ) : s i n ( s — b ) : s i n ( s — c ) . 6 . t a n r t a n r j t a n r 2 t a n ? 3 = n 2 . 7 . c o t r t a n r j t a n r 2 t a n r 3 = s i n 2 s . 8 t a n R — - c o s i a s ^ n 4 - c o s ^ 0 2 n 9 t a n R = 2 c o s b a c o s \ o s ™\ c , 3 ? i 1 0 . t a n R j : t a n B 2 : t a n R 3 = c o s ( S — A ) : c o s ( S — B ) : c o s ( S — C ) . 1 1 . c o t R c o t B , r c o t R 2 c o t R 3 = N 2 . 1 2 . t a n R c o t R j c o t R 2 c o t R 3 = c o s 2 S . A R E A S O F T R I A N G L E S . 2 1 9 . P r o b l e m . — T o f i n d t h e a r e a o f a s p h e r i c a l t r i a n g l e , h a v i n g g i v e n t h e t h r e e a n g l e s . L e t r = t h e r a d i u s o f t h e s p h e r e . E = t h e s p h e r i c a l e x c e s s = A + B + C — 1 8 0 ° . K = a r e a o f t r i a n g l e A B C . I t i s s h o w n i n G e o m e t r y ( A r t . 7 3 8 ) t h a t t h e a b s o l u t e a r e a o f a s p h e r i c a l t r i a n g l e i s t o t h a t o f t h e s u r f a c e o f t h e s p h e r e a s i t s s p h e r i c a l e x c e s s , i n d e g r e e s , i s t o 7 2 0 ° . . - . K : 4 T 9 - 2 = E : 7 2 0 ° . „ E . 3 3 0 S P H E R I C A L T R I G O N O M E T R Y . C o r . T h e a r e a s o f t h e c o l u n a r t r i a n g l e s a r e ( 2 A - E ) ^ ( 2 B - E ) , ( 2 C - E ) ^ 1 8 0 ° ' 1 8 0 ° , 1 8 0 ° 2 2 0 . P r o b l e m . — T o f i n d t h e a r e a o f a t r i a n g l e , h a v i n g g i v e n t h e t h r e e s i d e s . H e r e t h e o b j e c t i s t o e x p r e s s E i n t e r m s o f t h e s i d e s . I . C a g n o l V s T h e o r e m . s i n £ E = s i n | ( A + B + C -t t ) = s i n $ ( A + B ) s i n — c o s | ( A + B ) c o s = s i n £ C c o s $ C p c o g i ( a _ 6 ) _ c o s ¿ ( a + 6 ) ] . ( A r t . 1 9 8 ) c o s f c _ s i n ^ a s i n ^ 6 s i n C _ s i n ^ a s i n ^ 6 2 n ( A r t 1 9 5 ) ( 1 ) c o s \ c c o s £ c s i n a s i n 6 . - . s i n i E = ( 2 ) 2 c o s j ¡ a c o s £ 6 c o s I I . L h u i l i e r , s T h e o r e m . t a n j E = s i D K A + B + C - . ) ? c o s i ( A + B + C - i r ) _ s i n | ( A + B ) -s i n £ ( t t -C ) i A r t 4 g . c o s i ( A + B ) + c o s | ( i r - C ) ' ' _ s i n | ( A + B ) — c o s ~ ~ c o s | ( A + B ) + s i n ^ C _ c o s £ ( a — 6 ) — c o s £ c _ c o s ^ C , A r f c 1 9 g ; c o s £ ( a + 6 ) + c o s ^ c s i n £ C = 8 i n K « - 6 ) s m f r ( 8 - a ) c o H C ( A r t . 4 5 ) c o s £ s c o s £ ( s — c ) = V í a n | - s t a n £ ( s — a ) t a n £ ( s — 6 ) t a n £ ( s — c ) ( A r t . 1 9 5 ) ( 3 ) A R E A S O F T R I A N G L E S . 3 3 1 2 2 1 . P r o b l e m . — T o f i n d t h e a r e a o f a t r i a n g l e , h a v i n g g i v e n t w o s i d e s a n d t h e i n c l u d e d a n g l e . c o s £ E = c o s [ £ ( A + B ) - ( £ , r - £ C ) ] = c o s £ ( A + B ) s i n £ C + s i n £ ( A + B ) c o s £ C = c o s i ( a + 6 ) s i n 2 £ C + c o s | ( a - 6 ) c o s 2 ^ C ( A r t . 1 9 8 ) = [ c o s ^ a c o s J 6 + s i n £ « s i n ^ 6 c o s C ] s e c ^ c . . ( 1 ) D i v i d i n g ( 1 ) o f A r t . 2 2 0 b y t h i s e q u a t i o n , a n d r e d u c i n g , w e h a v e t a n | E = t a n £ « t a n | 6 s i n C _ ( 2 ) 1 + t a n £ a t a n £ b c o s C E X A M P L E S . 1 . G i v e n a = 1 1 3 ° 2 ' 5 6 " . 6 4 , b = 8 2 ° 3 9 ' 2 8 " . 4 , c = 7 4 ° 5 4 ' 3 1 " . 0 6 ; f i n d t h e a r e a o f t h e t r i a n g l e , t h e r a d i u s o f t h e s p h e r e b e i n g r . B y f o r m u l a ( 3 ) o f A r t . 2 2 0 , £ s = 6 7 ° 3 9 ' 1 4 " . 0 2 5 £ ( s - a ) = l l ° 7 ' 4 5 " . 7 0 5 £ ( s - 6 ) = 2 6 ° 1 9 ' 2 9 " . 8 2 5 b { s -c ) = 3 0 ° l l ' 5 8 " . 4 9 5 l o g t a n £ s = 0 . 3 8 6 0 8 4 0 l o g t a n $ ( s -f l ) = 9 . 2 9 3 8 5 8 3 l o g t a n \ ( s - b ) = 9 . 6 9 4 4 0 5 8 l o g t a n l ( s -c ) = 9 . 7 6 4 9 2 6 1 l o g t a n 2 ^ E = 9 . 1 3 9 2 7 4 2 l o g t a n = 9 . 5 6 9 6 3 7 1 . ^ E = 2 0 ° 2 1 ' 5 8 " . 2 5 . E = 8 1 ° 2 7 ' 5 3 " = 2 9 3 2 7 3 " . . - - K = m m X r > . . . . [ ( 1 ) o f A r t . 2 1 9 ] a = 1 1 3 ° 2 ' 5 6 " . 6 4 b = 8 2 ° 3 9 ' 2 8 " . 4 0 c = 7 4 ° 5 4 ' 3 1 " . 0 6 2 s = = 2 7 0 ° 3 6 ' 5 6 " . 1 0 s = 1 3 5 ° 1 8 ' 2 8 " . 0 5 s - a = 2 2 ° 1 5 ' 3 1 " . 4 1 , s - 6 = 5 2 ° 3 8 ' 5 9 " . 6 5 , s - c = 6 0 ° 2 3 ' 5 6 " . 9 9 . 3 3 2 S P H E R I C A L T R I G O N O M E T R Y . 2 . G i v e n A = 8 4 ° 2 0 ' 1 9 " , B = 2 7 ° 2 2 ' 4 0 " , C = 7 5 ° 3 3 ' ; f i n d E = 7 ° 1 5 ' 5 9 " . 3 . G i v e n a = 4 6 ° 2 4 ' , b = 6 7 ° 1 4 ' , c = 8 1 ° 1 2 ' ; f i n d K = M i t t s ' 4 . G i v e n a = 1 0 8 ° 1 4 ' , 6 = 7 5 ° 2 9 ' , c = 5 6 ° 3 7 ' ; f i n d E = 4 8 ° 3 2 ' 3 4 " . 5 . 5 . P r o v e c o s l E = 1 + c o s a + c o s 6 + c o s c 4 c o s \ a c o s ^ b c o s \ c _ c o s 2 ^ a + c o s 2 ^ 6 + c o s 2 ^ c — 1 2 c o s ^ a c o s ^ f t c o s _ J - c 6 . " s i n 1 E / s i H s s i n ^ ( s - a ) s m i ( s - 6 ) s i n | ( s - c ) 4 \ c o s ^ a c o s ^ 6 c o s ^ c . 7 . " e o s j E = / c o s l s c o H ( s - a ) c o s i ( s - 6 ) c o s ^ ( s - - c ) \ c o s ^ - a c o s ^ f r c o s ^ c 8 " c o t 1 E = c o H C T c o H 6 + c o s C J s i n C _ c o t ^ 6 c o t ^ c + c o s A s i n A c o t \ c c o t ^ a + c o s B s i n B E X A M P L E S . P r o v e t h e f o l l o w i n g : 1 . s i n ( s — a ) + s i n ( s — 6 ) + s i n ( s — c ) — s i n s = 4 s i n ^ r t s i n £ 6 s i n ^ e . 2 . s i n s + s i n ( s — b ) + s i n ( s — c ) — s i n ( s — a ) = 4 s i n ^ a c o s £ 6 c o s ^ c . E X A M P L E S . 3 3 3 3 . s i n ( s — £ > ) s i n ( . s — c ) + s i n ( s — c ) s i n ( s — a ) + s i n ( s — a ) s i n ( s — b ) + s i n s s i n ( s — a ) + s i n s s i n ( s — b ) + s i n s s i n ( s — c ) = s i n 6 s i n c + s i n c s i n a + s i n a s i n b . 4 . s i n ( s — 6 ) s i n ( s — c ) + s i n ( s — c ) s i n ( s — a ) — s i n ( s — < z ) s i n ( s — b ) + s i n s s i n ( s — a ) 4 - s i n s s i n ( s — b ) — s i n s s i n ( s — c ) = s i n 6 s i n c + s i n c s i n a — s i n a s i n b . 5 . s i n 2 s + s i n 2 ( s — o ) + s i n 2 ( s — b ) + s i n 2 ( s — c ) = 2 ( 1 — c o s a c o s f t c o s e ) . 6 . s i n 2 s + s i n 2 ( s — a ) — s i n 2 ( s — b ) — s i n 2 ( s — c ) = 2 c o s a s i n b s i n e . 7 . c o s 2 s + c o s 2 ( s — a ) + c o s 2 ( s — b ) + c o s 2 ( s — c ) = 2 ( 1 + c o s a c o s b c o s c ) . 8 . c o s 2 s + c o s 2 ( s — a ) — c o s 2 ( s — b ) — c o s 2 ( s — c ) = — 2 c o s a s i n b s i n c . 9 . t a n r c o t r j t a n r 2 t a n r 3 = s i n 2 ( s — a ) . 1 0 . t a n r t a n r , c o t r 2 t a n r 3 = s i n 2 ( s — b ) . 1 1 . t a n r t a n r j t a n r 2 c o t r 3 = s i n 2 ( s — c ) . 1 2 . c o t r s i n s = c o t \ A c o t c o t ^ C . 1 3 . t a n r j + t a n r 2 + t a n r 3 — t a n r = — — ^ N s i n S s i n A s i n B s i n C . , . . . 4 s i n X a s i n A 6 s i n A c 1 4 . c o t r , + c o t r 2 + c o t r 3 — c o t ? - = « — ■ , — 1 « » . < . i s i n f f l s i n 6 s i n c 1 0 . t a n r j : t a n r 2 : t a n r s = : : -1 + c o s A 1 + c o s B 1 + c o s C c t a n r , + t a n r , + t a n r , — t a n r , / H , , , , x 1 6 . ^ 2 = + ( l + c o s a + c o s 6 + c o s c ) . c o t r j + c o t r 2 + c o t r 3 — c o t r 3 3 4 S P H E R I C A L T R I G O N O M E T R Y . 1 7 . c o t 2 r , + c o t V 2 + c o t 2 r 3 + c o t 2 r = ^ ( l - c o s a c o s 6 c o s c ) ^ 1 9 . c o t r 2 c o t r 3 + c o t r s c o t r j + c o t r j c o t r 2 + c o t r ( c o t r y - f c o t r 2 + c o t r 3 ) _ s i n b s i n c + s i n c s i n a + s i n a s i n b 2 0 . t a n r 2 t a n r 3 + t a n r 3 t a n r , + t a n r 2 t a n r 2 + t a n r ( t a n 1 \ + t a n r 2 + t a n r 3 ) = s i n 6 s i n c + s i n c s i n a + s i n a s i n b . 2 1 . c o t R t a n R j c o t R 2 c o t R 3 = c o s 2 ( S — A ) . 2 2 . c o t R c o t R j t a n R 2 c o t R 3 = c o s 2 ( S — B ) . 2 3 . c o t R c o t R j c o t R 2 t a n R 3 = c o s 2 ( S -C ) . 2 4 . t a n R j - f t a n R 2 = c o t r + c o t r s . 2 5 . t a n R j + t a n R 2 + t a n R 3 — t a n R = 2 c o t r . 2 6 . t a n R — t a n R j + t a n R 2 + t a n R 3 = 2 c o t r v 2 7 . t a n R + t a n R j — t a n R 2 + t a n R 3 = 2 c o t r 2 . 2 8 . t a n R + t a n R j + t a n R 2 — t a n R 3 = 2 c o t ? 3 . 2 9 . c o t ^ + c o t r 2 + c o t r 3 — c o t r = 2 t a n R . 3 0 . c o t r — c o t r , + c o t r 2 + c o t 1 3 = 2 t a n R ^ 3 1 . c o t r + c o t r j — c o t + c o t r s = 2 t a n R 2 . 3 2 . c o t r + c o t 1 \ + c o t r 2 — c o t r 3 = 2 t a n R 3 . 3 3 . t a n R + c o t r = t a n R , + c o t r j = e t c . , = | ( c o t r 4 - c o t r , + c o t r , + c o t r 3 ) . 3 4 . t a n 2 R + t a n 2 R j + t a n 2 R 2 + t a n 2 R 3 i _ 2 ( 1 + c o s A c o s B c o s C ) n ' 1 8 . 3 2 c o s a s i n 6 s i n c E X A M P L E S . 3 3 5 g g t a n 2 R + t a n 2 R t + t a n 2 R 2 + t a n 2 R 3 _ ^ c o t 2 ? - + c o t 2 r , + c o t 2 ? - 2 + c o t 2 r 3 3 6 . t a n 2 R + t a n 2 R i — t a n 2 R 2 — t a n 2 R 3 _ 2 ( c o s A s i n B s i n C ) 3 7 . e o t 2 r + c o t 2 r , -c o t 2 r 2 -c o t 2 r 3 = 2 c o s « s i n 6 s i n c n 2 g g t a n 2 R + t a n 2 R t — t a n 2 R 2 — t a n 2 R 3 _ _ c o s A c o t 2 r + c o t 2 r r — c o t 2 r 2 — c o t 2 r 3 c o s a 3 9 . t a n R c o t R j = t a n ^ 6 t a n £ c . 4 0 . ( c o t r + t a n R ) 2 + 1 = ( s i n a + + s i n C J . 4 1 . ( c o t r , -t a n R ) 2 + 1 = ^ i n 6 + r i n c -s i n o j , N 4 2 . t a n £ A s i n ( s — a ) = 2 c o s J A c o s \ B c o s £ C £ 3 t a n r _ c o s ( S — A ) c o s ( S — B ) c o s ( S - C ) t a n R 2 c o s £ A c o s £ B c o s £ C 4 4 . c o t ( s — b ) c o t ( s — c ) + c o t ( s — c ) c o t ( s — a ) + c o t ( s — a ) c o t ( s — b ) = c o s e c 2 r . 4 5 . c o t ( s — 6 ) c o t ( s — c ) — c o t s c o t ( s — 6 ) — c o t s c o t ( s — c ) = c o s e c 2 r , -4 6 . c o t ( s — c ) c o t ( s — a ) — c o t s c o t ( s — c ) — c o t s c o t ( s — a ) = c o s e c 2 r 2 . 4 7 . c o t ( s — a ) c o t ( s — 6 ) — c o t s c o t ( s — a ) — c o t s c o t ( s — 6 ) = c o s e c 2 r 3 . c o t ( s — a ) c o t ( s — b ) _ | _ c o t ( 8 — c ) 2 c o t s s i n 2 ^ s i n 2 r 2 s i n 2 r 3 s i n 2 r = 3 c o t ( s — a ) c o t ( s — 6 ) c o t ( s — c ) . 3 3 6 S P H E R I C A L T R I G O N O M E T R Y . 4 9 . c o s e c 2 r , + c o s e c 2 r t + c o s e c 2 r a — c o s e c 2 r = — 2 c o t s [ c o t ( s — a ) + c o t ( s — b ) + c o t ( s — c ) ] . -— I I 1 " T " s i n 2 r s i n 2 ^ s i n 2 r 2 s i n 2 n — 2 2 t a n ( s — a ) t a n ( s — 6 ) t a n s t a n ( s — a ) t a n ( s — b ) t a n ( s — c ) 5 1 . c o t R — c o t R j -c o t B 2 -c o t R 3 -2 N c o s s . n 5 2 . s i n ( A - - E ) -n 2 c o s s i n | 6 s i n £ c 5 3 . s i n ( B - - E ) = n 2 s i n £ o c o s £ 6 s i n £ c 5 4 . s i n ( C - . E ) -M 2 s i n | a s i n £ 6 c o s | c 5 5 . c o s ( A -J E ) — s 1 n 2 £ ^ + s i n 2 | c — s i n 2 £ a 2 c o s £ a s i n £ 6 s i n £ c 5 6 . c o s ( B -i " E \ — s ^ n 2 £ c + s i n 2 ^ a — s i n 2 £ 6 2 s i n £ a c o s | 6 s i n £ c 5 7 . c o s ( C — ¿ E ) — s n 2 b a + s i n 2 £ 6 — s i n 2 £ c 2 s i n £ a s i n c o s \ c 5 8 . c o t ( A — , - n \ c o t ^ - a t a n ^ 6 — c o s C ; s i n C t a n % b t a n | - c + c o s A t a n £ a c o t £ 6 — c o s B s i n A s i n B 5 9 . t a n | ( A - - \ E ) = V c o t ^ - s c o t \ ( s — a ) t a n \ ( s — b ) t a n £ ( s — c ) . 6 0 . t a n £ ( B - - | E ) = V c o t | s t a n ^ ( s — a ) c o t | ( s — 6 ) t a n ^ ( s - - c ) . 6 1 . t a n > ( C - - j E ) = V c o t ^ s t a n i ( s — a ) t a n i ( s — 6 ) c o t ^ ( s - - c ) . 6 2 . I f S , S i , S 2 , S 3 d e n o t e t h e s u m s o f t h e a n g l e s o f a t r i a n g l e a n d i t s t h r e e c o l u n a r s , p r o v e t h a t S + S , + S 2 + S 3 = 3 i r . 6 3 . I n a n e q u i l a t e r a l t r i a n g l e , t a n R = 2 t a n r . E X A M P L E S . 3 3 7 6 4 . I f E b E 2 , E 3 d e n o t e t h e s p h e r i c a l e x c e s s e s o f t h e c o l u n a r s o n a , b , c , r e s p e c t i v e l y , s h o w t h a t E + E : + E 2 + E 3 = 2 i r ; a n d t h e r e f o r e t h e s u m o f t h e a r e a s o f a n y t r i a n g l e a n d i t s c o l u n a r s i s h a l f t h e a r e a o f t h e s p h e r e . 6 5 . G i v e n a = 1 0 8 ° 1 4 ' , b = 7 5 ° 2 9 ' , c = 5 6 ° 3 7 ' ; f i n d E = 4 8 ° 3 2 ' 3 4 " . 5 . 6 6 . G i v e n a = 6 3 ° 5 4 ' , 6 = 4 7 ° 1 8 ' , c = 5 3 ° 2 6 ' ; f i n d E = 2 4 ° 2 9 ' 4 9 £ " . 6 7 . G i v e n a = 6 9 ° 1 5 ' 6 " , 6 = 1 2 0 ° 4 2 ' 4 7 " , c = 1 5 9 ° 1 8 ' 3 3 " ; f i n d E = 2 1 6 ° 4 0 ' 2 3 " . 6 8 . G i v e n a = 3 3 ° l ' 4 o " , b = 1 5 5 ° 5 ' 1 8 " , C = 1 1 0 ° 1 0 ' ; f i n d E = 1 3 3 ° 4 8 ' 5 5 " . 6 9 . G i v e n a = b = c = 1 ° , o n t h e e a r t h , s s u r f a c e ; f i n d E = 2 7 " . 2 1 . 7 0 . G i v e n a = b = c = 6 0 ° , o n a s p h e r e o f 6 i n c h e s r a d i u s ; f i n d t h e a r e a o f t h e t r i a n g l e . A n s . 1 9 . 8 4 5 s q u a r e i n c h e s . 7 1 . I f a n d c = ^ , p r o v e s i n \ E = \ , a n d c o s E = \ . 3 2 7 2 . I f C = ^ , p r o v e s i n \ E = s i n ^ a s i n \ b s e c \ c , a n d c o s i E = c o s ^ a c o s ^ 6 s e c ^ c . 7 3 . I f a = 6 , a n d C = - , p r o v e t a n E = £ t a n a s e c a . 7 4 . I f A + B + C = 2 i r , p r o v e c o s ^ o + c o s ^ 6 + c o s 2 | c = l . 7 5 . I f a + b = i T , p r o v e t h a t E = C ; a n d i f E ' d e n o t e t h e s p h e r i c a l e x c e s s o f t h e p o l a r t r i a n g l e , p r o v e t h a t s i n ^ E ' = s i n a c o s ^ C . 7 6 . P r o v e s i n 2 £ E = V s i n j ^ E s i n ^ E t s i n ^ E ^ s i n ^ E , c o t £ a c o t \ b c o t £ c 3 3 8 S P H E R I C A L T R I G O N O M E T R Y . C H A P T E R X I I I . A P P L I C A T I O N S O F S P H E R I O A L T R I G O N O M E T R Y . S P H E R I C A L A S T R O N O M Y . 2 2 2 . A s t r o n o m i c a l D e f i n i t i o n s . T h e c e l e s t i a l s p h e r e i s t h e i m a g i n a r y c o n c a v e s u r f a c e o f t h e v i s i b l e h e a v e n s i n w h i c h a l l t h e h e a v e n l y b o d i e s a p p e a r t o b e s i t u a t e d . T h e s e n s i b l e h o r i z o n o f a p l a c e i s t h e c i r c l e i n w h i c h a p l a n e t a n g e n t t o t h e e a r t h , s s u r f a c e a t t h e p l a c e m e e t s t h e c e l e s t i a l s p h e r e . T h e r a t i o n a l h o r i z o n i s t h e g r e a t c i r c l e i n w h i c h a p l a n e t h r o u g h t h e c e n t r e o f t h e e a r t h p a r a l l e l t o t h e s e n s i b l e h o r i z o n m e e t s t h e c e l e s t i a l s p h e r e . B e c a u s e t h e r a d i u s o f t h e c e l e s t i a l s p h e r e i s s o g r e a t , i n c o m p a r i s o n w i t h t h e r a d i u s o f t h e e a r t h , t h e s e t w o h o r i z o n s w i l l s e n s i b l y c o i n c i d e , a n d f o r m a g r e a t c i r c l e c a l l e d t h e c e l e s t i a l h o r i z o n . T h e z e n i t h o f a p l a c e i s t h a t p o l e o f t h e h o r i z o n w h i c h i s e x a c t l y o v e r h e a d ; t h e o t h e r p o l e o f t h e h o r i z o n d i r e c t l y u n d e r n e a t h i s c a l l e d t h e n a d i r . V e r t i c a l c i r c l e s a r e g r e a t c i r c l e s p a s s i n g t h r o u g h t h e z e n i t h a n d n a d i r . T h e t w o p r i n c i p a l v e r t i c a l c i r c l e s a r e t h e c e l e s t i a l m e r i d i a n a n d t h e p r i m e v e r t i c a l . T h e c e l e s t i a l m e r i d i a n o f a p l a c e i s t h e g r e a t c i r c l e i n w h i c h t h e p l a n e o f t h e t e r r e s t r i a l m e r i d i a n m e e t s t h e c e l e s t i a l s p h e r e ; t h e p o i n t s i n w h i c h i t c u t s t h e h o r i z o n a r e c a l l e d t h e n o r t h a n d s o u t h p o i n t s . S P H E R I C A L A S T R O N O M Y . 3 3 9 T h e p r i m e v e r t i c a l i s t h e v e r t i c a l c i r c l e w h i c h i s p e r p e n d i c u l a r t o t h e m e r i d i a n ; t h e p o i n t s i n w h i c h i t c u t s t h e h o r i z o n a r e c a l l e d t h e e a s t a n d w e s t p o i n t s . T h e a x i s o f t h e e a r t h o r o f t h e c e l e s t i a l s p h e r e i s t h e i m a g i n a r y l i n e a b o u t w h i c h t h e e a r t h r o t a t e s . T h e c e l e s t i a l e q u a t o r , o r e q u i n o c t i a l , i s t h e g r e a t c i r c l e i n w h i c h t h e p l a n e o f t h e e a r t h , s e q u a t o r i n t e r s e c t s t h e c e l e s t i a l s p h e r e . T h e p o l e s o f t h e e q u i n o c t i a l a r e t h e p o i n t s i n w h i c h t h e a x i s p i e r c e s t h e c e l e s t i a l s p h e r e . H o u r c i r c l e s , o r c i r c l e s o f d e c l i n a t i o n , a r e g r e a t c i r c l e s p a s s i n g t h r o u g h t h e p o l e s o f t h e e q u i n o c t i a l . T h e e c l i p t i c i s a g r e a t c i r c l e o f t h e c e l e s t i a l s p h e r e , a n d t h e a p p a r e n t p a t h o f t h e s u n d u e t o t h e r e a l m o t i o n o f t h e e a r t h r o u n d t h e s u n . T h e e q u i n o x e s a r e t h e p o i n t s i n w h i c h t h e e c l i p t i c c u t s t h e e q u i n o c t i a l . T h e r e a r e t w o , c a l l e d t h e v e r n a l a n d t h e a u t u m n a l e q u i n o x , w h i c h t h e s u n p a s s e s o n M a r c h 2 0 a n d S e p t e m b e r 2 2 . T h e o b l i q u i t y o f t h e e c l i p t i c i s t h e a n g l e b e t w e e n t h e p l a n e s o f t h e e c l i p t i c a n d e q u a t o r , a n d i s a b o u t 2 3 ° 2 7 ' . C i r c l e s o f l a t i t u d e a r e g r e a t c i r c l e s p a s s i n g t h r o u g h t h e p o l e s o f t h e e c l i p t i c . 2 2 3 . S p h e r i c a l C o o r d i n a t e s . — T h e p o s i t i o n o f a p o i n t o n t h e c e l e s t i a l s p h e r e m a y b e d e n o t e d b y a n y o n e o f t h r e e s y s t e m s . I n e a c h s y s t e m t w o g r e a t c i r c l e s a r e t a k e n a s s t a n d a r d s o f r e f e r e n c e , a n d t h e p o i n t i s d e t e r m i n e d b y m e a n s o f t h e s e c i r c l e s , w h i c h a r e c a l l e d i t s s p h e r i c a l c o o r d i n a t e s , a s f o l l o w s : I . T h e h o r i z o n a n d t h e c e l e s t i a l m e r i d i a n o f t h e p l a c e . T h e a z i m u t h o f a s t a r i s t h e a r c o f t h e h o r i z o n i n t e r c e p t e d b e t w e e n t h e s o u t h p o i n t a n d t h e v e r t i c a l c i r c l e , 3 4 0 S P H E R I C A L T R I G O N O M E T R Y . p a s s i n g t h r o u g h t h e s t a r ; i t i s g e n e r a l l y r e c k o n e d f r o m t h e s o u t h p o i n t o f t h e h o r i z o n r o u n d b y t h e w e s t , f r o m 0 ° t o 3 6 0 ° . T h e a l t i t u d e o f a s t a r i s i t s a n g u l a r d i s t a n c e a b o v e t h e h o r i z o n , m e a s u r e d o n a v e r t i c a l c i r c l e . T h e c o m p l e m e n t o f t h e a l t i t u d e i s c a l l e d t h e z e n i t h d i s t a n c e . I I . H i e e q u i n o c t i a l a n d t h e h o u r c i r c l e t h r o u g h t h e v e r n a l e q u i n o x . T h e r i g h t a s c e n s i o n o f a s t a r i s t h e a r c o f t h e e q u i n o c t i a l i n c l u d e d b e t w e e n t h e v e r n a l e q u i n o x a n d t h e h o u r c i r c l e p a s s i n g t h r o u g h t h e s t a r ; i t i s r e c k o n e d e a s t w a r d f r o m 0 ° t o 3 6 0 ° , o r f r o m 0 h t o 2 4 h . T h e a n g l e a t t h e p o l e b e t w e e n t h e h o u r c i r c l e o f t h e s t a r a n d t h e m e r i d i a n o f t h e p l a c e i s c a l l e d t h e h o u r a n g l e o f t h e s t a r . T h e d e c l i n a t i o n o f a s t a r i s i t s d i s t a n c e f r o m t h e e q u i n o c t i a l , m e a s u r e d o n i t s h o u r c i r c l e ; i t m a y b e n o r t h o r s o u t h , a n d i s u s u a l l y r e c k o n e d f r o m 0 ° t o 9 0 ° . I t c o r r e s p o n d s t o t e r r e s t r i a l l a t i t u d e . T h e p o l a r d i s t a n c e o f a s t a r i s i t s d i s t a n c e f r o m t h e p o l e , a n d i s t h e c o m p l e m e n t o f i t s d e c l i n a t i o n . T h e r i g h t a s c e n s i o n a n d d e c l i n a t i o n o f c e l e s t i a l b o d i e s a r e g i v e n i n n a u t i c a l a l m a n a c s . I I I . T h e e c l i p t i c a n d t h e c i r c l e o f l a t i t u d e t h r o u g h t h e v e r n a l e q u i n o x . T h e l a t i t u d e o f a s t a r i s i t s a n g u l a r d i s t a n c e f r o m t h e e c l i p t i c m e a s u r e d o n a c i r c l e o f l a t i t u d e ; i t m a y b e n o r t h o r s o u t h , a n d i s r e c k o n e d f r o m 0 ° t o 9 0 ° . T h e l o n g i t u d e o f a s t a r i s t h e a r c o f t h e e c l i p t i c i n t e r c e p t e d b e t w e e n t h e v e r n a l e q u i n o x a n d t h e c i r c l e o f l a t i t u d e p a s s i n g t h r o u g h t h e s t a r . S P H E R I C A L C O O R D I N A T E S . 3 4 1 2 2 4 . G r a p h i c R e p r e s e n t a t i o n o f t h e S p h e r i c a l C o o r d i n a t e s . — T h e f i g u r e w i l l s e r v e t o i l l u s t r a t e t h e p r e c e d i n g d e f i n i t i o n s . O i s t h e e a r t h , P H P ' R i s t h e m e r i d i a n , P t h e n o r t h p o l e , H R t h e h o r i z o n , E Q t h e e q u i n o c t i a l , Z t h e z e n i t h . T h e n , o f a p l a c e w h o s e z e n i t h i s Z , Q Z i s t h e t e r r e s t r i a l l a t i t u d e ; a n d s i n c e Q Z = P R , . - . P R = t h e l a t i t u d e . B u t P R i s t h e e l e v a t i o n o f t h e p o l e a b o v e t h e h o r i z o n . H e n c e t h e e l e v a t i o n o f t h e p o l e a b o v e t h e h o r i z o n i s e q u a l t o t h e l a t i t u d e . L e t V b e t h e v e r n a l e q u i n o x , a n d l e t S b e a n y h e a v e n l y b o d y , s u c h a s t h e s u n o r a s t a r ; t h e n i t s p o s i t i o n i s d e n o t e d a s f o l l o w s : V K = r i g h t a s c e n s i o n o f t h e b o d y = « . K S = d e c l i n a t i o n " " t t = 8 , Z P S o r Q K = h o u r a n g l e " " u = t , P S = n o r t h p o l a r d i s t a n c e " " i t = P , H T = a z i m u t h « " i t = « , T S = a l t i t u d e " " i t = h , Z S = z e n i t h d i s t a n c e " " i t = z , Q Z = P R = l a t i t u d e o f t h e o b s e r v e r T h e t r i a n g l e Z P S i s c a l l e d t h e a s t r o n o m i c a l t r i a n g l e ; Z P = 9 0 ° — < f > = c o - l a t i t u d e o f t h e o b s e r v e r , P S = 9 0 ° - 8 , S Z = 9 0 ° - 7 i . 3 4 2 S P H E R I C A L T R l G O N O M E T l i T . L e t t h e s m a l l c i r c l e M M ' , p a s s i n g t h r o u g h S , a n d p a r a l l e l t o t h e e q u i n o c t i a l , r e p r e s e n t t h e a p p a r e n t d i u r n a l m o t i o n o f t h e h e a v e n l y b o d y S ( t h e d e c l i n a t i o n b e i n g s u p p o s e d c o n s t a n t ) ; t h e n t h e b o d y S w i l l a p p e a r t o r i s e a t A ( i f w e s u p p o s e t h e E a s t e r n h e m i s p h e r e i s r e p r e s e n t e d i n t h e d i a g r a m ) . I t w i l l b e a t B a t 6 o , c l o c k i n t h e m o r n i n g , a t M a t n o o n , a t M ' a t m i d n i g h t , a n d a t < o i t w i l l b e e a s t . 2 2 5 . P r o b l e m s . — B y m e a n s o f t h e f o r e g o i n g d e f i n i t i o n s a n d d i a g r a m w e m a y s o l v e s e v e r a l a s t r o n o m i c a l p r o b l e m s o f a n e l e m e n t a r y c h a r a c t e r a s f o l l o w s : ( 1 ) G i v e n t h e l a t i t u d e o f a p l a c e a n d t h e d e c l i n a t i o n o f a s t a r ; t o f i n d t h e t i m e o f i t s r i s i n g . L e t A b e t h e p o s i t i o n o f t h e s t a r i n t h e h o r i z o n . T h e n i n t h e t r i a n g l e A P R , r i g h t a n g l e d a t R , w e h a v e c o s R P A = -c o s Z P A = t a n R P c o t A P . . - . c o s t = — t a n t a n 8 ( 1 ) f r o m w h i c h t h e h o u r a n g l e i s f o u n d . S i n c e t h e h o u r l y r a t e a t w h i c h a h e a v e n l y b o d y a p p e a r s t o m o v e f r o m e a s t t o w e s t i s 1 5 ° , i f t h e h o u r a n g l e b e d i v i d e d b y 1 5 t h e t i m e w i l l b e f o u n d . I n t h e c a s e o f t h e s u n , f o r m u l a ( 1 ) g i v e s t h e t i m e f r o m s u n r i s e t o n o o n , a n d h e n c e t h e l e n g t h o f t h e d a y . E x . R e q u i r e d t h e a p p a r e n t t i m e o f s u n r i s e a t a p l a c e w h o s e l a t i t u d e i s 4 0 ° 3 0 ' 2 3 " . 9 , o n J u l y 4 , 1 8 8 1 , w h e n t h e s u n , s d e c l i n a t i o n i s 2 2 ° 5 2 ' 1 " . $ = 4 0 ° 3 6 ' 2 3 " . 9 , S = 2 2 ° 5 2 ' 1 " . l o g t a n = 9 . 9 3 3 1 3 5 2 l o g t a n 8 = 9 . 6 2 5 0 3 6 2 l o g c o s t = 9 . 5 5 8 1 7 1 4 -. - . t = 1 1 1 ° 1 1 ' 4 4 " = 7 h 2 4 m 4 7 s , n e a r l y , P R O B L E M S O F S P H E R I C A L A S T R O N O M Y . 3 4 3 w h i c h t a k e n f r o m 1 2 h , t h e t i m e o f a p p a r e n t n o o n , g i v e s 4 h 3 5 m 1 3 s , t h e t i m e o f a p p a r e n t s u n r i s e . ( 2 ) G i v e n t h e l a t i t u d e o f a p l a c e a n d t h e d e c l i n a t i o n o f a s t a r ; t o f i n d i t s a z i m u t h f r o m t h e n o r t h a t r i s i n g . L e t A = t h e a z i m u t h = A R . T h e n i n t h e t r i a n g l e A P R w e h a v e s i n A P = c o s A R c o s P R , o r s i n 8 = c o s A c o s < f > . . : c o s A = s i n 0 s e c < £ ( 2 ) E x . R e q u i r e d t h e h o u r a n g l e a n d a z i m u t h o f A r c t u r u s w h e n i t r i s e s t o a n o b s e r v e r i n N e w Y o r k , l a t . 4 0 ° 4 2 ' N . , t h e d e c l i n a t i o n b e i n g 1 9 ° 5 7 ' N . A n s . 7 h 1 2 m 4 0 " . 3 ; N . 0 3 ° 1 5 ' 1 1 " E . ( 3 ) G i v e n t h e l a t i t u d e o f t h e o b s e r v e r a n d t h e h o u r a n g l e a n d d e c l i n a t i o n o f a s t a r ; t o f i n d i t s a z i m u t h a n d a l t i t u d e . H e r e w e h a v e g i v e n , i n t h e t r i a n g l e Z P S , t w o s i d e s a n d t h e i n c l u d e d a n g l e ; t h a t i s , P Z = 9 0 ° -< £ , P S = 9 0 ° -8 , a n d Z P S = t . L e t A = t h e a z i m u t h f r o m t h e n o r t h = R T , p = t h e a n g l e Z S P , a n d z = Z S . T h e n b y D e l a m b r e , s A n a l o g i e s ( A r t . 1 9 8 ) , s i n £ ( p + A ) c o s \ z = c o s \ ( h — < f > ) c o s \ t , s i n — A ) s i n \ z = s i n \ ( h — < j > ) c o s £ £ , c o s + A ) c o s \ z = s i n £ ( 8 + < j > ) s i n \ t , c o s i ( p — A ) s i n \ z = c o s | ( 8 + < f > ) s i n - J . H e n c e , w h e n < £ , t , a n d 8 a r e g i v e n , t h a t i s , t h e l a t i t u d e o f t h e p l a c e , a n d t h e h o u r a n g l e a n d t h e d e c l i n a t i o n o f a h e a v e n l y b o d y , A , z , a n d p c a n b e f o u n d . I n a s i m i l a r m a n n e r m a y b e s o l v e d t h e c o n v e r s e p r o b l e m : G i v e n t h e l a t i t u d e o f t h e o b s e r v e r a n d t h e a z i m u t h a n d a l t i t u d e o f a s t a r ; t o f i n d U s h o u r a n g l e a n d . d e c l i n a t i o n . I n t h e s e e x a m p l e s n o c o r r e c t i o n s a r e a p p l i e d f o r r e f r a c t i o n , s e r a i - d i a m e t e r o f t h e s u n , c h a n g e i n d e c l i n a t i o n f r o m n o o n , e t c . 3 4 4 S P H E R I C A L T R I G O N O M E T R Y . ( 4 ) G i v e n t h e r i g h t a s c e n s i o n s a n d d e c l i n a t i o n s o f t w o s t a r s ; t o f i n d t h e d i s t a n c e b e t w e e n t h e m . P a - a ' L e t P b e t h e p o l e , S a n d S ' t h e t w o s t a r s . L e t a a n d « ' b e t h e r i g h t a s c e n s i o n s o f t h e s t a r s ; 8 a n d 8 ' t h e i r d e c l i n a t i o n s ; a n d d t h e r e q u i r e d d i s t a n c e . T h e n w e h a v e g i v e n , i n t h e t r i a n g l e P S S ' , t w o s i d e s a n d t h e i n c l u d e d a n g l e ; t h a t i s , P S = 9 0 ° - 8 = p , P S ' = 9 0 ° -8 ' = p > , a n d P = « -« ' . T h i s m a y b e s o l v e d b y A r t . 1 9 8 , o r b y t h e s e c o n d m e t h o d o f A r t . 2 0 9 , a s f o l l o w s : D r a w S D p e r p e n d i c u l a r t o P S ' p r o d u c e d ; l e t P D = w i . T h e n c o s P = t a n P D c o t P S . . - . t a n m = c o s P t a n p . A l s o c o s S S ' = c o s P S c o s S ' D s e c P D . - . c o s d = c o s p c o s S ' D s e c m . ( A r t . 2 0 9 ) E x . R e q u i r e d t h e d i s t a n c e b e t w e e n S i r i u s a n d A l d e b a r a n , t h e r i g h t a s c e n s i o n s b e i n g 0 h 3 8 m 3 7 ' . 6 a n d 4 h 2 7 m 2 5 ' . 9 , a n d t h e d e c l i n a t i o n s 1 6 ° 3 1 ' 2 " S . a n d 1 0 ° 1 2 ' 2 7 " N . , r e s p e c t i v e l y . l o g c o s P = 9 . 9 2 4 5 7 8 9 l o g t a n p = 0 . 5 2 7 9 1 6 1 -l o g t a n m = 0 . 4 5 2 4 9 5 0 -. - . m = 1 0 9 ° 2 5 ' 5 5 " . l o g c o s S ' D = 9 . 9 0 9 9 3 0 2 l o g c o s p = 9 . 4 5 3 7 8 2 3 -c o l o g c o s m = 0 . 4 7 7 9 6 4 3 -l o g c o s d = 9 . 8 4 1 6 7 6 8 H e r e P = 2 h l l r a l l ' . T = 3 2 ° 4 7 ' 5 5 " P 1 0 6 ° 3 1 ' 2 " m = 1 0 9 ° 2 5 ' 5 5 " p > = 7 3 ° 4 7 ' 3 3 " S ' D = 3 5 ° 3 8 ' 2 2 " P = 1 0 6 ° 3 1 ' 2 " , m = 1 0 9 ° 2 5 ' 5 5 " , S S ' = d 4 6 ° 0 ' 4 4 " . P R O B L E M S O F S P H E R I C A L A S T R O N O M Y . , 3 4 5 ( 5 ) G i v e n t h e r i g h t a s c e n s i o n a n d d e c l i n a t i o n o f a s t a r ; t o f i n d i t s l a t i t u d e a n d l o n g i t u d e . L e t V b e t h e v e r n a l e q u i n o x , S t h e s t a r , V D , V L t h e t h e e q u a t o r a n d t h e e c l i p t i c , S D , S L p e r p e n d i c u l a r t o V D , V L . T h e n V D = r i g h t a s c e n s i o n = « , S D = d e c l i n a t i o n = 8 , V L = l o n g i t u d e = X , S L = l a t i t u d e = I . D e n o t e t h e o b l i q u i t y o f t h e e c l i p t i c D V L b y « > , a n d t h e a n g l e . D V S b y 6 . F r o m t h e r i g h t t r i a n g l e s S V D , S V L w e g e t c o t 6 = s i n a c o t 8 . . . . t a n A . = c o s ( 6 — < d ) t a n « s e c 6 s i n I = s i n ( 0 — & > ) s i n 8 c o s e c 6 ( 1 ) ( 2 ) ( 3 ) F r o m ( 1 ) , 6 i s d e t e r m i n e d ; a n d f r o m ( 2 ) a n d ( 3 ) , A a n d I a r e d e t e r m i n e d . E x . G i v e n t h e r i g h t a s c e n s i o n o f a s t a r 5 h 6 m 4 2 a . 0 1 , a n d i t s d e c l i n a t i o n 4 5 ° 5 1 ' 2 0 " . 1 N . ; t o f i n d i t s l o n g i t u d e a n d l a t i t u d e , t h e o b l i q u i t y o f t h e e c l i p t i c b e i n g 2 3 ° 2 7 ' 1 9 " . 4 5 . a = 7 6 ° 4 0 ' 3 0 " . 1 5 8 = 4 5 ° 5 1 ' 2 0 " . l 6 = 4 6 ° 3 8 ' 1 1 " . 8 a > = 2 3 ° 2 7 ' 1 9 " . 4 5 6 -o > = 2 3 ° 1 0 ' 5 2 " . 3 o A . = 7 9 ° 5 8 ' 3 " . 4 4 . Z = 2 2 ° 5 1 ' 4 8 " . 4 . l o g s i n a l o g c o t 8 l o g c o t 6 l o g c o s ( 6 — i d ) l o g t a n a c o l o g c o s 6 l o g t a n A . l o g s i n ( 6 — w ) l o g s i n 8 c o l o g s i n 6 l o g s i n I = 9 . 9 8 8 1 4 7 9 = 9 . 9 8 7 0 2 7 7 = 9 . 9 7 5 1 7 5 6 = 9 . 9 6 3 4 4 0 1 = 1 0 . 6 2 5 5 2 6 6 = 0 . 1 6 3 2 8 1 6 = 1 0 . 7 5 2 2 4 8 3 = 9 . 5 9 5 0 9 9 6 = 9 . 8 5 5 8 7 4 3 = 0 . 1 3 8 4 5 7 5 = 9 . 5 8 9 4 3 1 4 3 4 6 S P H E R I C A L T M G O N O M E T R Y . E X A M P L E S . 1 . F i n d t h e a p p a r e n t t i m e o f s u n r i s e a t a p l a c e w h o s e l a t i t u d e i s 4 0 ° 4 2 ' , w h e n t h e s u n , s d e c l i n a t i o n i s 1 7 ° 4 9 ' N . A n s . 4 h 5 6 m . 2 . G i v e n t h e l a t i t u d e o f a p l a c e = 4 0 ° 3 6 ' 2 3 " . 9 , t h e h o u r a n g l e o f a s t a r = 4 6 ° 4 0 ' 4 " . 5 , a n d i t s d e c l i n a t i o n = 2 3 ° 4 ' 2 4 " . 3 ; t o f i n d i t s a z i m u t h a n d a l t i t u d e . A n s . A z i m u t h = 8 0 ° 2 3 ' 4 " . 4 7 , a l t i t u d e = 4 7 ° 1 5 ' 1 8 " . 3 . 3 . F i n d t h e a l t i t u d e a n d a z i m u t h o f a s t a r t o a n o b s e r v e r i n l a t i t u d e 3 8 ° 5 3 ' N . , w h e n t h e h o u r a n g l e o f t h e s t a r i s 3 h 1 5 m 2 0 ' W . , a n d t h e d e c l i n a t i o n i s 1 2 ° 4 2 ' N . A n s . A l t i t u d e = 3 9 ° 3 8 ' 0 " ; a z i m u t h = S . 7 2 ° 2 8 ' 1 4 " W . 4 . G i v e n t h e l a t i t u d e s o f N e w Y o r k C i t y a n d L i v e r p o o l 4 0 ° 4 2 ' 4 4 " N . a n d 5 3 ° 2 5 ' N . , r e s p e c t i v e l y , a n d t h e i r l o n g i t u d e s 7 4 ° 0 ' 2 4 " W . a n d 3 ° W . , r e s p e c t i v e l y ; t o f i n d t h e s h o r t e s t d i s t a n c e o n t h e e a r t h , s s u r f a c e b e t w e e n t h e m i n m i l e s , c o n s i d e r i n g t h e e a r t h a s a p e r f e c t s p h e r e w h o s e r a d i u s i s 3 9 5 6 m i l e s . N o t e . — T h i s i s e v i d e n t l y a c a s e o f ( 4 ) w h e r e t w o s i d e s a n d t h e i n c l u d e d a n g l e a r e g i v e n , t o f i n d t h e t h i r d s i d e . A n s . 3 3 0 5 m i l e s . 5 . T h e l a t i t u d e s o f P a r i s a n d P e k i n a r e 4 8 ° 5 0 ' 1 4 " K a n d 3 9 ° 5 4 ' 1 3 " N . , a n d t h e i r d i f f e r e n c e o f l o n g i t u d e i s 1 1 4 ° 7 ' 3 0 " ; f i n d t h e d i s t a n c e b e t w e e n t h e m i n d e g r e e s . A n s . 7 3 ° 5 6 ' 4 0 " . G E O D E S Y . 2 2 6 . T h e C h o r d a l T r i a n g l e . — G i v e n t w o s i d e s a n d t h e i n c l u d e d a n g l e o f a s p h e r i c a l t r i a n g l e ; t o f i n d t h e c o r r e s p o n d i n g a n g l e o f t h e c h o r d a l t r i a n g l e . G E O D E S Y . 3 4 7 T h e c h o r d a l t r i a n g l e i s t h e t r i a n g l e f o r m e d b y t h e c h o r d s o f t h e s i d e s o f a s p h e r i c a l t r i a n g l e . L e t A B C b e a s p h e r i c a l t r i a n g l e , O t h e / c e n t r e o f t h e s p h e r e , A ' B C t h e c o l u n a r t r i a n g l e , a n d M , N t h e m i d d l e p o i n t s o f \ t h e a r c s A ' B , A ' C . T h e n t h e c h o r d A B i s M p a r a l l e l t o t h e r a d i u s O M , s i n c e t h e y a r e b o t h p e r p e n d i c u l a r t o t h e c h o r d A ' B . S i m i l a r l y , A C i s p a r a l l e l t o O N " . I n t h e s p h e r i c a l t r i a n g l e A ' M j S T , w e h a v e c o s M N = c o s A ' N c o s A ' M + s i n A ' N s i n A ' M c o s A ' ( A r t . 1 9 1 ) D e n o t e t h e a n g l e B A C o f t h e c h o r d a l t r i a n g l e b y A j . T h e n a r c M N o r a n g l e M O N " = A u A ' N = 6 ) , A ' M = £ ( - < ; ) , a n d A ' = A . . - . c o s A , = s i n $ b s i n ^ - c + c o s ^ b c o s ^ c c o s A . ( 1 ) w i t h s i m i l a r v a l u e s f o r c o s l 5 j a n d c o s C ^ C o r . 1 . I f t h e s i d e s b a n d c a r e s m a l l c o m p a r e d w i t h t h e r a d i u s o f t h e s p h e r e , A , w i l l n o t d i f f e r m u c h f r o m A . L e t A j = A — 6 ; t h e n c o s A j = c o s A + 6 s i n A , n e a r l y . B u t s i n ^ 6 s i n ^ c = s i n 2 \ ( b + c ) — s i n 2 \ ( b — c ) , a n d c o s % b c o s £ c = c o s 2 £ ( 6 + c ) — s i n 2 \ ( b — c ) . S u b s t i t u t i n g i n ( 1 ) a n d r e d u c i n g , w e g e t 0 = t a n £ A s i n 2 ^ ( 6 + c ) - c o t £ A s i n 2 i ( 6 - c ) . ( 2 ) w h i c h i s t h e c i r c u l a r m e a s u r e o f t h e e x c e s s o f a n a n g l e o f t h e s p h e r i c a l t r i a n g l e o v e r t h e c o r r e s p o n d i n g a n g l e o f t h e c h o r d a l t r i a n g l e . T h e v a l u e i n s e c o n d s i s o b t a i n e d b y d i v i d i n g t h e c i r c u l a r m e a s u r e b y t h e c i r c u l a r m e a s u r e o f o n e s e c o n d , o r , a p p r o x i m a t e l y , b y t h e s i n e o f o n e s e c o n d . 3 4 8 S P H E R I C A L T R I G O N O M E T R Y . C o r . 2 . T h e a n g l e s o f t h e c h o r d a ! t r i a n g l e a r e , r e s p e c t i v e l y , e q u a l t o t h e a r c s j o i n i n g t h e m i d d l e p o i n t s o f t h e s i d e s o f t h e c o l u n a r t r i a n g l e s . 2 2 7 . L e g e n d r e ' s T h e o r e m . — I f t h e s i d e s o f a s p h e r i c a l t r i a n g l e b e s m a l l c o m p a r e d w i t h t h e r a d i u s o f t h e s p h e r e , t h e n e a c h a n g l e o f t h e s p h e r i c a l t r i a n g l e e x c e e d s b y o n e - t h i r d o f t h e s p h e r i c a l e x c e s s t h e c o r r e s p o n d i n g a n g l e o f t h e p l a n e t r i a n g l e , t h e s i d e s o f w h i c h a r e o f t h e s a m e l e n g t h a s t h e a r c s o f t h e s p h e r i c a l t r i a n g l e . L e t a , b , c b e t h e l e n g t h s o f t h e s i d e s o f t h e s p h e r i c a l t r i a n g l e , a n d r t h e r a d i u s o f t h e s p h e r e ; t h e n t h e c i r c u l a r m e a s u r e s o f t h e s i d e s a r e r e s p e c t i v e l y — , - -H e n c e , j r r r n e g l e c t i n g p o w e r s o f — a b o v e t h e f o u r t h , r a b e c o s c o s c o s -c o s A = — -— ( A r t . 1 9 1 ) b . c s i n s i n -r r ( l - a . , + \ 2 r 2 4 r 4 2 r 2 4 r V \ 2 r 2 4 r VJ ( A r t . 1 5 6 ) r \ 6 r y v _ ^ / y - + c 2 -a 2 a 4 -b -c 4 - 6 b 2 f \ A _ b 2 + c \ - 1 ~ b c \ 2 i 2 4 r 4 A 6 r J b 2 + c 2 -a 2 f q ^ - f t ^ - ^ - 0 f t 2 c ^ ^ + 6 2 + ( A ' --( 1 ) 2 6 c 2 4 f t C r 2 J \ 6 b 2 + c 3 -a ? 2 b - t r + 2 c 2 a 2 + 2 a 2 b 2 -a -b -c 2 b e 2 4 6 c r 2 T h e t e r m -i s t h e c i r c u l a r m e a s u r e o f t h e a n g l e w h i c h t h e a r c a s u b t e n d s a t t h e r b e c e n t r e o f t h e s p h e r e ; a n d s i m i l a r l y f o r _ a n d — G E O D E S Y . 3 4 9 N o w i f A ' , B ' , C d e n o t e t h e a n g l e s o f t h e p l a n e t r i a n g l e w h o s e s i d e s a r e a , b , c , r e s p e c t i v e l y , w e h a v e c o s A ' = 6 2 + 2 ^ ~ f t 2 . . . . ( A r t . 9 6 ) , - J A , 2 b 2 c 2 + 2 c 2 a ? + 2 a 2 b 2 - a - b - c 4 . 1 A n , a n d s m s A ' = ! ' . — ( A r t . 1 0 0 ) 4 b 2 c 2 K ' T h e r e f o r e ( 1 ) b e c o m e s c o s A = c o s A ' - 6 c S 1 " 2 A ' . . . . ( 2 ) L e t A = » A ' + 6 , w h e r e 6 i s a v e r y s m a l l q u a n t i t y ; t h e n c o s A = c o s A ' — 6 s i n A ' , n e a r l y . , = t e « a A ' = A ( A r t . 1 0 1 ) w h e r e A d e n o t e s t h e a r e a o f t h e p l a n e t r i a n g l e w h o s e s i d e s a r e a , b , c . W e h a v e t h e r e f o r e A = A ' 4 - — -3 r 2 S i m i l a r l y B = B ' + A , C = C + A 3 r 3 r . . . A + B + C -A ' - B ' - C ' = - o ; o r A + B + C — j r = — = s p h e r i c a l e x c e s s ( A r t . 2 1 9 ) . - . A - A ' = B - B ' = C - C ' = A _ i s p h e r i c a l e x c e s s . 3 r 3 r C o r . 1 . I f t h e s i d e s o f a s p h e r i c a l t r i a n g l e b e v e r y s m a l l c o m p a r e d w i t h t h e r a d i u s o f t h e s p h e r e , t h e a r e a o f t h e s p h e r i c a l t r i a n g l e i s a p p r o x i m a t e l y e q u a l t o Y i + + y + Y \ 2 4 r J 3 5 0 S P H E R I C A L T R I G O N O M E T R Y . F o r , t a n £ E = A / t a n — t a n — t a n — t a n — ( A r t . 2 2 0 ) 1 \ 2 r 2 r 2 r 2 r a n d t a n ± = ± f l + _ s - 1 ; t a n ^ = t ^ f l + e t c . 2 r 2 r \ _ 1 2 r ! J ' 2 r 2 r | _ 1 2 ^ J , ( A r t . 1 5 6 ) t a n j E = J - g - . g = g . £ = j . i = g r i + _ ^ ¥ i + ( s - » ) ^ T \ 2 r 2 r 2 r 2 r | _ 1 2 r ! J | _ 1 2 r 2 . . . i e = / 7 T « ' + ( » -g ) ! + ( 8 -f t p + ( a - c ) ! ( A r t s . 1 0 1 a n d 1 5 6 ) = A A a 2 + b 1 + c ° \ = A A g + y + c \ 4 ^ 1 2 r 3 j 4 r \ 2 4 r 2 / ^ 2 4 r 2 y T h a t i s : t h e a r e a o f t h e s p h e r i c a l t r i a n g l e e x c e e d s t h e a r e a o f t h e p l a n e t r i a n g l e b y a ^ ~ ~ 0 £ > a r £ o / < f t e f a t t e r . 2 4 r " C o r . 2 . I f w e o m i t t e r m s o f t h e s e c o n d d e g r e e i n - , w e h a v e r E r = A . H e n c e , i f t h e s i d e s o f a s p h e r i c a l t r i a n g l e b e v e r y s m a l l c o m p a r e d w i t h t h e r a d i u s o f t h e s p h e r e , i t s a r e a i s a p p r o x i m a t e l y e q u a l t o t h e a r e a o f t h e p l a n e t r i a n g l e h a v i n g s i d e s o f t h e s a m e l e n g t h . 2 2 8 . R o y ' s R u l e . — T h e a r e a o f a s p h e r i c a l t r i a n g l e o n t h e E a r t h , s s u r f a c e b e i n g k n o w n , t o e s t a b l i s h a f o r m u l a f o r c o m p u t i n g t h e s p h e r i c a l e x c e s s i n s e c o n d s . L e t A b e t h e a r e a o f t h e t r i a n g l e i n s q u a r e f e e t , a n d n t h e n u m b e r o f s e c o n d s i n t h e s p h e r i c a l e x c e s s . T h e n w e h a v e R O Y ' S R U L E . 3 5 1 A = E x 6 0 x 6 0 ^ 1 8 0 ° x 6 0 x 6 0 v ; _ n i r t - 2 _ n r 2 ~ 1 8 0 x 6 0 x 6 0 ~ ~ 2 0 6 2 6 5 > N o w , t h e l e n g t h o f a d e g r e e o n t h e E a r t h , s s u r f a c e i s f o u n d b y a c t u a l m e a s u r e m e n t t o b e 3 6 5 1 5 5 f e e t . . . . - ^ = 3 6 5 1 5 5 . . - . r = 1 8 0 x 3 6 5 1 5 5 . 1 8 0 ° . i r S u b s t i t u t i n g t h i s v a l u e o f r i n ( 1 ) , a n d r e d u c i n g , w e g e t l o g n = l o g A - 9 . 3 2 6 7 7 3 7 ( 2 ) T h i s f o r m u l a i s c a l l e d G e n e r a l R o y , s r u l e , a s i t w a s u s e d b y h i m i n t h e T r i g o n o m e t r i c S u r v e y o f t h e B r i t i s h I s l e s . H e g a v e i t i n t h e f o l l o w i n g f o r m : F r o m t h e l o g a r i t h m o f t h e a r e a o f t h e t r i a n g l e , t a k e n a s a p l a n e t r i a n g l e , i n s q r t a r e f e e t , s u b t r a c t t h e c o n s t a n t l o g a r i t h m 9 . 3 2 6 7 7 3 7 ; a n d t h e r e m a i n d e r i s t h e l o g a r i t h m o f t h e e x c e s s a b o v e 1 8 0 ° , i n s e c o n d s , n e a r l y . E x . I f t h e o b s e r v e d a n g l e s o f a s p h e r i c a l t r i a n g l e a r e 4 2 ° 2 ' 3 2 " , 6 7 ° 5 5 ' 3 9 " , 7 0 ° 1 ' 4 8 " , a n d t h e s i d e o p p o s i t e t h e a n g l e A i s 2 7 4 0 4 . 2 f e e t , r e q u i r e d t h e n u m b e r o f s e c o n d s i n t h e s u m o f t h e e r r o r s m a d e i n o b s e r v i n g t h e t h r e e a n g l e s . H e r e t h e a p p a r e n t s p h e r i c a l e x c e s s i s A + B + C -1 8 0 ° = -1 " . T h e a r e a o f t h e t r i a n g l e i s c a l c u l a t e d f r o m t h e e x p r e s s i o n a 2 s i n B s i n C 2 s i n A ( A r t . 1 0 1 ) a n d b y R o y , s R u l e t h e c o m p u t e d s p h e r i c a l e x c e s s i s f o u n d t o b e . 2 3 " . N o w s i n c e t h e c o m p u t e d s p h e r i c a l e x c e s s m a y b e s u p p o s e d t o b e t h e r e a l s p h e r i c a l e x c e s s , t h e s u m o f t h e o b s e r v e d a n g l e s o u g h t t o h a v e b e e n 1 8 0 ° + . 2 3 " . H e n c e i t a p p e a r s t h a t t h e s u m o f t h e e r r o r s o f t h e o b s e r v a t i o n s i s . 2 3 " — ( — 1 " ) = 1 " . 2 3 , w h i c h t h e o b s e r v e r m u s t 3 5 2 S P H E R I C A L T R I G O N O M E T R Y . a d d t o t h e t h r e e o b s e r v e d a n g l e s , i n s u c h p r o p o r t i o n s a s h i s j u d g m e n t m a y d i r e c t . O n e w a y i s t o i n c r e a s e e a c h o f t h e o b s e r v e d a n g l e s b y o n e - t h i r d o f 1 " . 2 3 , a n d t a k e t h e a n g l e s t h u s c o r r e c t e d f o r t h e t r u e a n g l e s . 2 2 9 . K e d u c t i o n o f a n A n g l e t o t h e H o r i z o n . — G i v e n t h e a n g l e s o f e l e v a t i o n o r d e p r e s s i o n o f t w o o b j e c t s , w h i c h a r e a t a s m a l l a n g u l a r d i s t a n c e f r o m t h e h o r i z o n , a n d t h e a n g l e w h i c h t h e o b j e c t s s u b t e n d , t o f i n d t h e h o r i z o n t a l a n g l e b e t w e e n t h e m . L e t a , b b e t h e t w o o b j e c t s , t h e a n g u l a r d i s t a n c e b e t w e e n w h i c h i s m e a s u r e d b y a n o b s e r v e r a t O ; l e t O Z b e t h e d i r e c t i o n a t r i g h t a n g l e s t o t h e o b s e r v e r , s h o r i z o n . D e s c r i b e a s p h e r e r o u n d O a s a c e n t r e , a n d l e t v e r t i c a l p l a n e s t h r o u g h O a , O b , m e e t t h e h o r i z o n a t O A , O B , r e s p e c t i v e l y ; t h e n t h e h o r i z o n t a l a n g l e A O B , o r A B , i s r e q u i r e d . L e t a b = 6 , A B = 6 + x , A a = h , ' B b = k . T h e n i n t h e t r i a n g l e a Z b w e h a v e c o s a b — c o s a Z c o s b Z c o s A B = c o s a Z b = -o r c o s ( 6 + x ) = s i n a Z s i n b Z c o s 6 — s i n h s i n k c o s h c o s k T h i s g i v e s t h e e x a c t v a l u e o f A B ; b y a p p r o x i m a t i o n w e o b t a i n , w h e r e x i s e s s e n t i a l l y s m a l l , n . a c o s 6 — h k . - . x s i n 6 = h k — % ( h 2 + k 2 ) c o s 6 , n e a r l y . x = 2 h k -( h 2 + k 2 ) ^ c o s 2 1 -s i n 2 1 ^ 2 s i n 0 = \ [ ( 7 i + k y t a . n $ 6 - ( h -k ) 2 c o t £ S M A L L V A R I A T I O N S I N P A R T S O F T R I A N G L E S . 3 5 3 E X A M P L E S . 1 . P r o v e t h a t t h e a n g l e s s u b t e n d e d b y t h e s i d e s o f a s p h e r i c a l t r i a n g l e a t t h e p o l e o f i t s c i r c u m c i r c l e a r e r e s p e c t i v e l y d o u b l e t h e c o r r e s p o n d i n g a n g l e s o f i t s c h o r d a l t r i a n g l e . 2 . I f A i , B „ d ; A 2 , B 2 , C 2 ; A 3 , B 3 , C 3 ; b e t h e a n g l e s o f t h e c h o r d a l t r i a n g l e s o f t h e c o l u n a r s , p r o v e t h a t o o s A ^ c o s J a s i n S , c o s B , = s i n J 6 s i n ( S — C ) , c o s C 1 = s i n J c s i n ( S — B ) , c o s A 2 = s i n 2 a s i n ( S — C ) , c o s B 2 = c o s 2 f t s i n S , c o s C 2 = s i n J c s i n ( S — A ) , c o s A 3 = s i n J o s i n ( S — B ) , c o s B 3 = s i n J 6 s i n ( S — A ) , c o s C 3 = c o s J c s i n S . 3 . P r o v e L e g e n d r e , s T h e o r e m f r o m e i t h e r o f t h e f o r m u l a e f o r s i n £ A , c o s \ A , t a n \ A , r e s p e c t i v e l y , i n t e r m s o f t h e s i d e s . 4 . I f C = A + B , p r o v e c o s C = — t a n \ a t a n \ b . 2 3 0 . S m a l l V a r i a t i o n s i n t h e F a r t s o f a S p h e r i c a l T r i a n g l e . I t i s s o m e t i m e s i m p o r t a n t i n G e o d e s y a n d A s t r o n o m y t o d e t e r m i n e t h e e r r o r i n t r o d u c e d i n t o o n e o f t h e c o m p u t e d p a r t s o f a t r i a n g l e f r o m a n y s m a l l e r r o r i n t h e g i v e n p a r t s . I f t w o p a r t s o f a s p h e r i c a l t r i a n g l e r e m a i n c o n s t a n t , t o d e t e r m i n e t h e r e l a t i o n b e t w e e n t h e s m a l l v a r i a t i o n s o f a n y o t h e r t w o p a r t s . S u p p o s e C a n d c t o r e m a i n c o n s t a n t . ( 1 ) R e q u i r e d t h e r e l a t i o n b e t w e e n t h e s m a l l v a r i a t i o n s o f a s i d e a n d t h e o p p o s i t e a n g l e ( a , A ) . T a k e t h e e q u a t i o n W e s u p p o s e a a n d A t o r e c e i v e v e r y s m a l l i n c r e m e n t s d a a n d d A ; t h e n w e r e q u i r e t h e r a t i o o f d a a n d d A w h e n b o t h a r e e x t r e m e l y s m a l l . T h u s s i n A s i n c = s i n C s i n a ( 1 ) s i n ( A + d A ) s i n c = s i n C s i n ( a - f d a ) , 3 5 4 S P H E R I C A L T R I G O N O M E T R Y . o r ( s i n A c o s d A + c o s A s i n d A ) s i n c = s i n C ( s i n a c o s d a + c o s a s i n d a ) ( 2 ) B e c a u s e t h e a r c s d A a n d d a a r e e x t r e m e l y s m a l l , t h e i r s i n e s a r e e q u a l t o t h e a r c s t h e m s e l v e s a n d t h e i r c o s i n e s e q u a l 1 : t h e r e f o r e ( 2 ) m a y b e w r i t t e n s i n A s i n c + c o s A s i n c d A = s i n C s i n a + s i n C c o s a d a ( 3 ) S u b t r a c t i n g ( 1 ) f r o m ( 3 ) , w e h a v e c o s A s i n c d A = s i n C c o s a d a . d a _ c o s A s i n c _ t a n a d A s i n C c o s a t a n A ( 2 ) R e q u i r e d t h e r e l a t i o n b e t w e e n t h e s m a l l v a r i a t i o n s o f t h e o t h e r s i d e s ( a , b ) . W e h a v e c o s c = c o s a c o s b + s i n a s i n 6 c o s C ( 1 ) . - . c o s c = c o s ( a + d a ) c o s ( b + d b ) + s i n ( a + d a ) s i n ( o + d b ) c o s C , o r = ( c o s a — s i n a d a ) ( c o s 6 — s i i i b d b ) + ( s i n a + c o s a d a ) ( s i n b + c o s b d b ) c o s C . ( 2 ) S u b t r a c t i n g ( 2 ) f r o m ( 1 ) a n d n e g l e c t i n g t h e p r o d u c t d a d b , w e h a v e 0 = ( s i n a c o s b — c o s a s i n b c o s C ) d a + ( c o s a s i n b — s i n a c o s b c o s C ) d b , Q r q _ ( c o t 6 s i n a — c o s a c o s C ) ^ a . ( c o t a s i n f t — c o s o c o s C ) ^ s i n a s i n 6 q _ c o t B s i n C d a c o t A s i n C d b ( A r t 1 9 3 ) s i n a s i n b d a _ _ _ c o s A d b c o s B ( 3 ) R e q u i r e d t h e r e l a t i o n b e t w e e n t h e s m a l l v a r i a t i o n s o f t h e o t h e r a n g l e s ( A , B ) . S M A L L V A R I A T I O N S I N P A R T S O F T R I A N G L E S . 3 5 5 B y m e a n s o f t h e p o l a r t r i a n g l e , w e m a y d e d u c e f r o m t h e r e s u l t j u s t f o u n d , t h a t d A _ _ c o s a d B c o s b ( 4 ) R e q u i r e d t h e r e l a t i o n b e t w e e n t h e s m a l l v a r i a t i o n s o f a s i d e a n d t h e a d j a c e n t a n g l e ( b , A ) . W e h a v e c o t c s i n b = c o t C s i n A + c o s b c o s A . . . ( A r t . 1 9 3 ) G i v i n g t o b a n d A v e r y s m a l l i n c r e m e n t s , a n d s u b t r a c t i n g , a s b e f o r e , w e g e t c o t c c o s b d b = c o t C c o s A d A — s i n b c o s A d b — c o s b s i n A d A . ( c o t e c o s b + s i n b c o s A ) d b = ( c o t C c o s A — c o s b s i n A ) d A . . . . 5 2 1 ^ d b = -^ ! J d a , A r t s 1 9 1 a n d 1 9 2 s s i n c s i n C d b _ c o s B s i n ? > _ s i n 6 c o t B d A c o s a s i n B c o s a E X A M P L E S . 1 . I f A a n d c a r e c o n s t a n t , p r o v e t h e f o l l o w i n g r e l a t i o n s b e t w e e n t h e s m a l l v a r i a t i o n s o f a n y t w o p a r t s o f t h e o t h e r e l e m e n t s : d a _ _ t a n a . d b _ s i n a d C ~ ~ t a n C , d ~ B ~ s i n C -d 6 t a n a r f C d C s i n C , d B = — c o s « . I f B a n d C r e m a i n c o n s t a n t , p r o v e t h e f o l l o w i n g : d b t a n b d A . - r , . n — = ; — = s i n B s i n C . d c t a n c d a d A , , , , d a . s i n a - = s i n A t a n o ; d c d c s i n c c o s 6 3 5 6 S P H E R I C A L T R I G O N O M E T R Y . P O L Y E D R O N S . 2 3 3 . T o f i n d t h e I n c l i n a t i o n o f T w o A d j a c e n t F a c e s o f a R e g u l a r F o l y e d r o n . L e t C a n d D b e t h e c e n t r e s o f t h e c i r c l e s i n s c r i b e d i n t h e t w o a d j a c e n t f a c e s w h o s e c o m m o n e d g e i s A B ; b i s e c t A B i n E , a n d j o i n C E a n d D E ; C E a n d D E w i l l b e p e r p e n d i c u l a r t o A B . . - . Z C E D i s t h e i n c l i n a t i o n o f t h e t w o a d j a c e n t f a c e s , w h i c h d e n o t e b y I . I n t h e p l a n e C E D d r a w C O a n d D O a t r i g h t a n g l e s t o C E a n d D E , r e s p e c t i v e l y , a n d m e e t i n g i n O . J o i n O A , O E , O B , a n d f r o m O a s c e n t r e d e s c r i b e a s p h e r e , c u t t i n g O A , O C , O E a t a , c , e , r e s p e c t i v e l y ; t h e n a c e i s a s p h e r i c a l t r i a n g l e . S i n c e A B i s p e r p e n d i c u l a r t o C E a n d D E , i t i s p e r p e n d i c u l a r t o t h e p l a n e C E D ; t h e r e f o r e t h e p l a n e A O B , i n w h i c h A B l i e s , i s p e r p e n d i c u l a r t o t h e p l a n e C E D . . - . Z a e c i s a r i g h t a n g l e . L e t m b e t h e n u m b e r o f s i d e s i n e a c h f a c e , a n d n t h e n u m b e r o f p l a n e a n g l e s i n e a c h s o l i d a n g l e . T h e n Z a c e = Z A C E = ^ = ^ , 2 m m a n d Z c a e = £ Z o f t h e p l a n e s O A C a n d O A D . , 2 i , r . - . Z c a e = — = — 2 n n I n t h e r i g h t t r i a n g l e c a e w e h a v e c o s c a e — c o s c e s i n a c e . B u t c o s c e = c o s C O E = c o s ( - — -| = s i n — ^ 2 2 J 2 T T I . 7 T . - . c o s - = s i n - s i n — n 2 m 1 I t i T . - . s i n - = c o s - c o s e c — 2 n m V O L U M E O F A P A R A L L E L O P I P E D . 3 5 7 C o r . 1 . I f r b e t h e r a d i u s o f t h e i n s c r i b e d s p h e r e , a n d a b e a s i d e o f o n e o f t h e f a c e s , t h e n a , i r , I r = - c o t — t a n — 2 m 2 F o r , r = O C = C E t a n C E O = A E c o t A C E t a n C E O a , i r . I = - c o t — t a n — 2 m 2 C o r . 2 . I f T l b e t h e c i r c u m r a d i u s o f t h e p o l y e d r o n , t h e n R = - t a n — t a n — 2 n 2 F o r , r = 0 A c o s a o c = E c o t e c a c o t e a c = R c o t — c o t — m n . : R — ^ t a n - t a n — C o r . 3 . T h e s u r f a c e o f a r e g u l a r p o l y e d r o n , F b e i n g t h e n u m b e r o f f a c e s , -m a — c o t — . 4 m F o r , t h e a r e a o f o n e f a c e = — m c o t — -. - . e t c . 4 m C o r . 4 . T / i e v o l u m e o f a r e g u l a r p o l y e d r o n m a V F , i r c o t — 1 2 m F o r , t h e v o l u m e o f t h e p y r a m i d w h i c h h a s o n e f a c e o f t h e p o l y e d r o n f o r b a s e a n d O f o r v e r t e x r m a 2 , i r . = c o t — -. - . e t c . 3 4 m 2 3 2 . V o l u m e o f a P a r a l l e l o p i p e d . — T o f i n d t h e v o l u m e o f a p a r a l l e l o p i p e d i n t e r m s o f i t s e d g e s a n d t h e i r i n c l i n a t i o n s t o o n e a n o t h e r . 3 5 8 S P H E R I C A L T R I G O N O M E T R Y . L e t t h e e d g e s b e O A = a , O B = b , O C = c , a n d l e t t h e i n c l i n a t i o n s b e B O C = a , G O A = P , A O B = y . D r a w C H p e r p e n d i c u l a r t o t h e f a c e A O B E . D e s c r i b e a s p h e r e r o u n d O a s c e n t r e , m e e t i n g O A , O B , O C , O E , i n a , b , c , e , r e s p e c t i v e l y . T h e v o l u m e o f t h e p a r a l l e l o -p i p e d i s e q u a l t o t h e a r e a o f t h e b a s e O A E B m u l t i p l i e d b y t h e a l t i t u d e C H ; t h a t i s , v o l u m e = a b s i n y . C H = a b c s i n y s i n c e w h e r e c e i s t h e p e r p e n d i c u l a r a r c f r o m c o n a b . . - . v o l u m e = a b c s i n y s i n a c s i n b a c . . . ( A r t . 1 8 6 ) 2 n = a b c s i n y s i n / ? —s i n / ? s i n y ( A r t . 1 9 5 ) = a b c V l — c o s 2 r e — c o s 2 / ? — c o s 2 y + 2 c o s a c o s ( 3 c o s y . C o r . 1 . T h e s u r f a c e o f a p a r a l l e l o p i p e d = 2 ( b e s i n a + c a s i n / ? + a b s i n y ) . C o r . 2 . T h e v o l u m e o f a t e t r a e d r o n = i a 6 c V 1 -3 0 s 2 / 3 — e o s 2 y + 2 c o s a c o s / J c o s y . F o r , a t e t r a e d r o n i s o n e - s i x t h o f a p a r a l l e l o p i p e d w h i c h h a s t h e s a m e a l t i t u d e a n d i t s b a s e d o u b l e t h a t o f t h e t e t r a e d r o n . 2 3 3 . D i a g o n a l o f a P a r a l l e l o p i p e d . — T o f i n d t h e d i a g o n a l o f a p a r a l l e l o p i p e d i n t e r m s o f i t s e d g e s , a n d t h e i r m u t u a l i n c l i n a t i o n s . L e t O D ( f i g u r e o f A r t . 2 3 2 ) b e a p a r a l l e l o p i p e d , w h o s e e d g e s O A = a , O B = b , O C = c , a n d t h e i r i n c l i n a t i o n s B O C = a , C O A = p , A O B = y ; l e t O D b e t h e d i a g o n a l r e q u i r e d , T A B L E O F F O R M U L A S . 3 5 9 a n d O E t h e d i a g o n a l o f t h e f a c e O A B . T h e n t h e t r i a n g l e O E D g i v e s O T 3 2 = O E 2 + E D 2 + 2 O E - E D c o s C O E = a 2 + 6 2 + 2 a 6 c o s y + c 2 + 2 c - O E c o s C O E ( 1 ) N o w , i t i s c l e a r t h a t O E c o s C O E i s t h e p r o j e c t i o n o f O E o n t h e l i n e O C , a n d t h e r e f o r e i t m u s t b e e q u a l t o t h e s u m o f t h e p r o j e c t i o n s o f O B a n d B E ( o r o f O B a n d O A ) , o n t h e s a m e l i n e . . - . O E c o s C O E = b c o s a + a c o s / ? , w h i c h i n ( 1 ) g i v e s O D 2 = a 2 + b - + c - + 2 6 c c o s « + 2 c a c o s / ? + 2 a b c o s y . ( 2 ) 2 3 4 . T a b l e o f F o r m u l a e i n S p h e r i c a l T r i g o n o m e t r y . — F o r t h e c o n v e n i e n c e o f t h e s t u d e n t , m a n y o f t h e p r e c e d i n g f o r m u l a e a r e s u m m e d u p i n t h e f o l l o w i n g t a b l e : 1 . c o s c = c o s a c o s 6 ( A r t . 1 8 5 ) 2 . s i n 6 = s i n B s i n c . 3 . s i n a = s i n A s i n c . 4 . c o s C = — c o s A c o s B ( A r t . 1 8 9 ) 5 . s i n B = s i n b s i n C . 6 . s i n A = s i n a s i n C . 7 ^ = s i n 6 = s m c ( A r t . 1 9 0 ) s i n A s i n B s i n C 8 . c o s a = « o s b c o s c + s i n b s i n c c o s A . . ( A r t . 1 9 1 ) 9 . c o s b = c o s c c o s a + s i n c s i n a c o s B . 1 0 . c o s c = c o s a c o s b + s i n a s i n b c o s C . 1 1 . c o s A = — c o s B c o s C + s i n B s i n C c o s a ( A r t . 1 9 2 ) 1 2 . c o s B = = — c o s C c o s A + s i n C s i n A c o s b . F r o m t h e n a t u r e o f p r o j e c t i o n s ( P l a n e a n d S o l i d G e o r a . , A r t . 3 2 6 ) . 3 6 0 S P H E R I C A L T R I G O N O M E T R Y . 1 3 . c o s C = — c o s A c o s B + s i n A s i n B c o s c . 1 4 . c o t a s i n b = c o t A s i n C + c o s C c o s b . . ( A r t . 1 9 3 ) 1 5 . c o t a s i n c = c o t A s i n B 4 - c o s B c o s c . 1 6 . c o t b s i n a = c o t B s i n C + c o s C c o s a . 1 7 . c o t b s i n c = c o t B s i n A + c o s A c o s c . 1 8 . c o t c s i n a = c o t C s i n B + c o s B c o s a . 1 9 . c o t c s i n 6 = c o t C s i n A + c o s A c o s b . 2 0 . s i n a c o s B = c o s b s i n c — s i n 6 c o s c c o s A ( A r t . 1 9 4 ) 2 1 . s i n a c o s C = s i n b c o s c — c o s b s i n c c o s A . 2 2 . s i n b c o s A = c o s a s i n c — s i n a c o s c c o s B . 2 3 . s i n b c o s C = s i n a c o s c — c o s a s i n c c o s B . 2 4 . s i n c c o s A = c o s a s i n b — s i n a c o s b c o s C . 2 5 . s i n c c o s B = s i n a c o s b — c o s a s i n 6 c o s C . 2 6 . s i n ^ A = J ^ s T 6 ) s i n ( s - ^ . . . ( A r t . 1 9 5 ) s i n s s i n ( s — a ) 2 9 . s i n A 2 V s i n s s i n ( s — a ) s i n ( s — 6 ) s i n ( s — c ) s i n b s i n c 2 n s i n 6 s i n c w h e r e n = V s i n s s i n ( s — a ) s i n ( s — 6 ) s i n ( s — c ) . ( A r t . 1 9 6 ) T A B L E O F F O R M U L A S . 3 6 1 3 2 . t a n | a = J c o 8 S c o s ( S - A l \ c o s ( S - B ) c o s ( S - C ) 3 3 s i n a — 2 v , ~ c o s t j c o s K S — A ) c o s ( S — B ) c o s ( S — C ) ' s i n B s i n C 3 4 . t a i a ( A + B ) = c o s H a - f r ) c o t ^ C . . ( A r t . 1 9 7 ) 3 5 . t a n | ( A -B ) = s i n ( a ~ 6 ) c o t A C . 5 V y s i n £ ( a + 6 ) 3 6 . t a n ( « + " 6 ) = c o s ~ B ^ t a n ¿ a 5 V ' c o s £ ( A + B ) 5 3 7 . t a n i ( « - 6 ) = S 1 " K A - B ) , 5 V ; s i n £ ( A + B ) 1 3 8 . s i n £ ( A + B ) e o s | c = c o s £ ( a - 6 ) c o s £ C ( A r t . 1 9 8 ) 3 9 . s i n £ ( A — B ) s i n £ c = s i n £ ( a — 6 ) c o s £ C . 4 0 . c o s ^ ( A + B ) c o s = c o s £ ( a + 6 ) s i n J C . 4 1 . c o s £ ( A — B ) s i n | c = s i n f ( a + b ) s i n £ C . 4 2 . t a n r = . / s i n ( s ~ a ) s i n [ ~ 6 ) s i n ( s " c ) \ s i n s . . . ( A r t . 2 1 5 ) s i n s n s i n s 4 3 . t a n R = - ? - [ s i n ( s — a ) + s i n ( s — b ) + s i n ( s — c ) — s i n s ] 2 n ( A r t . 2 1 7 ) 4 4 . K = a r e a o f A = ^ - - n - r 1 ( A r t . 2 1 9 ) 1 8 0 4 5 . s i n I E = -. . . ( A r t . 2 2 0 ) 2 c o s £ o c o s £ 6 c o s £ c 4 6 . t a n J E = V t a n J s t a n £ ( s — a ) t a n £ ( s — b ) t a n £ ( s — c ) . 3 6 2 S P H E R I C A L T R I G O N O M E T H Y . E X A M P L E S . 1 . F i n d t h e t i m e o f s u n r i s e a t a p l a c e w h o s e l a t i t u d e i s 4 2 ° 3 3 ' N . , w h e n t h e s u n , s d e c l i n a t i o n i s 1 3 ° 2 8 ' N . A n s . 5 h 9 m 1 3 ' . 2 . F i n d t h e t i m e o f s u n s e t a t C i n c i n n a t i , l a t . 3 9 ° 6 ' N . , w h e n t h e s u n , s d e c l i n a t i o n i s 1 5 ° 5 6 ' S . A n s . 5 h 6 , n . 3 . F i n d t h e t i m e o f s u n r i s e a t l a t . 4 0 ° 4 3 ' 4 8 " N . , i n t h e l o n g e s t d a y i n t h e y e a r , t h e s u n , s g r e a t e s t d e c l i n a t i o n b e i n g 2 3 ° 2 7 ' N . A n s . 4 h 3 2 m 1 6 ' . 4 . 4 . F i n d t h e t i m e o f s u n r i s e a t B o s t o n , l a t . 4 2 ° 2 1 ' N . , w h e n t h e s u n , s d e c l i n a t i o n i s 8 ° 4 7 ' S . A n s . 6 h 1 4 m . 5 . F i n d t h e l e n g t h o f t h e l o n g e s t d a y a t l a t . 4 2 ° 1 6 ' 4 8 " . 3 N . , t h e s u n , s g r e a t e s t d e c l i n a t i o n b e i n g 2 3 ° 2 7 ' N . A n s . 1 5 h 5 r a 5 0 ' . 6 . F i n d t h e l e n g t h o f t h e s h o r t e s t d a y a t N e w B r u n s w i c k , N . J . , l a t . 4 0 ° 2 9 ' 5 2 " . 7 N . , t h e s u n , s g r e a t e s t d e c l i n a t i o n b e i n g 2 3 ° 2 7 ' S . 7 . F i n d t h e h o u r a n g l e a n d a z i m u t h o f A n t a r e s , d e c l i n a t i o n 2 6 ° 6 ' S . , w h e n i t s e t s t o a n o b s e r v e r a t P h i l a d e l p h i a , l a t . 3 9 ° 5 7 ' N . A n s . 4 h 2 3 m 5 ' . 7 ; S . 5 4 ° 5 8 ' 4 4 " W . 8 . F i n d t h e h o u r a n g l e a n d a z i m u t h o f t h e N e b u l a o f A n d r o m e d a , d e c l i n a t i o n 4 0 ° 3 5 ' N . , w h e n i t r i s e s t o a n o b s e r v e r a t N e w B r u n s w i c k , N . J . , l a t . 4 0 ° 2 9 ' 5 2 " . 7 N . 9 . F i n d t h e a z i m u t h a n d a l t i t u d e o f R e g u l u s , d e c l i n a t i o n 1 6 ° 1 3 ' N . , t o a n o b s e r v e r a t N e w Y o r k , l a t . 4 0 ° 4 2 ' N . , w h e n t h e s t a r i s t h r e e h o u r s e a s t o f t h e m e r i d i a n . A n s . A z i m u t h = S . 7 1 ° 1 2 ' 3 0 " E . ; A l t i t u d e = 4 4 ° 1 0 ' 3 3 " . 1 0 . F i n d t h e a z i m u t h a n d a l t i t u d e o f F o m a l h a u t , d e c l i n a t i o n 3 0 ° 2 5 ' S . , t o a n o b s e r v e r i n l a t . 4 2 ° 2 2 ' N . , w h e n t h e s t a r i s 2 h 5 m 3 6 " e a s t o f t h e m e r i d i a n . A n s . A z i m u t h = S . 2 7 ° 1 8 ' 4 0 " E . ; A l t i t u d e = 1 1 ° 4 1 ' 3 7 " . E X A M P L E S . 3 6 3 1 1 . F i n d t h e a z i m u t h a n d a l t i t u d e o f a s t a r t o a n o b s e r v e r i n l a t . 3 9 ° 5 7 ' N . , w h e n t h e h o u r a n g l e o f t h e s t a r i s 5 h 1 7 r a 4 0 " e a s t , a n d t h e d e c l i n a t i o n i s 6 2 ° 3 3 ' N . A n s . A z i m u t h = N . 3 5 ° 5 4 ' E . ; A l t i t u d e = 3 9 ° 2 4 ' . 1 2 . F i n d t h e h o u r a n g l e ( t ) a n d d e c l i n a t i o n ( 8 ) o f a s t a r t o a n o b s e r v e r i n l a t . 4 0 ° 3 6 ' 2 3 " . 9 N . , w h e n t h e a z i m u t h o f t h e s t a r i s 8 0 ° 2 3 ' 4 " . 4 7 , a n d t h e a l t i t u d e i s 4 7 ° 1 5 ' 1 8 " . 3 . A n s . t = 4 6 ° 4 0 ' 4 " . 5 3 ; 8 = 2 3 ° 4 ' 2 4 " . 3 3 . 1 3 . F i n d t h e d i s t a n c e b e t w e e n R e g u l u s a n d A n t a r e s , t h e r i g h t a s c e n s i o n s b e i n g 1 0 h 0 m 2 9 M 1 a n d 1 6 h 2 0 m 2 0 ' . 3 5 , a n d t h e p o l a r d i s t a n c e s 7 7 ° 1 8 ' 4 1 " . 4 a n d 1 1 6 ° 5 ' 5 5 " . 5 . A n s . 9 9 ° 5 5 ' 4 4 " . 9 . 1 4 . F i n d t h e d i s t a n c e b e t w e e n t h e s u n a n d m o o n w h e n t h e r i g h t a s c e n s i o n s a r e 1 2 h 3 9 m 3 ' . 2 2 a n d 6 h 5 5 m 3 2 " . 7 3 , a n d t h e d e c l i n a t i o n s 9 ° 2 3 ' 1 6 " . 7 S . a n d 2 2 ° 5 0 ' 2 1 " . 9 N . A n s . 8 9 ° 5 2 ' 5 5 " . 5 . 1 5 . F i n d t h e s h o r t e s t d i s t a n c e o n t h e e a r t h , s s u r f a c e , i n m i l e s , f r o m N e w Y o r k , l a t . 4 0 ° 4 2 ' 4 4 " N . , l o n g . 7 4 ° 0 ' 2 4 " W . , t o S a n F r a n c i s c o , l a t . 3 7 ° 4 8 ' N . , l o n g . 1 2 2 ° 2 3 ' W . A n s . 2 5 6 2 m i l e s . 1 6 . F i n d t h e s h o r t e s t d i s t a n c e o n t h e e a r t h , s s u r f a c e f r o m S a n F r a n c i s c o , l a t . 3 7 ° 4 8 ' N . , l o n g . 1 2 2 ° 2 3 ' W . , t o P o r t J a c k s o n , l a t . 3 3 ° 5 1 ' S . , l o n g . 1 5 1 ° 1 9 ' E . A n s . 6 4 4 4 n a u t i c a l m i l e s . 1 7 . G i v e n t h e r i g h t a s c e n s i o n o f a s t a r 1 0 h l m 9 " . 3 4 , a n d i t s d e c l i n a t i o n 1 2 ° 3 7 ' 3 6 " . 8 N . ; t o f i n d i t s l a t i t u d e a n d l o n g i t u d e , t h e o b l i q u i t y o f t h e e c l i p t i c b e i n g 2 3 ° 2 7 ' 1 9 " . 4 5 . A n s . L a t i t u d e = ; L o n g i t u d e = 1 8 . G i v e n t h e o b l i q u i t y o f t h e e c l i p t i c < o , a n d t h e s u n , s l o n g i t u d e k ; t o f i n d h i s r i g h t a s c e n s i o n a a n d d e c l i n a t i o n 8 . A n s . t a n a = c o s w t a n A . ; s i n 8 = s i n a > s i n A . . 3 6 4 S P H E R I C A L T R I G O N O M E T R Y . 1 9 . G i v e n t h e o b l i q u i t y o f t h e e c l i p t i c 2 3 ° 2 7 ' 1 8 " . 5 , a n d t h e s u n , s l o n g i t u d e 5 9 ° 4 0 ' 1 " . 6 ; t o f i n d h i s r i g h t a s c e n s i o n ( a ) , a n d d e c l i n a t i o n ( 8 ) . A n s . a = 3 h 4 9 r a 5 2 ' . 6 2 ; 8 = 2 0 ° 5 ' 3 3 " . 9 K 2 0 . G i v e n t h e s u n , s d e c l m a t i o n 1 6 ° 0 ' 5 6 " . 4 N . , a n d t h e o b l i q u i t y o f t h e e c l i p t i c 2 3 ° 2 7 ' 1 8 " . 2 ; t o f i n d h i s r i g h t a s c e n s i o n ( « ) , a n d l o n g i t u d e ( X ) . A n s . a = 9 h 1 4 m 1 9 " . 2 ; X = 1 3 6 ° 7 ' 6 " . 5 . 2 1 . G i v e n t h e s u n , s r i g h t a s c e n s i o n 1 4 h 8 m 1 9 ' . 0 6 , a n d t h e o b l i q u i t y o f t h e e c l i p t i c 2 3 ° 2 7 ' 1 7 " . 8 ; t o f i n d h i s l o n g i t u d e ( X ) , a n d d e c l i n a t i o n ( 8 ) . A n s . X = 2 1 4 ° 2 0 ' 3 4 " . 7 ; 8 = 1 2 ° 5 8 ' 3 4 " . 4 S . 2 2 . G i v e n t h e s u n , s l o n g i t u d e 2 8 0 ° 2 3 ' 5 2 " . 3 , a n d h i s d e c l i n a t i o n 2 3 ° 2 ' 5 2 " . 2 S . ; t o f i n d h i s r i g h t a s c e n s i o n ( a ) . A n s . a = 1 8 " 4 5 m 1 4 " . 7 . 2 3 . I n l a t i t u d e 4 5 ° N . , p r o v e t h a t t h e s h a d o w a t n o o n o f a v e r t i c a l o b j e c t i s t h r e e t i m e s a s l o n g w h e n t h e s u n , s d e c l i n a t i o n i s 1 5 ° S . a s w h e n i t i s 1 5 ° N . 2 4 . G i v e n t h e a z i m u t h o f t h e s u n a t s e t t i n g , a n d a l s o a t 6 o , c l o c k ; f i n d t h e s u n , s d e c l i n a t i o n , a n d t h e l a t i t u d e . 2 5 . I f t h e s u n , s d e c l i n a t i o n b e 1 5 ° N . , a n d l e n g t h o f d a y f o u r h o u r s , p r o v e t a n = s i n 6 0 ° t a n 7 5 ° . 2 C . G i v e n t h e s u n , s d e c l i n a t i o n a n d t h e l a t i t u d e ; s h o w h o w t o f i n d t h e t i m e . w h e n h e i s d u e e a s t . 2 7 . I f t h e s u n r i s e n o r t h e a s t i n l a t i t u d e < j > , p r o v e t h a t c o t h o u r a n g l e a t s u n r i s e = — s i n < £ . 2 8 . G i v e n t h e l a t i t u d e s a n d l o n g i t u d e s o f t w o p l a c e s ; f i n d t h e s u n , s d e c l i n a t i o n w h e n h e i s o n t h e h o r i z o n o f b o t h a t t h e s a m e i n s t a n t . 2 9 . G i v e n t h e s u n , s d e c l i n a t i o n 8 , h i s a l t i t u d e h a t 6 o , c l o c k , a n d h i s a l t i t u d e h ' w h e n d u e e a s t ; p r o v e s i n 2 8 = s i n h s i n h , . E X A M P L E S . 3 6 5 3 0 . G i v e n t h e d e c l i n a t i o n o f ' a s t a r 3 0 ° ; f i n d a t w h a t l a t i t u d e i t s a z i m u t h i s 4 5 ° a t t h e t i m e o f r i s i n g . 3 1 . G i v e n t h e s u n , s d e c l i n a t i o n 8 , a n d t h e l a t i t u d e o f t h e p l a c e < f > ; f i n d h i s a l t i t u d e w h e n d u e e a s t . 3 2 . G i v e n t h e d e c l i n a t i o n s o f t w o s t a r s , a n d t h e d i f f e r e n c e o f t h e i r a l t i t u d e s w h e n t h e y a r e o n t h e p r i m e v e r t i c a l ; f i n d t h e l a t i t u d e o f t h e p l a c e . 3 3 . I f t h e d i f f e r e n c e b e t w e e n t h e l e n g t h s o f t h e l o n g e s t a n d s h o r t e s t d a y a t a g i v e n p l a c e b e s i x h o u r s , f i n d t h e l a t i t u d e . 3 4 . I f t h e r a d i u s o f t h e e a r t h b e 4 0 0 0 m i l e s , w h a t i s t h e a r e a o f a s p h e r i c a l t r i a n g l e w h o s e s p h e r i c a l e x c e s s i s 1 ° ? 3 5 . I f A " , B " , C " b e t h e c h o r d a l a n g l e s o f t h e p o l a r t r i a n g l e o f A B C , p r o v e c o s A " = s i n £ A c o s ( s — a ) , e t c . 3 6 . I f t h e a r e a o f a s p h e r i c a l t r i a n g l e b e o n e - f o u r t h t h e a r e a o f t h e s p h e r e , s h o w t h a t t h e b i s e c t o r o f a s i d e i s t h e s u p p l e m e n t o f h a l f t h a t s i d e . 3 7 . I f t h e a r e a o f a s p h e r i c a l t r i a n g l e b e o n e - f o u r t h t h e a r e a o f t h e s p h e r e , s h o w t h a t t h e a r c s j o i n i n g t h e m i d d l e p o i n t s o f i t s s i d e s a r e q u a d r a n t s . 3 8 . G i v e n t h e b a s e a n d a r e a ; s h o w t h a t t h e a r c j o i n i n g t h e m i d d l e p o i n t s o f t h e s i d e s i s c o n s t a n t ; a n d i f i t i s a q u a d r a n t , t h e n t h e a r e a o f t h e t r i a n g l e i s n - r 2 . 3 9 . T w o c i r c l e s o f a n g u l a r r a d i i , a a n d / ? , i n t e r s e c t o r t h o g o n a l l y o n a s p h e r e o f r a d i u s r ; f i n d i n a n y m a n n e r t h e a r e a c o m m o n t o t h e t w o . 4 0 . I f E b e t h e s p h e r i c a l e x c e s s o f a t r i a n g l e , p r o v e t h a t ^ E = t a n \ a t a n i b s i n C — \ ( t a n \ a t a n £ b ) 2 s i n 2 C + e t c . 4 1 . S h o w t h a t t h e s u m o f t h e t h r e e a r c s j o i n i n g t h e m i d d l e p o i n t s o f t h e s i d e s o f t h e c o l u n a r s i s e q u a l t o t w o 3 6 6 S P H E R I C A L T R I G O N O M E T R Y . r i g h t a n g l e s , t h e s i d e s o f t h e o r i g i n a l t r i a n g l e b e i n g r e g a r d e d a s t h e b a s e s o f t h e c o l u n a r s . 4 2 . P r o v e t h a t c o s 2 £ a s i n 2 S + s i n 2 1 6 s i n 2 ( S -C ) + s i n 2 | c s i n 2 " ( S -B ) + 2 c o s \ a s i n | b s i n \ c s i n S s i n ( S — B ) s i n ( S — C ) = 1 . 4 3 . H a v i n g g i v e n t h e b a s e a n d t h e a r c j o i n i n g t h e m i d d l e p o i n t s o f t h e c o l u n a r o n t h e b a s e , t h e c i r c u m c i r c l e i s f i x e d . 4 4 . P r o v e s i n \ b s i n | c s i n ( S — A ) + c o s \ b c o s ^ c s i n S = c o s ^ a . 4 5 . I f A + B + C = 2 i T , p r o v e t h a t c o s 2 \ a + c o s 2 £ 6 + c o s 2 \ c = 1 , a n d c o s C = — c o t \ a c o t \ b . 4 6 . S o l v e t h e e q u a t i o n s , s i n b c o s c s i n Z + s i n c c o s b s i n Y = s i n a , s i n e c o s a s i n X + s i n a c o s e s i n Z = s i n b , s i n a c o s b s i n Y + s i n 6 c o s a s i n X = s i n c , f o r s i n X , s i n Y , a n d s i n Z . 4 7 . I f b a n d c a r e c o n s t a n t , p r o v e t h e f o l l o w i n g r e l a t i o n s b e t w e e n t h e s m a l l v a r i a t i o n s o f a n y t w o p a r t s o f t h e o t h e r e l e m e n t s o f t h e s p h e r i c a l t r i a n g l e A B C : d V > t a n B d a d C t a n C d B = — s i n a t a n C : d a . - r , . d A s i n A . = s i n B s i n e ; ( J A ' d ' B s i n B c o s C , d a . 4 . - o d A s i n A = — s i n a t a n B ; c Z C d G c o s B s i n C 4 8 . I f A a n d c r e m a i n c o n s t a n t , p r o v e t h e f o l l o w i n g : d a s i n r t d a d B t a n C , d b - c o s C . E X A M P L E S . 3 6 7 4 9 . I f B a n d C r e m a i n c o n s t a n t , p r o v e t h e f o l l o w i n g : d A . . , d a s i n a = s i n A t a n c ; d b d b s i n b c o s c 5 0 . I f A a n d a r e m a i n c o n s t a n t , p r o v e t h e f o l l o w i n g : d b _ t a n b , d c _ t a n c _ d B ~ t a n B , d C ~ t a n C , d b c o s B d b s i n 6 d c c o s C , d ( J t a n B c o s c 5 1 . T w o e q u a l s m a l l c i r c l e s a r e d r a w n t o u c h i n g e a c h o t h e r ; s h o w t h a t t h e a n g l e b e t w e e n t h e i r p l a n e s i s t w i c e t h e c o m p l e m e n t o f t h e i r s p h e r i c a l r a d i u s . 5 2 . O n a s p h e r e w h o s e r a d i u s i s r a s m a l l c i r c l e o f s p h e r i c a l r a d i u s 6 i s d e s c r i b e d , a n d a g r e a t c i r c l e i s d e s c r i b e d h a v i n g i t s p o l e o n t h e s m a l l c i r c l e ; s h o w t h a t t h e l e n g t h o f t h e i r c o m m o n c h o r d i s V — c o s 2 6 . s i n 6 5 3 . G i v e n t h e b a s e c o f a t r i a n g l e , a n d t h a t t a n £ a t a n \ b = t a n 2 \ B , B b e i n g t h e b i s e c t o r o f t h e b a s e , f i n d a — b i n t e r m s o f c . 5 4 . I f C = A + B , s h o w t h a t 1 — c o s a — c o s 6 + c o s c = 0 . 5 5 . I f A d e n o t e o n e o f t h e a n g l e s o f a n e q u i l a t e r a l t r i a n g l e , a n d A ' a n a n g l e o f i t s p o l a r t r i a n g l e , s h o w t h a t c o s A c o s A ' = c o s A + c o s A ' . 5 6 . S h o w t h a t c o s a c o s B — c o s b c o s A _ c o s C + c o s c s i n a — s i n b s i n c K r j - r , „ „ „ „ „ . c o s a s i n b — s i n a c o s b c o s C 5 7 . J r r o v e c o s A = ; , s i n c j a , t > 2 s i n ( a + b ) s i n 2 A C a n d c o s A + c o s B = ^ — s i n e 3 6 8 S P H E R I C A L T R I G O N O M E T R Y , 5 8 . P r o v e L e g e n d r e , s T h e o r e m b y m e a n s o f t h e r e l a t i o n s s i n A _ s i n B _ s i n C s i n a s i n b s i n c 5 9 . T w o p l a c e s a r e s i t u a t e d o n t h e s a m e p a r a l l e l o f l a t i t u d e < j > ; f i n d t h e d i f f e r e n c e o f t h e d i s t a n c e s s a i l e d o v e r b y t w o s h i p s p a s s i n g b e t w e e n t h e m , o n e k e e p i n g t o t h e g r e a t c i r c l e c o u r s e , t h e o t h e r t o t h e p a r a l l e l ; t h e d i f f e r e n c e o f l o n g i t u d e o f t h e p l a c e s b e i n g 2 A . A n s . 2 r [ A . c o s < f > — s i n - 1 ( c o s $ s i n A . ) ] . 6 0 . I f t h e s i d e s o f a t r i a n g l e b e e a c h 6 0 ° , s h o w t h a t t h e c i r c l e s d e s c r i b e d , e a c h h a v i n g a v e r t e x f o r p o l e , a n d p a s s i n g t h r o u g h t h e m i d d l e p o i n t s o f t h e s i d e s w h i c h m e e t a t i t , h a v e t h e s i d e s o f t h e s u p p l e m e n t a l t r i a n g l e f o r c o m m o n t a n g e n t s . 6 1 . F i n d t h e v o l u m e a n d a l s o t h e i n c l i n a t i o n o f t w o a d j a c e n t f a c e s ( 1 ) o f a r e g u l a r t e t r a e d r o n , ( 2 ) o f a r e g u l a r o c t a e d r o n , ( 3 ) o f a r e g u l a r d o d e c a e d r o n , a n d ( 4 ) o f a r e g u l a r i c o s a ' e d r o n , t h e e d g e b e i n g o n e i n c h . A n s . ( 1 ) 1 1 7 . 8 5 c u . i n . , 7 0 ° 3 1 ' 4 3 " . 4 ; ( 2 ) . 4 7 1 4 c u . i n . , 1 0 9 ° 2 8 ' 1 6 " ; ( 3 ) 7 . 6 6 3 c u . i n . , 1 1 6 ° 3 3 ' 5 4 " ; ( 4 ) 2 . 1 8 1 7 c u . i n . , 1 3 8 ° 1 1 ' 2 2 " . 6 . 6 2 . I n t h e t e t r a e d r o n , p r o v e ( 1 ) t h a t t h e c i r c u m r a d i u s i s e q u a l t o t h r e e t i m e s i t s i n - r a d i u s , a n d ( 2 ) t h a t t h e r a d i u s o f t h e s p h e r e t o u c h i n g i t s s i x e d g e s i s a m e a n p r o p o r t i o n a l b e t w e e n t h e i n - r a d i u s a n d c i r c u m r a d i u s . 6 3 . P r o v e t h a t t h e r a t i o o f t h e i n - r a d i u s t o t h e c i r c u m r a d i u s i s t h e s a m e i n t h e c u b e a n d t h e o c t a e d r o n , a n d a l s o i n t h e d o d e c a e d r o n a n d i c o s a e d r o n .
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Skip to main content Lower bound for monochromatic triangles in Kn Ask Question Asked Modified 7 years, 11 months ago Viewed 3k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. Say Kn is a complete graph of n nodes, and every edge is either blue or red. I'm trying to find Tn, which is the lower bound for the number of monochromatic triangles in Kn (monochromatic meaning triangles whose 3 edges are of the same color). I'm trying this approaches: A. Let vi a vertex in Kn. This vertex belongs to (n−12) different triangles. Let r be the number of red edges that vi belongs b the number of blue edges. Of course, r+b=n−1 So, the number of non-monochromatic (dual/stereo-chromatic?) triangles is r⋅b, and so our bound would be (for vertex Vi) Tn≤(n−12)−r⋅b Question is - how do I know when r⋅b is the maximal possible number? Also, this would give me the upper bound, how do I find the lower bound? Thanks in advance! graph-theory ramsey-theory Share CC BY-SA 3.0 Follow this question to receive notifications edited Sep 18, 2017 at 0:35 Alex Ravsky 107k55 gold badges6565 silver badges202202 bronze badges asked Jan 24, 2012 at 15:49 yotamooyotamoo 2,87199 gold badges3636 silver badges4848 bronze badges Add a comment | 2 Answers 2 Reset to default This answer is useful 4 Save this answer. Show activity on this post. Here's a proof, from Wikipedia, that K6 has at least 2 monochromatic triangles. You may find that this method of proof generalizes to give you the lower bound you want. "An alternate proof works by double counting. It goes as follows: Count the number of ordered triples of vertices x,y,z such that the edge (xy) is red and the edge (yz) is blue. Firstly, any given vertex will be the middle of either 0×5=0 (all edges from the vertex are the same color), 1×4=4 (four are the same color, one is the other color), or 2×3=6 (three are the same color, two are the other color) such triples. Therefore there are at most 6×6=36 such triples. Secondly, for any non-monochromatic triangle (xyz), there exist precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore at least 2 of the 20 triangles in the K6 are monochromatic." On the other hand, if you want all the work done for you, there is an answer in A W Goodman, On sets of acquaintances and strangers at any party, Amer. Math. Monthly 66 (1959) 778-783, also in Frank Harary, The two-triangle case of the acquaintance graph, Math. Mag. 45 (1972) 130-135. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jan 25, 2012 at 1:27 answered Jan 25, 2012 at 1:08 Gerry MyersonGerry Myerson 186k1313 gold badges231231 silver badges404404 bronze badges 1 Also tabulated at oeis.org/A014557 – Gerry Myerson Commented Sep 4, 2022 at 0:48 Add a comment | This answer is useful -3 Save this answer. Show activity on this post. If you take G=Kn, n≥3 and colour each edge red with probability 1/2 and blue otherwise. Define X to be the random variable counting the number of monochromatic triangles. Then E[X]=((n2)3)(12)3 Hence, there is a colouring of G such that there are least this many. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 24, 2012 at 16:13 alext87alext87 2,9291818 silver badges2929 bronze badges 1 How does that help with a lower bound? – mjqxxxx Commented Jan 24, 2012 at 16:18 Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions graph-theory ramsey-theory See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 2 In every coloring in two colors of the edges of K10 there are at least 20 monochromatic triangles. 0 Graph with |G|=6, G and its complement G′ contains at least two triangles together Related 10 Showing that K7 contains at least 4 monochromatic triangles 0 Finding the maximal complete subgraph which contains no monochromatic triangles of a complete graph 3 2-coloring of K1,2,2,2,2 without monochromatic triangles 3 Upper bound for monochromatic edges in vertex coloring with k colors 2 Show that there exists a coloring of the edge set of Kn that has at most (n3)4 monochromatic triangles 1 Monochromatic vertex disjoint triangles in K6n Hot Network Questions How do computer models, languages, and algorithms relate to philosophy? 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https://en.wikipedia.org/?title=Inequality_of_arithmetic_and_geometric_means&redirect=no
Inequality of arithmetic and geometric means - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Inequality of arithmetic and geometric means [x] Add languages Add links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Add interlanguage links Print/export Download as PDF Printable version In other projects From Wikipedia, the free encyclopedia Redirect to: AM–GM inequality This page is a redirect. The following categories are used to track and monitor this redirect: From a page move: This is a redirect from a page that has been moved (renamed). This page was kept as a redirect to avoid breaking links, both internal and external, that may have been made to the old page name. When appropriate, protection levels are automatically sensed, described and categorized. Retrieved from " Hidden category: Redirects from moves This page was last edited on 19 January 2024, at 07:57(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search Inequality of arithmetic and geometric means Add languagesAdd topic
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https://aquascaperoom.ca/substrate-calculator/?srsltid=AfmBOorrp47a5oUATJVDfcAVVI5e0c0-ddwMKHwtiWKR_52MXtuIkWGi
Aquarium Substrate Calculator Canada | Aquascaperoom Have a question? More Home Deals! 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Input these measurements into the calculator fields above and click the Calculate button. The tool will estimate how many liters of substrate or aqua soil you need for your planted tank. This estimate helps Canadian aquascapers purchase the right amount of substrate to achieve their desired depth, whether it’s for carpeting plants, rooted species, or a showpiece aquascape. By adjusting your tank dimensions, you can customize the calculation to ensure accuracy. Choosing the Right Aqua Soil for Your Aquarium Aquarium substrate plays a vital role in the success of planted aquariums. The right aqua soil provides nutrients, supports healthy plant roots, improves water clarity, and enhances the look of your aquascape. In Canada, hobbyists often choose nutrient-rich substrates like ADA Aqua Soil for professional results. 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Q: How often should I replace aquarium soil? A: Quality aqua soil can last several years. Over time, nutrients may need to be replenished with fertilizers or root tabs rather than replacing the whole substrate. Q: Will aqua soil cloud my aquarium water? A: Some soils release particles initially. Rinse lightly if the brand recommends it, and fill your tank slowly to minimize disruption. Cloudiness usually clears in a few days. Q: Can I mix different substrates? A: Yes, but match nutrient levels and pH effects. Many aquascapers layer nutrient-rich soils on the bottom and cap them with decorative gravel or sand. Q: How deep should substrate be for planted tanks? A: 1–3 inches works for most aquariums. Shallow-rooted carpeting plants may need less, while rooted swords or crypts benefit from deeper layers. Q: Can I safely add aqua soil to an established tank? A: Yes — carefully layer the soil to avoid stirring debris. 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https://www.k-state.edu/counseling/services/resources/self_help/tipsformultiplechoiceexams.html
Tips for Multiple Choice Answers Jump to main contentJump to footer Kansas State University Canvas OrgCentral Navigate KSIS HRIS Webmail Sign in Browse A-Z Counseling and Psychological Services Search About Mission Meet the staff Annual report Parking information FAQ Services Need help now Types of services Specific concerns Attention Deficit Hyperactivity Disorder (ADHD) Additional resources Informed consent Concerned about a student? 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Friends Family/Parent(s) Faculty/Staff Student of concern report Suicide prevention Making a CAPS referral Professional training Internship in Health Service Psychology Practicum Training K-State home CAPS Services Additional Resources Self Help Tips for Multiple Choice Answers Tips for Multiple Choice Exams By Natalie Umberger Multiple choice questions usually include a phrase or stem followed by three to five options: Test strategies: • Read the directions carefully Know if each question has one or more correct option Know if you are penalized for guessing Know how much time is allowed (this governs your strategy) • Preview the test Read through the test quickly and answer the easiest questions first Mark those you think you know in some way that is appropriate • Read through the test a second time and answer more difficult questions You may pick up cues for answers from the first reading, or become more comfortable in the testing situation • If time allows, review both questions and answers It is possible you mis-read questions the first time. Answering options: Improve your odds, think critically • Cover the options, read the stem, and try to answer Select the option that most closely matches your answer • Read the stem with each option Treat each option as a true-false question, and choose the "most true" Strategies to answer difficult questions: • Eliminate options you know to be incorrect If allowed, mark words or alternatives in questions that eliminate the option • Give each option of a question the "true-false test:" This may reduce your selection to the best answer • Question options that grammatically don't fit with the stem • Question options that are totally unfamiliar to you • Question options that contain negative or absolute words. Try substituting a qualified term for the absolute one, like frequently for always; or typical for every to see if you can eliminate it • "All of the above:" If you know two of three options seem correct, "all of the above" is a strong possibility • Number answers: Toss out the high and low and consider the middle range numbers • "Look alike options" Probably one is correct; choose the best but eliminate choices that mean basically the same thing, and thus cancel each other out • Double negatives: Create the equivalent positive statement and consider • Echo options: If two options are opposite each other, chances are one of them is correct • Favor options that contain qualifiers The result is longer, more inclusive items that better fill the role of the answer • If two alternatives seem correct, Compare them for differences, Then refer to the stem to find your best answer Guessing: • Always guess when there is no penalty for guessing or you can eliminate options • Don't guess if you are penalized for guessing and if you have no basis for your choice • Use hints from questions you know to answer questions you do not. • Change your first answers when you are sure of the correction, or other cues in the test cue you to change. • Remember that you are looking for the best answer, not only a correct one, and not one which must be true all of the time, in all cases, and without exception. Resource www.studygs.net © All staff articles are used by permission of the respective author(s). Copyright belongs to the University Life Café. No part of this may be used without authorization. Need help now Types of services Goal-oriented individual therapy Workshops Group therapy Couple/polyamorous therapy Specific concerns Eating disorders Emotional support animals Career assessment Consultation for mental health concerns Attention Deficit Hyperactivity Disorder (ADHD) Additional resources K-State Resources Self-help Apps Manage your stress Presentations/outreach Global campus resources Community resources How to find a therapist Informed consent Lafene CAPS Counseling and Psychological Services Lafene Health Center 1105 Sunset Avenue, Room 101, Manhattan, KS 66502 785-532-6927| 866-793-8010 fax |counsel@k-state.edu Hours Monday-Friday: 8 a.m. - 5 p.m. Closed weekends and university holidays Contact Us Emergency Statements and Disclosures Accessibility KBOR Free Expression Statement © Kansas State University Updated 2/15/2024 Updated: 2/15/24
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https://en.wikipedia.org/wiki/Helly%27s_selection_theorem
Jump to content Helly's selection theorem Deutsch Français Italiano 日本語 Edit links From Wikipedia, the free encyclopedia On convergent subsequences of functions that are locally of bounded total variation In mathematics, Helly's selection theorem (also called the Helly selection principle) states that a uniformly bounded sequence of monotone real functions admits a convergent subsequence. In other words, it is a sequential compactness theorem for the space of uniformly bounded monotone functions. It is named for the Austrian mathematician Eduard Helly. A more general version of the theorem asserts compactness of the space BVloc of functions locally of bounded total variation that are uniformly bounded at a point. The theorem has applications throughout mathematical analysis. In probability theory, the result implies compactness of a tight family of measures. Statement of the theorem [edit] Let (fn)n ∈ N be a sequence of increasing functions mapping a real interval I into the real line R, and suppose that it is uniformly bounded: there are a,b ∈ R such that a ≤ fn ≤ b for every n ∈ N. Then the sequence (fn)n ∈ N admits a pointwise convergent subsequence. Proof [edit] Step 1. An increasing function f on an interval I has at most countably many points of discontinuity. [edit] Let , i.e. the set of discontinuities, then since f is increasing, any x in A satisfies , where ,, hence by discontinuity, . Since the set of rational numbers is dense in R, is non-empty. Thus the axiom of choice indicates that there is a mapping s from A to Q. It is sufficient to show that s is injective, which implies that A has a non-larger cardinity than Q, which is countable. Suppose x1,x2∈A, x1<x2, then , by the construction of s, we have s(x1)<s(x2). Thus s is injective. Step 2. Inductive Construction of a subsequence converging at discontinuities and rationals. [edit] Let , i.e. the discontinuities of fn, , then A is countable, and it can be denoted as {an: n∈N}. By the uniform boundedness of (fn)n ∈ N and B-W theorem, there is a subsequence (f(1)n)n ∈ N such that (f(1)n(a1))n ∈ N converges. Suppose (f(k)n)n ∈ N has been chosen such that (f(k)n(ai))n ∈ N converges for i=1,...,k, then by uniform boundedness, there is a subsequence (f(k+1)n)n ∈ N of (f(k)n)n ∈ N, such that (f(k+1)n(ak+1))n ∈ N converges, thus (f(k+1)n(ai))n ∈ N converges for i=1,...,k+1. Let , then gk is a subsequence of fn that converges pointwise in A. Step 3. gk converges in I except possibly in an at most countable set. [edit] Let , then, hk(a)=gk(a) for a∈A, hk is increasing, let , then h is increasing, since supremes and limits of increasing functions are increasing, and for a∈ A by Step 2. By Step 1, h has at most countably many discontinuities. We will show that gk converges at all continuities of h. Let x be a continuity of h, q,r∈ A, q<x<r, then ,hence Thus, Since h is continuous at x, by taking the limits , we have , thus Step 4. Choosing a subsequence of gk that converges pointwise in I [edit] This can be done with a diagonal process similar to Step 2. With the above steps we have constructed a subsequence of (fn)n ∈ N that converges pointwise in I. Generalisation to BVloc [edit] Let U be an open subset of the real line and let fn : U → R, n ∈ N, be a sequence of functions. Suppose that (fn) has uniformly bounded total variation on any W that is compactly embedded in U. That is, for all sets W ⊆ U with compact closure W̄ ⊆ U, : where the derivative is taken in the sense of tempered distributions. Then, there exists a subsequence fnk, k ∈ N, of fn and a function f : U → R, locally of bounded variation, such that fnk converges to f pointwise almost everywhere; and fnk converges to f locally in L1 (see locally integrable function), i.e., for all W compactly embedded in U, : : : 132 and, for W compactly embedded in U, : : : 122 Further generalizations [edit] There are many generalizations and refinements of Helly's theorem. The following theorem, for BV functions taking values in Banach spaces, is due to Barbu and Precupanu: Let X be a reflexive, separable Hilbert space and let E be a closed, convex subset of X. Let Δ : X → [0, +∞) be positive-definite and homogeneous of degree one. Suppose that zn is a uniformly bounded sequence in BV([0, T]; X) with zn(t) ∈ E for all n ∈ N and t ∈ [0, T]. Then there exists a subsequence znk and functions δ, z ∈ BV([0, T]; X) such that for all t ∈ [0, T], and, for all t ∈ [0, T], and, for all 0 ≤ s < t ≤ T, See also [edit] Bounded variation Fraňková-Helly selection theorem Total variation References [edit] ^ Jump up to: a b Ambrosio, Luigi; Fusco, Nicola; Pallara, Diego (2000). Functions of Bounded Variation and Free Discontinuity Problems. Oxford University Press. doi:10.1093/oso/9780198502456.001.0001. ISBN 9780198502456. Rudin, W. (1976). Principles of Mathematical Analysis. International Series in Pure and Applied Mathematics (Third ed.). New York: McGraw-Hill. 167. ISBN 978-0070542358. Barbu, V.; Precupanu, Th. (1986). Convexity and optimization in Banach spaces. Mathematics and its Applications (East European Series). Vol. 10 (Second Romanian ed.). Dordrecht: D. Reidel Publishing Co. xviii+397. ISBN 90-277-1761-3. MR 0860772 Retrieved from " Categories: Compactness theorems Theorems in mathematical analysis Hidden categories: Articles with short description Short description matches Wikidata
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https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:common-factor/a/taking-common-factors
Factoring polynomials by taking a common factor (article) | Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content Algebra 2 Course: Algebra 2>Unit 3 Lesson 3: Taking common factors Taking common factor from binomial Taking common factor from trinomial Taking common factor: area model Factoring polynomials by taking a common factor Factor polynomials: common factor Math> Algebra 2> Polynomial factorization> Taking common factors © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Factoring polynomials by taking a common factor AZ.Math: A1.A‑SSE.A.2, A2.A‑SSE.A.2, A2.A‑SSE.B.3.c Google Classroom Microsoft Teams Learn how to factor a common factor out of a polynomial expression. For example, factor 6x²+10x as 2x(3x+5). What you should be familiar with before this lesson The GCF (greatest common factor) of two or more monomials is the product of all their common prime factors. For example, the GCF of 6 x‍ and 4 x 2‍ is 2 x‍. If this is new to you, you'll want to check out our greatest common factors of monomials article. What you will learn in this lesson In this lesson, you will learn how to factor out common factors from polynomials. The distributive property: a(b+c)=a b+a c‍ To understand how to factor out common factors, we must understand the distributive property. For example, we can use the distributive property to find the product of 3 x 2‍ and 4 x+3‍ as shown below: 3 x 2(4 x+3)=3⏜x 2(4 x)+3⏜x 2(3)‍ Notice how each term in the binomial was multiplied by a common factor of 3 x 2‍. However, because the distributive property is an equality, the reverse of this process is also true! 3 x 2(4 x)+3 x 2(3)=3⏜⏜x 2(4 x+3)‍ If we start with 3 x 2(4 x)+3 x 2(3)‍, we can use the distributive property to factor out 3 x 2‍ and obtain 3 x 2(4 x+3)‍. The resulting expression is in factored form because it is written as a product of two polynomials, whereas the original expression is a two-termed sum. Check your understanding Problem 1 Write 2 x(3 x)+2 x(5)‍ in factored form. Choose 1 answer: Choose 1 answer: (Choice A) (2 x)(3 x)(5)‍ A (2 x)(3 x)(5)‍ (Choice B) 2 x(3 x+5)‍ B 2 x(3 x+5)‍ (Choice C) 6 x 2+10 x‍ C 6 x 2+10 x‍ Check Explain Factoring out the greatest common factor (GCF) To factor the GCF out of a polynomial, we do the following: Find the GCF of all the terms in the polynomial. Express each term as a product of the GCF and another factor. Use the distributive property to factor out the GCF. Let's factor the GCF out of 2 x 3−6 x 2‍. Step 1: Find the GCF 2 x 3=2⋅x⋅x⋅x‍ 6 x 2=2⋅3⋅x⋅x‍ So the GCF of 2 x 3−6 x 2‍ is 2⋅x⋅x=2 x 2‍. Step 2: Express each term as a product of 2 x 2‍ and another factor. 2 x 3=(2 x 2)(x)‍ 6 x 2=(2 x 2)(3)‍ I'd like to see how to find the other factors. We can find the missing factor by dividing each term by the greatest common factor (2 x 2)‍. The first will be 2 x 3 2 x 2=x‍. The second term will be 6 x 2 2 x 2=3‍. Notice that these expressions are comprised of the remaining factors in each monomial: 2 x 3=2⋅x⋅x⋅x‍ 6 x 2=2⋅3⋅x⋅x‍ So the polynomial can be written as 2 x 3−6 x 2=(2 x 2)(x)−(2 x 2)(3)‍. Step 3: Factor out the GCF Now we can apply the distributive property to factor out 2 x 2‍. 2 x 2(x)−2 x 2(3)=2⏜⏜x 2(x−3)‍ Verifying our result We can check our factorization by multiplying 2 x 2‍ back into the polynomial. 2 x 2(x−3)=2⏜x 2(x)−2⏜x 2(3)‍ Since this is the same as the original polynomial, our factorization is correct! Check your understanding Problem 2 Factor out the greatest common factor in 12 x 2+18 x‍. Choose 1 answer: Choose 1 answer: (Choice A) 6 x(2 x+18 x)‍ A 6 x(2 x+18 x)‍ (Choice B) 6 x(2 x+3)‍ B 6 x(2 x+3)‍ (Choice C) 6 x(3+12 x 2)‍ C 6 x(3+12 x 2)‍ Check Explain Problem 3 Factor out the greatest common factor in the following polynomial. 10 x 2+25 x+15=‍ Check Explain Problem 4 Factor out the greatest common factor in the following polynomial. x 4−8 x 3+x 2=‍ Check Explain Can we be more efficient? If you feel comfortable with the process of factoring out the GCF, you can use a faster method: Once we know the GCF, the factored form is simply the product of that GCF and the sum of the terms in the original polynomial divided by the GCF. See, for example, how we use this fast method to factor 5 x 2+10 x‍, whose GCF is 5 x‍: 5 x 2+10 x=5 x(5 x 2 5 x+10 x 5 x)=5 x(x+2)‍ Factoring out binomial factors The common factor in a polynomial does not have to be a monomial. For example, consider the polynomial x(2 x−1)−4(2 x−1)‍. Notice that the binomial 2 x−1‍ is common to both terms. We can factor this out using the distributive property: x(2 x−1)−4(2 x−1)=(x−4)(2 x−⏜⏜1)‍ I'm still confused. Is there another explanation? Sometimes it is easier to visualize this factorization if we make a substitution. The distributive property tells us that A B−C B=B(A−C)‍. But if A=x‍, B=2 x−1‍ and C=4‍, we can substitute to justify the factorization. x(2 x−1)−4(2 x−1)=(2 x−1)(x−4)‍ Why is 2x-1 in parentheses? If we wrote the factorization as 2 x−1(x−4)‍ this would imply that only the −1‍ gets distributed to x−4‍. We know that the entire factor of 2 x−1‍ must be distributed, and so we must put parentheses around the factor. Check your understanding Problem 5 Factor out the greatest common factor in the following polynomial. 2 x(x+3)+5(x+3)=‍ Check Explain Different kinds of factorizations It may seem that we have used the term "factor" to describe several different processes: We factored monomials by writing them as a product of other monomials. For example, 12 x 2=(4 x)(3 x)‍. We factored the GCF from polynomials using the distributive property. For example, 2 x 2+12 x=2 x(x+6)‍. We factored out common binomial factors which resulted in an expression equal to the product of two binomials. For example: x(x+1)+2(x+1)=(x+1)(x+2)‍ While we may have used different techniques, in each case we are writing the polynomial as a product of two or more factors. So in all three examples, we indeed factored the polynomial. Challenge problems Problem 6 Factor out the greatest common factor in the following polynomial. 12 x 2 y 5−30 x 4 y 2=‍ Check Explain Problem 7 A large rectangle with an area of 14 x 4+6 x 2‍ square meters is divided into two smaller rectangles with areas 14 x 4‍ and 6 x 2‍ square meters. 14 x 4‍6 x 2‍Length‍Width‍ The width of the rectangle (in meters) is equal to the greatest common factor of 14 x 4‍ and 6 x 2‍. What is the length and width of the large rectangle? Width=‍ meters Length=‍ meters Check Explain Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Avery Baker 5 years ago Posted 5 years ago. Direct link to Avery Baker's post “I have the following prob...” more I have the following problem 6x^3 + 8x^2 - 4x but I can't get it right can someone help me, please. Answer Button navigates to signup page •Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer A/V 5 years ago Posted 5 years ago. Direct link to A/V's post “So let's begin with GCF. ...” more So let's begin with GCF. Looking at the polynomial, it seems that 2x is the GCF of that. Let's take that out: 2x(3x²+4x-2) Noticing that there is a trinomial that might factor, we use the technique: a c = -6 a + c = 4 Noticing that all factors of 6 cannot add up to 4, we leave it at that. In some cases you cannot factor a trinomial, and this is an example of such. Answer: 2x(3x²+4x-2) 1 comment Comment on A/V's post “So let's begin with GCF. ...” (12 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Revika 4 years ago Posted 4 years ago. Direct link to Revika's post “factors of (a+b)^3 - (a-b...” more factors of (a+b)^3 - (a-b)^3 Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jerry Nilsson 4 years ago Posted 4 years ago. Direct link to Jerry Nilsson's post “Using the binomial theore...” more Using the binomial theorem, we get (𝑎 + 𝑏)³ − (𝑎 − 𝑏)³ = 𝑎³ + 3𝑎²𝑏 + 3𝑎𝑏² + 𝑏³ − (𝑎³ − 3𝑎²𝑏 + 3𝑎𝑏² − 𝑏³) Combining like terms, we get 6𝑎²𝑏 + 2𝑏³ Finally, we factor out 2𝑏, which gives us 2𝑏(3𝑎² + 𝑏²) Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... BethanyVemulapalli 7 years ago Posted 7 years ago. Direct link to BethanyVemulapalli's post “So this isn't really on t...” more So this isn't really on this page but I don't know where else to ask it. I have some homework that I don't fully understand a similar problem to the one I am confused on is 15x(x+6)^2+45x(x+6)+35 Answer Button navigates to signup page •3 comments Comment on BethanyVemulapalli's post “So this isn't really on t...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer that1guy1234 9 months ago Posted 9 months ago. Direct link to that1guy1234's post “To factor the expression ...” more To factor the expression ( 15x(x+6)^2 + 45x(x+6) + 35 ): Factor out the greatest common factor (GCF): [ 15x(x+6)^2 + 45x(x+6) + 35 = 5(3x(x+6)^2 + 9x(x+6) + 7) ] Simplify the expression inside the parentheses: [ 3x(x+6)^2 + 9x(x+6) + 7 ] Factor the quadratic expression (if possible): [ 5(3x(x+6)^2 + 9x(x+6) + 7) ] The expression is now factored as ( 5(3x(x+6)^2 + 9x(x+6) + 7) ). Further simplification may not be possible without additional context or constraints Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... kashishthakur24 a year ago Posted a year ago. Direct link to kashishthakur24's post “how to solve algebra” more how to solve algebra Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “You might be in the wrong...” more You might be in the wrong lesson rn, try the Algebra 1 course! Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more ndzupin 10 months ago Posted 10 months ago. Direct link to ndzupin's post “Tutors, please help :) I...” more Tutors, please help :) I solved an equation using two different methods and generated two DIFFERENT results. I asked ChatGPT to solve the same equation using the same two methods and it also generated the SAME DIFFERENT results as me. Please take a look at the link (below) to my Drive with the screenshot of the equation, and tell me where is the mistake :) Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer TheReal3A 10 months ago Posted 10 months ago. Direct link to TheReal3A's post “20x^2/(2x^2) = 20x^2 1...” more20x^2/(2x^2) = 20x^2 1/(2x^2) = 10`. Make sure the entire denominator is in groupings, which is why we get the 1/2 in the first place! Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... 小企飞弹🐧 a year ago Posted a year ago. Direct link to 小企飞弹🐧's post “For problem 5, do I have ...” more For problem 5, do I have to write the answer in order? For example, one of the equations for distributed property tells us that AB-CB = B(A-C). I wrote my answer for problem 5 as (2x+5)(x+3) when it is supposed to be written as (x+3)(2x+5), but I still got it correct. Answer Button navigates to signup page •1 comment Comment on 小企飞弹🐧's post “For problem 5, do I have ...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “The answers are equivalen...” more The answers are equivalent. The commutative property of multiplication applies. Just like 2(3) is the same as 3(2), your version (2x+5)(x+3) is the same as (x+3)(2x+5). Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Grace Muriithi 9 years ago Posted 9 years ago. Direct link to Grace Muriithi's post “On question 4, I typed in...” more On question 4, I typed in the wrong answer, then it said it was wrong. Then, I put in the right answer and it says it is still wrong! Why? Please fix! Answer Button navigates to signup page •1 comment Comment on Grace Muriithi's post “On question 4, I typed in...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer sarra 2 years ago Posted 2 years ago. Direct link to sarra's post “someone do this for me ( ...” more someone do this for me ( 𝑥^2 − 1 ) ( 𝑥^2 + 𝑥 + 3) + (𝑥2 + 1 )^2 and does it have anything to do with this lesson? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “You should first try it y...” more You should first try it yourself. That's the only true way to learn. You need to follow order of operations (PEMDAS). Here are the basic steps. 1) Multiply the first 2 polynomials & simplify: ( 𝑥^2 − 1 ) ( 𝑥^2 + 𝑥 + 3) 2) Multiply the last 2 binomials. Remember, (𝑥^2 + 1 )^2 = (𝑥^2 + 1 ) (𝑥^2 + 1 ) and will create a trinomial when you FOIL. 3) Combine like terms Here are videos that might help you 1) Squaring binomials: 2) Multiply polynomials: Hope these help. Comment back if you have questions. 4 comments Comment on Kim Seidel's post “You should first try it y...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Diane Ray 5 years ago Posted 5 years ago. Direct link to Diane Ray's post “Example 3 trinomial facto...” more Example 3 trinomial factors to two binomials! Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 5 years ago Posted 5 years ago. Direct link to Kim Seidel's post “There are quite a few vid...” more There are quite a few videos on that topic. See lessons at this link: FYI - You can find lessons yourself by using the search bar at the top of any KA screen. 1 comment Comment on Kim Seidel's post “There are quite a few vid...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more LGP25 5 years ago Posted 5 years ago. Direct link to LGP25's post “How would you solve this ...” more How would you solve this cubic equation by factoring out the GCF first? x^3-10x^2+24x Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Victor 5 years ago Posted 5 years ago. Direct link to Victor's post “The GCF in this equation ...” more The GCF in this equation x³ - 10x² + 24x is x. Extracting x will give you x (x² - 10x + 24). Then you can factor your expression (if you've learned how to) to x (x - 6)(x - 4). I hope this helped! 1 comment Comment on Victor's post “The GCF in this equation ...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. 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https://www.quora.com/What-is-the-probability-that-at-least-one-head-appears-in-two-tosses-of-a-fair-coin
What is the probability that at least one head appears in two tosses of a fair coin? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Coin Toss Solution Counting Statistics Sources Probability of an Event Probability Theory Coin Flipping Probability (general) Mathematical Sciences 5 What is the probability that at least one head appears in two tosses of a fair coin? All related (68) Sort Recommended Jered M. Mathematics and statistics educator · Author has 3K answers and 5.5M answer views ·5y There are four possible equally-likely outcomes from two coin tosses: HH, HT, TH, TT. Of these, three contain at least three heads. Therefore, the probability is 3 4.3 4. If that approach seems too “naive”, there is a more arithmetic approach: find the probability of the complement. The complement of “getting at least one head” is “getting no heads”. On each toss, the probability of getting no head is 1 2,1 2, and we have to do that twice in a row to have no heads after two flips. This gives us P(no heads)=1 2×1 2=1 4.P(no heads)=1 2×1 2=1 4. Finally, use the fact that P(\text{at least one h P(\text{at least one h Continue Reading There are four possible equally-likely outcomes from two coin tosses: HH, HT, TH, TT. Of these, three contain at least three heads. Therefore, the probability is 3 4.3 4. If that approach seems too “naive”, there is a more arithmetic approach: find the probability of the complement. The complement of “getting at least one head” is “getting no heads”. On each toss, the probability of getting no head is 1 2,1 2, and we have to do that twice in a row to have no heads after two flips. This gives us P(no heads)=1 2×1 2=1 4.P(no heads)=1 2×1 2=1 4. Finally, use the fact that P(at least one head)=1−P(no heads)P(at least one head)=1−P(no heads) to obtain the desired probability. (Of course, it is the same as the result from the first approach.) Upvote · 9 5 9 2 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 619 Related questions More answers below A fair coin is tossed three times. what is the probability that at least one head appears? A fair coin is tossed seven times. What is the probability that at least one head appears? If two coins are tossed 3 times, what is the probability of at least one head? A fair coin is tossed 5 times. What is the probability of getting at least one head on consecutive tosses? What is the probability of getting one head while tossing a coin twice? Assistant Bot · 1y To find the probability that at least one head appears in two tosses of a fair coin, we can use the complementary probability approach. Calculate the total possible outcomes: When tossing a coin twice, the possible outcomes are: HH (two heads) HT (head and tail) TH (tail and head) TT (two tails) So, there are 4 possible outcomes. Calculate the probability of the complementary event: The complementary event to getting at least one head is getting no heads, which occurs only when both tosses result in tails (TT). The probability of getting tails in one toss is 1 2 1 2. T Continue Reading To find the probability that at least one head appears in two tosses of a fair coin, we can use the complementary probability approach. Calculate the total possible outcomes: When tossing a coin twice, the possible outcomes are: HH (two heads) HT (head and tail) TH (tail and head) TT (two tails) So, there are 4 possible outcomes. Calculate the probability of the complementary event: The complementary event to getting at least one head is getting no heads, which occurs only when both tosses result in tails (TT). The probability of getting tails in one toss is 1 2 1 2. Therefore, the probability of getting tails in both tosses is: P(TT)=P(T)×P(T)=1 2×1 2=1 4 P(TT)=P(T)×P(T)=1 2×1 2=1 4 2. Calculate the probability of at least one head: Now, we can find the probability of getting at least one head by subtracting the probability of getting no heads from 1: P(at least one head)=1−P(TT)=1−1 4=3 4 P(at least one head)=1−P(TT)=1−1 4=3 4 Thus, the probability that at least one head appears in two tosses of a fair coin is 3 4 3 4 or 75%. Upvote · Samuel Pickering C,D in A-level Mathematics&Physics A Level, Notre Dame Catholic Sixth Form College (Graduated 2019) ·4y Originally Answered: A fair coin is tossed three times. what is the probability that at least one head appears? · At least one head can be quite tedious to figure out as there can be either 1, 2 or 3 heads appearing out of the many different scenarios. So, what we can do instead is find out the probability where no heads occur. This way, we know that from the possible scenarios, only 1 can occur. This being three tails in a row. since we know that a fair coin has two sides and therefore, each side has a 1/2 chance, we can use the formula for AND scenarios in probability. This gives us this: P(3Tails) = 1/21/21/2 (1/2 times 1/2 times 1/2) From here, we can then do fraction multiplication and get a value fo Continue Reading At least one head can be quite tedious to figure out as there can be either 1, 2 or 3 heads appearing out of the many different scenarios. So, what we can do instead is find out the probability where no heads occur. This way, we know that from the possible scenarios, only 1 can occur. This being three tails in a row. since we know that a fair coin has two sides and therefore, each side has a 1/2 chance, we can use the formula for AND scenarios in probability. This gives us this: P(3Tails) = 1/21/21/2 (1/2 times 1/2 times 1/2) From here, we can then do fraction multiplication and get a value for the probability of P(3tails) = 1/8. Since we now have the probability of no heads occurring, we can reverse it to have every scenario where at least 1 head occurs. Which happens to be every other scenario. Since we know that all probabilities all must be equal to 1, 1–1/8 will give us P(at least 1 heads) Which is equal to 7/8. Upvote · 9 5 Doug O'Brien Former Volunteer, Peace Corps Panama Fluent in Spanish · Author has 185 answers and 262.8K answer views ·5y Originally Answered: A fair coin is tossed three times. what is the probability that at least one head appears? · Ch, this would be easier if you specified what you mean by “at least one”. At least one what? I’ll assume you forgot to finish your question and that you meant to say “at least one head”. The probability of at least one head is 1 - probability of all tails. The probability of all 3 tails - these being independent events - is 1/2 x 1/2 x 1/2 = 1/8. Therefore, the probability of at least on head is 1 - 1/8 = 7/8. Upvote · 9 4 Related questions More answers below What is the probability of having at least one head if a fair coin is tossed four times? What's the probability that a head appears on tossing a coin? What is the probability of getting at most 2 heads in 7 tosses of a fair coin? What is the probability of getting 5 consecutive heads in 10 tosses of a fair coin? What is the probability of getting at least one head, one head, no head, and two heads, when two coins are tossed simultaneously? Aram Abrahamyan BS in Data Science Student ·1y Originally Answered: What is the possibility of having at least a head when a fair coin is tossed 2 times? · If you want to know the possibility, then it is possible to have at least 1 head tossing a fair coin 2 times. Possibility refers to the potential occurrence or existence of something. It indicates that an event or outcome could happen, but it does not provide information about the likelihood or chance of it happening. If you want to know the probability, then you need to do some calculations. We can consider the complementary event, which is not getting any heads, which means getting tails on both flips. Getting tails on one flip has the probability of 0.5 since it is a fair coin. Since the tos Continue Reading If you want to know the possibility, then it is possible to have at least 1 head tossing a fair coin 2 times. Possibility refers to the potential occurrence or existence of something. It indicates that an event or outcome could happen, but it does not provide information about the likelihood or chance of it happening. If you want to know the probability, then you need to do some calculations. We can consider the complementary event, which is not getting any heads, which means getting tails on both flips. Getting tails on one flip has the probability of 0.5 since it is a fair coin. Since the tosses are independent we need to raise 0.5 to the second power (0.5^2) to get the probability of getting 2 tails on 2 flips of a fair coin. 0.5^2 = 0.25. As we were calculating the complementary event, now we need to substract the result from one to get the answer for our initial problem. 1–0.25=0.75 So the final result is 0.75 or 75% Upvote · Sponsored by LPU Online Career Ka Turning Point with LPU Online. 100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support. Apply Now 999 262 Nigel Parsons Puzzler, gamer, Sci-fi nut. · Author has 2.2K answers and 1.9M answer views ·2y Originally Answered: If two coins are tossed, what will be the probability of getting at least one head? · If two coins are tossed, what will be the probability of getting at least one head? With one fair toss of two fair coins, or two fair tosses of one fair coin, there are two possible outcomes for each coin/toss. There are 4 (22) possible outcomes for 2 tosses. These are: Heads Heads Tails Tails Heads Tails Tails Heads Of these, all except Tails Tails contain ‘at least one head’. Probability =3/4 = 75% Upvote · 9 4 Dave Wade-Stein I am a cashless society · Author has 1.9K answers and 8.2M answer views ·5y Well, what are the outcomes of two tosses? tails + tails tails + heads heads + tails heads + heads So in 3 out of 4 of those cases, a head appears, therefore the likelihood is 75%. Upvote · 9 1 Sponsored by RedHat Know what your AI knows, with open source models. Your AI should keep records, not secrets. Learn More 99 20 Peggy Lucas Solver of puzzles, prob/stat instructor, calculus instructor ·2y Originally Answered: If two coins are tossed, what will be the probability of getting at least one head? · P(at least 1 head) + P(no heads) = 1 because those two sets compose the whole sample space and don't overlap. So P(at least 1 head)= 1- P(no heads)=1-tails tails =1-(1/2)(1/2)=3/4 Upvote · 9 3 Harvinder Kumar 3y Originally Answered: When two fair coins are tossed, what is the probability of occurrence of at least one head? · The answer is. 3/4 because we are asked of at least one head so outcomes are , HH,TT,HT,TH we can see that 3 options comes at least one head so answer is 3/4 Hope it will help you Upvote · 9 2 Sponsored by JetBrains Write better C++ code with less effort. Boost your efficiency with refactorings, code analysis, unit test support, and an integrated debugger. Download 999 897 Nigel Parsons Studied at Bishop of Llandaff Church in Wales High School · Author has 2.2K answers and 1.9M answer views ·4y Originally Answered: What is the probability that at least one head appears in 3 tosses of a fair coin? · What is the probability that at least one head appears in 3 tosses of a fair coin? With 3 tosses of a fair coin there are 8 (2^3) possible outcomes. The only outcome which does not contain ‘at least one head’ is all tails. The probability of this is 1/8 So the probability of at least 1 head is the complement of this, 1 - 1/8 = 7/8 = 87.5% Upvote · 9 3 Colinjivadi Mahadevan Former Executive Director(Retired) at Life Insurance Corporation of India (LIC) (1965–2002) · Author has 1.4K answers and 982.8K answer views ·2y Originally Answered: If two coins are tossed, what will be the probability of getting at least one head? · When two coins are tossed,the possible combinations of outcome are HH,HT,TT & TH.i.e 4 total combinations. The combinations with at least one head are HH,HT, & TH, i.e 3 combinations. The required probability =3/4 Upvote · 9 2 Archit Gupta Software Engineer at Google DeepMind ·Updated 9y Originally Answered: When two fair coins are tossed, what is the probability of occurrence of at least one head? · Imagine yourself to be out there tossing 2 coins.you have 1 coin in left hand and the other in right. In one toss,the outcome of the coin in left hand is head and in the right hand is tail. In the next outcome,the result is opposite ie in left hand is tail and in right hand is head. The 2 events are different right? Upvote · 9 3 Arthur Fisher Former Administration and Logistics, now retired · Author has 9K answers and 3.4M answer views ·4y Originally Answered: What is the probability that at least one head appears in 3 tosses of a fair coin? · Well the only chance of there NOT being at least one head, is a sequence of three tails in three tosses. The chances of that happening is 0.5^3 = 0.125. And so the answer you are looking for is 1–0.5^3 = 0.875 or seven-eights or 87½%. Upvote · 9 2 Related questions A fair coin is tossed three times. what is the probability that at least one head appears? A fair coin is tossed seven times. What is the probability that at least one head appears? If two coins are tossed 3 times, what is the probability of at least one head? A fair coin is tossed 5 times. What is the probability of getting at least one head on consecutive tosses? What is the probability of getting one head while tossing a coin twice? What is the probability of having at least one head if a fair coin is tossed four times? What's the probability that a head appears on tossing a coin? What is the probability of getting at most 2 heads in 7 tosses of a fair coin? What is the probability of getting 5 consecutive heads in 10 tosses of a fair coin? What is the probability of getting at least one head, one head, no head, and two heads, when two coins are tossed simultaneously? If you toss two fair coins, what is the probability of getting at least one head in four tosses? What is the conditional problem of getting two heads given that at least one head white tossing two coins simultaneously? What is the probability that at least one head appears in 3 tosses of a fair coin? What is the probability of getting at least one tail when two coins are tossed? A coin is tossed three times, what is the probability of tossing at least two heads? Related questions A fair coin is tossed three times. what is the probability that at least one head appears? A fair coin is tossed seven times. What is the probability that at least one head appears? If two coins are tossed 3 times, what is the probability of at least one head? A fair coin is tossed 5 times. What is the probability of getting at least one head on consecutive tosses? What is the probability of getting one head while tossing a coin twice? 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https://pubchem.ncbi.nlm.nih.gov/compound/Menaquinone-7
Menaquinone 7 | C46H64O2 | CID 5287554 - PubChem An official website of the United States government Here is how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. NIH National Library of Medicine NCBI PubChem About Docs Submit Contact Search PubChem compound Summary Menaquinone 7 PubChem CID 5287554 Structure Primary Hazards Laboratory Chemical Safety Summary (LCSS) Datasheet Molecular Formula C 46 H 64 O 2 Synonyms Menaquinone 7 Menaquinone-7 2124-57-4 Vitamin K2(35) Menaquinone K7 View More... Molecular Weight 649.0 g/mol Computed by PubChem 2.2 (PubChem release 2021.10.14) Dates Create: 2005-03-27 Modify: 2025-08-30 Description Menaquinone-7 is a menaquinone whose side-chain contains seven isoprene units in an all-trans-configutation. It has a role as a Mycoplasma genitalium metabolite, a bone density conservation agent, an Escherichia coli metabolite, a human blood serum metabolite and a cofactor. ChEBI Menaquinone 7 is under investigation in clinical trial NCT00402974 (The Effect of Vitamin K Supplementation on Osteocalcin Carboxylation in Healthy Children). DrugBank Menaquinone-7 has been reported in Brevibacillus brevis with data available. LOTUS - the natural products occurrence database View More... 1 Structures 1.1 2D Structure Structure Search Get Image Download Coordinates Chemical Structure Depiction Full screen Zoom in Zoom out PubChem 1.2 3D Status Conformer generation is disallowed since too flexible PubChem 2 Names and Identifiers 2.1 Computed Descriptors 2.1.1 IUPAC Name 2-[(2 E,6 E,10 E,14 E,18 E,22 E)-3,7,11,15,19,23,27-heptamethyloctacosa-2,6,10,14,18,22,26-heptaenyl]-3-methylnaphthalene-1,4-dione Computed by Lexichem TK 2.7.0 (PubChem release 2021.10.14) PubChem 2.1.2 InChI InChI=1S/C46H64O2/c1-34(2)18-12-19-35(3)20-13-21-36(4)22-14-23-37(5)24-15-25-38(6)26-16-27-39(7)28-17-29-40(8)32-33-42-41(9)45(47)43-30-10-11-31-44(43)46(42)48/h10-11,18,20,22,24,26,28,30-32H,12-17,19,21,23,25,27,29,33H2,1-9H3/b35-20+,36-22+,37-24+,38-26+,39-28+,40-32+ Computed by InChI 1.0.6 (PubChem release 2021.10.14) PubChem 2.1.3 InChIKey RAKQPZMEYJZGPI-LJWNYQGCSA-N Computed by InChI 1.0.6 (PubChem release 2021.10.14) PubChem 2.1.4 SMILES CC1=C(C(=O)C2=CC=CC=C2C1=O)C/C=C(\C)/CC/C=C(\C)/CC/C=C(\C)/CC/C=C(\C)/CC/C=C(\C)/CC/C=C(\C)/CCC=C(C)C Computed by OEChem 2.3.0 (PubChem release 2024.12.12) PubChem 2.2 Molecular Formula C 46 H 64 O 2 Computed by PubChem 2.2 (PubChem release 2021.10.14) PubChem 2.3 Other Identifiers 2.3.1 CAS 2124-57-4 CAS Common Chemistry; ChemIDplus; DrugBank; EPA DSSTox; European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); Hazardous Substances Data Bank (HSDB) 2.3.2 European Community (EC) Number 870-176-9 European Chemicals Agency (ECHA) 2.3.3 UNII 8427BML8NY FDA Global Substance Registration System (GSRS) 2.3.4 ChEBI ID CHEBI:44245 ChEBI 2.3.5 ChEMBL ID CHEMBL1230575 ChEMBL 2.3.6 DrugBank ID DB13075 DrugBank 2.3.7 DSSTox Substance ID DTXSID801317474 EPA DSSTox 2.3.8 Lipid Maps ID (LM_ID) LMPR02030041 LIPID MAPS 2.3.9 Metabolomics Workbench ID 56546 Metabolomics Workbench 2.3.10 Nikkaji Number J141.747I Japan Chemical Substance Dictionary (Nikkaji) J16.269H Japan Chemical Substance Dictionary (Nikkaji) 2.3.11 RXCUI 2001758 NLM RxNorm Terminology 2.3.12 Wikidata Q27120546 Wikidata 2.4 Synonyms 2.4.1 MeSH Entry Terms menaquinone 7 menaquinone K7 1,4-naphthalenedione, 2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-, (all-E)- menaquinone-7 vitamin K2(35) vitamin MK 7 Medical Subject Headings (MeSH) 2.4.2 Depositor-Supplied Synonyms Menaquinone 7 Menaquinone-7 2124-57-4 Vitamin K2(35) Menaquinone K7 Vitamin MK 7 Menakinon 7 8427BML8NY 2-((2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-Heptamethyloctacosa-2,6,10,14,18,22,26-heptaen-1-yl)-3-methylnaphthalene-1,4-dione CHEBI:44245 1,4-Naphthalenedione, 2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-, (all-E)- 2-[(2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-heptamethyloctacosa-2,6,10,14,18,22,26-heptaenyl]-3-methylnaphthalene-1,4-dione DTXSID801317474 2-[(2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-heptamethyloctacosa-2,6,10,14,18,22,26-heptaen-1-yl]-3-methylnaphthalene-1,4-dione 2-((2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-heptamethyloctacosa-2,6,10,14,18,22,26-heptaenyl)-3-methylnaphthalene-1,4-dione RefChem:925473 DTXCID90908566 MFCD06200757 MK-7 (all-E)-2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-1,4-naphthalenedione 13425-62-2 2-Methyl-3-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptenyl)-1,4-naphthoquinone Mk7 (all-E)-2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosahept-aenyl)-3-methyl-1,4-naphthoquinone Menaquinone 7, Synthetic (90%) Menaquinone K Menlaquinone 7 HSDB 1040 menaquinone(7) Menaquinone MK-7 C46H64O2 trans-menaquinone 7 Vitamin K2 (MK-7) Menaquinone-7 (Standard) UNII-8427BML8NY MENAQUINONE 7 [MI] MENAQUINONE-7 [DSC] MENAQUINONE 7 [HSDB] SCHEMBL436068 CHEMBL1230575 MENAQUINONE-7 [USP-RS] MENAQUINONE-7 [WHO-DD] SCHEMBL29359122 MSK1597 Vitamin K2 Menaquinone-7 (MK-7) AKOS027326821 DB13075 FM25056 HY-112499R 1,4-Naphthalenedione, 2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosahept-aenyl)-3-methyl-, (all-E)- 1,4-Naphthoquinone, 2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosahept-aenyl)-3-methyl-, (all-E)- 2-[(2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-heptamethyloctacosa-2,6,10,14,18,22,26-heptaenyl]-3-methyl-naphthalene-1,4-dione AC-36642 AS-78195 DA-68610 SY353428 HY-112499 CS-0046153 NS00072810 124M574 Q27120546 (all-E)-2-(3,7,11,15,19,23,27-Heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-1,4-naphthalenedione;Vitamin K2(35) (All-ee)-2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-1,4-naphthalenedione 1,4-NAPHTHALENEDIONE, 2-((2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-HEPTAMETHYL-2,6,10,14,18,22,26-OCTACOSAHEPTAENYL)-3-METHYL- 1,4-NAPHTHOQUINONE, 2-(3,7,11,15,19,23,27-HEPTAMETHYL-2,6,10,14,18,22,26-OCTACOSAHEPTAENYL)-3-METHYL-, (ALL-E)- 2-[(2E,6E,10E,14E,18E,22E)-3,7,11,15,19,23,27-Heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl]-3-methylnaphthoquinone # PubChem 3 Chemical and Physical Properties 3.1 Computed Properties Property Name Property Value Reference Property Name Molecular Weight Property Value 649.0 g/mol Reference Computed by PubChem 2.2 (PubChem release 2021.10.14) Property Name XLogP3 Property Value 14.5 Reference Computed by XLogP3 3.0 (PubChem release 2021.10.14) Property Name Hydrogen Bond Donor Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2021.10.14) Property Name Hydrogen Bond Acceptor Count Property Value 2 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2021.10.14) Property Name Rotatable Bond Count Property Value 20 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2021.10.14) Property Name Exact Mass Property Value 648.49063128 Da Reference Computed by PubChem 2.2 (PubChem release 2021.10.14) Property Name Monoisotopic Mass Property Value 648.49063128 Da Reference Computed by PubChem 2.2 (PubChem release 2021.10.14) Property Name Topological Polar Surface Area Property Value 34.1 Ų Reference Computed by Cactvs 3.4.8.18 (PubChem release 2021.10.14) Property Name Heavy Atom Count Property Value 48 Reference Computed by PubChem Property Name Formal Charge Property Value 0 Reference Computed by PubChem Property Name Complexity Property Value 1310 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2021.10.14) Property Name Isotope Atom Count Property Value 0 Reference Computed by PubChem Property Name Defined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Defined Bond Stereocenter Count Property Value 6 Reference Computed by PubChem Property Name Undefined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Covalently-Bonded Unit Count Property Value 1 Reference Computed by PubChem Property Name Compound Is Canonicalized Property Value Yes Reference Computed by PubChem (release 2021.10.14) PubChem 3.2 Experimental Properties 3.2.1 Physical Description Light yellow solid; [Merck Index] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 3.2.2 Color / Form Light yellow microcrystalline plates from petroleum ether O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) Crystals Lide, D.R. CRC Handbook of Chemistry and Physics 88TH Edition 2007-2008. CRC Press, Taylor & Francis, Boca Raton, FL 2007, p. 3-324 Hazardous Substances Data Bank (HSDB) LIGHT-YELLOW SOLIDS OR OILS Osol, A. and J.E. Hoover, et al. (eds.). Remington's Pharmaceutical Sciences. 15th ed. Easton, Pennsylvania: Mack Publishing Co., 1975., p. 942 Hazardous Substances Data Bank (HSDB) 3.2.3 Boiling Point 200 °C at 0.0002 mm Hg (some decomposition) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) 3.2.4 Melting Point 54 °C Lide, D.R. CRC Handbook of Chemistry and Physics 88TH Edition 2007-2008. CRC Press, Taylor & Francis, Boca Raton, FL 2007, p. 3-324 Hazardous Substances Data Bank (HSDB) Yellow crystals from alcohol; MW: 444.65; MP: 35 °C; uv max: 248 nm (E(1%)(1 cm) 439) /Menaquinone 4/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) Yellow crystals from acetone + alcohol or petroleum ether. MW: 580.88; MP: 50 °C; uv max (petroleum ether): 243, 248, 261, 325-328 nm (E(1%)(1 CM) 304, 320, 290, 292, 53) /Menaquinone 6/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) 3.2.5 Solubility Slightly less soluble than vitamin K1 in the same organic solvents O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) INSOL IN WATER; SOL IN FAT SOLVENTS Osol, A. and J.E. Hoover, et al. (eds.). Remington's Pharmaceutical Sciences. 15th ed. Easton, Pennsylvania: Mack Publishing Co., 1975., p. 942 Hazardous Substances Data Bank (HSDB) SOL IN OILS Rossoff, I.S. Handbook of Veterinary Drugs. New York: Springer Publishing Company, 1974., p. 658 Hazardous Substances Data Bank (HSDB) SOL IN FATS Goodman, L.S., and A. Gilman. (eds.) The Pharmacological Basis of Therapeutics. 5th ed. New York: Macmillan Publishing Co., Inc., 1975., p. 1592 Hazardous Substances Data Bank (HSDB) 3.3 Chemical Classes Biological Agents -> Vitamins and Derivatives Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 3.3.1 Drugs Pharmaceuticals -> Listed in ZINC15 S55 | ZINC15PHARMA | Pharmaceuticals from ZINC15 | DOI:10.5281/zenodo.3247749 NORMAN Suspect List Exchange 3.3.2 Lipids Lipids -> Prenol Lipids [PR] -> Quinones and hydroquinones [PR02] -> Vitamin K [PR0203] LIPID MAPS 4 Spectral Information 4.1 Mass Spectrometry 4.1.1 GC-MS 1 of 2 items MoNA ID JP008164 MS Category Experimental MS Type GC-MS MS Level MS1 Instrument HITACHI M-80 Instrument Type EI-B Ionization Mode positive Top 5 Peaks 69 99.99 225 83.88 81 57.75 135 29.86 187 27.94 SPLASH splash10-016r-8961000000-9f4b60b646775c88a22c Thumbnail License CC BY-NC-SA MassBank of North America (MoNA) 2 of 2 items NIST Number 33145 Library Main library Total Peaks 93 m/z Top Peak 69 m/z 2nd Highest 225 m/z 3rd Highest 648 Thumbnail NIST Mass Spectrometry Data Center 4.2 UV Spectra UV max absorption (petroleum ether): 243, 248, 261, 270, 325-328 nm at epsilon = 278, 195, 266, 267, 48, respectively, (epsilon = 1%, 1 cm) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) 5 Related Records 5.1 Related Compounds with Annotation Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here. PubChem 5.2 Related Compounds Same Connectivity Count 12 Same Stereo Count 3 Same Isotope Count 10 Same Parent, Connectivity Count 18 Same Parent, Stereo Count 9 Same Parent, Isotope Count 16 Same Parent, Exact Count 7 Mixtures, Components, and Neutralized Forms Count 14 Similar Compounds (2D) View in PubChem Search Similar Conformers (3D) View in PubChem Search PubChem 5.3 Substances 5.3.1 PubChem Reference Collection SID 516571381 PubChem 5.3.2 Related Substances All Count 132 Same Count 115 Mixture Count 17 PubChem 5.3.3 Substances by Category PubChem 5.4 Entrez Crosslinks PubMed Count 1 Protein Structures Count 2 Taxonomy Count 1 Gene Count 2 PubChem 5.5 NCBI LinkOut NCBI 6 Chemical Vendors PubChem 7 Drug and Medication Information 7.1 Drug Indication Open Targets 7.2 FDA National Drug Code Directory National Drug Code (NDC) Directory 7.3 Drug Labels Active ingredient and drug DailyMed 7.4 Clinical Trials 7.4.1 ClinicalTrials.gov ClinicalTrials.gov 7.5 Therapeutic Uses There is no typical dosage for vitamin K. Some multivitamin preparations contain vitamin K as vitamin K1 (phylloquinone or phytonadione) or vitamin K2 (menaquinones) at doses of 25 to 100 ug. PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 711 Hazardous Substances Data Bank (HSDB) Vitamin K is used to treat anticoagulant-induced prothrombin deficiency caused by warfarin, hyporprothrombinemia secondary to antibiotic therapy and hypoprothrombinemia secondary to vitamin C deficiency from various causes, including malabsorption syndromes. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 709 Hazardous Substances Data Bank (HSDB) Because hemorrhagic disease of the newborn can be effectively prevented by administrating vitamin K, infants born in the US and Canada routinely receive 0.5-1 mg pf phylloquinone intramuscularly or 2.0 mg orally within 6 hours of birth. This practice is supported by both US and Canadian pediatric societies. /Phylloquinone/ National Academies of Sciences Institute of Medicine; Vitamin K. In: Dietary Reference Intakes. p.255-61 (2006) National Academies Press, Washington, DC Hazardous Substances Data Bank (HSDB) The current recommendations of the American Academy of Pediatrics advise that "vitamin K (phylloquinone) should be given to all newborns as a single, intramuscular dose of 0.5-1 mg" and if this advice is followed, the disease /Vitamin K deficiency bleeding/ is effectively prevented. /Vitamin K/ Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) For more Therapeutic Uses (Complete) data for Vitamin K2 (8 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 7.6 Drug Warnings ... MK-7 induced more complete carboxylation of osteocalcin, and hematologists should be aware that preparations supplying 50 ug/d or more of MK-7 may interfere with oral anticoagulant treatment in a clinically relevant way. PMID:17158229 Schurgers LJ et al; Blood 109 (8): 3279-83 (2007) Hazardous Substances Data Bank (HSDB) It has been suggested that vitamin K may have roles in osteoporosis and vascular health. However, this is difficult to establish on the basis of the studies performed thus far. /Vitamin K/ National Academies of Sciences Institute of Medicine; Vitamin K. In: Dietary Reference Intakes. p.255-61 (2006) National Academies Press, Washington, DC Hazardous Substances Data Bank (HSDB) Pregnant women and nursing mothers should avoid supplemental intakes of vitamin K greater than RDA amounts (65 ug daily) unless higher amounts are prescribed by their physicians. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 711 Hazardous Substances Data Bank (HSDB) Individuals on chronic warfarin therapy may require dietary counseling on how to maintain steady vitamin K intake levels. Because habitual vitamin K intake may modulate warfarin dosage in patients using this anticoagulant, these individuals should maintain their normal dietary and supplementation patterns once an effective dose of warfarin has been established. /Vitamin K/ National Academies of Sciences Institute of Medicine; Vitamin K. In: Dietary Reference Intakes. p.255-61 (2006) National Academies Press, Washington, DC Hazardous Substances Data Bank (HSDB) For more Drug Warnings (Complete) data for Vitamin K2 (6 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 8 Food Additives and Ingredients 8.1 FDA Generally Recognized as Safe - GRAS Notices 1 of 2 items GRAS Notice Number(GRN) 245 Substance Menaquinone-7 Intended Use As a nutrient supplement in dairy products, margarine substitutes, and dairy products including yoghurt, cottage cheese, hard cheeses, milk drinks and substitute products at levels up to 50 micrograms per day Basis Scientific procedures FDA Letter At the notifier's request, FDA ceased to evaluate this notice Notifier NattoPharma ASA Date of Filing 04/02/2008 FDA Generally Recognized as Safe (GRAS) 2 of 2 items GRAS Notice Number(GRN) 887 Substance Menaquinone-7 Intended Use Ingredient and as a nutrient according to 21 CFR 170.3(o)(20) in nutritional beverage products intended for consumers ages 1-13 years at a use level of 4 g/serving. Basis Scientific procedures FDA Letter FDA has no questions (in PDF) Notifier Synergia Life Sciences Pvt. Ltd. FDA Generally Recognized as Safe (GRAS) 9 Pharmacology and Biochemistry 9.1 MeSH Pharmacological Classification Hemostatics Agents acting to arrest the flow of blood. Absorbable hemostatics arrest bleeding either by the formation of an artificial clot or by providing a mechanical matrix that facilitates clotting when applied directly to the bleeding surface. These agents function more at the capillary level and are not effective at stemming arterial or venous bleeding under any significant intravascular pressure. Medical Subject Headings (MeSH) 9.2 Bionecessity Vitamin K functions as a coenzyme for biological reactions involved in blood coagulation and bone metabolism. It also plays an essential role in the conversion of certain residues in proteins into biologically active forms. These proteins include plasma prothrombin (coagulation factor II) and the plasma procoagulants, factors VII, IX, and X. /Vitamin K/ National Academies of Sciences Institute of Medicine; Vitamin K. In: Dietary Reference Intakes. p.255-61 (2006) National Academies Press, Washington, DC Hazardous Substances Data Bank (HSDB) Vitamin K deficiency can occur under certain conditions. These include, inadequate dietary intake, malabsorption syndromes (cystic fibrosis, Crohn's disease, ulcerative colitis, Whipple's disease, celiac sprue, short bowel syndrome) and loss of storage sites due to hepatocellular disease. Vitamin K deficiency frequently occurs in those with chronic liver disease, such as primary biliary cirrhosis. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 708 Hazardous Substances Data Bank (HSDB) Symptoms of vitamin K deficiency include easy bruisability, epistaxis, gastrointestinal bleeding, menorrhagia and hematuria. Chronic vitamin K deficiency may also result in osteoporosis and increased risk of fractures. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 708 Hazardous Substances Data Bank (HSDB) The classical sign of a vitamin K deficiency has been the development of a hemorrhagic syndrome, and the first proteins identified as requiring vitamin K for their synthesis were plasma clotting factors. ... Prothrombin (clotting factor II) is the circulating zymogen of the procoagulant thrombin, and was the first protein shown to be dependent on vitamin K for its synthesis. Clotting factors VII, IX, and X were all initially identified because their activity was decreased in the plasma of a patient with a hereditary bleeding disorder and were subsequently shown to depend on vitamin K for their syntheses. ... Proteins C and S were discovered after it had been shown that prothrombin contained Gla residues. They were subsequently shown to have an anticoagulant, rather that a procoagulant, role in hemostasis. The seventh vitamin K-dependent plasma protein, protein Z, is not a protease zymogen and also exhibits an anticoagulant role under some conditions. /Vitamin K/ Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) For more Bionecessity (Complete) data for Vitamin K2 (16 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 9.3 Absorption, Distribution and Excretion Vitamin K, mainly in the form of vitamin K1, is principally absorbed from the jejunum and ileum. ... Vitamin K is delivered to the enterocytes in micelles formed from bile salts and other substances. Vitamin K is secreted by enterocytes into the lymphatics in the form of chylomicrons. It enters the circulation via the thoracic duct and is carried in the circulation to various tissues including hepatic, bone and spleen, in the form of chylomicron remnants. In the liver, some vitamin K is stored, some is oxidized to inactive end products and some is secreted with VLDL (very low density lipoprotein). Approximately 50% of vitamin K is carried in the plasma in the form of VLDL, about 25% in LDL (low-density lopoprotein) and about 25% in HDL (high-density lipoprotein). /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 709 Hazardous Substances Data Bank (HSDB) Excretion of vitamin K and its metabolites is mainly via the feces. Some urinary excretion of vitamin K also occurs. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 710 Hazardous Substances Data Bank (HSDB) ...Menaquinones are adequately absorbed from the GI tract only if bile salts are present. Menaquinones and its water-soluble derivatives, however, are absorbed even in the absence of bile. ... Menaquinones are absorbed almost entirely by way of the lymph. /Menaquinones/ Gilman, A.G., T.W. Rall, A.S. Nies and P. Taylor (eds.). Goodman and Gilman's The Pharmacological Basis of Therapeutics. 8th ed. New York, NY. Pergamon Press, 1990., p. 1565 Hazardous Substances Data Bank (HSDB) Menaquinone forms of vitamin K are produced by bacteria in the lower bowel, where the forms appear in large amounts. However, their contributuion to the maintenance of vitamin K status has been difficult to assess. Although the content is extremely variable, the human liver contains about 10 times as much vitamin K as a mixture of menaquinones than as phylloquinone. /Menaquinones/ National Academies of Sciences Institute of Medicine; Vitamin K. In: Dietary Reference Intakes. p.255-61 (2006) National Academies Press, Washington, DC Hazardous Substances Data Bank (HSDB) 9.4 Metabolism / Metabolites A major pathway of vitamin K metabolism is that which is involved in the reduction and recycling of the epoxide formed by the carboxylase. /Vitamin K/ Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) Vitamin K undergoes some oxidative metabolism. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 710 Hazardous Substances Data Bank (HSDB) 9.5 Mechanism of Action In vivo and in vitro studies have shown that vitamin K may directly act on bone metabolism. In vitro studies have demonstrated that vitamin K2 inhibits bone resorption by, in part, inhibiting the production of bone resorbing substances such as prostaglandin E2 and interleukin-6. Vitamin K2 has been reported to enhance human osteoblast-induced mineralization in vitro and to inhibit bone loss in steroid-treated rats and ovariectomized rats. PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 709 Hazardous Substances Data Bank (HSDB) Certain naphthoquinones, in particular the synthetic vitamin K menadione, have been found to have antitumor activity in vitro and in vivo. Vitamin K2 has been found to induce the in vitro differentiation of myeloid leukemic cell lines. The mechanism of the possible anticarcinogenic activity of vitamin K is not well understood. Menadione is an oxidative stress inducer and its possible anticarcinogenic activity may, in part, be explained by induction of apoptotic cell death. One study suggested that the induction of apoptosis by menadione is mediated by the Fas/Fas ligand system. Another study reported that menadione induces cell cycle arrest and cell death by inhibiting Cda 25 phosphatase. PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 709 Hazardous Substances Data Bank (HSDB) Vitamin K is involved as a cofactor in the posttranslational gamma-carboxylation of glutamic acid residues of certain proteins in the body. These proteins include the vitamin K-dependent coagulation factors II (prothrombin), VII (proconvertin), IX (Christmas factor), X (Stuart factor), protein C, protein S, protein Zv and a growth-arrest-specific factor (Gas6). In contrast to the other vitamin K-dependent proteins in the blood coagulation cascade, protein C and protein X serve anticoagulant roles. The two vitamin K-dependent proteins found in bone are osteocalcin, also known as bone G1a (gamma-carboxyglutamate) protein or BGP, and the matrix G1a protein or MGP. Gamma-carboxylation is catalyzed by the vitamin K-dependent gamma-carboxylases. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 708 Hazardous Substances Data Bank (HSDB) The primary gene product of the vitamin K-dependent proteins contains a very homologous domain between the amino terminus of the mature protein and the signal sequence that targets the polypeptide for the secretory pathway. This "propeptide" region appears to be both a "docking" or "recognition " site for the enzyme and a modulator of the activity of the enzyme by decreasing the apparent Km of the Glu site substrate. ... A key finding essential to a complete understanding of the detailed mechanism of action of this enzyme has been the identification of an intermediate chemical form of vitamin K, which could be sufficiently basic to abstract the gamma-hydrogen of the glutamyl residue. It has been proposed that the initial attack of O(2) at the naphthoquinone carbonyl carbon adjacent to the methyl group results in the formation of a dioxetane ring, which generates an alkoxide intermediate. /Vitamin K/ Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) For more Mechanism of Action (Complete) data for Vitamin K2 (8 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 9.6 Biochemical Reactions Rhea - Annotated Reactions Database 10 Use and Manufacturing 10.1 Uses Sources/Uses Menaquinones (vitamin K2) have vitamin K activity; Produced by intestinal bacteria and used as a vitamin (prothrombogenic); [Merck Index] Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Menadione is used as a source of vitamin K activity in poultry and swine rations. /Menadione/ Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) The nutritional supplement forms of vitamin K are vitamin K1 and vitamin K2. PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 709 Hazardous Substances Data Bank (HSDB) Forms of vitamin K commonly incorporated into nonhuman-primate diets include the water-soluble derivatives menadione dimethylpyrimidinol bisulfite (MPB), menadione sodium bisulfite (MSB), and menadione sodium bisulfite complex (MSBC). Vitamins K1 and K2 and menadione also have been used, but they are fat-soluble, so it is difficult to distribute them uniformly in dry feeds. Committee on Animal Nutrition, Ad Hoc Committee on Nonhuman Primate Nutrition, National Research Council; Nutrient Requirements of Nonhuman Primates: Second Revised Edition (2003). Available from, as of March 17, 2011: Hazardous Substances Data Bank (HSDB) THERAPEUTIC CATEGORY: Vitamin (prothrombogenic) /Menaquinones/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) 10.2 Methods of Manufacturing A ... short synthesis of vitamin K1 and vitamins K2 has been patented by F. Hoffmann-LaRoche. Menadione is first subjected to a Diels-Alder reaction with cyclopentadiene, producing the crystalline endo-Diels-Alder adduct in 93% yield. Subsequently, the proton alpha to the carbonyl group can be abstracted with a strong base (e.g., potassium tertbutoxide) and the resulting enolate alkylated with phytyl bromide (E/Z greater than or equal to 98:2). The alkylation product is obtained in a yield of greater than or equal to 90%. As a result of the basic alkylation conditions, the trans geometry of the double bond in phytyl bromide (greater than or equal to 98% E) is completely retained during alkylation. The alkylation product is thermally unstable and undergoes a retro-Diels-Alder reaction. On heating in toluene (110 °C), it decomposes rapidly into vitamin K1 (greater than or equal to 98% E) and cyclopentadiene. The yield of vitamin K1 is practically quantitative. The cyclopentadiene can be recovered by distillation and reused. The vitamins K2 (menaquinones) and ubiquinones can be synthesized according to the same scheme. /Vitamin K1 and Vitamins K2/ Eggersdorfer M et al; Ullmann's Encyclopedia of Industrial Chemistry 7th ed. (2010). NY, NY: John Wiley & Sons; Vitamins. Online Posting Date: June 15, 2000 Hazardous Substances Data Bank (HSDB) Isolation from putrefied fish meal: McKee et al, J Biol Chem 131, 327 (1939); From cultures of Bacillus brevis: Tishler, Sampson, Proc Soc Exp Biol Med, 136 (1948). Structure and synthesis: Isler et al, Helv Chim Acta 41, 786 (1958). /Menaquinones/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) 10.3 General Manufacturing Information In the early 1930s, Henrik Dam observed that chicks consuming very low lipid diets developed subdural or muscular hemorrhages and that blood taken from these animals clotted slowly. This hemorrhagic disease could not be cured by supplementation with any other known dietary factor, and Dam proposed the existence of a new fat-soluble factor, vitamin K. ... The term vitamin K is now used as a generic descriptor of 2-methyl-1,4-naphthoquinone (menadione) and all derivatives of this compound that exhibit antihemorrhagic activities in animals fed a vitamin K-deficient diet. /Vitamin K/ Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) The compound, 2-methyl-3-farnesyl-geranylgeranyl-1,4-naphthaquinone, first isolated from putrefied fish meal, is one of a series of vitamin K compounds with unsaturated side chains called multiprenylmenaquinones, which are produced by a limited number of anaerobic bacteria and are present in large quantities in the lower bowel. This particular menaquinone has 7 isoprenoid units in the side chain and was once called vitamin K2. That term is currently used to describe any of the vitamers with an unsaturated side chain, and this compound is more correctly identified as menaquinone-7 (MK-7). The predominant menaquinones found in the gut are MK-7 through MK-9, but smaller amounts of others are also present. Suttie JW; Vitamin K. In: Encyclopedia of Dietary Supplements, ed. Coates PM et al; Marcel Dekker, New York, NY pp. 771-82 (2005) Hazardous Substances Data Bank (HSDB) Nomenclature is based on the number of isoprene residues comprising the side chain. /Menaquinones/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) Vitamin K is a generic term for a group of substances which contain the 2-methyl-1,4-naphthoquinone ring structure and which possess hemostatic activity. ... Vitamin K1 or phylloquinone is the principal dietary source of vitamin K and its predominant circulating form. ... Vitamin K1 is a fat-soluble substance. Vitamin K2, which is also fat soluble, is the collective term for a number of substances known as menaquinones. The dietary contribution of vitamin K2 is much less than that of vitamin K1. The amount of vitamin K contributed to the body by the intestinal microflora remains unclear. Vitamin K3 or menadione is a fat-soluble synthetic compound which is used in animal feed and dog and cat food. It is metabolized to vitamin K2. PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 708 Hazardous Substances Data Bank (HSDB) For more General Manufacturing Information (Complete) data for Vitamin K2 (6 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 11 Identification 11.1 Clinical Laboratory Methods As a result of its high selectivity and sensitivity, high performance liquid chromatography (HPLC) is the method of choice for the determination of phylloquinone and menaquinones in the blood, tissues, milk, and in foods. Various procedures for extraction and preliminary purification, internal standardization, HPLC conditions (with normal or reversed phase) and possibilities for UV, electrochemical, and fluorescence detection (both after electrochemical or chemical reduction and after photochemical decomposition) of the vitamins K have been described. /Phylloquinone and Menaquinones/ Eggersdorfer M et al; Ullmann's Encyclopedia of Industrial Chemistry 7th ed. (2010). NY, NY: John Wiley & Sons; Vitamins. Online Posting Date: June 15, 2000 Hazardous Substances Data Bank (HSDB) A sensitive and highly selective high-performance liquid chromatography (HPLC) method was developed for the determination of vitamin K homologues including phylloquinone (PK), menaquinone-4 (MK-4) and menaquinone-7 (MK-7) in human plasma using post-column peroxyoxalate chemiluminescence (PO-CL) detection following on-line ultraviolet (UV) irradiation... PMID:17481401 Ahmed S et al; Anal Chim Acta 591 (2): 148-54 (2007) Hazardous Substances Data Bank (HSDB) 12 Safety and Hazards 12.1 Hazards Identification 12.1.1 GHS Classification Note This chemical does not meet GHS hazard criteria for 100% (105 of 105) of all reports. GHS Hazard Statements Not Classified Reported as not meeting GHS hazard criteria by 105 of 105 companies. For more detailed information, please visit ECHA C&L website. ECHA C&L Notifications Summary Aggregated GHS information provided per 105 reports by companies from 1 notifications to the ECHA C&L Inventory. Reported as not meeting GHS hazard criteria per 105 of 105 reports by companies. There are 0 notifications provided by 0 of 105 reports by companies with hazard statement code(s). Information may vary between notifications depending on impurities, additives, and other factors. The percentage value in parenthesis indicates the notified classification ratio from companies that provide hazard codes. Only hazard codes with percentage values above 10% are shown. For more detailed information, please visit ECHA C&L website. European Chemicals Agency (ECHA) 12.1.2 Hazard Classes and Categories Not Classified European Chemicals Agency (ECHA) 12.2 Accidental Release Measures 12.2.1 Disposal Methods SRP: Criteria for land treatment or burial (sanitary landfill) disposal practices are subject to significant revision. Prior to implementing land disposal of waste residue (including waste sludge), consult with environmental regulatory agencies for guidance on acceptable disposal practices. Hazardous Substances Data Bank (HSDB) 13 Toxicity 13.1 Toxicological Information 13.1.1 Interactions ... MK-7 induced more complete carboxylation of osteocalcin, and hematologists should be aware that preparations supplying 50 ug/d or more of MK-7 may interfere with oral anticoagulant treatment in a clinically relevant way. PMID:17158229 Schurgers LJ et al; Blood 109 (8): 3279-83 (2007) Hazardous Substances Data Bank (HSDB) Broad specturm antibiotics may sterilize the bowel and decrease the vitamin K contribution to the body by the intestinal microflora. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 711 Hazardous Substances Data Bank (HSDB) Cephalosporins containing side chains of N-methylthiotetrazole (cefmenoxime, cefoperazone, cefotetan, cefamandole, latamoxef) or methylthiadiazole (cefazolin) can cause vitamin K deficiency and hypoprothombinemia. These cephalosporins are inhibitors of hepatic vitamin K epoxide reductase. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 711 Hazardous Substances Data Bank (HSDB) Concomitant intake of cholestyramine and vitamin K may reduce the absorption of vitamin K. /Vitamin K/ PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 711 Hazardous Substances Data Bank (HSDB) For more Interactions (Complete) data for Vitamin K2 (16 total), please visit the HSDB record page. Hazardous Substances Data Bank (HSDB) 13.1.2 Antidote and Emergency Treatment /SRP:/ Immediate first aid: Ensure that adequate decontamination has been carried out. If patient is not breathing, start artificial respiration, preferably with a demand valve resuscitator, bag-valve-mask device, or pocket mask, as trained. Perform CPR if necessary. Immediately flush contaminated eyes with gently flowing water. Do not induce vomiting. If vomiting occurs, lean patient forward or place on the left side (head-down position, if possible) to maintain an open airway and prevent aspiration. Keep patient quiet and maintain normal body temperature. Obtain medical attention. /Poisons A and B/ Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3Rd edition, Elsevier Mosby, St. Louis, MO 2005, p. 160 Hazardous Substances Data Bank (HSDB) /SRP:/ Basic treatment: Establish a patent airway (oropharyngeal or nasopharyngeal airway, if needed). Suction if necessary. Watch for signs of respiratory insufficiency and assist ventilations if needed. Administer oxygen by nonrebreather mask at 10 to 15 L/min. Monitor for pulmonary edema and treat if necessary ... . Monitor for shock and treat if necessary ... . Anticipate seizures and treat if necessary ... . For eye contamination, flush eyes immediately with water. Irrigate each eye continuously with 0.9% saline (NS) during transport ... . Do not use emetics. For ingestion, rinse mouth and administer 5 mL/kg up to 200 mL of water for dilution if the patient can swallow, has a strong gag reflex, and does not drool ... . Cover skin burns with dry sterile dressings after decontamination ... . /Poisons A and B/ Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3Rd edition, Elsevier Mosby, St. Louis, MO 2005, p. 160 Hazardous Substances Data Bank (HSDB) /SRP:/ Advanced treatment: Consider orotracheal or nasotracheal intubation for airway control in the patient who is unconscious, has severe pulmonary edema, or is in severe respiratory distress. Positive-pressure ventilation techniques with a bag valve mask device may be beneficial. Consider drug therapy for pulmonary edema ... . Consider administering a beta agonist such as albuterol for severe bronchospasm ... . Monitor cardiac rhythm and treat arrhythmias as necessary ... . Start IV administration of D5W /SRP: "To keep open", minimal flow rate/. Use 0.9% saline (NS) or lactated Ringer's if signs of hypovolemia are present. For hypotension with signs of hypovolemia, administer fluid cautiously. Watch for signs of fluid overload ... . Treat seizures with diazepam or lorazepam ... . Use proparacaine hydrochloride to assist eye irrigation ... . /Poisons A and B/ Currance, P.L. Clements, B., Bronstein, A.C. (Eds).; Emergency Care For Hazardous Materials Exposure. 3Rd edition, Elsevier Mosby, St. Louis, MO 2005, p. 160-1 Hazardous Substances Data Bank (HSDB) 13.1.3 Human Toxicity Excerpts /HUMAN EXPOSURE STUDIES/ The supplemental forms of vitamin K, vitamin K1 and vitamin K2 are well tolerated. In one study, doses of 90 mg daily of vitamin K2 were given for 24 weeks. Few adverse effects were noted. Reversible elevations of some liver tests were noted in a few subjects in the study. PDR for Nutritional Supplements 2nd ed. Thomson Reuters, Montvale, NJ 2008, p. 711 Hazardous Substances Data Bank (HSDB) /HUMAN EXPOSURE STUDIES/ ... The synthetic short-chain vitamin K(1) is commonly used in food supplements, but recently the natural long-chain menaquinone-7 (MK-7) has also become available as an over-the-counter (OTC) supplement. The purpose of this paper was to compare in healthy volunteers the absorption and efficacy of K(1) and MK-7. Serum vitamin K species were used as a marker for absorption and osteocalcin carboxylation as a marker for activity. Both K(1) and MK-7 were absorbed well, with peak serum concentrations at 4 hours after intake. A major difference between the 2 vitamin K species is the very long half-life time of MK-7, resulting in much more stable serum levels, and accumulation of MK-7 to higher levels (7- to 8-fold) during prolonged intake. MK-7 induced more complete carboxylation of osteocalcin, and hematologists should be aware that preparations supplying 50 ug/d or more of MK-7 may interfere with oral anticoagulant treatment in a clinically relevant way. PMID:17158229 Schurgers LJ et al; Blood 109 (8): 3279-83 (2007) Hazardous Substances Data Bank (HSDB) /SIGNS AND SYMPTOMS/ No adverse effects have been reported with high intakes of vitamin K from food or supplements in healthy individuals who are not intentionally blocking vitamin K activity with anticoagulation medications. /Vitamin K/ National Academies of Sciences Institute of Medicine; Vitamin K. In: Dietary Reference Intakes. p.255-61 (2006) National Academies Press, Washington, DC Hazardous Substances Data Bank (HSDB) /OTHER TOXICITY INFORMATION/ ... This study ... measured circulating levels of vitamin K1, menaquinone-4 (MK-4) and menaquinone-7 (MK-7) in 23 normal healthy women aged 52-93 years (mean +/- SD: 80.1 +/- 3.5), 13 female patients with vertebral fractures aged 66-93 years (80.3 +/- 7.8) and 38 female patients with hip fractures aged 76-87 years (79.8 +/- 9.2), (all Japanese) ... . Serum circulating levels of MK-4 was undetectable in most subjects (only one out of 74). Appreciable numbers from these three groups had undetectable levels of MK-7 (52% of the control group, 23% of the vertebral fracture group and 24% of the hip fracture group). Eight subjects from the normal control group (35%) and five patients from the vertebral group (38%) had undetectable levels of vitamin K1. ... A significant difference in the measurable levels of vitamin K1, MK-4 and MK-7 in patients with vertebral fractures or patients with hip fractures /was not found/ compared to age-matched normal controls... PMID:11678581 Kawana K et al; Endocr Res 27 (3): 337-43 (2001) Hazardous Substances Data Bank (HSDB) 13.1.4 Non-Human Toxicity Excerpts /OTHER TOXICITY INFORMATION/ Although inadequate dietary vitamin K alters bone osteocalcin /in chickens/, symptoms associated with the skeletal system are not as apparent as blood clotting problems. Hemorrhaging may occur subcutaneously, intermuscularly, and internally and may lead to anemia and the appearance of hypoplastic bone marrow. A greatly extended blood clotting time may result in death from exsanguination. ... Inadequate vitamin K under practical circumstances is most likely to occur during the starting period, and supplementation of the feed at this time is advantageous. Starting feeds seldom contain forage meals, and a poorly developed intestinal microflora together with the use of antimicrobials further reduces access to the vitamin. ... Adults usually have a well-developed intestinal microflora, and vitamin K inadequacies are unusual. ... Use of vitamin K by embryos parallels that by adults. A deficiency with the embryo alters bone metabolism, but no physical deformities occur. Adverse effects on blood clotting are not apparent until after hatching, when hemorrhaging and mortality occur should trauma be encountered. Subcommittee on Poultry Nutrition, National Research Council; Nutrient Requirements of Poultry: Ninth Revised Edition, 1994. Available from, as of March 17, 2011: Hazardous Substances Data Bank (HSDB) /OTHER TOXICITY INFORMATION/ ...Menaquinones are nontoxic to animals, even when given in huge amounts. /Menaquinones/ Gilman, A.G., T.W. Rall, A.S. Nies and P. Taylor (eds.). Goodman and Gilman's The Pharmacological Basis of Therapeutics. 8th ed. New York, NY. Pergamon Press, 1990., p. 1564 Hazardous Substances Data Bank (HSDB) 13.2 Ecological Information 13.2.1 Natural Pollution Sources Group of prenylated naphthoquinone derivatives having the physiological activity of vitamin K1 ... Compounds of varying chain length are produced by bacteria, particularly by normal intestinal flora, and may serve as a source of vitamin K for humans. Birds and mammals are capable of alkylating mendione ... to produce menaquinone 4 /Menaquinones/ O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1008 Hazardous Substances Data Bank (HSDB) Vitamin K functions as a coenzyme during the synthesis of the biologically active form of a number of proteins involved with blood coagulation and bone metabolism. /Vitamin K/ Institute of Medicine; p. 162 in Dietary Reference Intakes. Washington, DC: Natl Acad Press (2001) Hazardous Substances Data Bank (HSDB) 13.2.2 Food Survey Values Vitamin K content of milk and meat products(1). /Vitamin K/ Foods Vitamin K content (mg/100 g) Foods Cow's milk Vitamin K content (mg/100 g) 0.5-1 Foods Cheese Vitamin K content (mg/100 g) 10-50 Foods Butter Vitamin K content (mg/100 g) 10 Foods Skeletal meats Vitamin K content (mg/100 g) <1-5 Foods Liver Vitamin K content (mg/100 g) 20-100 (1) Eggersdorfer M et al; Ullmann's Encyclopedia of Industrial Chemistry. 7th ed. (1999-2011). New York, NY: John Wiley & Sons; Vitamins. Online Posting Date: 15 Jun 2000 Hazardous Substances Data Bank (HSDB) 13.2.3 Plant Concentrations Plants with highest amount of phylloquinone(1), the plant form of Vitamin-K(SRC). Genus species Common name Concentration Plant part Genus species Vaccinium corymbosum L Common name Blueberry Concentration 300 ppm Plant part Fruit Genus species Triticum aestivum L. Common name Wheat Plant part Leaf (1) USDA; Dr. Duke's Phytochemical and Ethnobotanical Databases. Plants with a chosen chemical. Phylloquinone. Washington, DC: US Dept Agric, Agric Res Service. Available from, as of Jan 11, 2011 Hazardous Substances Data Bank (HSDB) 13.2.4 Average Daily Intake The average intake for men and women is 120 and 90 ug/day, respectively. /Vitamin K/ Institute of Medicine; p. 182 in Dietary Reference Intakes. Washington, DC: Natl Acad Press (2001) Hazardous Substances Data Bank (HSDB) 14 Associated Disorders and Diseases Open Targets 15 Literature 15.1 Consolidated References PubChem 15.2 NLM Curated PubMed Citations Medical Subject Headings (MeSH) 15.3 Chemical Co-Occurrences in Literature PubChem 15.4 Chemical-Gene Co-Occurrences in Literature PubChem 15.5 Chemical-Disease Co-Occurrences in Literature PubChem 15.6 Chemical-Organism Co-Occurrences in Literature PubChem 16 Patents 16.1 Depositor-Supplied Patent Identifiers PubChem Link to all deposited patent identifiers PubChem 16.2 WIPO PATENTSCOPE Patents are available for this chemical structure: PATENTSCOPE (WIPO) 16.3 Chemical Co-Occurrences in Patents PubChem 16.4 Chemical-Disease Co-Occurrences in Patents PubChem 16.5 Chemical-Gene Co-Occurrences in Patents PubChem 16.6 Chemical-Organism Co-Occurrences in Patents PubChem 17 Interactions and Pathways 17.1 Protein Bound 3D Structures RCSB Protein Data Bank (RCSB PDB) View 2 proteins in NCBI Structure PubChem 17.1.1 Ligands from Protein Bound 3D Structures PDBe Ligand Code MQ7 PDBe Structure Code 1PRC PDBe Conformer Protein Data Bank in Europe (PDBe) 17.2 Chemical-Target Interactions Comparative Toxicogenomics Database (CTD) 17.3 Drug-Drug Interactions DrugBank 17.4 Pathways PubChem 18 Taxonomy LOTUS - the natural products occurrence database; Natural Product Activity and Species Source (NPASS); The Natural Products Atlas 19 Classification 19.1 MeSH Tree Medical Subject Headings (MeSH) 19.2 ChEBI Ontology ChEBI 19.3 LIPID MAPS Classification LIPID MAPS 19.4 ChemIDplus ChemIDplus 19.5 UN GHS Classification GHS Classification (UNECE) 19.6 NORMAN Suspect List Exchange Classification NORMAN Suspect List Exchange 19.7 EPA DSSTox Classification EPA DSSTox 19.8 The Natural Products Atlas Classification The Natural Products Atlas 19.9 LOTUS Tree LOTUS - the natural products occurrence database 19.10 MolGenie Organic Chemistry Ontology MolGenie 19.11 Chemicals in PubChem from Regulatory Sources PubChem 20 Information Sources Filter by Source CAS Common ChemistryLICENSE The data from CAS Common Chemistry is provided under a CC-BY-NC 4.0 license, unless otherwise stated. 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ECHA permits and encourages organisations and individuals to create links to the ECHA website under the following cumulative conditions: Links can only be made to webpages that provide a link to the Legal Notice page. (all-E)-2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-1,4-naphthalenedione (all-E)-2-(3,7,11,15,19,23,27-heptamethyl-2,6,10,14,18,22,26-octacosaheptaenyl)-3-methyl-1,4-naphthalenedione (EC: 870-176-9) FDA Global Substance Registration System (GSRS)LICENSE Unless otherwise noted, the contents of the FDA website (www.fda.gov), both text and graphics, are not copyrighted. They are in the public domain and may be republished, reprinted and otherwise used freely by anyone without the need to obtain permission from FDA. Credit to the U.S. Food and Drug Administration as the source is appreciated but not required. 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https://www.kaptest.com/study/gre/gre-vocabulary-using-word-roots/?srsltid=AfmBOoqJIrydFmcfTjoFAGJ6ntGoAkkwTaa2XQGsmxIlbmAUd7KEx32x
GRE Vocabulary: Using Word Roots – Kaplan Test Prep ☰ About the GRE® ▼ GRE Exam Info What's Tested on the GRE? What's a Good GRE Score? Taking the GRE Studying for the GRE ▼ GRE Study Plans GRE Tips and Tricks GRE Books Applying to Grad School Free GRE Practice Questions ▼ GRE Question of the Day GRE Pop Quiz GRE 20-Minute Workout Free GRE Practice Test GRE Prep ▼ GRE Qbank GRE Practice Packs GRE Classes GRE Vocabulary: Using Word Roots August 9, 2023/in GRE/by admin Vocabulary is tested in many ways on the GRE which is why it’s important to build your GRE-level vocabulary as you study for the exam. During the Verbal section, your knowledge of vocabulary will be directly tested with Sentence Equivalence, Text Completion, and Reading Comprehension questions. Many students find GRE Verbal to be one of the most challenging sections on the test because of the breadth of vocabulary tested. Good old fashioned memorization can be an effective strategy, but it’s time-consuming and may not be the most efficient course of action, especially if you only have a month or two until your GRE test date. One way to increase your GRE vocabulary knowledge is to study word roots. Learning word roots will help you recognize the basic meaning of GRE-level words, enhancing your test prep. [ KEEP READING: How to Study for the GRE in Two Months] How do word roots help you build your GRE vocabulary? Word roots can help you in two major ways on the GRE. First, instead of learning one word at a time, you can learn a whole group of words that contain a certain root. They’ll be related in meaning, so if you remember one, it will be easier for you to remember others. Second, roots can often help you decode an unknown GRE word. If you recognize a familiar root, you could get a good enough grasp of the word to answer the question. Keep reading for a list of word roots that will give you the component parts of many typical GRE vocabulary words. This list is a starting point and a quick review, not an exhaustive guide (though it is lengthy!). Roots are given in their most common forms, with their most common or broadest definitions; often, other forms and meanings exist. Similarly, the definitions for the words given as examples may be incomplete, and other senes of those words may exist. To boost your GRE Verbal prep, get in the habit of looking up unfamiliar words in a good, current dictionary—whether on paper or on the Internet—and be sure to check their etymologies while you’re there. [ RELATED:Top 52 GRE Vocabulary Words] FREE PREMIUM CONTENT GRE Vocabulary Root List Unlock access to our free GRE Vocabulary Root List. Get Access GRE Vocabulary: Kaplan’s Root List A A/AN: not, without agnostic: one who believes the existence of God is not provable amoral:neither moral nor immoral; having no relation to morality atrophy:the wasting away of body tissue AB: off, away from, apart, down abdicate:to renounce or relinquish a throne abject:cast down; degraded abstract: conceived apart from concrete realities, specific objects, or actual instances abstruse: hard to understand; secret, hidden ABLE/IBLE: capable of, worthy of changeable: able to be changed combustible:capable of being burned; easily inflamed AC/ACR:sharp, bitter, sour acerbic: sour or astringent in taste; harsh in temper acid: something that is sharp, sour, or ill-natured acrimonious: caustic, stinging, or bitter in nature acumen: mental sharpness; quickness of wit ACT/AG: to do, to drive, to force, to lead agile:quick and well-coordinated in movement; active, lively pedagogue:a teacher AD:to, toward, near (often the d is dropped and the first letter to which a is prefixed is doubled.) accede: to yield to demand; to enter office addict: to give oneself over, as to a habit or pursuit adjoin:to be close or in contact with AL/ALI/ALTER:other, another alias: an assumed name allegory:figurative treatment of one subject under the guise of another altruist: a person unselfishly concerned for the welfare of others AM: love amateur:a person who engages in an activity for pleasure rather than financial or professional gain amity: friendship; peaceful harmony amorous:inclined to love, esp. sexual love AMBI/AMPHI: both, on both sides, around ambidextrous:able to use both hands equally well ambient: moving around freely; circulating ambiguous:open to various interpretations AMBL/ABUL: to go, to walk ambulance:a vehicle equipped for carrying sick people (from a phrase meaning “walking hospital”) ambulatory: of, pertaining to, or capable of walking perambulator: one who makes a tour of inspection on foot ANIM: of the life, mind, soul, breath animosity: a feeling of ill will or enmity equanimity: mental or emotional stability, especially under tension magnanimous: generous in forgiving an insult or injury ANNUI/ENNI: year annals:a record of events, esp. a yearly record annual: of, for, or pertaining to a year; yearly perennial: lasting for an indefinite amount of time ANT/ANTE:before antecedent:existing, being, or going before antedate:precede in time antediluvian: belonging to the period before the biblical flood; very old or old-fashioned ANTRHO/ANDR: man, human androgen:any substance that promotes masculine characteristics androgynous:being both male and female misanthrope:a person who hates humans or humanity APO: away apocalypse:revelation; discovery; disclosure apogee: the highest or most distant point apostasy:a total desertion of one’s religion, principles, party, cause, etc. ARCH/ARCHI/ARCHY: chief, principal, ruler anarchy:a state or society without government or law monarchy:a government in which the supreme power is lodged in a sovereign oligarchy:a state or society ruled by a select group ARD: to burn ardent:burning; fierce; passionate ardor:flame; passion arson:the crime of setting property on fire AUTO:self autonomy:independence or freedom autocrat:an absolute ruler B BE: about, to make, to surround, to affect(often used to transform words into transitive verbs) belie:to misrepresent; to contradict belittle:to make small; to make something appear smaller bemoan:to lament; to moan for BEL/BELL:beautiful belle:a beautiful woman embellish:to make beautiful; to ornament BEN/BENE: good benediction:act of uttering a blessing benefit:anything advantageous to a person or thing benign:having a kindly disposition BI/BIN: two biennial:happening every two years bilingual:able to speak one’s native language and another with equal facility binocular:involving two eyes BON/BOUN:good, generous bona fide:in good faith; without fraud bountiful:generous BREV/BRID: short, small abbreviate:to shorten breviloquent:laconic; concise in one’s speech brevity:shortness BURS:purse, money bursar:treasurer bursary:treasury disburse:to pay C CAD/CID: to fall, to happen by chance cascade:a waterfall descending over a steep surface coincidence:a striking occurrence of two or more events at one time, apparently by chance recidivist:one who repeatedly relapses, as into crime CAP/CIP/CEPT: to take, to get capture: to take by force or stratagem percipient: having perception; discerning; discriminating precept: a commandment or direction given as a rule of conduct CAP/CAPIT/CIPIT: to head, headlong capital:the city or town that is the official seat of government capitulate:to surrender unconditionally or on stipulated terms precipice:a cliff with a vertical face CARD/CORD/COUR: heart concord:agreement; peace, amity concordance:agreement, concord, harmony discord:lack of harmony between persons or things CED/CEED/CESS: to go, to yield, to stop accede:to yield to a demand; to enter office cessation: a temporary or complete discontinuance incessant:without stop CELER: speed accelerant:something used to speed up a process celerity:speed; quickness CERN/CERT/CRET/CRIM/CRIT:to separate, to judge, to distinguish, to decide certitude:freedom from doubt discrete:detached from others, separate hypocrite:a person who pretends to have beliefs that she does not CHROM: color chromatic:having to do with color chromosome:genetic material that can be studied by coloring it with dyes CHRON: time anachronism:something that is out-of-date or belonging to the wrong time chronic:constant, habitual chronometer:a highly accurate clock or watch CIRCU/CIRCUM: around circuitous:roundabout, indirect circumspect:cautious; watching all sides CIS: to cut exorcise:to seek to expel an evil spirit by ceremony incisive:penetrating, cutting CLA/CLO/CLU: to shut, to close claustrophobia: an abnormal fear of enclosed places cloister:a courtyard bordered with covered walks, esp. in a religious institution preclude:to prevent the presence, existence, or occurrence of CLAIM/CLAM: to shout, to cry out clamor:a loud uproar disclaim:to deny interest in or connection with proclaim:to announce or declare in an official way CLI: to lean toward climax:the most intense point in the development of something disinclination:aversion, distaste Proclivity:inclination, bias CO/COL/COM/CON: with, together commensurate:suitable in measure, proportionate conciliate:to placate, win over connect:to bind or fasten together CONTRA/CONTRO/COUNTER: against contrary:opposed to; opposite countermand:to retract an order CORP/CORS: body corps:a body (an organized group) of troops corpulent:obese; having a lot of flesh incorporation:combining into a single body COSM: order, universe, world cosmology: a theory of the universe as a whole cosmopolitan:worldly microcosm:a small system that reflects a larger whole COUR/CUR: running, a course concur:to agree cursory:going rapidly over something; hasty; superficial incursion:a hostile entrance into a place, esp. suddenly CER/CRESC/CRET: to grow accretion:an increase by natural growth accrue:to be added a matter of periodic gain excrescence:an outgrowth CRED: to believe, to trust credo:any formula of belief credulity:willingness to believe or trust too readily CULP: fault, blame culpable:deserving blame or censure inculpate:to charge with fault mea culpa:through my fault; my fault D DAC/DOC: to teach didactic:intended for instruction indoctrinate:to imbue a person with learning DE: away, off, down, completely, reversal defame:to attack the good name or reputation of deferential:respectful; to yield to judgment defile:to make foul, dirty, or unclean DEM: people democracy:government by the people endemic:peculiar to a particular person or locality pandemic:general, universal DI: day diurnal:daily quotidian:everyday; ordinary DI/DIF/DIS: away from, apart, reversal, not diffuse:to pour out and spread, as in a fluid dilatory:inclined to delay or procrastinate dissipate:to scatter wastefully DIGN: worth condign:well deserved; fitting; adequate dignitary:a person who holds a high rank or office disdain:to look upon or treat with contempt DOL: to suffer, to pain, to grieve doleful:sorrowful, mournful dolorous:full of pain or sorrow, grievous indolence:a state of being lazy or slothful DUB: doubt dubiety:doubtfulness dubious:doubtful indubitable:unquestionable DULC: sweet dulcet: sweet; pleasing dulcified:sweetened; softened dulcimer:a musical instrument DYS: faulty, abnormal dysfunctional:poorly functioning dyspepsia:impaired digestion dystrophy:faulty or inadequate nutrition or development E E/EX: out, out of, from, former, completely efface:to rub or wipe out; surpass, eclipse extricate:to disentangle, release EPI: upon epidermis:the outer layer of the skin epigram:a witty or pointed saying tersely expressed epilogue:a concluding part added to a literary word EQU: equal, even equation:the act of making equal iniquity:gross injustice; wickedness ERR: to wander errant:wandering or traveling, especially in search of adventure erratic:deviating from the proper or usual course in conduct ESCE: becoming convalescent: recovering from illness obsolescent:becoming obsolete EU: good, well euphemism:pleasant-sounding term for something unpleasant euphony:pleasantness of sound F FAB/FAM: to speak affable:friendly, courteous defame:to attack the good name of ineffable:too great for description in words; that which much not be uttered FATU:foolish fatuity:foolishness; stupidity fatuous:foolish; stupid infatuated:swept up in a fit of passion impairing one’s reason FI/FID: faith, trust affidavit:a written statement on oath fiduciary:of a trust; held or given in trust infidel:disbeliever in the supposed true religion FLU/FLUX:to flow confluence:merging into one effluence:flowing out of (light, electricity, etc.) mellifluous:pleasing, musical FULG: to shine effulgent: shining forth refulgent:radiant; shining FUS: to pour diffuse: to spread widely or thinly profuse:lavish, extravagant, copious suffuse:to spread throughout or over from within G GEN: birth, creation, race, kind congenital:existing or as such from birth progeny:offspring, descendants GNI/GNO: to know ignoramus:a person lacking knowledge, uninformed prognosis:to forecast, especially of disease GRAT: pleasing gratuity:money given for good service ingratiate:to bring oneself into favor GREG: flog aggregate:a number of things considered a collective whole egregious:remarkably bad; standing out from the crowd gregarious:sociable; enjoying spending time with others H HAP: by chance haphazard:at random hapless:without luck HER/HES: to stick adherent:able to adhere; believer or advocate of a particular thing coherent:logically consistent; having waves in phase and of one wavelength inherent:involved in the constitution or essential character of something HOL: whole catholic:universal holograph:a document written entirely by the person whose name it’s in holistic:considering something as a unified whole HYPER: over, excessive hyperactive:excessively active hyperbole:purposeful exaggeration for effect HYPO: under, beneath, less than hypochondriac:one affected by extreme depression of mind or spirits, often centered on imaginary physical ailments hypocritical:pretending to have beliefs one does not hypodermic:relating to the parts beneath the skin I ICON: image, idol iconic:being representative of a culture or movement iconoclast:one who attacks established beliefs; one who tears down images iconology:symbolism IN/IM: in, into (often the m is dropped and the first letter to which i is prefixed is doubled) incarnate:given a bodily, esp. a human, form influx:the act of flowing in intrinsic:belonging to a thing by its very nature IN/IM: not, without indigent: poor, needy, lacking in what is needed indolence:showing a disposition to avoid exertion; slothful innocuous:not harmful or injurious J JECT: to throw, to throw down abject:utterly hopeless, humiliating, or wretched conjecture:formation of opinion on incomplete information eject:to throw out, expel JOC: joke jocose:given to joking; playful jocular:in a joking manner; funny jocund:merry; cheerful JUD: to judge abjudicate:to act as a judge judicious:having good judgment JUR: law, to swear abjure:to renounce an oath adjure:to beg or command perjury:willful lying under oath L LAUD: praise, honor laudable:praiseworthy laudatory:expressing praise LAV/LAU/LU: to wash ablution:act of cleansing antediluvian:before the biblical flood; extremely old deluge:a great flood of water LEC/LEG/LEX: to read, to speak legible:readable lexicographer:a writer of dictionaries lexicon:a dictionary LECT/LEG: to select, to choose eclectic:selecting ideas, etc. from various sources predilection:preference, liking LI/LIG: to tie, to bind liable:legally responsible; bound by law lien:the right to hold a property due to an outstanding debt ligature:a connection between two letters; a bond LIBER: free liberality:generosity libertine:one who follows one’s own path, without regard for morals or other restrictions LITH: stone acrolith:a statue with a stone head and limbs (but a wooden body) lithography:a printing process that originally involved writing on a flat stone lithology:the study of rocks and stones megalith:a very big stone LOC/LOG/LOQU: word, speech, though colloquial:of ordinary or familiar conversation elocution:art of clear and expressive speaking grandiloquent:pompous or inflated in language loquacious: talkative LUD/LUS: to play allude:to refer casually or indirectly delude:to mislead the mind or judgment of, deceive elude:to avoid capture or escape defection by M MACRO: great, long macrobiotics:a system intended to prolong life macrocephalous:having a large head macrocosm:the universe; a large system that is reflected in at least one of its subsets MAG/MAJ/MAX: big, great magnanimous:generous in forgiving an insult or injury magnate: a powerful or influential person maxim:an expression of general truth or principle MAL/MALE: bad, ill, evil, wrong maladroit:clumsy; tactless malapropism:humorous misuse of a word malfeasance:a misconduct or wrongdoing often committed by a public official malign:to speak harmful untruths about, to slander MEGA: large, great megapolis:a very large city megaton:explosive power equal to 1000 tons of TNT MICRO: very small microbe:a very small organism micron:a millionth of a meter MIN: small diminution: the act or process of diminishing minutiae:small or trivial details MIS: bad, wrong, to hate misadventure:bad luck; an unlucky accident misanthrope:one who hates people or humanity mischance:bad luck; an unlucky accident MIS/MIT: to send emissary:a messenger or agent sent to represent the interests of another remit:to send money MOLL: soft emollient:something that softens or soothes (e.g., a lotion) mollify:sooth; soften; calm MON/MONIT: to remind, to warn admonish:to counsel against something; caution premonition:forewarning, presentiment remonstrate:to say or please in protest, object, or reproof MON/MONO: one monarchy:rule by a single person monograph:a scholarly study of a single subject monomania:an obsession with a single subject MOR/MORT: death immortal:not subject to death moribund:dying, decaying MORPH: shape amorphous:without definite form; lacking a specific shape anthropomorphism:attribution of human characteristics to inanimate objects, animals, or natural phenomena MULT: many multiplex:having many parts multitudinous:very many; containing very many; having very many forms MUT: to change commute:to substitute; exchange; interchange immutable:unchangeable, invariable N NAT/NAS/NAI/GNA: birth cognate:related by blood; having a common ancestor nascent:starting to develop NIHIL: nothing, none annihilate: wipe out; reduce to nothing nihilism:denial of all moral beliefs; denial that existence has any meaning NOC/NOX: harm innocuous: not harmful or injurious noxious:injurious or harmful to health or morals NOCT/NOX: night noctambulant:walking at night; sleepwalking nocturne:a dreamlike piece of music; a painting set at night NOM/NYM/NOUN/NOWN: name nomenclature: a system of names; systematic naming nominal:existing in name only; negligible renown:fame; reputation NON: not nonentity:something that doesn’t exist; something that isn’t important nonpareil:something with no equal NOV/NEO/NOU: new neologism:a newly coined word, phrase, or expression neophyte:a beginner; a new convert; a new worker neoplasm:a new growth in the body; a tumor O OB: toward, to, against, over obfuscate:to render indistinct or dim; darken obsequious:overly submissive obstinate:stubbornly adhering to an idea, inflexible obstreperous:noisily defiant, unruly OMNI: all omnibus:an anthology of the works of one author or of writings on related subjects omnipotent:all powerful omnipresent:everywhere at one time ONER: burden onerous:burdensome; difficult onus:a burden; a responsibility P PAC/PEAC: peace appease:to bring peace to pacify:to ease the anger or agitation of PALP: to feel palpable:capable of being felt; tangible palpate:to feel; to examine by feeling palpitate:to beat quickly, as the heart; to throb PAN/PANT: all, everyone panegyric:formal or elaborate praise at an assembly panoply:a wide-ranging and impressive array or display PAR: equal disparage:to belittle, speak disrespectfully about parity:equally, as in amount, status, or character PARA: next to, beside paragon:a model of excellence parody:to imitate for purpose of satire PAU/PO/POV/PU: few, little, poor paucity:smallness of quantity; scarcity; scantiness pauper:a person without any personal means of support puerile:childish, immature pusillanimous:lacking courage or resolution PEC: money impecunious:having no money; penniless peculation:embezzlement pecuniary:relating to money PED/POD: foot antipodes: places that are diametrically opposite each other on the globe impede:to retard progress by means of obstacles or hindrances PEL: to drive, to push dispel:to drive away; to disperse impel:to force; to drive forward PER: completely perforate:to make a way through or into something perfunctory:performed merely as routine duty pertinacious:resolute, persistent PET/PIT: to go, to seek, to strive impetuous:characterized by sudden or rash action or emotion petulant:showing sudden irritation, esp. over some annoyance PHIL: love bibliophile:one who loves or collects books philatelist:one who loves or collets postage stamps PHON: sound euphony:the quality of sounding good polyphony:the use of simultaneous melodic lines to produce harmonies in musical compositions PLAC: to please complacent:self-satisfied, unconcerned complaisant:inclined or disposed to please implacable:unable to be pleased POLY: many polyandry:the practice of having multiple husbands polyglot:someone who speaks many languages POT: to drink potable:drinkable; safe to drink; a drink potation:drinking; a drink PRI/PRIM: first primal:original; most important primeval:ancient; going back to the first age of the world PROB: to prove, to test approbation:praise, consideration opprobrium:the disgrace incurred by shameful conduct probity:honesty, high-mindedness PROP/PROX: near proximate:nearby; coming just before or just after proximity:nearness; distance PROT/PROTO: first protagonist:the main character in a play or story prototype:the first version of an invention, on which later models are based PUG: to fight impugn:to challenge as false pugilist:a fighter or boxer pugnacious:to quarrel or fight repeatedly PUNC/PUNG/POIGN: to point, to prick, to pierce compunction:a feeling of uneasiness for doing wrong expunge:to erase, eliminate completely punctilious:strict or exact in the observance of formalities PYR: fire pyre:a bonfire, usually for burning a dead body pyrosis:heartburn Q QUAD/QUAR/QUAT: four quadrille:a square dance involving four couples quart:one-fourth of a gallon QUIE/QUIT: quiet, rest acquiesce:to comply, give in quiescence:the condition of being at rest, still, inactive QUIN/QUINT: five quinquennial:a five-year period quintuple:five times as many R RACI/RADI: root deracinate:to uproot radish:a root vegetable RAMI: branch ramiform:branchlike RECT: straight, right erect: upright; starting up straight rectitude:moral uprightness; moral straightness REG: king, rule interregnum: a period between kings regent:one who serves on behalf of a king; one who rules regicide:killing a king; one who kills a king RETRO: backward retroactive:extending to things that happened in the past retrofit:to install newer parts into an older device or structure RUB/RUD: red rubella:German measles; a disease marked by red spots rubicund:reddish; rosy-cheeked RUD: crude rude:uncivilized; impolite rudimentary: undeveloped S SACR/SANCT: holy sacrament: something regarded as possessing sacred character sacrilege: the violation of anything sacred sanctify: to make holy SAG/SAP/SAV: taste, thinking, discerning sagacious: perceptive; discerning; insightful sage: wise sapient: wise savant: a learned person SAL/SIL/SAULT/SULT: to leap, to jump assault: a sudden or violent attack exult: to show or feel triumphant joy SALU: health salubrious: healthful salutary: healthful SALV: to save salvage: to save; something saved or recovered salvation: being saved savior: one who saves SAN: healthy sane: mentally healthy sanitarium: a place of healing SANG: blood consanguinity: being related by blood sanguinary: bloody; bloodthirsty SAT: enough dissatisfied: feeling that one does not have enough sate: to fill saturate: to fill completely; to entirely satisfy SCI: to know conscience: the inner sense of what is right or wrong, impelling one toward right action omniscient: knowing everything prescient: having knowledge of things before they happen SE: apart, away secede: to withdraw formally from an association sedition: incitement of discontent or rebellion against a government SEC/SEQU/SUE/SUI: to follow non sequitur: an inference or a conclusion that does not follow from the premises obsequious: fawning SED/SESS/SID: to sit, to settle assiduous: diligent, persistent, hardworking (literally, “sitting down” to business) insidious: intended to entrap or beguile; lying in wait to entrap SEMI: half semicircle: half a circle semiconscious: only partly conscious; half awake SEN: old senate: the highest legislative body (from “council of elders”) senescent: getting old senile: relating to old age; experiencing memory loss or other age-related mental impairments SENS/SENT: to feel, to be aware dissent: to differ in opinion, esp. from the majority insensate: without feeling or sensitivity presentiment: a feeling that something is about to happen SOL: alone desolate: deserted; laid waste; left alone isolate: to set apart from others soliloquize: talk to oneself; talk onstage as if to oneself solipsism: the belief that the only thing that really exists, or can really be known, is oneself SOL: to loosen, to free absolution: forgiveness for wrongdoing dissolute: indifferent to moral restraints SOL: sun parasol:an umbrella that protects from the sun solarium: a sunroom; a room with windows for taking in the sun SOMN: sleep somnambulist: a sleepwalker somniferous: sleep-inducing somniloquist: one who talks while asleep somnolent: sleep-inducing; sleepy; drowsy SOPH: wisdom philosopher: one who studies logic, beauty, truth, etc.; one who seeks wisdom sophisticated: complex; worldly; experienced SOURC/SURG/SURRECT: to rise insurgent: rising up in revolution; rushing in insurrection: rising up in armed rebellion surge: to rise up forcefully, as ocean waves SPEC/SPIC: to look, to see circumspect: watchful and discreet, cautious conspicuous: easily seen or noticed; readily observable specious: deceptively attractive SUA: sweet, pleasing, to urge assuage: to make less severe, ease, relieve suave: smoothly agreeable or polite; sweet SUB/SUP: below, under subliminal: existing or operating below the threshold of consciousness subsidiary: serving to assist or supplement suppose: to put down as a hypothesis; to use as the underlying basis of an argument; to assume SUMM: highest, total consummate: highly qualified; complete; perfect summit: highest point SUPER/SUR: over, above supercilious: arrogant, haughty, condescending superfluous: extra, more than necessary SYM/SYN: together symbiosis: living together in a mutually beneficial relationship symposium: a meeting at which ideas are discussed T TAC/TIC: to be silent reticent: disposed to be silent or not to speak freely tacit: unspoken understanding taciturn: uncommunicative TEND/TENS/TENT/TENU: to stretch, to thin attenuate: to weaken or reduce in force extenuating: making less serious by offering excuses THEO: god apotheosis: glorification, glorified ideal theocracy: a form of government in which a deity is recognized as the supreme ruler THERM: heat thermal: relating to heat; retaining heat thermonuclear: relating to a nuclear reaction that takes place at high temperatures TOR/TORQ/TORT: to twist contort: to twist; to distort distort: to pull out of shape, often by twisting; to twist or misrepresent facts extort: to wring money, property, or services out of somebody using threats or force TORP: stiff, numb torpid: numbed; sluggish torpor: numbness; listlessness; apathy TOX: poison antitoxin: an antibody that counteracts a given poison toxic: poisonous TRANS: across, beyond intransigent: refusing to agree or compromise transcendent: going beyond ordinary limits transgress: to violate a law, command, or moral code U ULT: last, beyond penultimate: second-to-last ulterior: beyond what is immediately present; future; beyond what is stated; hidden ultimate: last; final UMBR: shadow adumbrate: to foreshadow; to sketch; to overshadow penumbra: a shaded area between pure shadow and pure light umbrage: shade; shadow; displeasure; resentment UN: not unseen: not seen unusual: not usual; exceptional; strange UND: wave abound: to be plentiful; to overflow (from water flowing in waves) inundate: to flood undulate: to move in a wavelike way UNI/UN: one unanimous: of one mind; incomplete accord uniform: of one kind; consistent US/UT: to use abuse: to use wrongly or improperly usurp: to seize and hold utilitarian: efficient, functional, useful V VAIL/VAL: strength, use, worth ambivalent: being caught between contradictory feelings of equal power or worth avail: to have force; to be useful; to be of value convalescent: recovering strength; healing VEN/VENT: to come or to move toward adventitious: accidental contravene: to come into conflict with convene: to assemble for some public purpose VER: truth aver: to affirm, to declare to be true veracious: habitually truthful verisimilitude: the appearance or semblance of truth VERB: word proverb: an adage; a byword; a short, commonly known saying verbatim: exactly as stated; word-for-word verbiage: excessive use of words; diction VERD: green verdant: green with vegetation; inexperienced verdure: fresh, rich vegetation VI: life convivial: sociable viable: capable of living vivacity: the quality of being lively, animated, spirited VIL: base, mean revile: to criticize with harsh language vile: loathsome, unpleasant vilify: to slander, to defame VIRU: poison virulent: acrimonious; very bitter; very poisonous viruliferous: containing a virus VOC/VOK: call, word advocate: to support or urge by argument convoke: to call together vocabulary: the stock of words used by or known to a particular person or group vociferous: crying out noisily VOL: wish benevolent: characterized by or expressing goodwill malevolent: characterized by or expressing bad will volition: free choice, free will; act of choosing VOLU/VOLV: to roll, to turn convolution: a twisting or folding voluble: easily turning; fluent; changeable VOR: to eat carnivorous: meat-eating omnivorous: eating or absorbing everything voracious: having a great appetite Get The Test Take a free practice test How would you do if you took the GRE® today? Get The Test Take a free practice test How would you do if you took the GRE® today? Tags:GRE, gre verbal, gre vocab, GRE vocabulary Share this entry Share on Facebook Share on Twitter 441 1356 admin admin 2023-08-09 12:30:55 2024-01-31 17:08:55 GRE Vocabulary: Using Word Roots You might also like GRE Sentence Equivalence: Tips and Strategies What's on the GRE: GRE Quantitative Section How to Reschedule Your GRE GRE Quantitative: Combinations and Permutations How to Study for the GRE in 3 Months Should I Take the GRE as an Undergrad? Take The Quiz How would you score if you took the GRE today? Learn More About the GRE The Shorter GRE: What’s Changing? How Do I Register for the GRE? How to Study for the GRE Start Studying for the GRE GRE 1-Month Study Plan GRE 2-Month Study Plan GRE 3-Month Study Plan What’s the Best GRE® Prep For You? How Difficult is the GRE? How is the GRE scored? What’s a good GRE score? What GRE score do I need? 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https://study.com/skill/learn/using-ksp-to-calculate-the-solubility-of-a-compound-explanation.html
Using Ksp to Calculate the Solubility of a Compound | Chemistry | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Using Ksp to Calculate the Solubility of a Compound High School Chemistry Skills Practice An error occurred trying to load this video. Try refreshing the page, or contact customer support. You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 Provide the… 03:51 Find the molar… 05:28 Find the molar… Jump to a specific example Wesley Smith, Jamie Ludwig Instructors Wesley Smith Wesley D. Smith has a Ph.D. in theoretical chemistry from Brigham Young University. He has taught freshman chemistry a total 30 years at five different universities. View bio Jamie Ludwig Jamie Ludwig has taught chemistry courses at the college level for more than 7 years. She has a BA in Chemistry from Illinois Wesleyan University and a PhD in Chemistry from UT Southwestern Medical Center in Dallas. View bio Example SolutionsPractice Questions Steps for Using Ksp to Calculate the Solubility of a Compound Step 1: Write the equation for the compound's solubility reaction. Step 2: Tabulate the initial conditions. Step 3: Tabulate the equilibrium conditions in terms of x, taking into account the stoichiometry of the reaction. Step 4: Plug these values into the K s p expression and solve for x. Step 5: Attach the units m o l L to the value of x. This will be the solubility of the compound. Formulas and Definitions for Using Ksp to Calculate the Solubility of a Compound The Solubility Product.The solubility product, K s p, is the equilibrium constant for the reaction in which an ionic solid dissociates into its ions as it dissolves in water. For example, nickel(II) hydroxide dissolves very slightly in water forming an equilibrium: N i(O H)2(s)⇌N i 2++2 O H 1− and the solubility product is K s p=[N i 2+][O H 1−]2 where the bracket notation means the molar concentration of the ion at equilibrium but without the units. The the numerical values for the solubility products of slightly soluble ionic solids are tabulated in reference works. One such table is probably in the back of your textbook. For nickel(II) hydroxide, K s p=5.48×10−16. Solubility. Solubility is a physical property of a substance. It is a measure of how much of it dissolves in a given volume of water at the standard temperature of 298 K. The solubility of a substance is related to its solubility product. The smaller it solubility product, the less soluble the substance is. Here are three example problems to help you calculate a compound's solubility from its solubility product. Example Problem 1 - Using Ksp to Calculate the Solubility of a Compound The molar solubility of nickel(II) hydroxide turns out to be 5.16×10−6 m o l L. How was the molar solubility of nickel(II) hydroxide calculated? Step 1: Write the equation for the compound's solubility reaction. N i(O H)2(s)⇌N i 2++2 O H 1− Step 2: Tabulate the initial conditions. Before any solute dissolves, the initial values of all dissolved ions are zero. (This may seem to be a superfluous step, but in more advanced problems, some common ions from a separate source may be present. Thus, completing this step is good practice for the future.) Step 3: Tabulate the equilibrium conditions in terms of x, taking into account the stoichiometry of the reaction. Let the equilibrium value for N i 2+ be represented by the unknown x. Then, stoichiometry says that the equilibrium value for O H 1− will be double that, or 2 x. Step 4: Plug these values into the K s p expression and solve for x. K s p=[N i 2+][O H 1−]2=(x)(2 x)2=5.48×10−16 4 x 3=5.48×10−16 x=5.48×10−16 4 3=5.16×10−6 Step 5: Attach the units m o l L to the value of x. This will be the solubility of the compound. The molar solubility of nickel(II) hydroxide is5.16×10−6 mol L. Example Problem 2 - Using Ksp to Calculate the Solubility of a Compound What is the molar solubility of lead(II) sulfide, PbS? (K s p=3×10−28) Step 1: Write the equation for the compound's solubility reaction. P b S(s)⇌P b 2++S 2− Step 2: Tabulate the initial conditions. The initial conditions are zero. Step 3: Tabulate the equilibrium conditions in terms of x, taking into account the stoichiometry of the reaction. Let [P b 2+]=[S 2−]=0+x Step 4: Plug these values into the K s p expression and solve for x. K s p=[P b 2+][S 2−]=x 2=3×10−28 x=3×10−28=1.7×10−14 Step 5: Attach the units m o l L to the value of x. This will be the solubility of the compound. 1.7×10−14 mol Lis the molar solubility of lead(II) sulfide. Example Problem 3 - Using Ksp to Calculate the Solubility of a Compound What is the molar solubility of silver phosphate, A g 3 P O 4? (K s p=8.89×10−17) Step 1: Write the equation for the compound's solubility reaction. A g 3 P O 4(s)⇌3 A g 1++P O 4 3− Step 2: Tabulate the initial conditions. The initial conditions are zero. Step 3: Tabulate the equilibrium conditions in terms of x, taking into account the stoichiometry of the reaction. Let [P O 4 3−]=0+x. Then [A g 1+]=0+3 x. Step 4: Plug these values into the K s p expression and solve for x. K s p=[A g 1+]3[P O 4 3−]=(3 x)3(x)=8.89×10−17 27 x 4=8.89×10−17 x=8.89×10−17 27 4=4.26×10−5 Step 5: Attach the units m o l L to the value of x. This will be the solubility of the compound. 4.26×10−5 mol Lis the molar solubility of silver phosphate. Get access to thousands of practice questions and explanations! Create an account Table of Contents Steps Formulas and Definitions Example Problem 1 Example Problem 2 Example Problem 3 Test your current knowledge Practice Using Ksp to Calculate the Solubility of a Compound Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources Time Period of Pride and Prejudice by Jane Austen Melquiades in One Hundred Years of Solitude Rasselas by Samuel Johnson | Summary, Characters & Analysis Nancy Clutter in In Cold Blood | Family, Analysis & Quotes Bronx Masquerade by Nikki Grimes | Summary, Style & Analysis Angela Lee Duckworth's Concept of Grit | Overview &... 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https://www.sciencedirect.com/science/article/pii/S0307904X15001407
Comparing heuristics for the product allocation problem in multi-level warehouses under compatibility constraints - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Abstract Keywords 1. Introduction 2. Related works 3. The multi-level PAP with compatibility constraints 4. A rollout-based heuristic for multi-level PAP with compatibility constraints 5. Computational experiments 6. Conclusions and future works References Show full outline Cited by (33) Figures (2) Tables (11) Table 1 Table 2 Table 3 Table 4 Table 5 Table 6 Show all tables Applied Mathematical Modelling Volume 39, Issues 23–24, December 2015, Pages 7375-7389 Comparing heuristics for the product allocation problem in multi-level warehouses under compatibility constraints Author links open overlay panelF.Guerriero a, O.Pisacane b, F.Rende a Show more Outline Add to Mendeley Share Cite rights and content Under an Elsevier user license Open archive Abstract One of the most significant activities in warehouse management concerns the allocation of products to the storage positions. This problem is known in the literature as the Product Allocation Problem (PAP). It mainly aims to optimize both the warehouse space utilization and the products handling costs (at least 40% of the total logistics cost). This paper addresses the PAP in a multi-layer warehouse, with compatibility constraints among the product classes. It has already been addressed from a modeling point of view in the literature and it has been formulated as a Mixed Integer Linear Programming model. However, solving the problem to optimality becomes impracticable in real-life settings. To this purpose, an Iterated Local Search-based Heuristic (ILS) and a Cluster-based Heuristic (CH) have already been proposed in the literature. This paper presents a Rollout-based heuristic whose performances are evaluated on the basis of a detailed computational phase, including also a real case study and compared with those of both the ILS and the CH, in terms of the computational times and the quality of the final solutions. Previous article in issue Next article in issue Keywords Warehouse management Clustering Logistics Iterated Local Search Rollout 1. Introduction Warehouse managers are usually interested in providing high quality services to their customers at minimum cost. From a tactical, strategic and operational point of view, the main issues concern both the Warehouse and the Inventory Management. Although these decisions are linked, they can be solved independently by implementing a top–down approach. Firstly, the strategic decisions are considered and then, some limits and restrictions are imposed on those to take at the tactical level. Therefore, the decisions taken at the operational level are affected by the two previous states. Focusing attention on warehouse management, the main activities include the receiving, storage, order picking accumulation, sorting and shipping. Indeed, a warehouse is usually divided into three parts: the Receiving, Storage and Shipping area . In the first area, the ingoing flows are unloaded from the entering vehicles and then, they are sent to the storage zone where they are properly managed to be allocated to the storage positions (slots) and, eventually, packed to compose the final orders (picking orders operation). In order to organize the storage zone adequately, two main issues should be addressed , : the storage allocation, aimed at determining the dimension of each Stock-Keeping Unit (SKU) and the storage assignment aimed at determining the most convenient locations for the SKUs. Finally, in the shipping zone, the products are loaded into the outgoing vehicles in order to be distributed. It is worth noting that the orders picking operations are 65% and 50% of the total cost and of the workforce respectively and they involve the allocation of products and the routing strategy. In particular, the former aims to assign the products optimally to the slots, in such a way that the most required items are placed in positions next to the I/O doors . The routing strategy, instead, aims to determine the best routes for the operators employed, inside the warehouse, during the picking operations. These operations considerably affect the times required to manage the products in the storage area and to handle the orders. This paper focuses attention on the Product Allocation Problem (PAP) in a multi-layer warehouse under compatibility constraints, taking into consideration the hypothesis already described in where a first Mixed Integer Linear Programming (MILP) formulation of the problem is proposed. It is worth observing that, for real world problems, this MILP model becomes mathematically intractable. In fact, although the preliminary grouping phase of the products in homogeneous classes tends to reduce the number of items considerably (i.e., the total number of variables and constraints), the problem hardness remains a critical issue in realistic settings. In order to overcome this drawback, both an Iterated Local Search heuristic (ILS) and a Cluster-based Heuristic (CH) are described in the literature, as detailed in Section 2. In this paper, a rollout-based approach to solve the problem under investigation is designed. The Rollout Heuristics (RH), originally introduced in , , can be used to solve NP-hard combinatorial optimization problems. The basic idea is to use the cost obtained by applying a heuristic method, called base heuristic H, to discriminate among several search options at each step. These algorithms are very appealing from the practical point of view. Indeed, RH s have been successfully used to solve both deterministic and stochastic problems in sequential and parallel computing systems (see , , , . Experimental results have shown that RH significantly improves the performances of H in terms of solution quality. Moreover, it is more robust than other techniques (e.g.,Tabu Search and Simulated Annealing, as shown in ) since it does not require to fix any input parameter a priori. The RH, here proposed, is a hybrid procedure, in the sense that H is obtained by combining a constructive heuristic with a local search method. To the best of our knowledge this is the first work that proposes rollout approaches for solving the PAP in multi-layer warehouses with compatibility constraints. For the sake of completeness, the developed solution strategy is also compared with ILS and CH on a large set of instances. The numerical comparisons are discussed in terms of both computational times and solution quality. It is worth noting that the quality of a solution is evaluated with reference to both the handling costs and the penalties. The proposed approach is also evaluated on a real case. The rest of the paper is organized as follows: Section 2 revises the previous related work from both the modeling and the methodological point of view; Section 3 gives the problem statement; while Section 4 outlines the proposed RH; Section 5 shows the numerical comparisons; finally, Section 6 concludes the work. 2. Related works The Storage Location Assignment Problem (SLAP) aims at allocating the products to the slots in a warehouse, minimizing both the handling costs and penalties and maximizing the space utilization. The allocation mechanism follows the principle that the highly demanded products are allocated in slots next to the I/O doors in order to reduce the total handling times (especially during the picking activity). Among the several proposed strategies, the dedicated storage policy always assigns a fixed number of slots to the same product type. It is very easy to be implemented, but it has the disadvantage that an empty slot cannot be used for a different product type. For this reason, it leads to a waste of space especially in cases where the goods are subjected to the seasonality factor . To overcome this limit, in , a heuristic based on the Duration of Stay (DOS) concept is proposed. In particular, DOS(i) represents how long the item i stays in the warehouse. The items with the lowest DOS (i.e., highly demanded) are then assigned to slots next to the I/O doors. When the information on the products is available, SLAP can be heuristically solved through local searches. For example, a Random storage (RS) approach allocates each item unit to a slot, randomly chosen among the free ones. Thus, each free slot is equally likely to be chosen. This allows for a more uniform warehouse space utilization with the possibility of operators sometimes taking long routes. Alternatively, in , a Closest Open Location Storage approach is described in which the slots are chosen directly by the operators among the free ones closer to them. This strategy does not allow uniformly distributing goods in the warehouse since it usually tends to allocate the units to the slots next to the I/O doors. On the contrary, a Dedicated Storage (DS) strategy places each item unit in a specific limited area inside the warehouse. On the one hand, a free slot cannot be used for assigning items of different types; however, a fixed assignment policy (slot,item) could help the operators in picking activities. This paper takes into consideration a Class-based Storage (CS) location strategy in which the products are assumed to be grouped into different classes (according to a specific criterion). The approach firstly ranks the classes and then, it allocates them by starting from the most critical ones. Finally, it assigns the more profitable slots to the groups. However, the criterion followed for grouping the products into classes still remains a critical issue . The storage assignment policies here described do not take into consideration the possible relations among the items. In some cases, in fact, the customers could require a set of items of different types and thus, it could be convenient to allocate different classes next to each other (i.e., the Family Grouping policy). Alternative approaches are both the Complementary- Based Method and Contact-Based Method. The former consists of a grouping phase of the products, according to the demand complementarity degree and of an allocation step in which the items of the same group are assigned to the adjacent slots , , . The second approach is quite similar to the first one except for the grouping phase , . It is based on the contact frequency of each pair of items (i,j), i.e., the number of times that an operator picks up product type j immediately after product type i. Larson et al. propose a class-based storage warehouse layout procedure with the aim of optimizing the space utilization and the products handling operations. In , the authors propose a linear mathematical model for PAP, taking into account some constraints. Among them, for example, each product type has to be allocated in a specific stocking zone to simplify the picking activity, the inventory management and the equipment utilization, assuming that the quantity to store and the capacity required by the products are known. The authors also fix some allocation priorities. A product, in fact, is firstly assigned to adjacent positions; secondly, to the opposite ones; then, to the posterior positions and, finally, to the generic positions. The goal is to minimize the penalties related to both the assignments and the handling costs. For handling real world problems, the authors propose a branch-and-bound based heuristic. They firstly find a feasible initial solution and then, try to improve it by applying some local adjustments. The solution space search is performed by changing the neighborhoods for the products allocated into more than one slot. In , a multiple-level warehouse shelf configuration is designed in order to minimize the annual carrying costs, without taking into account the compatibility constraints. For this purpose, the aim is different from the one addressed in this paper. Finally, the authors also propose a particle swarm optimization algorithm to determine the optimal layout. In , the PAP is formulated as a MILP model, by introducing specific restrictions, according also to the same allocation priorities described in . In particular, the authors take into consideration the compatibility constraints among the classes and the multi-layers in the warehouse layout, guaranteeing a three-dimensional aspect to the problem. To the best of our knowledge only two heuristic approaches have been proposed in the literature to solve this version of the PAP, i.e., CH and ILS. In particular, the former, described in , basically consists of five consecutive steps. In the first step, the product classes are properly grouped in clusters according to their compatibilities. It is worth noting that a product class could belong to more than one cluster. In the second step, an auxiliary graph G=〈V,E〉 is built where the set of the vertexes V represents the cluster set and an edge (i,j),i∈V,j∈V exists if and only if the two clusters (i and j) share at least one product class. In the third step, the connected components of G are properly determined. Then, for each connected component, a tree is built by performing a depth visit (the fourth step) and finally, the product classes are allocated according to the information on the tree (the fifth step). The ILS, described in , starting from an initial randomly generated solution, explores the current solution’s neighborhood by applying, up to a fixed number of iterations, two different types of local moves (i.e., switch and remove move). In addition, to avoid becoming stuck in a local optimum, a perturbation phase is applied. In particular, a new solution is randomly generated if, after a certain number of consecutive iterations, no improvement on the current best solution is obtained. According to the numerical results presented in the above cited work, this approach detects good solutions for the PAP in multi-layer warehouses with compatibility constraints, in a reasonable amount of time. In this paper, an alternative heuristic approach, based on the RH, is designed and compared to both CH and ILS. 3. The multi-level PAP with compatibility constraints This section aims to give the problem statement with the related assumptions. In particular, the allocation of units of different product types, distinguished according to their physical features and grouped into classes for a better management in the warehouse, is taken into account. Although the compatibility among the products of the same class is always satisfied, the one among items of different groups could not be verified. For example, foodstuffs and detersives can be allocated neither in the same slots nor in the adjacent ones. In addition, the warehouse also presents a multi-layer layout that assures a three-dimensional aspect to the problem. Each slot, in fact, has a width, a length and a height. Thus, a block indicates a set of overlapped slots. In order to reduce the product decentralization and consequently, assure, if possible, that the units of a class are allocated in the same slot, the allocation priorities given in are taken into account. The notation used hereafter is reported in the following: N={1,…,n} is the set of product classes, while m and w are the number of the vertical and the horizontal slots, respectively. The number of levels is h and S is the set of the slots belonging to the first layer such that |S|=m×w. C is a N×N binary matrix such that c ij is equal to 1 if the class i is compatible with the class j,0 otherwise. P is an array of 4 discounts. In particular, p 1,p 2,p 3 and p 4 are discounts applied in the case in which all the units of a class are assigned to the same block, to adjacent slots in the same aisle, to opposite slots and to posterior ones, respectively. Instead, a penalty p 5 is considered if a class is assigned to more than one slot. K is the set of the I/O doors, while D is the distance matrix, where d slk denotes the distance (in meters) from the slot (s,l) to the door k. F is the handling matrix such that f ik indicates how many daily handling operations (measured in load units) of the class i are performed from the door k, while r i is the number of load units of the class i. Cap is the slot capacity, expressed in load units. T 1,T 2 and T 3 denote the set of the slots of the first level with one adjacent slot in the same aisle, an opposite slot and a posterior one, respectively, while ∊ is the cost for moving one load unit for one meter. In the MILP model proposed in , the objective function presents three components: the total handling cost Cost f, related to the flows, the total penalty Cost p due to p 5 and finally, the total saving Sav p associated with the discounts p 1,p 2,p 3 and p 4. The goal is to minimize both Cost f and the product decentralization, mathematically expressed by the difference between Cost p and Sav p. The constraints of the model: (a) state the demand satisfaction for each product class; (b) guarantee that in both the same and the adjacent blocks of the same aisle only compatible classes are allocated and (c) assure that the capacity of each slot is not exceeded. 4. A rollout-based heuristic for multi-level PAP with compatibility constraints As already remarked in Section 1, owing to the problem hardness, a RH is proposed in this work and it is detailed in the following. It is worth noting that a RH is mainly suitable to solve problems whose solution s can be represented by m(⩾1) components: s=(s 1,s 2,…,s m). Thus, the related optimization problem can be sequentially solved by choosing one component at a time. In the following, the k-state (S k) denotes a partial solution in which the first k components are fixed (S k=(s 1,s 2,…,s k-1,s k) with k<m). Starting from S k, the base heuristic H is applied (m-k) times in order to determine a complete solution. Hereafter, L={π 1,π 2,…,π n} denotes the list of the n product classes to allocate in the warehouse. Moreover, it is assumed that the instance to solve is feasible, i.e., the total warehouse capacity is such that all the classes can be allocated. Therefore, there is at least one feasible allocation under the compatibility constraints. It is worth observing that this assumption does not limit the applicability of our solution approach that instead remains valid in most of the real industrial cases. The designed RH builds a configuration (i.e., a feasible allocation) by starting from a specific product class and by iteratively adding a new one. The choice of the product class to add is done through H. This procedure is repeated until all the classes are allocated in the warehouse (i.e., L=∅). Therefore, S t=(π 1,π 2,π 3,…,π t) represents a partial class allocation (i.e., a partial solution) obtained at the iteration t, whereas V(S t) indicates its neighborhood such as the set of product classes that are not allocated yet. At the iteration t+1, the algorithm performs the following steps: 1.(n-t) new solutions are determined applying H to the solution S t∪π̃ obtained by adding to S t the product class π̃,∀π̃∈V(S t). The cost of the complete solution obtained by applying H to S t∪π̃ is denoted by c(π̃). Therefore, the aim is to determine the product class to allocate in the position t+1 and that is not fixed yet; 2.the product class π Best is determined such that π Best=arg min π̃∈V(S t)c(π̃); 3.π Best is added to S t in order to determine the solution S t+1=(π 1,π 2,π 3,…,π t,π Best); 4.π Best is removed from L. The solution quality is determined by the theoretical properties of H. More specifically, if H is sequentially consistent (e.g., a greedy algorithm), it is possible to show that RH remarkably outperforms H in terms of solution quality. In addition, it has been shown that better performances can be achieved by designing H as a hybrid approach, i.e., combining a sequentially consistent heuristic with a deterministic local search method . In this work, a hybrid rollout approach is proposed and outlined in Fig. 1, where a local search is applied to each solution generated at Step 1 of RH. 1. Download: Download high-res image (246KB) 2. Download: Download full-size image Fig. 1. A rollout-based heuristic for PAP. Both H and the local search method designed are detailed in the following sub-sections. 4.1. Base heuristic H In the proposed RH,H is a constructive approach, i.e., it determines a feasible solution by selecting and allocating a product class at a time. In particular, it is assumed that L is sorted according to a specific ordering policy. At each step t of H, a product class π m is selected from L and each product π j∈V(S t) is allocated next to π m (in the same block, if possible). Then, the other classes in V(S t) are assigned by looking for the first free slots adjacent to the ones already occupied by π j. In the developed base heuristic, three different ordering policies have been implemented: •the product flow sorting policy (FP): the most handled product class is firstly selected; •the demand-based sorting policy (DP): the product classes are sorted by increasing value of the demand; •the random sorting policy (RP): the product classes are randomly sorted. 4.2. Local search A local search is performed in order to improve the quality of the determined solution and it mainly consists of two steps: • Switch move: let be (π i,π j) a pair of two distinct product classes and (s,q) their slots. The move Switch(π i s,π j q) assigns π i to q and π j to s. From all the possible changes, the procedure selects only the feasible moves with reference to the following conditions: 1.Capacity constraints: π i can be assigned to q only if the residual capacity of q (i.e., its residual capacity before switching the two classes plus the units of π j to remove) is greater or equal to the units of π i to add. 2.Compatibility constraints: π i can be allocated to q only if it is compatible with the product types already assigned to q. From all the feasible moves, the procedure selects the least cost one. •Remove move: let be π i a product type and s its slot. Remove(π i,s) eliminates the units of π i allocated to s and tries to assign them to a free slot. It is worth noting that this procedure performs only the moves that satisfy the compatibility constraints for the adjacent slots in both the same block and same aisle. Then, from all the feasible moves, it selects the least cost configuration. During the local search phase, the new configuration is evaluated in terms of gaps in cost with respect to the current solution in order to accelerate the neighborhood exploration. Let S be the current solution, the new solution S′, obtained from S by applying the move Switch(π i s,π j q), is evaluated as in the following:(1)Cost f(S′)=Cost f(S)-Cost f(π i,s)-Cost f(π j,q)+Cost f(π i,q)+Cost f(π j,s)where Cost f(S) returns the total handling cost of S, while Cost f(π i,s) determines the handling cost related to a product class π i when it is assigned to a slot s. Same considerations are applied to both Cost p(S′) and to Sav p(S′) as detailed in (2), (3):(2)Cost p(S′)=Cost p(S)-Cost p(π i,s)-Cost p(π j,q)+Cost p(π i,q)+Cost p(π j,s),where Cost p(S) returns the total penalty of S, while Cost p(π i,s) determines the penalty related to a product class π i when it is assigned to a slot s and(3)Sav p(S′)=Sav p(S)+Sav p(π i,s)+Sav p(π j,q)-Sav p(π i,q)-Sav p(π j,s),where Sav p(S) returns the total saving of S, while Sav p(π i,s) determines the saving related to a product class π i when it is assigned to a slot s. 5. Computational experiments The RH for PAP in a multi-layer warehouse with compatibility constraints was coded in Java. The numerical results were compared with the ones found by solving the MILP model through the Mixed Integer Linear Programming solver of Cplex (release 10.1). Both the heuristic and the model were tested on an Intel core2 Duo P7350 processor (2.00 GHz, 3 GB RAM and 250 GB HD). The experiments were carried out on five groups of instances, generated by varying: n (parameter VarN), Cap (parameter VarCap) and the layout parameters m,w,h (VarM,VarW and VarH, respectively). For each of them, the compatibility degree (ρ) was varied generating simple scenarios (S) with all compatible products, middle scenarios (M) with a few of incompatibilities and finally, complex scenarios (C) with many incompatibilities. In order to reduce the computational overhead of RH, two different strategies were considered. The first is aimed to reduce the size of the neighborhood of each partial solution explored on each iteration. In particular, the following two cases were considered: (1) the neighborhood is fully explored; (2) only 10% of all the possible combinations are evaluated. Instead, the second strategy is aimed to reduce the number of times in which the local search is executed. Therefore, two versions of RH were implemented: in the former, the local search is executed on each solution determined by H (Step 1 of RH) while in the latter, the local search is executed only once at the end of RH. In our tests, the following codes are compared: RH all 100 such as RH with a full exploration of the neighborhoods and in which the local search is executed on each solution; RH all 10 as RH all 100 but only the 10% of all the possible options are evaluated; RH end 100 as RH all 100 but the local search is executed only once at the end; RH end 10 as RH all 10 but the local search is executed only once at the end. In addition, RH and ILS were implemented and tested taking into consideration the sorting policies described in Section 4.1. The numerical results found by solving the MILP model were obtained with ∊=0.6 and CPU time limit of 2 h. The performances of both the proposed RH and the approaches considered for making comparisons were analyzed along the following two dimensions : •Solution Quality: to evaluate the gap between the value of the objective function obtained by RH and the one provided by the MILP model. In particular, the percentage gap in cost was computed as in the following:(4)g c=(c H-c∗)c∗×100 where c H and c∗ denote the cost given by a heuristic and the one found by solving the MILP model, respectively. • Computational Effort: since the computational time is a key factor, the following information was considered to assess the behavior of the proposed approach: –Total run time such as the algorithm execution time prior to termination by its stopping rule; –Percentage gap in time such as a measure of the percentage gap between the time required by the heuristic compared to the one needed to solve the MILP model:(5)g t=(t H-t∗)t∗×100 where t H and t∗ denote the computational time required by the heuristic and for solving the MILP model, respectively. Computational experiments were carried out to assess the influence of the considered sorting strategies, in terms of solution quality, on both the performances of RH and ILS. The numerical results are reported in Table 1, where the average cost values of the solutions obtained by the proposed strategies on all the test problems are highlighted. Table 1. Average cost obtained by the proposed approaches according to the sorting policies used. | Empty Cell | FP | DP | RP | --- | ILS | 69607.65 | 70862.85 | 70311.14 | | RH all 100 | 69153.59 | 69585.11 | 68966.56 | | RH all 10 | 70165.42 | 71197.91 | 70245.77 | | RH end 100 | 69021.47 | 70166.99 | 69167.03 | | RH end 10 | 69892.19 | 70848.28 | 69993.74 | Table 1 clearly shows that, on average, the best results (highlighted in bold) are obtained by using FP. For this reason, hereafter attention will be focused only on the numerical results obtained applying this sorting strategy. 5.1. Numerical results on the VarN test problems In this data set, the number of product classes (n) is varied in the range {5,10,20,50,80,100}. For each of these values, the three different compatibility degrees are taken into consideration. The computational results obtained on this set are given in Table 2, Table 3. In particular, Table 2 shows the percentage gaps in cost, whereas the results related to the computational effort are reported in Table 3. Table 2. Computational results obtained on the VarN test problems: solution quality. | n | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 5 | S | 2.02 | 1.93 | 2.10 | 2.10 | 2.10 | 0.00 | | 10 | S | 1.80 | 1.80 | 1.80 | 2.82 | 1.80 | 1.80 | | 20 | S | 2.64 | 1.73 | 2.69 | 1.07 | 2.69 | 1.01 | | 50 | S | 3.75 | 0.84 | 3.77 | 1.45 | 2.63 | 2.59 | | 80 | S | 2.08 | 0.54 | 2.14 | 0.75 | 1.17 | 2.97 | | 100 | S | 2.23 | 0.78 | 2.28 | 0.71 | 1.04 | 2.08 | | | | Average | | 2.46 | 1.37 | 2.50 | 1.64 | 2.08 | 1.67 | | | | 5 | M | 1.73 | 0.00 | 1.73 | 1.90 | 1.73 | 0.19 | | 10 | M | 2.21 | −0.39 | 2.21 | 0.64 | 2.21 | −0.35 | | 20 | M | 2.54 | −0.08 | 2.54 | 0.52 | 1.83 | 1.15 | | 50 | M | 3.13 | 0.88 | 3.23 | 0.87 | 2.10 | 2.40 | | 80 | M | 0.55 | −0.75 | 0.57 | −0.08 | −0.37 | 1.76 | | 100 | M | 1.26 | −0.28 | 1.43 | −0.16 | −0.34 | 1.93 | | | | Average | | 2.03 | −0.07 | 2.06 | 0.77 | 1.50 | 1.03 | | | | 5 | C | 1.99 | 0.05 | 1.99 | 0.05 | 1.99 | 5.57 | | 10 | C | 3.82 | 1.90 | 3.82 | 2.80 | 3.82 | 2.02 | | 20 | C | 2.59 | 1.31 | 2.71 | 1.36 | 1.95 | 1.61 | | 50 | C | 2.79 | 1.07 | 2.85 | 1.45 | 1.98 | 3.11 | | 80 | C | 1.70 | 0.33 | 1.72 | 0.42 | 0.45 | 1.56 | | 100 | C | −14.45 | −16.01 | −14.42 | −15.98 | −15.20 | −13.26 | | | | Average | | 2.58 | 0.93 | 2.62 | 1.21 | 2.04 | 2.77 | Table 3. Computational results obtained on the VarN test problems: computational effort. | n | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 5 | S | −57.03 | −71.43 | −69.94 | −82.29 | −84.11 | 25.83 | | 10 | S | −95.13 | −99.53 | −93.85 | −89.02 | −93.82 | −51.56 | | 20 | S | −99.62 | −97.25 | −99.33 | −99.10 | −99.68 | −95.47 | | 50 | S | −99.99 | −99.79 | −99.97 | −99.98 | −99.99 | −99.90 | | 80 | S | −99.99 | −99.19 | −99.93 | −99.94 | −99.99 | −99.80 | | 100 | S | −99.99 | −98.41 | −99.85 | −99.88 | −99.99 | −99.74 | | | | Average | | −90.35 | −93.44 | −92.61 | −94.06 | −95.52 | −64.18 | | | | 5 | M | −97.67 | −99.27 | −99.32 | −99.27 | −99.51 | −75.83 | | 10 | M | −99.81 | −99.72 | −99.90 | −99.91 | −99.95 | −97.99 | | 20 | M | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.96 | | 50 | M | −100.00 | −99.87 | −99.98 | −99.98 | −100.00 | −99.90 | | 80 | M | −99.99 | −99.51 | −99.93 | −99.94 | −99.99 | −99.81 | | 100 | M | −99.99 | −98.56 | −99.87 | −99.88 | −99.99 | −99.75 | | | | Average | | −99.49 | −99.67 | −99.83 | −99.82 | −99.89 | −94.70 | | | | 5 | C | −96.91 | −98.21 | −99.35 | −99.27 | −99.43 | −67.89 | | 10 | C | −99.09 | −98.30 | −99.42 | −99.50 | −99.93 | −89.00 | | 20 | C | −99.85 | −99.85 | −99.88 | −99.85 | −99.97 | −98.43 | | 50 | C | −100.00 | −99.81 | −99.98 | −99.98 | −100.00 | −99.90 | | 80 | C | −99.99 | −99.20 | −99.93 | −99.93 | −100.00 | −99.83 | | 100 | C | −99.99 | −98.56 | −99.87 | −99.88 | −99.99 | −99.75 | | | | Average | | −99.17 | −99.07 | −99.71 | −99.71 | −99.86 | −91.01 | Table 2 shows that the RH approaches find good quality solutions. On all the test problems, RH all 100 and RH end 100 outperform both ILS and CH. In particular, on the simple compatibility cases, the average gap in cost is 1.37% and 1.64% for RH all 100 and RH end 100, respectively while the percentage worsening in the solution quality is equal to 1.67% and 2.46% for CH and ILS, respectively. Better results are obtained on the middle complex scenarios, in which RH all 100 outperforms the optimizer in terms of solution quality. In fact, a negative average gap in cost equal to -0.07% is observed. In addition, both RH end 100 and RH end 10 outperform ILS, whereas RH end 100 finds also better solutions than those given by CH. It is worth highlighting that RH all 10 behaves the worst. A similar trend is observed on the complex scenarios, in which all the RH approaches behave better than CH while RH end 100,RH end 10 and RH all 100 significantly outperform both ILS and RH all 10 with an average improvement of 1.21%, 2.04% and 0.93%, respectively, against 2.58% and 2.62%, respectively. Regarding the computational effort, Table 3 clearly shows that all the considered heuristic approaches outperform Cplex in terms of execution time. In addition, the RH approaches turn out to be more efficient than the other solution strategies considered for comparisons. 5.2. Numerical results on the VarCap test problems The following data set is obtained varying the slot capacity (Cap) in the range {11,20,30,50,80,100,150,200} and ρ, as described in Section 5. All these results have been detected by fixing the number of product classes at 20. In order to assess the behavior of the RH approaches, the results reported in Table 4 show the average percentage gaps in cost, whereas Table 5 reports the computational times. Table 4. Computational results obtained on the VarCap test problems: solution quality. | Cap | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 11 | S | −3.02 | 0.05 | 0.06 | 0.06 | 0.05 | −0.03 | | 20 | S | −0.93 | −0.03 | −0.02 | −0.05 | −0.05 | −0.12 | | 30 | S | −0.78 | 0.06 | 0.09 | 0.06 | 0.09 | 0.03 | | 50 | S | 0.32 | 0.12 | 0.38 | 0.29 | 0.35 | 0.17 | | 80 | S | 0.84 | 0.35 | 0.79 | 0.53 | 0.82 | 0.59 | | 100 | S | 2.64 | 1.73 | 2.69 | 1.07 | 2.69 | 1.01 | | 150 | S | 3.19 | 1.46 | 3.19 | 3.19 | 3.19 | 0.69 | | 200 | S | 2.75 | 2.70 | 2.91 | 2.91 | 2.91 | 0.64 | | | | Average | | 0.62 | 0.80 | 1.26 | 1.01 | 1.26 | 0.37 | | | | 11 | M | −7.58 | −12.75 | −7.59 | −15.28 | −7.59 | −7.82 | | 20 | M | −0.75 | −0.06 | −0.03 | −0.05 | −0.05 | −0.11 | | 30 | M | −0.17 | −0.06 | −0.08 | −0.08 | −0.08 | −0.12 | | 50 | M | 0.23 | −0.04 | 0.23 | −0.05 | 0.23 | 0.14 | | 80 | M | 0.28 | −0.24 | 0.28 | 0.03 | 0.32 | 0.60 | | 100 | M | 2.54 | −0.08 | 2.54 | 2.54 | 1.83 | 1.15 | | 150 | M | 2.49 | 0.38 | 2.49 | 2.49 | 2.49 | 0.71 | | 200 | M | 1.45 | 1.10 | 2.23 | 0.85 | 2.35 | 0.47 | | | | Average | | −0.19 | −1.47 | 0.01 | −1.19 | −0.06 | −0.62 | | | | 11 | C | −4.50 | −3.74 | −1.25 | −5.13 | −1.25 | −2.67 | | 20 | C | −0.40 | −0.04 | −0.02 | −0.03 | −0.03 | −0.07 | | 30 | C | −0.87 | −0.11 | −0.08 | −0.10 | −0.08 | −0.12 | | 50 | C | 0.39 | 0.17 | 0.44 | 0.44 | 0.44 | 0.16 | | 80 | C | 0.65 | 0.06 | 0.68 | 0.68 | 0.75 | 0.15 | | 100 | C | 2.59 | 1.31 | 2.71 | 1.36 | 1.95 | 1.61 | | 150 | C | 2.92 | 0.66 | 2.92 | 1.73 | 2.58 | 0.62 | | 200 | C | 4.14 | 2.22 | 4.14 | 4.14 | 2.65 | 0.92 | | | | Average | | 0.61 | 0.07 | 1.19 | 0.39 | 0.88 | 0.07 | Table 5. Computational results obtained on the VarCap test problems: computational effort. | Cap | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 11 | S | −99.99 | −99.98 | −100.00 | −100.00 | −100.00 | −99.90 | | 20 | S | −99.99 | −99.96 | −100.00 | −100.00 | −100.00 | −99.93 | | 30 | S | −100.00 | −99.97 | −100.00 | −100.00 | −100.00 | −99.94 | | 50 | S | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.96 | | 80 | S | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.97 | | 100 | S | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.97 | | 150 | S | −99.98 | −99.87 | −99.98 | −100.00 | −100.00 | −99.73 | | 200 | S | −99.94 | −99.68 | −99.95 | −99.99 | −99.98 | −99.48 | | | | Average | | −99.99 | −99.93 | −99.99 | −100.00 | −100.00 | −99.86 | | | | 11 | M | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.93 | | 20 | M | −100.00 | −99.99 | −100.00 | −100.00 | −100.00 | −99.95 | | 30 | M | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.95 | | 50 | M | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.96 | | 80 | M | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.97 | | 100 | M | −99.72 | −97.77 | −99.76 | −99.99 | −99.95 | −95.97 | | 150 | M | −99.68 | −97.54 | −99.69 | −99.91 | −99.89 | −96.17 | | 200 | M | −99.70 | −98.56 | −99.73 | −99.65 | −99.93 | −95.49 | | | | Average | | −99.89 | −99.22 | −99.90 | −99.94 | −99.97 | −98.43 | | | | 11 | C | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.92 | | 20 | C | −100.00 | −100.00 | −100.00 | −100.00 | −100.00 | −99.94 | | 30 | C | −100.00 | −99.99 | −100.00 | −100.00 | −100.00 | −99.95 | | 50 | C | −100.00 | −99.99 | −100.00 | −100.00 | −100.00 | −99.96 | | 80 | C | −100.00 | −99.98 | −100.00 | −100.00 | −100.00 | −99.97 | | 100 | C | −100.00 | −100.00 | −100.00 | −100.00 | −100.00 | −99.96 | | 150 | C | −99.98 | −99.85 | −99.99 | −99.97 | −100.00 | −99.78 | | 200 | C | −99.94 | −99.65 | −99.96 | −100.00 | −99.99 | −99.39 | | | | Average | | −99.99 | −99.93 | −99.99 | −99.99 | −100.00 | −99.86 | From Table 4, it is evident that, on simple scenarios, CH provides the best results (with an average gap in cost equal to 0.37%). However, on middle complex scenarios, RH all 100 behaves the best. Finally, on complex scenarios, RH all 100 and CH give the same results (with an average gap in cost equal to 0.07%). As shown in Table 5, all the considered heuristic approaches outperform Cplex in terms of computational times. 5.3. Numerical results on the VarM test problems The data set is generated varying the number of vertical slots (m) in the range {4,5,6,7,8,9,10} and ρ as described in Section 5. The number of product classes n is varied as reported in Section 5.1. The computational results are reported in Table 6, Table 7, where the average gaps in cost and in time are given. The results of Table 6 clearly show that, on this set of test problems, Cplex does not find a feasible solution, within 2 h (in more than 42% of the cases), whereas the compared heuristic approaches always find a feasible solution within the same time limit. The best performances are obtained by RH all 100, whereas RH all 10 shows the worst behavior in terms of solution quality. In addition, when the compatibility constraints become more stringent, the superiority of the rollout approaches in comparison to the other heuristics is more evident. In particular, on both middle and complex scenarios, RH end 100 outperforms both ILS and CH. Table 6. Computational results obtained on the VarM test problems: solution quality. The symbol - means that Cplex is not able to find a feasible solution within the CPU time limit. | m | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 4 | S | 2.42 | 1.27 | 2.46 | 1.75 | 1.90 | 1.58 | | 5 | S | 2.44 | 1.17 | 2.58 | 1.99 | 2.02 | 2.31 | | 6 | S | 2.37 | 1.14 | 2.50 | 1.57 | 2.20 | 0.89 | | 7 | S | 2.41 | 1.41 | 2.59 | 1.46 | 1.91 | 1.68 | | 8 | S | 0.87 | −0.18 | 1.05 | 0.20 | 0.84 | −0.46 | | 9 | S | −1.08 | −2.05 | −0.87 | −1.82 | −1.47 | −2.67 | | 10 | S | −6.84 | −7.91 | −6.65 | −7.45 | −7.25 | −7.62 | | | | Average | | 0.37 | −0.74 | 0.52 | −0.33 | 0.02 | −0.61 | | | | 4 | M | 1.91 | −0.10 | 1.95 | 0.61 | 1.19 | 1.18 | | 5 | M | – | – | – | – | – | – | | 6 | M | – | – | – | – | – | – | | 7 | M | – | – | – | – | – | – | | 8 | M | – | – | – | – | – | – | | 9 | M | – | – | – | – | – | – | | 10 | M | – | – | – | – | – | – | | | | 4 | C | −0.26 | −1.89 | −0.22 | −0.93 | −0.84 | 0.10 | | 5 | C | −1.70 | −3.05 | −1.68 | −2.38 | −2.11 | −2.10 | | 6 | C | −1.22 | −2.58 | −1.13 | −1.86 | −1.42 | −1.53 | | 7 | C | −8.23 | −9.57 | −8.18 | −9.13 | −8.54 | −8.69 | | 8 | C | – | – | – | – | – | – | | 9 | C | – | – | – | – | – | – | | 10 | C | – | – | – | – | – | – | | | | Average | | −2.85 | −4.27 | −2.80 | −3.57 | −3.23 | −3.05 | Table 7. Computational results obtained on the VarM test problems: computational effort. The symbol - means that Cplex is not able to find a feasible solution in the imposed time limit. | m | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 4 | S | −98.85 | −98.98 | −99.54 | −99.11 | −99.88 | −86.03 | | 5 | S | −98.95 | −98.45 | −99.58 | −99.67 | −99.88 | −86.94 | | 6 | S | −99.17 | −98.53 | −99.66 | −99.60 | −99.84 | −84.25 | | 7 | S | −98.94 | −98.34 | −99.64 | −99.76 | −99.84 | −84.85 | | 8 | S | −99.05 | −98.90 | −99.55 | −99.57 | −99.79 | −79.38 | | 9 | S | −99.16 | −98.41 | −99.60 | −99.76 | −99.83 | −78.32 | | 10 | S | −99.13 | −98.10 | −99.64 | −99.27 | −99.86 | −81.27 | | | | Average | | −99.04 | −98.53 | −99.60 | −99.53 | −99.85 | −83.01 | | | | 4 | M | −99.73 | −99.55 | −99.89 | −99.87 | −99.92 | −96.45 | | 5 | M | – | – | – | – | – | – | | 6 | M | – | – | – | – | – | – | | 7 | M | – | – | – | – | – | – | | 8 | M | – | – | – | – | – | – | | 9 | M | – | – | – | – | – | – | | 10 | M | – | – | – | – | – | – | | | | 4 | C | −99.41 | −99.27 | −99.79 | −99.88 | −99.91 | −93.91 | | 5 | C | −99.60 | −99.23 | −99.82 | −99.83 | −99.93 | −95.15 | | 6 | C | −99.58 | −99.10 | −99.81 | −99.84 | −99.93 | −95.05 | | 7 | C | −99.51 | −98.61 | −99.72 | −99.59 | −99.91 | −92.31 | | 8 | C | – | – | – | – | – | – | | 9 | C | – | – | – | – | – | – | | 10 | C | – | – | – | – | – | – | | | | Average | | −99.53 | −99.05 | −99.78 | −99.92 | −94.10 | −87.83 | Regarding the computational times, the results in Table 7 underline that, also on this set of test problems, the considered heuristics are faster than Cplex. 5.4. Numerical results on the VarW test problems The data set provided in this section is generated varying the number of horizontal slots (w) in the range {6,8,10,12,14,16,18} and ρ as described in Section 5. Again, the number of product classes n is fixed according to Section 5.1. The average percentage gains in cost and in time, in VarW test problems, are given in Table 8, Table 9, respectively. Table 8. Computational results obtained on the VarW test problems: solution quality. The symbol - means that Cplex is not able to find a feasible solution in the imposed time limit. | w | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 6 | S | 2.42 | 1.27 | 2.46 | 1.58 | 1.90 | 1.58 | | 8 | S | 2.09 | 1.12 | 2.18 | 1.22 | 1.80 | 1.54 | | 10 | S | 0.01 | −0.94 | 0.10 | −0.66 | −0.27 | −0.56 | | 12 | S | −0.05 | −1.00 | 0.03 | −0.72 | −0.33 | −0.62 | | 14 | S | −2.74 | −3.67 | −2.67 | −3.41 | −3.02 | −3.36 | | 16 | S | −6.28 | −7.16 | −6.20 | −6.86 | −6.51 | −6.88 | | 18 | S | −11.01 | −11.84 | −10.94 | −11.56 | −11.22 | −11.67 | | | | Average | | −2.22 | −3.18 | −2.15 | −2.92 | −2.52 | −2.85 | | | | 6 | M | 1.91 | −0.10 | 1.95 | 1.61 | 1.38 | 1.18 | | 8 | M | 1.23 | −0.72 | 1.34 | 0.45 | 0.99 | 0.56 | | 10 | M | −9.19 | −10.97 | −9.11 | −10.26 | −9.51 | −9.98 | | 12 | M | – | – | – | – | – | – | | 14 | M | – | – | – | – | – | – | | 16 | M | – | – | – | – | – | – | | 18 | M | – | – | – | – | – | – | | | | Average | | −2.02 | −3.93 | −1.94 | −2.74 | −2.38 | −2.75 | | | | 6 | C | −0.26 | −1.89 | −0.22 | −1.10 | −0.84 | 0.10 | | 8 | C | 1.89 | 0.39 | 1.99 | 1.33 | 1.43 | 2.64 | | 10 | C | 1.44 | −0.05 | 1.54 | 0.32 | 0.99 | 2.19 | | 12 | C | −9.20 | −10.57 | −9.14 | −10.16 | −9.58 | −8.60 | | 14 | C | – | – | – | – | – | – | | 16 | C | – | – | – | – | – | – | | 18 | C | – | – | – | – | – | – | | | | Average | | −1.53 | −3.03 | −1.46 | −2.40 | −2.00 | −0.92 | Table 9. Computational results obtained on the VarW test problems: computational effort. The symbol - means that Cplex is not able to find a feasible solution in the imposed time limit. | w | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 6 | S | −98.54 | −98.91 | −99.51 | −99.75 | −99.72 | −81.75 | | 8 | S | −98.99 | −98.86 | −99.64 | −99.79 | −99.80 | −82.34 | | 10 | S | −99.07 | −98.68 | −99.56 | −99.56 | −99.81 | −84.03 | | 12 | S | −99.07 | −98.49 | −99.63 | −99.55 | −99.81 | −78.91 | | 14 | S | −99.06 | −98.45 | −99.50 | −99.62 | −99.83 | −73.40 | | 16 | S | −99.28 | −98.20 | −99.63 | −99.56 | −99.83 | −77.94 | | 18 | S | −99.28 | −98.29 | −99.63 | −99.66 | −99.86 | −75.10 | | | | Average | | −99.04 | −98.55 | −99.59 | −99.64 | −99.81 | −79.07 | | | | 6 | M | −99.68 | −99.53 | −99.90 | −99.90 | −99.89 | −95.90 | | 8 | M | −99.76 | −99.33 | −99.90 | −99.83 | −99.96 | −95.50 | | 10 | M | −99.78 | −99.15 | −99.89 | −99.92 | −99.96 | −94.05 | | 12 | M | – | – | – | – | – | – | | 14 | M | – | – | – | – | – | – | | 16 | M | – | – | – | – | – | – | | 18 | M | – | – | – | – | – | – | | | | Average | | −99.74 | −99.34 | −99.90 | −99.88 | −99.93 | −95.15 | | | | 6 | C | −99.39 | −99.16 | −99.78 | −99.85 | −99.87 | −94.39 | | 8 | C | −99.57 | −99.13 | −99.78 | −99.84 | −99.88 | −94.33 | | 10 | C | −99.54 | −98.98 | −99.79 | −99.81 | −99.89 | −93.82 | | 12 | C | −99.71 | −98.86 | −99.82 | −99.81 | −99.94 | −94.04 | | 14 | C | – | – | – | – | – | – | | 16 | C | – | – | – | – | – | – | | 18 | C | – | – | – | – | – | – | | | | Average | | −99.55 | −99.03 | −99.79 | −99.83 | −99.89 | −94.14 | Table 8 clear highlights that the proposed RH approaches allow obtaining good quality solutions and, in all the considered instances, RH all 100 behaves the best. In particular, the average gain in cost obtained by RH all 100 is equal to -3.18%,-3.93% and -3.03%, on the simple, middle and complex scenarios, respectively. In addition, on the simple and middle compatibility cases, RH all 10 behaves very poorly, whereas on the complex cases, CH shows the worst performance. From Table 9, it is evident that all the considered heuristics always outperform the solver in terms of computational effort. 5.5. Numerical results on the VarH test problems This data set is generated varying the number of layers (h) in the range {4,5,6,7,8,9,10} and ρ, as described in Section 5. The number of product classes n is varied according to Section 5.1. Table 10 gives the average percentage gain in cost over the number of product classes obtained on VarH instances. The computational times are instead reported in Table 11 where, on each test problem, the average percentage gain is provided. Table 10. Computational results obtained on the VarH test problems: solution quality. The symbol - means that Cplex is not able to find a feasible solution in the imposed time limit. | h | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 4 | S | 2.42 | 1.27 | 2.46 | 1.74 | 1.48 | 1.58 | | 5 | S | 1.84 | 1.05 | 2.26 | 1.71 | 1.16 | 2.07 | | 6 | S | −0.74 | −1.60 | −0.28 | −0.91 | −0.87 | −0.52 | | 7 | S | −7.07 | −7.87 | −6.63 | −7.18 | −7.41 | −6.86 | | 8 | S | 0.72 | −0.24 | 1.18 | 0.46 | 0.33 | 0.98 | | 9 | S | −5.96 | −6.78 | −5.51 | −6.09 | −6.59 | −5.75 | | 10 | S | −4.55 | −5.40 | −4.10 | −4.71 | −4.85 | −4.33 | | | | Average | | −1.91 | −2.80 | −1.52 | −2.14 | −2.39 | −1.83 | | | | 4 | M | 1.91 | −0.10 | 1.95 | 1.04 | 0.92 | 1.18 | | 5 | M | 0.88 | −0.80 | 0.96 | 0.43 | −0.19 | 0.30 | | 6 | M | – | – | – | – | – | – | | 7 | M | – | – | – | – | – | – | | 8 | M | – | – | – | – | – | – | | 9 | M | – | – | – | – | – | – | | 10 | M | – | – | – | – | – | – | | | | Average | | 1.39 | −0.45 | 1.46 | 0.73 | 0.36 | 0.74 | | 4 | C | −0.26 | −1.89 | −0.22 | −0.95 | −1.65 | 0.10 | | 5 | C | 1.83 | 0.44 | 1.93 | 1.41 | 1.13 | 2.53 | | 6 | C | – | – | – | – | – | – | | 7 | C | – | – | – | – | – | – | | 8 | C | – | – | – | – | – | – | | 9 | C | – | – | – | – | – | – | | 10 | C | – | – | – | – | – | – | | | | Average | | 0.79 | −0.73 | 0.85 | 0.23 | −0.26 | 1.32 | Table 11. Computational results obtained on the VarH test problems: computational effort. The symbol – means that Cplex is not able to find a feasible solution in the imposed time limit. | h | ρ | ILS (%) | RH all 100 (%) | RH all 10 (%) | RH end 100 (%) | RH end 10 (%) | CH (%) | --- --- --- --- | | 4 | S | −98.87 | −98.88 | −99.58 | −99.28 | −99.82 | −81.93 | | 5 | S | −98.91 | −98.72 | −99.62 | −99.38 | −99.78 | −79.08 | | 6 | S | −98.78 | −98.45 | −99.52 | −99.68 | −99.77 | −73.98 | | 7 | S | −98.64 | −98.21 | −99.54 | −99.35 | −99.78 | −75.15 | | 8 | S | −98.88 | −98.12 | −99.53 | −99.57 | −99.86 | −75.70 | | 9 | S | −98.98 | −98.20 | −99.60 | −99.58 | −99.87 | −75.98 | | 10 | S | −98.92 | −97.97 | −99.56 | −99.61 | −99.84 | −67.19 | | | | Average | | −98.85 | −98.37 | −99.56 | −99.49 | −99.82 | −75.57 | | | | 4 | M | −99.67 | −99.54 | −99.90 | −99.92 | −99.96 | −96.36 | | 5 | M | −99.74 | −99.37 | −99.89 | −99.83 | −99.97 | −96.04 | | 6 | M | – | – | – | – | – | – | | 7 | M | – | – | – | – | – | – | | 8 | M | – | – | – | – | – | – | | 9 | M | – | – | – | – | – | – | | 10 | M | – | – | – | – | – | | | Average | | −99.71 | −99.45 | −99.89 | −99.87 | −99.96 | −96.20 | | | | 4 | C | −99.38 | −99.26 | −99.79 | −99.64 | −99.92 | −94.47 | | 5 | C | −99.62 | −99.17 | −99.83 | −99.87 | −99.94 | −94.75 | | 6 | C | – | – | – | – | – | – | | 7 | C | – | – | – | – | – | – | | 8 | C | – | – | – | – | – | – | | 9 | C | – | – | – | – | – | – | | 10 | C | – | – | – | – | – | – | | | | Average | | −99.50 | −99.21 | −99.81 | −99.76 | −99.93 | −94.61 | From Table 10, it is evident that, on this set of test problems, Cplex does not find a feasible solution within the imposed time limit (in 10 out of 21 (48%) of the cases), whereas the heuristic approaches always determine a feasible solution. Regarding the simple compatibility scenarios, RH all 100 behaves the best, followed by RH end 10,RH end 100,ILS,CH and RH all 10. A similar trend is also observed on the middle and complex cases, where RH all 100 outperforms all the other approaches. RH all 10, on average, shows the worst performances in both the simple and the middle scenarios, while it is CH with regard to the complex cases. Table 11 shows that, in all the considered instances, the heuristic procedures outperform the solver in terms of computational efforts. 5.6. Real case study In order to assess the behavior of the proposed solution approaches, computational experiments were also carried out on a real case study. In particular, a company in the South of Italy, that operates in the distribution of sports equipment and sportswear, provided the data used in the computational phase. The layout configuration, given in Fig. 2, is characterized by m=17 vertical and w=12 horizontal slots. There are 4 levels with 2 I/O doors and the slot capacity Cap is 650 load units. 1. Download: Download high-res image (364KB) 2. Download: Download full-size image Fig. 2. Warehouse layout of the real case study. There are 289 product classes and for each category, the demands related to a period of 353 days (a whole working year) were considered. The computational results collected by applying ILS and the different versions of RH are described in the following. Since CH has shown, on the average, the worst performances, the related results are not considered in the comparison. With reference to the real case study, the performances of the considered approaches are evaluated not only in terms of total cost but also with reference to the total penalty cost. This type of analysis is carried out with the main aim of showing how, in a real case study, the versions of RH outperform ILS, by detecting solutions in which a lower product decentralization in the warehouse is obtained. We think that it represents a significant aspect to highlight since, in almost all the real scenarios, the number of product classes can be very high, thus the product decentralization can become a key factor for the solution quality. About the total cost, the percentage gaps (i.e., the RH approaches with regard to the ILS) are: -6.61%,-6.58%,-6.52% and -6.34% for RH all 100,RH all 10,RH end 100 and RH end 10, respectively. About the total penalty cost, the percentage gaps are -8.37% for all the four versions of RH. This last result clearly highlights the better performances of the proposed RH approaches with regard to ILS, i.e., the product decentralization found by RH is lower than the one shown by ILS. Also, regarding the total computational cost, the gaps are always less than 2% for all the four versions of RH. 6. Conclusions and future works In this paper, several variants of a rollout-based heuristic approach were proposed to solve the product allocation problem in a multi-layer warehouse with compatibility constraints among the classes. The numerical results, carried out on different types of instances, showed that they are suitable to find good quality solutions, compared to the ones detected by Cplex, on simple and middle complex scenarios. Instead, they find better solutions on complex scenarios, not solved to optimality by Cplex. In all the considered cases, the heuristics outperformed the optimizer in terms of computational effort. In addition, the proposed rollout-based heuristics were compared to two alternative approaches: an iterative local search and a cluster-based algorithm. The numerical results, evaluated with reference to both the solution quality and the computational overhead, remarked that RH all 100 outperforms all the other approaches on average. In addition, a detailed analysis carried out on a real case study concluded the contribution. The collected computational results underlined the good performances of the four versions of the proposed RH compared to ILS, not only in terms of total cost but also with regard to the total penalty, key factor in real world settings, characterized by a large number of products. It is worth observing that the model proposed in and the solution approaches described in this paper can be adapted to solve the problem considered when significantly variable SKUs for volume and shape are taken into account. The easiest strategy to handle this specific situation is to approximate the dimension of each product as either a multiple or a fraction of the load unit. Then, the slot capacity constraints need to be defined by taking into account this approximation. Another possibility is to approximate each product with a three-dimensional item, to represent each slot as a three-dimensional bin and to introduce the classical 3-bin packing capacity constraints (see , ). Future work will concern the extension of the MILP model in order to take into consideration other realistic constraints such as the possibility of dividing a warehouse into zones according to the rotation index of the product classes. Consequently, the heuristics will be modified. Finally, a multi-objective version of the problem will be also addressed. Recommended articles References J.J. Bartholdi, S.T. Hackman Warehouse & Distribution Science: Release 0.89 The Supply Chain and Logistics Institute (2008) Google Scholar G. Ghiani, G. Laporte, R. Musmanno Introduction to Logistics Systems Management (second ed.), 978-1-119-94338-9, Wiley (2013) 478p Google Scholar R. Accorsi, R. Manzini, M. Bortolini A hierarchical procedure for storage allocation and assignment within an order-picking system. A case study Int. J. Logist. Res. Appl.: A Leading J. Supply Chain Manage., 15 (6) (2012), pp. 351-364 CrossrefView in ScopusGoogle Scholar R. Accorsi, R. Manzini, F. Maranesi A decision-support system for the design and management of warehousing systems Comput. Ind., 65 (1) (2014), pp. 175-186 View PDFView articleView in ScopusGoogle Scholar G.P. Broulias, E.C. Marcoulaki, G.P. Chondrocoukis, L.G. Laios Warehouse Management for Improved Order Picking Performance: An Application Case Study from the Wood Industry Department of Industrial Management & Technology, University of Piraeus (2005) Google Scholar F. Guerriero, R. Musmanno, O. Pisacane, F. Rende A mathematical model for the multi-levels product allocation problem in a warehouse with compatibility constraints Appl. Math. Modell., 37 (2013), pp. 4385-4398 View PDFView articleView in ScopusGoogle Scholar D.P. Bertsekas, J.N. Tsitsiklis Neuro-Dynamic Programming Athena Scientific, Belmont, MA (1996) Google Scholar D.P. Bertsekas, J.N. Tsitsiklis, C. Wu Rollout algorithms for combinatorial optimization J. Heuristics, 3 (1997), pp. 245-262 View in ScopusGoogle Scholar N. Secomandi A rollout policy for the vehicle routing problem with stochastic demands Oper. Res., 49 (2001), p. 796802 Google Scholar F. Guerriero, M. Mancini Parallelization strategies for rollout algorithms Comput. Optim. Appl., 31 (2) (2005), pp. 221-244 CrossrefView in ScopusGoogle Scholar F. Guerriero Hybrid rollout approaches for the job shop scheduling problem J. Optim. Theory Appl., 139 (2) (2008), pp. 419-438 CrossrefView in ScopusGoogle Scholar L. Bertazzi, A. Bosco, F. Guerriero, D. Laganá A stochastic inventory routing problem with stock-out Transp. Res. Part C: Emerg. Technol., 27 (2013), pp. 89-107 View PDFView articleView in ScopusGoogle Scholar M.K. Lee Optimization of warehouse storage capacity under a dedicated storage policy Int. J. Prod. Res., 43 (9) (2005), pp. 1785-1805 CrossrefView in ScopusGoogle Scholar M. Goetschalckx, H.D. Ratliff Shared storage policies based on the duration stay of unit loads Manage. Sci., 36 (9) (1990), pp. 1120-1132 CrossrefView in ScopusGoogle Scholar R.Z. Farahani, M. Hekmatfar Facility Location: Concepts, Models, Algorithms and Case Studies, Production and Operations Management, Physica-Verlag Heidelberg (2009) Google Scholar M.B. Rosenwein An application of cluster analysis to the problem of locationing items within a warehouse IEEE Trans., 22 (1994), pp. 101-103 CrossrefGoogle Scholar G. Wascher Order picking: a survey of planning problems and methods Supply Chain Manage. Reserve Logist. (2004), pp. 323-347 CrossrefGoogle Scholar V.R. Muppani, G.K. Adil Efficient formation of storage classes for warehouse storage location assignment: a simulated annealing approach Omega, 36 (4) (2007), pp. 609-618 Google Scholar D.L. Oudheusden, Y.J. Van Tzen, H. Ko Improving storage and order picking in a person-on-board AS/RS system Eng. Costs Prod. Econ., 13 (1988), pp. 273-283 Google Scholar D.L. Oudheusden, W. Van Zhu Storage layout of a AS/RS racks based on recurrent orders Eur. J. Oper. Res., 58 (1992), pp. 48-56 Google Scholar T.N. Larson, H. March, A. Kusiak A heuristic approach to warehouse layout with class-based storage IIE Trans., 29 (1997), pp. 337-348 View in ScopusGoogle Scholar O. Sanei, V. Nasiri, M.R. Marjani, S.M. Moattar Husseini, A heuristic algorithm for the warehouse space assignment problem considering operational constraints: with application in a case study, in: Proceedings of the 2011 International Conference on Industrial Engineering and Operations Management, Kuala Lumpur, Malaysia, January 22–24, 2011. Google Scholar S. Önüt, U.R. Tuzkaya, B. Doğaç A particle swarm optimization algorithm for the multiple-level warehouse layout design problem Comput. Ind. Eng., 54 (4) (2008), pp. 783-799 View PDFView articleView in ScopusGoogle Scholar F. Guerriero, F. Mari, R. Musmanno, O. Pisacane, F. Rende, (2012). A cluster-based heuristic for allocating products in multi-level warehouses, in: Proceedings of LOGMS 2012 – The 2012 International Conference on Logistics and Maritime Systems, Bremen, Germany, 22–24 August 2012, pp. 273–284. Google Scholar F. Guerriero, M. Mancini, R. Musmanno New rollout algorithms for combinatorial optimization problems Optim. Methods Software, 17 (4) (2002), pp. 627-654 View in ScopusGoogle Scholar R.S. Barr, B.L. Golden, J.P. Kelly, M.G.C. Resende, W.R. Stewart Jr. Designing and reporting on computational experiments with heuristic methods J. Heuristics, 1 (1995), pp. 9-32 View in ScopusGoogle Scholar S. Martello, D. Pisinger, D. Vigo The three-dimensional bin packing problem Oper. Res., 48 (2) (2000), pp. 256-267 View in ScopusGoogle Scholar Y. Wu, W. Li, M. Goh, R. de Souza Three-dimensional bin packing problem with variable bin height Eur. J. Oper. Res., 202 (2) (2010), pp. 347-355 View PDFView articleView in ScopusGoogle Scholar Cited by (33) Hybrid data mining and data-driven algorithms for a green logistics transportation network in the post-COVID era: A case study in the USA 2024, Systems and Soft Computing Show abstract This study examines the problem of item allocation in a post-COVID environment with various products and a large customer base. The number of customers has increased due to the rise of internet access and the growing willingness to shop online. Problems such as the timely delivery of goods or services, the selection and destination of orders in decentralized warehouses, and the allocation of warehouses to customers are difficult to overcome with a large variety of items and many customers. It has been proposed that mathematical modeling in combination with meta-heuristic solution techniques solve these problems. However, solving mathematical models is very time-consuming and labor-intensive because there are many different location situations. Due to computing power and memory capacity advances, researchers have been looking at data-driven solutions to these problems. This study aims to tackle the diversity of commodities and the number of consumers in the post-COVID era by proposing a hybrid data-driven approach that combines data mining and mathematical modeling to solve mathematical location models with high accuracy in less time. This paper was implemented based on data from real cases in the USA. ### Solving the multiple level warehouse layout problem using ant colony optimization 2020, Operational Research ### A Polling-Based Dynamic Order-Picking System considering Priority Orders 2020, Complexity ### The storage location assignment problem: A literature review 2019, International Journal of Industrial Engineering Computations ### Multi-objective optimization of electronic product goods location assignment in stereoscopic warehouse based on adaptive genetic algorithm 2018, Journal of Intelligent Manufacturing ### Product allocation of warehousing and cross docking: A genetic algorithm approach 2017, International Journal of Services and Operations Management View all citing articles on Scopus Copyright © 2015 Elsevier Inc. All rights reserved. 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https://www.youtube.com/watch?v=QWjoZlbkTjw
IB MAI HL - Largest Empty Circle Voronoi Pearson 7 IBvodcasting ibvodcasting 5300 subscribers 5 likes Description 901 views Posted: 2 Jun 2020 IB Math Video Transcript: okay so we're working on Voronoi diagrams and we are going to consider we want to construct a new branch at H which is 66 28 is the point so here is 66 is up here and 28 is going to be about let me just see here so here's 30 going across here so I believe here is going to be our H okay and so they're enough and so now what I want to do is I want to create my perpendicular bisectors and I know so I'm gonna estimate these and I'm just gonna sketch them out here just kind of dot on it up between H&G J here's the middle and I'm gonna perpendicular it until I hit the boundary now that I'm on this boundary it's with I so now I look for the middle between I and I want to make it perpendicular so it's going to be straight across because it's there both this is 66:12 as well so it's going to be a horizontal line here now I've hit the boundary with e so E and H I'm gonna find the midpoint which we about here and we perpendicular which will bring me about there and so now this is the boundary with G and H and so if I look between G and H here is the midpoint and about right there I think is gonna be the perpendicular and so here is my region as such something like that and so if I could erase the lines that are there I would erase let me I would raise this one here this one all these in here I should erase them away okay so that is a part so now b part says the bank wants to add one more branch and the area bounded by e G H and I so here is G and so when I look at that I want it to be bounded by here so I'm looking at this vertex and this vertex and that's the only one bounded by because this is bounded by J this one's bounded by E and H e IH this is e G H and so I'm looking at these two values here so if I call this one L and this one M I'm looking for the distance now I recognize that they are closest all closest to H right so L is here and E then J so this is the one they have in common so I'm gonna look for these distances here and so I'm gonna calculate by the distance formula L H and M H and so I would have to find these lines of intersection which is quite intensive I can estimate this circle here and let me see what I can come up with and so there's the one circle let's see if here's the other circle those circles are pretty darn close to each other but I think it's going to be L and so in order to find it I'm gonna have to actually find these perpendicular bisectors find this coordinate point and then do the distance formula that is a lot of tedious work but that's the process you have to go through and so I'm going to say L o L is the largest circle for this problem here you'll also note that this is a different answer than the back than the back they talked about F being in there so I think there's a mistake in the books answer teeth
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https://api.pageplace.de/preview/DT0400.9781351447898_A30865448/preview-9781351447898_A30865448.pdf
FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS FLOW RESISTANCE A DESIGN GUIDE FOR ENGINEERS Erwin Fried General Electric Company I E. Idelchik Industrial Research Institute, Moscow Taylor & Francis Publishers since 1798 Distribution Center: USA Publishing Office: Taylor & Francis Ltd. 325 Chestnut St., Suite 800 Philadelphia, Pa 19 106 Tel: 215-625-8900 Fax: 215-625-2940 Taylor & Francis Ltd. 47 Runway Road, Suite G Levittown, Pa 19057 Tel: 21 5-269-0400 Fax: 21 5-269-0363 Taylor & Francis Ltd. 11 New Fetter Lane London EC4P 4EE Tel: +44 (0) 171 583 9855 Fax: +44 (0) 171 842 2298 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS 3 4 5 6 7 8 9 0 G P G P 9 8 This book was set in Times Roman by Hemisphere Publishing Corporation. The editor was Jerry A. Orvedahl; the production supervisor was Peggy Rote; and the typesetter was Sandra F. Watts. Cover design by Sharon M. DePass. Library of Congress Cataloging-in-Publication Data Idel'chik, I. E. Flow resistance: a design guide for engineers 1 I.E. Idelchik : editor, Erwin Fried. p. cm. "Based almost exclusively on material presented in the recently published Handbook of hydraulic resistance, second edition, by I.E. Idelchik, a translation from the RussianM--Pref. Includes index. 1. Fluid dynamics. 2. Frictional resistance (Hydrodynamics) I. Fried, Erwin. 11. Idel'chik, I. E. Spavochnik po gidravlicheskim soprotivleniiam 1 1 1 . Title. TA357.133 1989 620.1 '064--dc20 ISBN 1-56032-487-2 (paperback) 89-7569 CIP CONTENTS Preface Nomenclature Useful Conversions of Units 1 General Information 1 - 1 General Guidelines General References 2 Flow in Straight Tubes and Conduits: Friction Coefficients and Roughness 2-1 General Guidelines 2-2 Diagrams of Friction Coefficients References 3 Flow at the Entrance Into Tubes and Conduits: Resistance Coefficients of Inlet Sections 3-1 General Guidelines 3-2 Diagrams of Resistance Coefficients References 4 Flow through Orifices with Sudden Change in Velocity and Flow Area: Resistance Coefficients of Sections with Sudden Expansion, Sudden Contraction, Orifices, Diaphragms, and Apertures 4- 1 General Guidelines 4-2 Diagrams of Resistance Coefficients References vii ix . . . Xlll vi CONTENTS 5 Flow with a Smooth Change in Velocity: Resistance Coefficients of Diffusers and Converging and other Transition Sections 5-1 General Guidelines 5-2 Diagrams of the Resistance Coefficients References 6 Flow with Changes of the Stream Direction: Resistance Coefficients of Curved Segments-Elbows and Bends 6-1 General Guidelines 6-2 Diagrams of the Resistance Coefficients References 7 Merging of Flow Streams and Division into Flow Streams: Resistance Coefficients of Wyes, Tees, and Manifolds 7-1 General Guidelines 7-2 Diagrams of the Resistance Coefficients References 8 Flow through Barriers Uniformly Distributed over the Channel Cross Section: Resistance Coefficients of Grids, Screens, Porous Layers, and Packings 8- 1 General Guidelines 8-2 Diagrams of the Resistance Coefficients References 9 Flow through Pipe Fittings and Labyrinth Seals: Resistance Coefficients of Throttling Devices, Valves, Plugs, and Labyrinth Seals 9- 1 General Guidelines 9-2 Diagrams of the Resistance Coefficients References 10 Flow Past Obstructions in a Tube: Resistance Coefficients of Sections with Protuberances, Trusses, Girders, and other Shapes 10- 1 General Guidelines 10-2 Diagrams of the Resistance Coefficients References 11 Flow at the Exit from Tubes and Channels: Resistance Coefficients of Exit Sections 1 1 - 1 General Guidelines 11-2 Diagrams of the Resistance Coefficients References PREFACE This book provides practical, applications-oriented data necessary for the design and evaluation of internal fluid system pressure losses. It was prepared for the practicing engineer, consultant, or designer who understands engineering and fluid flow fundamen- tals, but who needs an easy-to-use compilation of flow resistance coefficients in graphical or tabular form without the distraction of voluminous theory and text. It is based almost exclusively on material presented in the recently published Handbook o f Hydraulic Resis- tance, Second Edition, by I. E. Idelchik, a translation from the Russian, which contains the most extensive compilation of pressure loss coefficients currently available in one volume. The material in this book has been arranged in a convenient guidebook format so that it can be applied easily. The extensive coverage becomes self-evident when one reviews the hundreds of illustrations of flow passages and flow configurations. Most of these are sufficiently basic as to allow application to any shape of flow passage encountered in engineering practice. Each of the illustrations and flow coefficient graphs shows its limits of applicability. Source references are also shown, to allow for further data verification, if desired. In any compilation of empirical data, the accuracy decreases with increasing com- plexity of the component, due to analysis of experimental uncertainties. This book is no exception. Thus, a good rule to follow is to check more than one source, if possible. Since this guidebook is based on a Russian sourcebook, the symbols and nomencla- ture differ somewhat from U.S. practice. This, however, should provide no impediment to using the material, because the pictorial representations are quite clear and easy to follow. The nondimensionality of the pressure loss coefficients and the governing param- eters allow their use in any suitable system of units. Any work of this nature is subject to editorial or translation errors as well as data viii PREFACE reporting errors. The publisher and I would be most grateful to the readers and users of this book for information on such items. I would like to express my gratitude to Professor I. E. Idelchik, who passed away in 1987 after spending a lifetime in the theoretical and experimental investigation of fluid mechanics. For myself and many of my colleagues in the engineering profession, his name is synonymous with the concept of hydraulic resistance. Erwin Fried NOMENCLATURE Symbol Name of quantity Abridged notation in SI units a a a, b cp and c, speed of sound critical speed of sound sides of a rectangle specific heats of gases at constant pressure and constant volume, respectively coefficient of drag cross-section diameters hydraulic or equivalent diameter (4 X hydraulic radius) cross-sectional areas area ratio of a grid, orifice, perforated plate, etc. mass flow rate of liquid (gas) gravitational acceleration height specific heat ratio length of flow segment, depth of channel, or thickness of orifice Mach number coefficient of momentum (Boussinesq coefficient) wetting intensity exponent coefficient of kinetic energy power area ratio (degree of enlargement or reduction of cross section); polytropic exponent; number of elements static pressure total pressure or flow stagnation pressure x NOMENCLATURE Abridged notation Symbol Name of quantity in SI units Pex 4 ' Pdr Q R Rh R 7 (p in US literature) rln excess pressure overall pressure difference drag force volumetric flow rate gas constant hydraulic radius (4 Dh) radii of cross sections of a circular pipe or curved pipe length Reynolds number spacing (distance between rods in a bundle of pipes, between grid holes, etc.) length of a free jet surface area frontal area of a body in a flow thermodynamic temperature thermodynamic flow stagnation temperature (total temperature) internal energy specific volume; side discharge (inflow) velocity stream velocity longitudinally fluctuating stream velocity dust content dust capacity central angle of divergence or convergence; angle of a wye or tee branching; angle of stream incidence angle of turning (of a branch, elbow); angle of valve opening thickness of a wall, boundary layer, or wall layer; height of joint equivalent uniform roughness of walls mean height of wall roughness protuberances relative roughness of walls coefficient of jet contraction porosity (void fraction) degree of turbulence coefficient of fluid resistance (pressure loss coefficient), K in the US literature coefficient of local fluid resistance coefficient of friction resistance of the segment of length I dynamic viscosity cleaning coefficient Jlkg m2 /kg; mls mls mls glm2 kg/m2 degrees NOMENCLATURE xi Abridged notation Symbol Name of quantity in SI units = Sfrl(llDh) friction coefficient [friction resistance of - the segment of relative unit length (l/D, = 11; i.e., friction factor X = fin US literature friction factor - X = w/a, relative (reduced) stream velocity - P discharge coefficient: mass concentration - of suspended particles in flow Y kinematic viscosity mZ/s P density of liquid (gas) kg/m2 n cross-sectional (wetted) perimeter m (b velocity coefficient - SUBSCRIPTS Subscripts listed for the quantities F , f, D, d, I I , a, b, w, p, Q, and p refer to the following cross sections or pipe segments: con or gr br, st, ch out m governing cross section or minimum area larger cross section in the case of expansion or contraction of the flow segment larger cross section after equalization of the stream velocity intermediate cross section of curved channel (elbow, branch) or the working chamber of the apparatus contracted jet section at the discharge from an orifice (nozzle) orifice or a single hole in the perforated plate or screen front of the perforated plate, screen, orifice side branch, straight passage, and common channel of a wye or tee, respectively outlet velocity at infinity Subscripts 0, 1, 2, k, and g at 1 refer, respectively, to the straight inlet, straight outlet, intermediate (for a curved channel), and diffuser pipe lengths, I. Subscripts at Ap and 5 refer to the following forms of the fluid resistances: I fr ov tot out int exp sh b and st local friction overall total resistance of an impedance in the network total resistance of a diffuser or a branch at the outlet from the network internal resistance of a diffuser resistance to flow expansion in a diffuser shock resistance at sudden enlargement of the cross section resistance of a branch and straight passage of a wye or tee (for the resistance co- efficients reduced to the velocity in respective branch pipes) resistance coefficients of the side branch and of the straight passage of a wye or tee reduced to the velocity in a common channel of a wye or tee USEFUL CONVERSIONS OF UNITS Approximate or useful relationship 3 t f t z z l m 1 in -- 25 mm Given in ----, Multiplied by - Gives Physical quantity Gives r ---- Divided by Given in Length ft 0.3048 m in 25.4 (exact) mm mil 0.0254 mm yard 0.9144 m mile (mi) 1 609.3 m km 0.621388 mi ft2 in2 acre 100 ft" 9 m2 1 in2 -- 650 mm2 Area Volume ft3 U.S. gal U.S. gal L (liter) Brit gal U.S. gal barrel (U.S. pet.) barrel (U.S. pet.) m3 m3 liter (L) U.S. gal m3 ft3 m3 U.S. gal 35 ft3 1 1 m3 260 gal = 1 m3 1 gal--3$L 1 L =- 0.26 gal Velocity ftlP mls ftlmin mi/h km/h knots Mass 1 Ib, -- .45 kg 1 kg - 2.2 Ib, metric ton = lo3 kg Ibm kg metric ton ton (2 000 I b , ) Ib f Ibf kgf kgf dyne 4.44822 N = kg m/sZ 0.45359 kgf 2.2046 Ib f 9.80665 N 0.00001 (exact) N Force Amount of substance ibm-mol g-mol kg-moi mol kmol mol kmol kmol Mass flow rate lbmlh kgls Ibmls Ibm/min Reprinted f r o m International System of Units (SI), J. Taborek, in Heat Exchanger Design Hand- book, pp. xxvii-xxix, Hemisphere, Washington, D.C., 1984. xiv USEFUL CONVERSIONS OF UNITS Approximate Given in - Multiplied by - Gives or useful Physical quantity Gives - Divided by 4--- Given in relationship Volume flow rate U.S. gal/min 6.309 x lo- m3/s U.S. bbl/day 0.15899 m3 /day U.S. bbl/day 1.84 x m3/s ft3/s 0.02832 m3 1s ft3 lmin 0.000472 m3 /s Mass velocity (mass flux) kg/s m' Ibm/h ft' ~ t u ~ Btu Btu kcal ft Ibf Wh 1 055.056 J = N m = W s 0.2520 kcal 778.28 ft Ibf 4 186.8 J 1.3558 I 3 600 I Energy (work) (heat) 1 B t u z 1 0001 1 kcai -- 4 Btu Power Btu/h 0.2931 W = l/s W 3.41 18 Btu/h kcal/h 1.163 W ft Ibf/s 1.3558 W hp (metric) 735.5 W Btulh 0.2520 kcal/h tons refrig. 3 516.9 W Heat flux Btu/h ft' W/m2 kcal/cm2 s 3.1546 W/m2 0.317 Btu/h ft' 41.868 W/ma Heat transfer coefficient Btu/h ftl ' I : 5.6784 W/m2 K 1 000 Btu/h ft' O F = W/ml K 0.1761 Btu/h ft' O F 5 600 W/ml K kcal/cmZ s "C 41.868 W/m2 K Heat transfer resistance (Btulh ft' OF)-' 0.1761 (W/m2 K)-' 0.001 (Btu/h ft2 OF)-' z (Wlm' K)-I 5.6784 (Btu/h ft' OF)-' 0.000 18 (W/m2 K)-' Pressure Ibf/in2 (psi) kPa kN/mz = psi kPa kPa kPa kPa Pa Pa kPa kPa kPa 1 psi -- 7 kPa 14.5 psi =. 100 kPa bar Ibf/ftz mm Hg (torr) in Hg mmH,O inH,O at (kgf/cm2) atm (normal) 1 000 kPa = 1 MPa - 150 psi atm = 760 mmHg Mass flux Physical and Transport Properties Thermal conductivity Btu/ft h 'F W/m K kcal/m h "C Density Ibm/ft3 kg/m3 lbm/U.S. gal Specific heat capacity B tu/lbm "F kcallkg 'C Enthalpy Btu/lbm kcal/k&, Dynamic (absolute) centipoise (cP) viscosity poise (P) CP CP Ib,/ft h lbm/ft h CP Ibm/ft s steel =- 50 W/m K water (20°C) 4: 0.6 W/m K air (STP) 24 mW/m K water (10O0C), 0.31 cP air (lOO°C), 0.021 CP USEFUL CONVERSIONS OF UNITS xv Approximate Given in - Multiplied by Gives or useful Physical quantity Gives - Divided by - Given in relationship Kinematic vtscosity stoke (St), cm2 s 0.0001 m2 1s cent~stoke (cSt) mz /s ft2/s 0.092903 m2 1s Diffusivity ft2/s 0.092903 m' /s Thermal diffusivity ml/h ft'ls ft2/h Surface tension dynelcm dynejcm lbf/ft Temperature relations: "C = $ ["F - 32) OC = (OF + 40)g - 40 AT(OC) = ; aT(OF) K = "C + 273.15 " F = p c C ) + 3 2 " F = ("C + 40); - 40 aT("F) = $ AT("C) R = " F + 459.67 Miscellaneous: Acceleration of gravity (standard): g = 9.806 65 m/s2 Gas constant: R = 8 314.3 m N/K kmol Stefan-Boltzmann constant: 5.669 7 X Wlm' K4 1.714 x lo-' Btu/ft2 h R4 i ~ v e n though the abbreviations s and h were introduced only with the SI, they are used here throughout for consistency. Note: the calorie and Btu are based on theInternationalStandard Table values. The thermochemical calorie equals 4.184 J (exact) and is used in some older texts. CHAPTER ONE GENERAL INFORMATION 1-1 GENERAL GUIDELINES This design guide is intended to enable the practicing engineer to analyze and evaluate the flow resistance or pressure loss coefficient for most flow passage types, devices, and components. In keeping with the assumption that the user of this design guide has some understanding of engineering fundamentals, only that material necessary to use the charts and graphs presented herein will be provided. Should the user want to delve into the subject in greater depth, the source book for this design guide, Handbook of Hydraulic Resistance by I. E. Idelchik, published by Hemisphere Publishing Corporation, 1986, should be con- sulted. This source book will be cited throughout this text as "Idelchik." All references cited in Idelchik are shown herein, to allow the user access to the original sources of the data. For basic fluid mechanics information the user is referred to any convenient or familiar fluid mechanics text. For completeness and convenience, a list of recent fluid mechanics texts is included in the bibliography of this chapter. Following are some general guidelines to get the most usefulness from this book. 1. All sketches, diagrams, and graphs are self-explanatory, with flow direction, areas, and other features indicated. 2. Particular attention should be paid to the limits of applicability shown on each of the tables and graphs. These are usually expressed in terms of Reynolds number or in terms of geometric parameters. 3. It is assumed that the inlet and exit conditions are ideal, i.e., there are no flow profile distortions, unless otherwise indicated. There exists only a very limited amount of data on the effect of the inlet flow distortion or inlet swirl for most flow devices. Since each 2 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS application involving distorted flow is unique, it is recommended that experimental methods be considered when such conditions exist and pressure loss is of importance. 4. Unless otherwise indicated, the data shown herein apply to Newtonian fluids consid- ered homogeneous, incompressible, and involving neither work nor energy addition. The pipe or duct walls are considered rigid. 5. For graphs dealing with components involving a change in area, particular attention should be paid to the graph, whether the value of the pressure loss coefficient is based on the inlet, minimum, or exit area. 6. The nondimensionality of the parameters of most of the graphs allows their use in any convenient system of units. 7. The basic reference data given in this book are the static pressure loss coefficients, or K-factor as used in the US literature. This term can be considered the overall static pressure loss coefficient for the component of interest. It includes the nonrecoverable losses within the component as well as the frictional and the recoverable losses. The frictional losses are usually considered negligible when compared to the nonrecovera- ble losses and generally are neglected unless stated otherwise in the graphs. 8. If one considers how the pressure loss coefficient 3- is evaluated experimentally, this becomes evident. It is the measured static pressure drop Ap, divided by the dynamic or velocity head, p 4 , for the component. Thus, 9. The basic pressure loss equation to be used with the data given in this book is ~2 Ap = 3- - , in consistent units 2 10. The overall static pressure drop is considered a positive quantity if the sign convention used in this book is followed. Therefore, a static pressure rise, such as in a diffuser, will show up as negative quantity. 11. The effect of Reynolds number on the pressure loss coefficient is most pronounced at low values (Re < 10'). At higher values of Re it can be assumed as independent of Re, unless otherwise stated. 12. When there is no indication of the Reynolds number at which the value of 3- was obtained, it may be assumed that the given value of ! : is virtually independent of Re. However, in the case of purely laminar flow (Re < 2. lo3), the value of 3- is only an approximation. 13. For the determination of Reynolds numbers in noncircular ducts, an equivalent or hydraulic diameter must be used. It is defined as four times the cross-sectional flow area divided by the wetted perimeter n, with both measured in a direction perpendicu- lar to the flow. If as usual, the fluid fills the entire cross-section of the duct, this definition is equivalent to the relation 14. For a few simple configurations we have the following hydraulic diameter D,, GENERAL INFORMATION 3 Circle of Diameter D D Square with side a a Rectangle with sides a, b 2abl(a + b) Parallel plates separated a distance a 2a Annular duct of cylinders, Dl, D2 01 - D 2 15. Property data, such as viscosity, density, etc., can be obtained from any consistent source available to the user. It is purposely omitted here to keep the size of this book to a minimum. 16. For gases and steam, the variation of density is sometimes very important. If the calculation shows that the resulting pressure drop is such as to change the density, then the piping system can be subdivided and the calculation can be done on a section-by- section basis. In that method, the exit conditions of one section become the inlet conditions of the next section. For condensing steam the density can change quite rapidly and the segmentation method becomes important. It should be noted that the segmentation method is only an approximation, but with judicious selection of seg- ments it can provide acceptable engineering results. 17. Most values of the pressure loss coefficient shown in this book are valid for Mach numbers of less than 0.3 unless otherwise stated. 18. The value of the overall pressure losses in a piping network can be evaluated by use of electrical resistance network methods or by use of one of the several computer pro- grams currently available. This book will provide the necessary pressure loss coeffi- cients 3., or K-factors. 19. In a piping or ducting network, the pressure losses in each segment can be calculated as if the others did not exist and the pressure losses added. However, if the components are close to one another, the exit conditions of one may affect the entry conditions of the following component. Engineering judgement must be applied in such a case. 20. When a system is analyzed for pressure losses, it is often convenient to use the entry or similar dimension as the reference dimension, because the loss coefficient 5- depends on the velocity, which is a function of the cross section. In general, with variable density along the flow, the resistance coefficient 3., based on the velocity in any given section (area F,), is calculated for another section (area F2) using the relation For the case of no change in density, the usual case, this is simplified and becomes a most useful relation, which can be used to normalize any system. 21. A few comments need to be made about the calculation of friction losses in a system. When the straight runs of pipe are significant in relation to the flow obstructions or components, then it is advisable to calculate the friction losses for these straight runs and add them to the other section losses. The friction loss, rfr or K , , can be treated like another loss coefficient, r or K-factor, by use of the following relation. 4 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS In American practice this becomes Ap = (C K, + C K;) - 2 where Kfr = 3;, = f ( 1 l D ) is the friction loss coefficient, and Ki = 5;. is the static pressure loss coefficient from this book. 22. When friction factors are required for solution of an overall system, the graphs and tables allow the use of any friction factor sources familiar to the user, such as Moody or Fanning charts. It should be noted that the value of the Moody friction factor is 4 times that of the Fanning friction factor. This is due to the way the hydraulic diameter is defined. A convenient way to tell which of these two friction factors is given is by inspection of the laminar friction factor. Iff is 16/~e, then it is Fanning. If it is 6 4 / ~ e , it is Moody. GENERAL REFERENCES 1. Idelchik, I. E., Handbook of Hydraulic Resistance, 2nd Ed., Hemisphere Publ. Corp., 1986. 2. White, F. M., Fluid Mechanics, 2nd Ed., McGraw-Hill, New York, 1985. 3. Streeter, V. L., and Wylie, E. B., Fluid Mechanics, 8th Ed., McGraw-Hill, New York, 1985. 4. Blevins, R. D., Applied Fluid Dynamics Handbook, Van Nostrand Renhold, New York, 1984. 5. Miller, D. S., Inteml Flow Systems, British Hydromech. Research Assn., Cranfield, U.K., 1978. 6. Flow of Fluids Through 'Valves, Finings, and Pipe, Crane Co. Technical Paper No. 410, Chicago, 1957. 7. Ward-Smith, A. J., Inteml Fluid Flow, Clarendon Press, Oxford, U.K., 1980. 8. Olson, R. M., Essentials of Engineering FZuid Mechanics, 4th Ed., Harper & Row, New York, 1980. 9. Panton, R. L., Incompressible Flow, Wiley, New York, 1984. 10. White, F. M., Viscous Fluid Flow, McGraw-Hill, New York, 1974. 11. Marks Mechanical Engineers Handbook, Ed. by Th. Baumeister, McGraw-Hill, New York, 1978. 12. Rouse, H., Elementary Mechanics of Fluids, Wiley, New York, 1946. 13. Binder, R. C., Fluid Mechanics, 2nd Ed., Prentice Hall, 1950. 14. Kays, W. M., and London, A. L., Compact Heat Exchangers, McGraw-Hill, New York, 1964. CHAPTER TWO FLOW IN STRAIGHT TUBES AND CONDUITS Friction Coefficients and Roughness 2-1 GENERAL GUIDELINES 1. The pressure losses along a straight tube of constant cross section are calculated from the Darcy-Weisbach equation: where so is the area of the friction surface. Note that X corresponds to the friction factor f in US literature. This book uses the Moody [I271 friction factor. 2. The hydraulic diameter D, was discussed in Chapter 1 briefly. It is equal to the pipe diameter for circular pipe and varies for other shapes of ducts as shown in the tables and diagrams of this chapter. It should be noted that the concept of hydraulic diameter will provide values of pressure losses of acceptable engineering accuracy for most shapes, where there are no secondary flows of significance. Triangular ducts, and shapes where laminar flows can persist locally, are the exception. Empirically derived values are best for such cases, where available. For parallel plates and flat ducts it has been shown that the concept of effective laminar diameter, D,, [151, 1521, is a method that will yield slightly better pressure loss predic- 6 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS tions. Typical values of Deff/Dh for concentric annuli are shown below. Inside Diam. Outside Diam. 0 0.01 0.1 1 .o Reference [I521 provides a useful discussion of the above concept. Also see Diagram 2-7. 3. In laminar flow, due to the overriding effects of viscosity, even flow past surface asperities appears to be smooth. Therefore the roughness of the wall, unless it is very significant, does not affect the flow resistance. Under these conditions of flow the friction coefficient is always a function of the Reynolds number alone. 4. As the Reynolds number increases, the inertia forces, which are proportional to the velocity squared, begin to dominate. Turbulent motion is then initiated, which is character- ized by the development of transverse velocity components giving rise to agitation of the fluid throughout the entire stream and to momentum exchange between randomly moving masses of fluid. All this causes a significant increase in the resistance to motion in turbulent flow as compared with the case for laminar flow. When the surface of the walls is rough, separation occurs in the flow past roughness asperities and the resistance coefficient becomes a function not only of the Reynolds num- ber but also of the relative roughness - A , = - E , or - in U.S. practice D h Dh 5. Pipes and channels can be either smooth or rough, with the roughness being either uniform or nonuniform. These two types of roughness differ according to the shape of such protuberances, their dimensions, the spaces between them, etc. The majority of commercial pipes and tubes have nonuniform roughness. 6. The averaged height A, of asperities, in terms of the absolute length units, is called the absolute geometric roughness. The ratio of the average height of asperities to the tube diameter, that is, i , = A,/Dh or dDh, is called the relative roughness. In view of the fact that the geometric characteristics of the roughness cannot adequately determine the flow resistance of the tube, the concept of a derived hydraulic or equivalent roughness A is introduced, which is determined by measuring the resistance. 7. Although the resistance coefficient for smooth tubes should decrease with increas- ing Re, rough tubes show an increase in the coefficient A with increase of this number with constant geometric roughness. This is explained by the effect of a viscous sublayer. When the thickness of the viscous sublayer is larger than roughness protuberances (6 > A, Fig. 2- la), the latter are entirely covered with this layer. At low velocities, typical of a laminar Figure 2-1 Flow past roughness asperities for different modes of flow: (a) 6 > A; (b) 6 < A. FLOW IN STRAIGHT TUBES AND CONDUITS 7 Figure 2-2 Dependence of the resistance coefficient X on Re for tubes with uniform grain roughness . sublayer, the fluid moves smoothly past surface irregularities and they have no effect on the character of the flow. In this case X decreases with a rise in Re. 8. With an increase in the Reynolds number, the laminar sublayer becomes thinner and, at Re attaining a certain value, it can become smaller than the height of the asperities (6 < A, Fig. 2-lb). The asperities enhance the formation of vortices and hence increase the pressure losses, which result in the rise of X with increasing Re. Thus, tubes can be considered smooth as long as the height of asperities is smaller than the thickness of the laminar sublayer. 9. The equivalent roughness A depends on: The material of tubular products and the method by which they were manufactured. For example, iron pipes manufactured by centrifugal casting are smoother than welded tubes. Tubes manufactured by the same method have, as a rule, the same equivalent roughness irrespective of their diameter. The properties of the fluid flowing in a tube; liquids may cause corrosion on the inner surface of the tube, resulting in formation of protuberances and deposition of scale. The service life and history of the tubes. 10. The dependence of the frictional resistance coefficient X on Re and &, as deter- mined by the experiments of Nikuradse for a stabilized flow in tubes with uniform roughnesst (Fig. 2-2), suggests the existence of three principal regimes of flow. 11. The first regime, called the laminar regime, involves small values of the Reynolds number (up to Re = 2000) and is characterized by X being independent of roughness. From the Hagen-Poiseuille law [I171 +A form of artificial sand roughness is meant here, as obtained by Nikuradse. The curves for other forms can differ somewhat [loo]. 8 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS 12. The second regime, called the transition regime, consists of three segments of the resistance curves for uniform roughness: The segment related to the transition region between laminar and turbulent flow (approx- imately within Re = 2000-4000). The resistance coefficient X in this region increases rapidly with Re. However, this coefficient remains independent of the value of relative roughness. The segment for which the resistance curves of tubes with different roughness coincide with the Blasius curve for smooth tubes According to this equation, the resistance law is valid for the lower range of Reynolds numbers. For the larger values the relative roughness dominates. The segment for which the resistance curves of tubes with different roughness diverge from each other, departing from the straight line obtained from Eq. (2-4). Here, the resistance coefficients for certain ranges of Re increase with increasing relative roughness. 13. The third regime is called turbulent or square-law region. It is characterized by the resistance coefficients for each value of the relative roughness becoming constant, indepen- dent of Re. 14. For a stabilized flow and the region of purely turbulent flow, the friction coeffi- cient X of commercial circular tubes (with nonuniform roughness of walls), except for special cases for which the values of A are given separately, can be determined from the curves of Diagram 2-4 plotted on the basis of the Colebrook-White formula : or for engineering calculations, from Altshul's approximate formula 15. Using Colebrook's formula , Moody [I271 developed the now widely used Moody Chart (Fig. 2-3), which covers laminar as well as turbulent flow and relative roughness. It can be used for circular as well as noncircular ducts, provided the proper hydraulic diameter is used. The Moody chart may be used in preference to Diagrams 2-1 through 2-4. 16. The resistance coefficient of noncircular tubes depends on the shape of the cross section. It can be expressed in terms of the resistance coefficient of circular tubes through the use of a correction factor which allows for the effect of the shape of the tube cross section: a 3 SS3NH9nOtl 3AIlWl3ki V) - c o w N - 8 0 0 - 8 8 B 5888 8 8 8 0 - 0- 0 0 8 3 % s s s s s s s s s s s s 10 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS Figure 2-4 Dependence of the friction coefficient X on Re for a short starting length ([,,/Do - 2) with smooth walls: (1) test section is installed immediately downstream of a smooth inlet (lo/Do - 0); (2) upstream straight section of length lo/Do = 0.4 is installed between the smooth inlet and the test section; (3) relative length of the upstream section is lo/Do = 4.3; (4) three rows of paper bands are passed on the inner surface at the end of the upstream section of length lo/Do = 3.4; (5) the resistance curve is according to Blasius; (6) Hagen-Poiseuille curve. where X is the friction coefficient of circular tubes at the same Reynolds numbers Re = w,,llh/v = w,,DO/v; &on., is X for noncircular tubes; and k,,., is the correction factor allow- ing for the effect of tube cross-sectional shape (see Diagram 2-6). 17. When a fluid enters a straight duct, the newly formed boundary layer is quite thin and requires some distance before the boundary layer thickens and the flow becomes fully developed. This entrance length or nonstabilized flow region depends on the inlet shape, turbulence level, pre-existing conditions, etc. and results in higher flow resistance than is the case in the developed or stabilized region. This entrance length is usually expressed in terms of duct diameters. 18. Creation of conditions under which the flow becomes turbulent in the boundary layer at the inlet into the tube leads to an increase in the coefficient Lon,, for short lengths as well (see Fig. 2-4). Therefore, at relatively small Reynolds numbers (Re,, < Re < 5 x 10~-10~) for short tubes in real devices (in which the flow at the inlet is very much per- turbed as a rule), one may, with a certain factor of safety, assume that &,,, = A, until more detailed data are obtained. For a nonstabilized turbulent flow at larger values of Re where knon,, > 1.0 is the correction factor which compensates for the nonstabilized behav- ior of the flow and which is determined from the curve know,, = f(x/Dh) of Diagram 2-16. FLOW IN STRAIGHT TUBES AND CONDUITS 11 For nonstabilized laminar flow, the friction coefficient of the starting length is calcu- lated from Eq. (2-7), in which n , , , , , a function of the parameter Re (x/D,), is determined by use of graph b of Diagram 2-16. Table 2-1 Equivalent roughness of tubes and channels Group Type of tubes, material State of tube surface and conditions of use A, mm A. Metal tubes I Seamless tubes made Commercially smooth [122,129, 1391 from brass, copper, lead Aluminum tubes The same I1 Seamless steel tubes 1) New, unused (22,99,127] (commercial) 2) Cleaned after many years of use [I291 3) Bituminized [I201 4) Superheated steam pipes of heating systems and water pipes of heating systems with deaeration and chemical treatment of running water 5) After one year of use in gas pipelines 6) After several years of use as tubing in gas wells under various conditions 7) After several years of use as casings in gas wells under different conditions 8) Saturated steam ducts and water pipes of heating systems with minor water leakage (up to 0.5%) and deaeration of water supplied to balance leakage 9) Pipelines of water heating systems inde- pendent of the source of supply [ 131 10) Oil pipelines for intermediate operating conditions 1531 11) Moderately corroded 12) Small depositions of scale [I391 13) Steam pipelines operating periodically and condensate pipes with the open system of condensate 14) Compressed air pipes from piston- and turbocompressors 15) After several years of operation under different conditions (corroded or with small amount of scale) [4,84,129] 16) Condensate pipelines operating periodically and water heating pipes with no deaeration and chemical treatment of water and with substantial leakage from the system (up to 1.5-3%) 17) Water pipelines previously used 18) With large depositions of scale 19) Poor condition; nonuniform overlapping of joints [I191 I11 Welded steel tubes 1) New or old, but in good condition; welded or riveted joints [122, 1391 2) New, bituminized [I281 (See foomote on p. 13.) 12 FLOW RESISTANCE: A DESIGN GUIDE FOR ENGINEERS T P a b l e 2-1 Equivalent roughness of tubes and channels (Continued) Group Type of tubes, material State of tube surface and conditions of use A, mm A. Metal tubes (Cont.) IV Riveted steel tubes 3) Used previously, corroded, bitumen -0.10 partially dissolved [I391 4) Used previously, uniformly corroded -0.15 5) Without noticeable unevenness at joints 0.3-0.4 [ 1391 ; lacquered on the inside layer (10 mm thick); adequate state of surface 6) Gas mains after many years of use -0.5 7) With simple or double transverse riveted 0.6-0.7 joints; lacquered 10 mm thick on the inside or with no lacquer but not corroded [I221 8) Lacquered on the inside but rusted; soiled when transporting water but not corroded I1221 9) Layered deposits; gas mains after 20 years of use [I391 10) With double transverse riveted joints, not corroded; soiled during transport of water [99, 1391 11) Small deposits [ 1391 12) With double transverse riveted joints, heavily corroded [l22] 13) Appreciable deposits [I391 14) Used for 25 years in municipal gas mains, nonuniform deposits of resin and naphtha- lene [I391 15) Poor condition, nonuniform overlapping of joints 1) Lateral and longitudinal riveting with one 0.3-0.4 line of rivets; 10 mm thick lacquered on the inside; adequate state of the surface 2) With double longitudinal riveting and simple 0.6-0.7 lateral riveting; 10 mm thick lacquered on the inside, or without lacquer but not corroded [I221 3) With simple lateral and double longitudinal 1.2-1.3 riveting; from 10 to 20 mm thick lacquered or torred on the inside [I221 4) With four to six longitudinal rows of rivets; 2.0 long period of use 5) With four lateral and six longitudinal rows of 4.0 rivets; joints overlapped on the inside [I221 6) Very poor condition; uneven overlapping of b5.0 joints V Roofing steel sheets 1) Oiled 2) Not oiled VI Galvanized steel tubes 1) Bright galvanization; new [I391 0.07-0.10 2) Ordinary galvanization [ 1391 0.1-0.15 VII Galvanized sheet steel 1) New 2) Used previously VIII Cast-iron tubes 1) New 2) New, bituminized 3) Asphalt-coated 0271 (See foolnote on p. 13.) FLOW IN STRAIGHT TUBES AND CONDUITS 13 'Pable 2-1 Equivalent roughness of tubes and channels (Continued) Group Type of tubes, material State of tube surface and conditions of use A, mm - A. Metal tubes (Cont.) 4) Water pipelines, used previously 1.4 5) Used previously, corroded [I391 1.0-1.5 6) With deposits [127,139] 1.0-1.5 7) Appreciable deposits [129, 1391 2.0-4.0 8) Cleaned after use for many years [I391 0.3-1.5 9) Heavily corroded Up to 3.0 B. Concrete, Cement, and Other Tubes and Conduits Concrete tubes 1) Good surface, plaster finish [I391 2) Average conditions 3) Coarse (rough) surface [841 Reinforced concrete tubes Asbestos-cement tubes 1) New 2) Average Cement tubes 1) Smoothed 2) Nonprocessed [84, 1291 3) Mortar at joints not smoothed [I221 Conduit with a cement- mortar plaster 1) Good plaster made of pure cement with smoothed joints; all asperities removed; metal casing I1221 2) Steel-troweled [I31 Plaster over a metallic screen VII Ceramic salt-glazed conduits VIII IX Slag-concrete slabs [I31 Carefully made slabs [13, 1141 Slag and alabaster- filling slabs - - C. Wood, Plywood, and Glass Tubes I Wooden tubes 1) Boards very thoroughly dressed 2) Boards well dressed 3) Boards undressed but well-fitted 4) Boards undressed [ 1391 5) Staved I1 Plywood tubes 1) Of goodquality birch plywood with trans- verse grain [ 1 1 2) Of good-quality birch plywood with longi- tudinal grain [ 1 1 111 Glass tubes Pure glass [I271 a~epending on how long these were stored. 2.2 DIAGRAMS OF FRICTION COEFFIClENTS Circular tube with smooth walls; stabilized flow [6,118,135] Diagram 2-1 1. Laminar regime (Re 4 2000): A = 64 = =/'(Re) see graph a. 2. Transition regime (2000 4 Re < 4000): A = f (Re) see graph b. 3. Turbulent regime (4000 < Re < lo): 0.3164 A = - ReO.w See graph c. 4. Turbulent regime (Re > 4000): 1 A = (1.8 1 g Re - 1.64)' See graph '' Circular tube with walls of uniform roughness; Diagram stabilized flow; Re > 2000 [56,132] 2-2 A = Ap - 1 (pw~/2)(1/~,)- la, + b, lg (Re 6) + c, lg iil A = f (Re) see graph; the values of a, , b, , and c, are given below: for A see Table 2-1. At A < & l i m ~ o , for the values of A, see Diagram 2-1, where ilim = 17.85 Circular tube with walls of uniform roughness; stabilized flow; Re > 2000 [56, 1321 Diagram 2-2 Values of h Values of h Circular tube with walls of nonuniform roughness; stabilized flow; Diagram critical zone (Re, < Re < 4000) [66,69] 2-3 1. Re, < Re < Re, ; A 2 0.007 h = 4.4 Re-0-59J exp - - = f(Re, h) ( 0 . , , 7 5 > 2. Re, < Re < Re, A = (A2 - A) exp {-[0.0017 (Re, -Re)] + A = f(Re, B) at h < 0.007, A = h, = 0.032, and h, = = 7.244 Re-0.643 0 0109 0.145 at h > 0.007, A = A, - 0.0017 = 0.0758 - %, and ;4 = A ; = Re, = 754 exp t ~ ) Re, = 1160 (krl Re, = 2090 (iy'0635 where for A, see Table 2-1. Values of A Circular tube with walls of nonuniform roughness; stabilized flow; critical zone (Reo < Re < 4000) [66, 691 Diagram 2-3 Values of h -- - - - - R e x - A 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 to 1 . 4 22 2 . 6 3 . 0 3,4 R ~ X I O - ~ Intermediate values of Re and h Circular tube with walls of nonuniform roughness; stabilized flow; Re > Rq (for Rq, see Diagram 2-3) . Also see Fig. 2-3 Diagram 2-4 A = A P - - 1 (pw;/2)(//Dh) [2 ig (2.51/Re Ji + A/3.7)] or within the limits of = 0 . 0 0 0 0 8 -0 . 0 1 2 5 : A = 0 . 1 1 ( Z +325 see graph a for A, see Table 2-1. At A < BlimDn, for A see Diagram 2-1; for slim see graph b as a function of Re. Circular tube with walls of nonuniform roughness; stabilized flow; Re > Re2 (for Rez, see Diagram 2-3) . Also see Fig. 2-3 Diagram 2-4 Values of h A Re A = - Dh 3 x lo3 4 x lo3 6 x lo3 lo4 2 X lo4 4 X lo4 6 X lo4 los 2 X lo5 Values of h Circular tube with rough walls; stabilized flow; regime of quadratic resistance law (Reli, > 5601s) [65, 1321 Diagram 2-5 for A, see Table 2-1. Tubes of rectangular, elliptical, and other types of cross section; Diagram stabilized flow [56, 1051 2-6 where A is determined as for circular tubes from Diagrams 2-1 through 2-5 Shape of tube (conduit) cross section and schematic ( Laminar regime (Re < 2000, curve 1) Rectangle: Correction factor k , , , , Turbulent regime (Re > 2000, curve 2) knon~=k,,,l.10 1.08 1.06 1.04 1.02 1.01 1.0 Trapezoid: k,,, is determined in approximately the same way as for a rectangle Tubes of rectangular, elliptical, and other types of cross section; Diagram stabilized flow [56, 1051 2-6 -- - knon.,= k,,, = kSt = 1.0 Shape of the tube (conduit) cross section and schematic Circle with one or two recesses. Star-shaped circle -@ --@ - -@ + -. / / Ellipse -e 4a0bo Dh 1.5(a,, + b , ) - a More precisely: Dh = naobo 0.983a0 + 0.311b0 + 0.287bi/ao Correction factor knon+ Laminar regime (Re i 2000): knOn-. = k . 1 1 = 8 e y + ( $ 1 1 see graph b 3 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 a0 kell 1.21 1.16 1.11 1.08 1.05 1.03 1.02 1.01 1.01 1.0 kell fl Turbulent regime (Re > 2000); kell = 1.0 sug lnoyj!M snlnuue ~ P ~ U ~ O U O ~ e JO IZq aas ' ;fy IOJ 'aulqg - ('alp - I)(u/E) + ( Oa/p + I ) , I,OI x E < a x 1 " aqZa1 1uaInqIn.L (7=Ya 9E.I = fY = 3-u0uy (Oaulsg - ' a l p + I)( ' a l p - I ) ,OI x E > a x le a W a 1 n u ! -? s q 1eu!pn$!%uo1 ST3 I"PW@~O? ql!m (6'0 = Oa/p) snlnuue ~ o s n u 3p)ua3uo3 o n - /t ZO? - Z F 901 - #J -- SO'I SO'I PO'I 90'1 CO'I EO'I €0'1 ZO'I I O'T '01: 90'1 SO'I SO'I SO'I SO'I PO'I PO'I €0.1 20'1 O'I 90 I 90'1 90'1 90'1 90'1 SO'I SO'I VOI €0.1 ZO'I O'T so1 LO'I LO'I LO'T 90'1 90'1 SO'I SO'I PO'I EO'T O'I Or O' I 8'0 L'O 9'0 S-0 C'O E'O Z'O I.0 0 all ' a l p ' Z y 30 san@A e qdeB JO saun3 aas 'IZy = 23-U0U Y v :(oooz < ax) aunZa1 1uaInqIn.L OS'I - - 6P'T - 8P.I I SP.1 OP'I 0.1 " Y 0.1 8.0 L'O 9.0 s'o P' o E-o z'o 1-0 o ' a l p .e yded JO a m aas ul)/ z('alp) - I I + C( ' a l p ) + I = ,ry = 0-uouy :(OOOz > 3x1 a q ~ a r remure? ,(Oa/p) - I S-z y 2 n 0 1 ~ 1-2 smBe!a WOIJ saqq 1ern3q3 IOJ se hem aures ayl tq pau!urIalap s ! Y alaym 2-uou ( Y a / ~ ) t z ~ ! ~ d ) - - 3-uouy - Ya j-uouy = e d = L - O LI Y = - a - = Y a dv 1 dv ya"M OdP L-2 weBela [LPI '9L ' £ 9 'OP '61 'SI ' P I ] MOI3 paz~qelsfsaqnl EIn3Q3 v N Circular tubes; stabilized f l o w [14, 15, 19,40,63, 76, 1471 Diagram 2-7 Shape of the tube (channel) cross section I Schematic Spiral fins 2(Tlnd)(d/Do Dh = DO - a [ 1 - d / ~ . (A - B) - ? I , / nDo Eccentric annulus
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4.4.1: Auxin - Biology LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 4.4: Hormones 4: Plant Physiology and Regulation { } { "4.4.01:_Auxin" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.02:_Cytokinins" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.03:_Gibberellins" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.04:_Abscisic_Acid" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.05:_Ethylene" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.06:_Summary_Table" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.07:_Other_Signaling_Molecules" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.4.08:_Chapter_Summary" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "4.01:_Photosynthesis_and_Respiration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.02:_Environmental_Responses" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.03:_Nutrition_and_Soils" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.04:_Hormones" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.05:_Transport" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.06:_Development" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 28 Jul 2025 22:09:57 GMT 4.4.1: Auxin 32025 32025 Jennifer Rogers { } Anonymous Anonymous 2 false false [ "article:topic", "auxin", "phototropism", "gravitropism", "indole-3-acetic acid (IAA)", "plant hormones", "Apical Dominance", "showtoc:no", "license:ccbyncsa", "source-bio-5799", "source-bio-5799", "program:oeri", "cid:biol155", "authorname:haetal", "licenseversion:40" ] [ "article:topic", "auxin", "phototropism", "gravitropism", "indole-3-acetic acid (IAA)", "plant hormones", "Apical Dominance", "showtoc:no", "license:ccbyncsa", "source-bio-5799", "source-bio-5799", "program:oeri", "cid:biol155", "authorname:haetal", "licenseversion:40" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Botany and Horticulture 4. Botany (Ha, Morrow, and Algiers) 5. 4: Plant Physiology and Regulation 6. 4.4: Hormones 7. 4.4.1: Auxin Expand/collapse global location Botany (Ha, Morrow, and Algiers) Front Matter 1: Introduction to Botany 2: Biodiversity (Organismal Groups) 3: Plant Structure 4: Plant Physiology and Regulation 5: Ecology and Conservation Back Matter 4.4.1: Auxin Last updated Jul 28, 2025 Save as PDF 4.4: Hormones 4.4.2: Cytokinins Page ID 32025 Melissa Ha, Maria Morrow, & Kammy Algiers Yuba College, College of the Redwoods, & Ventura College via ASCCC Open Educational Resources Initiative ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. The Discovery of Auxin 3. Auxin Transport 1. Polar Transport 2. Nonpolar Transport Actions of Auxin Tropisms Growth and Development Embryo Development Vascular Tissue Differentiation Leaf Development and Arrangement Root Initiation and Development Shade Avoidance Interactions with Other Growth-Regulating Hormones Apical Dominance Flowering and Fruit Development Prevention of Abscission Mechanisms of Auxin Action Commercial Applications of Auxins Attributions Learning Objectives Explain the mechanism of polar auxin transport. Identify locations of synthesis and actions of auxin. Define apical dominance and explain the role of auxin in maintaining it. Describe the commercial applications of auxin. Interpret and predict outcomes of experiments that demonstrate the action of auxin. The term auxin is derived from the Greek word auxein, which means "to grow." While many synthetic auxins are used as herbicides, indole-3-acetic acid (IAA) is the only naturally occurring auxin that shows physiological activity (Figure 4.4.1.1). Auxin is synthesized in apical meristems, young leaves, and developing seeds. Figure 4.4.1.1: Chemical structure of indole-3-acetic acid (IAA) The Discovery of Auxin Recall from the Tropisms section that the Boysen-Jensen experiment showed a chemical signal must be downward from the tip of the coleoptile tip along the shaded side, resulting in phototropism. Went extracted the chemical signal involved in the Boysen-Jensen experiment. He removed the tips of several coleoptiles of oat, Avena sativa, seedlings. He placed these on a block of agar for several hours. At the end of this time, the agar block itself was able to initiate resumption of growth of the decapitated coleoptile. The growth was vertical because the agar block was placed completely across the stump of the coleoptile and no light reached the plant from the side (Figure 4.4.1.2). The unknown substance that had diffused from the agar block was named auxin. The amount of auxin in coleoptile tips was far too small to be purified and analyzed chemically. Therefore, a search was made for other sources of auxin activity. Figure 4.4.1.2: Went's experiment showing that the coleoptile transferred a chemical signal (now know to be auxin) into an agar block. The agar block could then stimulate cell elongation in the absence of the coleoptile tip. This search was aided by a technique called the Avena test developed by Went for determining the relative amount of auxin activity in a preparation. The material to be assayed is incorporated into an agar block, and the block is placed on one edge of a decapitated Avena coleoptile. As the auxin diffuses into that side of the coleoptile, it stimulates cell elongation and the coleoptile bends away from the block (Figure 4.4.1.3). The degree of curvature, measured after 1.5 hours in the dark, is proportional to the amount of auxin activity (e.g., number of coleoptile tips used). The use of living tissue to determine the amount of a substance, such as in the Avena test, is called a bioassay. Figure 4.4.1.3: In the Avena test, the coleoptile tip is removed. (The primary leaf pierces the coleoptile.) Plant tissue is then allowed to transfer the chemical signal (now know to be auxin) into an agar block. The agar block is then placed on one side of the coleoptile, resulting in bending in the absence of the coleoptile tip. The Avena test soon revealed that substances with auxin activity occur widely in nature. One of the most potent was first isolated from human urine. It was indole-3-acetic acid (IAA) and turned out to be the auxin actually used by plants. Auxin Transport Auxin moves through the plant by two mechanisms, called polar and nonpolar transport. Polar Transport In contrast to the other major plant hormones, auxins can be transported in a specific direction (polar transport) through parenchyma cells. The cytoplasms of parenchyma cells are neutral (pH = 7), but the region outside the plasma membranes of adjacent cells (the apoplast) is acidic (pH = 5). When auxin is in the cytoplasm, it releases a proton and becomes an anion (IAA-). It cannot pass through hydrophobic portion of the plasma membrane as an anion, but it does pass through special auxin efflux transporters called PIN proteins.Eight different types of these transmembrane proteins have been identified so far. When IAA- enters the acidic environment of the apoplast, it is protonated, becoming IAAH. This uncharged molecule can then pass through the plasma membrane of adjacent cells through diffusion or via influx transporters. Once it enters the cytoplasm, it loses its proton, becoming IAA- again. PIN proteins can be unevenly distributed around the cell (for example, only occurring on the bottom of the cell), which directs the flow of auxin (Figure 4.4.1.4). Figure 4.4.1.4: The polar transport of auxin. Auxin (IAAH) enters the cell through influx transporters (orange) or passes directly through the plasma membrane (not shown). In the cytoplasm (pH = 7), which has a higher pH than the apoplast (pH = 5), auxin dissociates to release a proton (H+) and anion (IAA-). It can no longer exit the cell through the way that it entered, but it can exit through PIN proteins (purple). In this example, PIN proteins are only on the lower side of the cell. Once IAA- leaves the cell and enters the acidic apoplast, it binds to a proton becoming IAAH again. It then enters the next cells (below the first cell). This process continues, resulting in polar (unidirectional) transport. Image by Jen Valenzuela (CC-BY-NC). Nonpolar Transport Auxins can also be transported nondirectionally (nonpolar transport) through the phloem. It passes in the assimilate that translocates through the phloem from where it is synthesized (its "source", usually the shoot) to a "sink" (e.g., the root). Actions of Auxin Tropisms Auxins are the main hormones responsible for phototropism and gravitropism. The auxin gradients that are required for these tropisms are facilitated by the movement of PIN proteins and the polar transport of auxin in response to environmental stimuli (light or gravity). Note that higher auxin concentration on one side of the stem typically causes that side of the stem to elongate; however, the effect is opposite in roots with higher auxin concentration inhibiting elongation (Figure 4.4.1.5). Figure 4.4.1.5: The graph (based on the work of K. V. Thimann) shows the effect of auxin concentration on root and stem growth. The x-axis shows the concentration of auxin in parts per million (ppm) on the log scale. The y-axis shows percent stimulation (positive) or inhibition (negative).The difference between the behavior of roots and stems lies in the difference in the sensitivity of their cells to auxin. Auxin concentrations high enough to stimulate stem growth (represented by a peak) inhibit root growth (represented by a dip). Growth and Development Embryo Development Auxins play a role in embryo development. From the very first mitotic division of the zygote, gradients of auxin guide the patterning of the embryo into the parts that will become the organs of the plant, including the shoot apex, primary leaves, cotyledon(s), stem, and root. Vascular Tissue Differentiation They also control cell differentiation of vascular tissue. Leaf Development and Arrangement The formation of new leaves in the apical meristem is initiated by the accumulation of auxin. Already-developing leaves deplete the surrounding cells of auxin so that the new leaves do not form too close to them. In this way, the characteristic pattern of leaves in the plant is established. Auxin also controls the precise patterning of the epidermal cells of the developing leaf. Root Initiation and Development The localized accumulation of auxin in epidermal cells of the root initiates the formation of lateral or secondary roots. Auxin also stimulates the formation of adventitious roots in many species. Adventitious roots grow from stems or leaves rather than from the regular root system of the plant. Once a root is formed, a gradient of auxin concentration develops highest at the tip promoting the production of new cells at the meristem, and lowest in the region of differentiation, thus promoting the elongation and differentiation of root cells. The drop in auxin activity in the regions of elongation and differentiation is mediated by cytokinin — an auxin antagonist. Shade Avoidance Auxins stimulate cell elongation parts of the plants that have access to light as part of the shade-avoidance response (see Etiolation and Shade Avoidance). Interactions with Other Growth-Regulating Hormones Auxin is required for the function of other growth-regulating hormones such as cytokinins; cytokinins promote cell division, but only in the presence of auxin. Apical Dominance Apical dominance—the inhibition of axillary bud (lateral bud) formation—is triggered by downward transport of auxins produced in the apical meristem. Many plants grow primarily at a single apical meristem and have limited axillary branches (Figure 4.4.1.6). Growth of the shoot apical meristem (terminal shoot) usually inhibits the development of the lateral buds on the stem beneath. If the shoot apical meristem of a plant is removed, the inhibition is lifted, and axillary buds begin growth. However, if the apical meristem is removed and IAA applied to the stump, inhibition of the axillary buds is maintained (Figure 4.4.1.7). Gardeners exploit this principle by pruning the terminal shoot of ornamental shrubs, etc. The release of apical dominance enables lateral branches to develop and the plant becomes bushier. The process usually must be repeated because one or two laterals will eventually outstrip the others and reimpose apical dominance. Figure 4.4.1.6: The auxin produced by the shoot apical meristem (apical bud) inhibits the growth of the axillary (lateral) buds, maintaining apical dominance. Image by Doctor Smart (CC BY-SA). Figure 4.4.1.7: The shoot apical meristem (terminal shoot) produces auxin and inhibits the growth of the axillary (lateral) buds, maintaining apical dominance (left). If the shoot apical meristem is removed, the axillary buds will growth into axillary (lateral) shoots (middle). If the shoot apical meristem is removed and replaced with an agar block containing axuin, apical dominance is maintained (right). The common white potato also illustrates the principle of apical dominance. Note that a potato is a tuber, which is an underground stem modified for starch storage. As with an ordinary shoot, the potato has a terminal bud (containing the shoot apical meristem) or "eye" and several axillary (lateral) buds. After a long period of storage, the terminal bud usually sprouts but the other buds do not. However, if the potato is sliced into sections, one bud to a section, the axillary buds develop just as quickly as the terminal bud (Figure 4.4.1.8). Figure 4.4.1.8: The eyes of the potato illustrate apical dominance. Each eye of the potato is a bud. In the top image, the shoot apical meristem of the terminal shoot produces auxin that inhibits the growth of axillary (lateral) buds. In the bottom image, the potato is sliced into sections, apical dominance does not occur, and each axillary bud (lateral bud/eye) grows. As will be discussed in the Abscisic Acid section, abscisic acid in the lateral buds inhibits production of auxin, and removal of the apical bud will release this inhibition of auxin, allowing the lateral buds to begin growing. Flowering and Fruit Development Auxins promote flowering and fruit setting and ripening. Pollination of the flowers of angiosperms initiates the formation of seeds. As the seeds mature, they release auxin to the surrounding flower parts, which develop into the fruit that covers the seeds. Prevention of Abscission Some plants drop leaves and fruits in response to changing seasons (based on temperatures, photoperiod, water, or other environmental conditions). This process is called abscission, and is regulated by interactions between auxin and ethylene. During the growing season, the young leaves and fruits produce high levels of auxin, which blocks activity of ethylene; they thus remain attached to the stem. As the seasons change, auxin levels decline and permit ethylene to initiate senescence, or aging (see Ethylene). Figure 4.4.1.9 demonstrates the role of auxin in abscission. If the blade of the leaf is removed, as shown in the figure, the petiole remains attached to the stem for a few more days. The removal of the blade seems to be the trigger as an undamaged leaf at the same node of the stem remains on the plant much longer, in fact, the normal length of time. If, however, auxin is applied to the cut end of the petiole, abscission of the petiole is greatly delayed. Figure 4.4.1.9: This experiment shows that removal of the leaf blades from a Mimosa plant results in abscission of the petiole. Leaf blades produce auxin, which prevents abscission. When an auxin paste is applied to the petiole, abscission does not occur. Mechanisms of Auxin Action Auxin effects are mediated by two different pathways: immediate, direct effects on the cell and turning on of new patterns of gene expression. The arrival of auxin in the cytosol initiates such immediate responses as changes in the concentration of and movement of ions in and out of the cell and reduction in the redistribution of PIN proteins. At the cellular level, auxin generally increases the rate of cell division and longitudinal cell expansion. Some of the direct effects of auxin may be mediated by its binding to a cell-surface receptor designated ABP1 ("Auxin-binding protein 1"). Many auxin effects are mediated by changes in the transcription of genes. Auxin enters the nucleus and binds to its receptor, a protein called TIR1 ("transport inhibitor response protein 1") which now can bind to proteins responsible for attaching ubiquitin to one or another of several Aux/IAA proteins. This triggers the destruction of the Aux/IAA proteins by proteasomes. Aux/IAA proteins normally bind transcription factors called auxin response factors (ARF) preventing them from activating the promoters and other control sequences of genes that are turned on (or off) by auxin. Destruction of the Aux/IAA proteins relieves this inhibition, and gene transcription begins. This mechanism is another of the many cases in biology where a pathway is turned on by inhibiting the inhibitor of that pathway (a double-negative is a positive). The presence in the cell of many different Aux/IAA proteins (29 in Arabidopsis), many different ARFs (23 in Arabidopsis) and several (~4) TIR1-like proteins provides a logical basis for mediating the different auxin effects that are described here, but how this is done remains to be discovered. Commercial Applications of Auxins Commercial use of auxins is widespread in for propagation in nurseries, crop production, and killing weeds. Horticulturists may propagate desirable plants by cutting pieces of stem and placing them base down in moist soil. Eventually adventitious roots grow out at the base of the cutting. The process can often be hastened by treating the cuttings with a solution or powder containing a synthetic auxin. Applying synthetic auxins to tomato plants in greenhouses promotes normal fruit development. Fruit growers often apply auxin sprays to cut down the loss of fruit from premature dropping. Additionally, outdoor application of auxin promotes synchronization of fruit setting and dropping to coordinate the harvesting season. Fruits such as seedless cucumbers can be induced to set fruit by treating unfertilized plant flowers with auxins. Synthetic auxins are widely used as herbicides. Examples include 2,4-dichlorophenoxy acetic acid (2,4-D) and 2,4,5-trichlorophenoxy acetic acid (2,4,5-T), shown in Figure 4.4.1.10. 2,4-D and its many variants are popular because they are selective herbicides, killing broad-leaved eudicots but not narrow-leaved monocots. (No one knows the basis of this selectivity). Why should a synthetic auxin kill the plant? It turns out that the auxin influx transporter works fine for 2,4-D, but that 2,4-D cannot leave the cell through the efflux transporters. Perhaps it is the resulting accumulation of 2,4-D within the cell that kills it. A mixture of 2,4,-D and 2,4,5-T was the "agent orange" used by the U.S. military to defoliate the forest in parts of South Vietnam. Because of health concerns, 2,4,5-T is no longer used in the U.S. Figure 4.4.1.10: Chemical structures of the synthetic auxins 2,4-D (top) and 2,4,5-T (bottom). As the formulas show, 2,4,5-T is 2,4-D with a third chlorine atom, instead of a hydrogen atom, at the #5 position in the benzene ring (circled and in blue). Attributions Curated and authored by Melissa Ha from the following sources: 30.6 Plant Sensory Systems and Responses from _\\_Biology 2e\_\_ by OpenStax (licensed CC-BY). Access for free at openstax.org. 16.2F Tropisms and 16.5B Auxin from Biology by John. W. Kimball (licensed CC-BY) Plant Hormones and Sensory Systems by Biology 1520 Introduction to Organismal Biology (licensed CC BY-NC-SA) This page titled 4.4.1: Auxin is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Melissa Ha, Maria Morrow, & Kammy Algiers (ASCCC Open Educational Resources Initiative) . Back to top 4.4: Hormones 4.4.2: Cytokinins Was this article helpful? Yes No Recommended articles 39.5.1: AuxinThe most important auxin produced by plants is indole-3-acetic acid (IAA). It undergoes both polar (unidirectional) and nonpolar transport. Key functi... 11.1: AuxinThe most important auxin produced by plants is indole-3-acetic acid (IAA). It undergoes both polar (unidirectional) and nonpolar transport. Key functi... 16.5B: AuxinThis page discusses auxins, particularly indole-3-acetic acid (IAA), which are vital plant hormones impacting growth, development, and responses to en... 16.2F: TropismsThis page explains tropisms in plants, which are growth movements influenced by external stimuli. There are positive (toward the stimulus) and negativ... 11: Plant HormonesHormones are long-distance chemical signals in plants. They coordinate many responses including growth, reproduction, dormancy, and stress responses. ... Article typeSection or PageAuthorMelissa Ha, Maria Morrow, and Kammy AlgiersLicenseCC BY-NC-SALicense Version4.0OER program or PublisherASCCC OERI ProgramShow TOCno Tags Apical Dominance auxin cid:biol155 gravitropism indole-3-acetic acid (IAA) phototropism plant hormones source-bio-5799 source-bio-5799 © Copyright 2025 Biology LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? 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https://www.statisticshowto.com/chance-vs-probability-vs-odds/
Skip to content Chance vs Probability vs Odds Statistics Definitions > Chance vs Probability vs Odds Probability is the study of the chance something will happen. The two terms—chance and probability—are often used as synonyms. If you’re talking about the chance something will happen, then that is an informal way of saying probability. For example, you might say you have a 10% chance of winning a raffle or you could also say there is a 10% probability you’ll win; they both mean the same thing. Many other terms are used in the same way: when we say that an event is “likely”, “unlikely”, or “almost definitively” to happen, we are also talking about probability. Chance vs Probability vs Odds: The Math Probability is expressed as a percentage from 0% (will never happen) to 100% (will certainly happen). These percentages can also be expressed as a decimal from 0 to 1; to convert a percentage to a decimal, move the decimal point two places to the right. For example, 100.00% becomes 1.00 as a decimal. You can look at a probability as a number that represents the chance or likelihood an event will happen. Chances are almost always expressed as a percentage; you can think of them as probabilities expressed in a scale “by the hundred”. In other words, the measurement scale for chances and probabilities is different. For example, if the probability of an event happening is 1/4, then the chance of the event happening is 25% (1/4 times 100 percent). Therefore, a chance of 25% literally means that 25 of every one hundred of this particular event would likely happen . Odds is the probability an event will happen, divided by the probability an event will not happen. As a formula: Odds = p / (1 – p), where p is the probability (“chance”) of an event happening. As an example, let’s say you buy a scratch off lottery ticket with stated odds of 1:5 that you’ll win a prize. That means you have one chance of winning vs. five chances of not winning. To convert odds to probability: Place the odds of winning in the numerator of a fraction. Place the odds of winning and losing in the denominator. Using the above example, odds of 1/5 becomes a probability of 1/(5+1) – 1/6. You can also convert more without using the formula, if you understand what the odds represents. Odds of 1 to 5 means “one for and five against”. In other words, out of six possible chances (1 + 5), you’ll win once. As a percentage, that’s 1/6; this is our probability. If we put that probability into the formula we get:Odds = (1/6) / (1 – (1/6)) = (1/6) / (5/6) = 1/5.Odds are usually not stated in fractional form though; it’s more common to say 1:5 or 1 to 5. Chance vs Probability vs Odds: References Fulton, L. et al. (2012). Confusion Between Odds and Probability, a Pandemic? Journal of Statistics Education, Volume 20, Number 3 Comments? Need to post a correction? Please Contact Us.
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https://study.com/academy/lesson/marginal-rate-of-substitution-definition-formula-example.html
Marginal Rate of Substitution | MRS Definition, Formula & Example - Lesson | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Business Courses Business 100: Intro to Business Marginal Rate of Substitution | MRS Definition, Formula & Example Contributors: Sarah Sagal, Toni Bonton Author Author: Sarah Sagal Show more Instructor Instructor: Toni Bonton Show more Learn how to calculate the marginal rate of substitution and its application in economics. View examples of the formula in use with real world application. Updated: 11/21/2023 Table of Contents Marginal Rate of Substitution (MRS): Definition Marginal Rate of Substitution Formula Marginal Rate of Substitution and Utility Function Lesson Summary Show FAQ What is the MRS formula? The marginal rate of substitution is equal to the amount the consumer is willing to sacrifice of good A divided by the amount they will gain from good B in exchange. The amount the consumer is willing to sacrifice is always negative, resulting in a negative MRS. How do you find the MRS of a utility function? A utility function can be denoted as U(Xa, Xb). To find the MRS, you can use (Xa, Xb) as a point along the indifference curve to calculate MRS. For example, MRS = -Xa / Xb. What is the marginal rate of substitution with example? The marginal rate of substitution is the amount of one good that a consumer is willing to sacrifice in exchange for some amount of another good. For example, if a consumer is willing to give up 6 bananas in exchange for 3 apples, the MRS = -6 / 3 = -3. Create an account LessonTranscript VideoQuizCourseGames Click for sound 5:29 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? Log in here for access Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. 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The marginal rate of substitution (MRS) is the rate at which a consumer is willing to substitute one good for some amount of another good, given that the new good brings the same level of satisfaction. The marginal rate of substitution is always negative because it is measuring the rate at which someone is willing to give up some amount of one good in exchange for gaining some amount of another good. Since the equation contains the amount that is willing to be given up (negative) in the numerator and the amount to be gained (positive) in the denominator, the result is always negative. The marginal rate of substitution will always illustrate how much of one item we are willing to give up for one unit of another item. For example, if the MRS is -3, then we are willing to give up 3 units of one item in exchange for 1 unit of another item. The MRS in this case can also be written as 3:1. Slope of Indifference Curve: MRS An indifference curve shows multiple combinations of two goods that result in the same satisfaction for the consumer, hence making them indifferent about any of the possible combinations on the curve. The marginal rate of substitution is equal to the slope of the indifference curve. It is important to note that the indifference curve is actually a curve, so the slope and the MRS are constantly changing along the curve. The indifference curve is used to explain consumer preferences within the limits of budget constraints. For example, say someone has a budget of $10, and they can buy any combination of hot dogs and fries. Hot dogs cost $1 and bring 10 units of satisfaction while fries cost $0.50 and bring 5 units of satisfaction. The combinations that would fall on the indifference curve are where the level of satisfaction is equal. If they decide to get 10 hot dogs for $10, they gain 100 units of satisfaction; the same applies to purchasing 5 hot dogs and 10 fries for $10, which also brings 100 units of satisfaction. In this case, they are willing to give up 5 hot dogs in exchange for 10 fries because either amount of each respective good brings the same amount of satisfaction. This makes the marginal substitution rate for hot dogs at this point on the curve -0.5 since they will give up 5 hot dogs to gain 10 fries (-5 / 10). One way to interpret this is that 0.5 units of a hot dog brings the same satisfaction as 1 unit of fries. This is the theoretical application of this concept, and in theory, the same satisfaction from either combination of the two goods would be received. However, the law of diminishing marginal utility plays its part and can vary the amount of satisfaction gained from each additional unit consumed. The law of diminishing marginal utility states that as each additional unit of a good or service is consumed, the marginal utility gained from each additional unit decreases. For example, the first hot dog consumed might bring 10 units of satisfaction, but the second hot dog might bring only 7 units. However, for the purpose of learning about this concept, we will pretend that the marginal utility for each unit of the same good is constant. To unlock this lesson you must be a Study.com memberCreate an account Marginal Rate of Substitution Formula ------------------------------------- The marginal rate of substitution formula is the change in good X (dx) divided by the change in good Y (dy). The amount of the good being given up will be good X since it will always be negative. The amount gained in exchange is always good Y. The negative value always goes in the numerator, and the positive value always goes in the denominator. MRS formula. Take this indifference curve, for example. The coordinates for Point A are (10 units, 20 units). To find the marginal rate of substitution, first determine which good is being given up and which good is being gained. It can work either way depending on what is being looked for. For example, say we are giving up 20 units of Good A to gain 10 units of Good B. The equation is -20/10 and results in an MRS of -2 for Good A at this point on the curve. The way to interpret this result is that we are willing to give up 2 units of Good A to gain 1 unit of Good B. But what if we are giving up Good B to gain Good A? In this case, we give up 10 units of Good B to gain 20 units of Good A. The resulting MRS is -0.5 for Good B at this point on the curve (-10 Good B / 20 Good A). Indifference curve of goods A and B. Thinking about this in practice makes sense; if Good A represents apples, and Good B represents bananas, we will most likely want some variety in the amount of fruit we have. We will probably not want 50 apples or 50 bananas, but rather some combination in between. At either end of the indifference curve, we have all of one good and none of the other. Since we have so much of one item, we will be willing to give up a larger amount of that item in order to gain a smaller amount of the other item. This seemingly unfair trade actually brings the same amount of satisfaction as the initial amount of the same good. It is important to note that indifference curves will vary by each person based on their preferences for one good over another, as well as their budgetary constraints. Marginal Rate of Substitution Examples Now, we will do some examples to illustrate further how to calculate the marginal rate of substitution. Using the graph below, we will find the MRS for mangoes at point C. To do this, we must first recognize that mangoes are measured along the X-axis, so the amount of mangoes the consumer is willing to give up for oranges at this point on the indifference curve is equal to 3 mangoes. The consumer will sacrifice 3 mangoes in exchange for 6 oranges, which are measured along the Y-axis. To find the MRS for mangoes at point C, the amount of mangoes the consumer is willing to give up (-3) is divided by the oranges they will gain (6) to find that the MRS for mangoes at point C is -0.5 What is the marginal rate of substitution for oranges at point C? We must reverse the equation to find this. The consumer is willing to give up 6 oranges in order to gain 3 mangoes. In this situation, MRS = -6 / 3 = -2. At this point, the consumer will give up 2 oranges for every mango they gain. Indifference curve for oranges and mangoes. To unlock this lesson you must be a Study.com memberCreate an account Marginal Rate of Substitution and Utility Function -------------------------------------------------- Utility function measures the comparative amount of satisfaction gained as a function of the consumption of goods. Utility functions are expressed as a function of a bundle of goods and services denoted as U(X 1, X 2, X N). A practical way to think about this is to imagine a basket full of groceries. It might have apples (X 1), chicken (X 2), crackers (X 3), and cookies (X 4). The aim of the utility function is to measure the preferences that a consumer applies to their choice of goods or services. Essentially, the order of the items shows the preference for the items through ranking; in this case, the consumer prefers apples over chicken since apples come first. When used to compare one bundle of goods to another bundle, the utility function is denoted as U(X A, X B). The marginal rate of substitution can be derived from the utility function. Since we are measuring utilities along the indifference curve, the utility for every combination will be the same. Because of this, U(X A, X B) = X A, X B and the MRS = - X A / X B. Thus, when given a utility function of U(1,8), the MRS = -8/1 = -8. To unlock this lesson you must be a Study.com memberCreate an account Lesson Summary -------------- The marginal rate of substitution is the rate at which a consumer is willing to substitute one good for some amount of another good, given that the new good brings the same level of satisfaction. The marginal rate of substitution (MRS) is equal to the slope of the indifference curve which shows multiple combinations of two goods that result in the same satisfaction for the consumer, hence making them indifferent about any of the possible combinations on the curve. The utility function can be used to derive the MRS, and it measures the comparative amount of satisfaction gained as a function of the consumption of goods. The law of diminishing marginal utility states that for each additional unit consumed of the same good or service, the marginal utility gained decreases. This is important to keep in mind for real life applications. To unlock this lesson you must be a Study.com memberCreate an account Video Transcript Marginal Rate of Substitution Brandy loves to shop for shoes and bags. In fact, she spends most of her free time and allowance on shopping sprees for more shoes and bags. While loading up on handbags during a sale at the mall, Brandy hears the store manager's voice over the intercom announcing an impromptu discount on shoes, but there's a catch. Customers may only receive a maximum of $500 in discounts per purchase. Brandy is now faced with dilemma. She must decide how many handbags she is willing to give up in exchange for each pair of shoes and still be satisfied with her purchase. This combination exchange is called the marginal rate of substitution. The marginal rate of substitution is the number of units a consumer is willing to give up of one good in exchange for units of another good and remain equally satisfied. The substitution doesn't indicate a preference in goods, only that the consumer is willing to give up units of one good for additional units of another good. In this case, Brandy is willing to give up a certain amount of handbags for each additional pair of shoes she would like to buy, as long as she doesn't have to compromise her total satisfaction. If her chosen combinations were graphed, it would look like this: Each combination of handbags and shoes will generate the same amount of satisfaction for Brandy. The Exchange Now that Brandy is willing to give up a certain amount of handbags for additional pairs of shoes that she would like to purchase, we can figure out the marginal rate of substitution. To calculate the marginal rate of substitution, the change in good x is divided by the change in good y: MRS(x,y) = the marginal rate of substitution between both goods dx = the change in good x, the number of units a consumer is willing to give up dy = the change in good y, the number of units a consumer gains by giving up units of good x If that sounds like a foreign language, let's bring it all together so that it makes sense. Listed below is the combination of handbags and shoes Brandy is willing to accept to be satisfied and still fall within her allowed discounts: The marginal rate of substitution begins at Combination B because it shows what Brandy had to give up in order to purchase an additional pair of shoes. At Combination B, Brandy had to give up three handbags (7 - 4 = 3) to gain one additional pair of shoes (2 - 1 = 1). Let's calculate the marginal rate of substitution: MRS(x,y) = 3 (the change in good x) / 1 (the change in good y) MRS(x,y) = 3 / 1 MRS(x,y) = 3 The marginal rate of substitution is 3, or 3:1. When the marginal rate of substitution is written as a ratio, it points out how many of good x were given up for good y. Now, Brandy has four handbags and two pair of shoes, but she has her eyes on another pair of shoes that she would love to have in her collection. This time Brandy's willing to give up two more handbags in order to buy one additional pair of shoes and still remain completely satisfied. Let's figure out the marginal rate of substitution for Combination C: MRS(x,y) = 2 (the change in good x) / 1 (the change in good y) MRS(x,y) = 2 / 1 MRS(x,y) = 2 The marginal rate of substitution is 2, or 2:1, meaning Brandy is willing to give up two more handbags so that she can purchase an additional pair of shoes. What Is Satisfaction? You might be wondering, do economists have to ask every customer how satisfied they are to calculate the marginal rate of substitution for a set of goods? Of course, satisfaction is a subjective measurement, but economists measure it by looking at the behavior of the market. For example, during the summer months there may be fluctuation (or change) in how many hot dogs and hamburgers are purchased. Say you're planning a BBQ, and you've gone to the store to get some meat for the grill. As you're shopping, you may decide to put back a few packs of hot dogs in order to buy another pack of hamburgers and still be satisfied. Since it would be time consuming to ask each individual consumer what combination of meat would keep them equally satisfied, economists analyze the market fluctuations of different purchases alongside the income elasticity of demand (in other words, consumers buying what they can afford) to monitor the rate of substitutions. Lesson Summary Let's review what we've learned… The marginal rate of substitution describes the rate at which a consumer is willing to give up units of one good in order to receive additional units of another good, as long as the level of satisfaction remains the same. The willingness to give up units of one good does not mean there is a preference of either good, just that a consumer is willing to give a certain number of good x in order to gain a certain number of good y. The formula for the marginal rate of substitution is as follows: It can also be seen as: MRS(x,y) = the marginal rate of substitution between both goods When calculated, the marginal rate of substitution is usually written out as a ratio, like X:1. What this translates to is that X units of the first good were given up of one unit for the other good. Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. 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Try it now Business 100: Intro to Business 26 chapters 137 lessons 12 flashcard sets Chapter 1 The Dynamic Business Environment How the Economic and Legal Environment Affects Business 5:45 min How The Technological Environment Affects Business: Examples & Effects 5:15 min How The Competitive Environment Affects Business: Examples & Importance 5:59 min Social Environment of a Business | Definition, Effects & Examples 9:04 min How The Global Business Environment Affects Business: Explanation & Examples 6:45 min Chapter 2 Practicing Social Responsibility and Ethical Behavior in Business Business Ethics | Theories, Principles & Examples 5:58 min Code of Ethics | Definition, Types & Examples 7:19 min Social Responsibility in Organizations | Overview & Examples 4:27 min Chapter 3 Economics and Business What is Economics? - Definition & Types 6:46 min Circular Flow of Economic Activity | Overview & Models 6:03 min Competition Within Free Markets: Types & Summary 5:01 min Command, Socialist & Mixed Economies | Definition & Examples 5:42 min Gross Domestic Product | GDP Definition, Components & Examples 4:56 min Business Cycle Definition, Parts & Example 12:00 min Consumer Price Index | Definition, Example & Calculation 8:41 min Chapter 4 Forms of Business Ownership Organizational Structure & Ownership of a Business 5:45 min Corporate Expansion: Mergers and Acquisitions 5:32 min Franchises: Opportunities and Challenges 5:56 min Chapter 5 Entrepreneurship and Small Business Entrepreneur | Definition & Examples 5:51 min Small Business in US Economy | Definition, Impact & Importance 7:57 min Small Business in a Global Market 5:54 min Chapter 6 Managing and Leading in Business Good Management Skills, Importance & Examples 6:27 min Functions of Management | Definition & Roles 6:37 min Planning Definition & Types 9:24 min SWOT Analysis | Definition, Purpose & Examples 5:35 min Internal Organizational Change | Definition & Examples 6:41 min Leadership Roles in Management | Overview & Types 4:30 min What Is Management by Objective? - Defining the MBO Process 3:42 min Chapter 7 Leadership Styles in Business Leadership Orientation | Definition, Styles & Examples 6:16 min Leadership Theory | Overview & Types 3:29 min What is New Leadership Theory? 3:39 min Leadership Style & Fit in the Workplace 5:46 min Using the 5Ps Leadership Analysis 5:22 min Chapter 8 Workplace Productivity & Motivation Motivation in Management | Definition, Process & Types 4:43 min What Is Staff Motivation? - Theories & Strategies 3:45 min Employee Motivation in the Workplace | Theories, Ways & Examples 4:38 min Motivation Theory: Needs-Based & Behavior-Based 5:50 min Chapter 9 Organizational Management Organizational Behavior Theories | Importance & Examples 7:25 min Organizational Management | Definition & Principles 8:18 min Organizational Design: Theory, Principles & Definition 4:19 min Organizational Design | Definition & Factors 7:43 min Types of Organizational Structure | Overview & Examples 7:46 min Types of Contemporary Organizational Designs: Matrix, Team & Network Designs 8:41 min Total Quality Management | TQM Overview, Key Features & Examples 7:23 min Modern Management Theories | Definition & Types 8:46 min Chapter 10 Product Development and Retailing Product Development and Business Growth: Process & Strategies 7:26 min Types of New Products: New Product Lines, Product Improvements & More 7:28 min Packaging in Marketing | Importance, Types & Reasons 5:37 min Product Adoption in Marketing | Definition & Process 5:20 min Product Life Cycles: Development, Design and Beyond 4:27 min Chapter 11 Business Production and Operations Operations Management: Focusing on Production Efficiency & Customer Satisfaction 8:01 min Productivity, Quality, Profitability and the Role of Managers 7:42 min Operations Management Planning | Definition & Examples 5:15 min PERT Analysis & Project Completion 6:17 min Henry Gantt's Chart | History, Invention & Elements 6:08 min Management Information Systems | MIS Definition, Scope & Careers 7:17 min Chapter 12 Product Distribution & Supply Chain Management Channel Intermediaries: Definition and Function in Business 5:30 min Wholesale Intermediaries | Definition, Types & Examples 6:05 min Distribution Strategy in Marketing | Definition, Types & Channels 4:25 min Non-Store Retailing | Types, Advantages & Examples 5:43 min Supply Chain Management: Technology, Measurement, Relationship & Material Integration 6:21 min Logistics Activities, Management & Services 5:31 min Chapter 13 Basics of Human Resources Human Resource Management: Hiring and Staffing 9:25 min Human Resources | Laws, Rights & Regulations 5:54 min Performance Appraisal and 360 Feedback 5:17 min Understanding Employee Compensation 6:20 min Chapter 14 Managing the Employer-Worker Relationship A Historical Outline of Organized Labor in the United States 7:12 min Function of Collective Bargaining 7:56 min Third-Party Interventions in Organizations | Methods & Types 5:09 min Chapter 15 Business Marketing Basics Competitive Advantage | Definition, Types & Examples 6:41 min Target Marketing Strategies | Definition & Examples 7:13 min Market Research Definition, Strategies & Examples 9:45 min B2B vs. B2C | Definition, Marketing Strategies & Examples 5:55 min Demographic vs Psychographic Segmentation in Marketing 6:51 min Influences on Consumer Buying Decisions: Cultures, Values & More 9:23 min Chapter 16 Pricing Strategy in Marketing Pricing Decisions: Profit-Oriented, Sales & Status Quo 5:26 min Pricing Objectives: How Firms Decide on a Pricing Strategy 4:47 min Pricing Strategy and Consumer Perception 5:58 min Chapter 17 Product Promotion in Business Promotion and the Consumer Communication Process 6:03 min Goals of Promotion and the Marketing Mix 5:00 min Integrated Marketing Communication | IMC Definition & Examples 5:00 min Effects of Advertising on Consumer Buying Behavior 5:08 min Advertising Strategies | Definition, Types & Comparison 5:04 min Advertising Media Definition, Types & Evaluation 6:46 min Trade Sales Promotion | Types, Goals & Examples 5:03 min Chapter 18 Business in Global Markets The Importance of the Global Market and Global Trade: Role & Advantages 9:21 min Import & Export Market Overview & Examples | What is an Import & Export? 7:24 min Strategies for Reaching Global Markets: Examples & Types 6:18 min Forces that Affect Trade in Global Markets 7:48 min The Changing Landscape of the Global Market 6:51 min Chapter 19 MIS Basics in Business Business Strategies: Market Advantages Provided by Information Systems 5:16 min Internet Communication: Social Media, Email, Blog, & Chat 4:32 min Internet in Business Collaboration: Communication & Platforms 5:03 min Business Intelligence (BI): Organizing, Categorizing and Accessing Data 8:13 min Management Information Systems (MIS): Manager Decision-Making Tools 5:10 min Chapter 20 Implications of Information Technology Information Technology in Business: Benefits & Limitations 6:04 min How Technology Helps Achieve Business Success 5:24 min E-Business Enhanced and E-Business Enabled Organizations 5:24 min Ethical Issues in Managing Technology in Business 6:51 min Chapter 21 Risk Management in Business Ways to Manage Risk: Insurable and Uninsurable Risk 7:17 min Understanding Insurance Policies and Risk Management 7:14 min Insurance Coverage for Various Types of Risk 6:00 min Chapter 22 Accounting Basics Purpose of Accounting | ALOE Equation & Examples 11:20 min External and Internal Users and Uses of Accounting 8:22 min Generally Accepted Accounting Principles | GAAP Overview & Rules 10:44 min Financial Statement | Definition, Types & Importance 4:47 min Accounting Cycle | Definition, Process & Examples 6:20 min Technology in Accounting: The Growing Role of Technology in Accounting 6:37 min Chapter 23 Money and Financial Institutions The Basic Functions of Money | Properties, Roles & Uses 5:38 min Multiplier Effect & Money Multiplier | Overview & Calculation 12:55 min Private Investment and Real Interest Rates 12:15 min Federal Reserve History, Role & Goals 8:58 min Reserve Requirement | Definition, History & Examples 11:47 min Institutions of the U.S. Banking System 4:26 min How Technology Makes Banking More Efficient 4:37 min International Banking | Definition, Services & Types 6:41 min The World Bank, IMF & Other International Banking Organizations 8:21 min Chapter 24 Financial Management in Business Financial Manager | Role, Responsibilities & Requirements 3:33 min Debt Capital vs. Equity Capital | Definition, Types & Examples 5:07 min Short-Term Financing | Definition, Purpose & Types 3:37 min Sources of Long-Term Financing 4:17 min The Financial Planning Process 7:57 min Chapter 25 Securities Markets and Business Investment Banking Overview & Types 6:27 min Selling Stock | Overview, Types & Benefits 7:36 min The Advantages of Bond Financing 5:19 min Chapter 26 Studying for Business 100 Dynamic Business Environment Flashcards Social Responsibility & Ethics in Business Flashcards Economics & Business Flashcards Business in Global Markets Flashcards Entrepreneurship & Forms of Business Flashcards Business Management & Leadership Flashcards Product Development & Distribution Flashcards Human Resources in the Workplace Flashcards Marketing Strategies in Business Flashcards Information Technology in Business Flashcards Risk Management Flashcards Finances in Business Flashcards Related Study Materials Marginal Rate of Substitution | MRS Definition, Formula & Example LessonsCoursesTopics ##### Consumption in Economics | Overview, Types & Example 5:11 ##### Income Effect in Economics | Definition & Examples 3:10 ##### Indifference Curves: Use & Impact in Economics 5:21 ##### Marginal Rate of Substitution: Definition, Formula & Examples 5:00 ##### Normal vs. Inferior Goods | Definition, Examples & Demand Curve 3:44 ##### Disposable Income | Definition, Importance & Examples 2:44 ##### Consumer Preference Concept & Assumptions | What is Consumer Preference? 5:12 ##### Who Is the Consumer in Microeconomics? 3:31 ##### Consumer Theories in Economics: Decision Making, Incentives & Preferences 5:21 ##### Indifference Curve, Substitute & Complement | Meaning & Examples 5:01 ##### Consumer Sovereignty | Definition, Advantages & Disadvantages 4:15 ##### Substitution Effect | Definition & Examples 3:27 ##### Expenditure Patterns of Older Adults 5:48 ##### Inferior Good in Economics | Definition & Examples 4:39 ##### Needs & Wants in Economics | Definition & Differences 4:57 ##### Consumer Income Meaning, Sources & Statistics 3:43 ##### The Motivation to Produce, Distribute & Consume within Society 6:48 ##### Economic Needs and Wants: Definition & Concept 3:13 ##### The Importance of Consumer Choice in Economics 5:36 ##### Engel Curve Definition, Mechanics & Examples 5:02 ##### Holt McDougal Economics - Concepts and Choices: Online Textbook Help ##### Business 100: Intro to Business ##### Business 110: Business Math ##### Intro to Excel: Essential Training & Tutorials ##### UExcel Introduction to Macroeconomics: Study Guide & Test Prep ##### UExcel Organizational Behavior: Study Guide & Test Prep ##### UExcel Business Law: Study Guide & Test Prep ##### UExcel Business Ethics: Study Guide & Test Prep ##### UExcel Workplace Communications with Computers: Study Guide & Test Prep ##### Business Strategy: Help & Review ##### Business Calculus: Help & Review ##### Leadership Study Guide ##### Investing: Help & Tutorials ##### Business Writing: Help & Review ##### Internet & Social Media Marketing: Help & Review ##### E-Commerce: Help & Tutorials ##### Intermediate Excel Training: Help & Tutorials ##### Microsoft Excel Certification: Practice & Study Guide ##### Intro to Email Marketing ##### Money Management: Help & Review Browse by Courses Marketing for Teachers: Professional Development DSST Principles of Public Speaking Study Guide and Test Prep Effective Communication in the Workplace: Certificate Program Supervision Principles for Teachers: Professional Development DSST Business Mathematics Study Guide and Test Prep Intro to PowerPoint: Essential Training & Tutorials GED Social Studies: Civics & Government, US History, Economics, Geography & World Business 103: Introductory Business Law Workplace Communications with Computers AP Macroeconomics Study Guide and Exam Prep Business 107: Organizational Behavior Business 100: Intro to Business CLEP Principles of Macroeconomics Study Guide and Exam Prep CSET Business Study Guide and Test Prep AEPA Economics (AZ035) Study Guide and Test Prep Browse by Lessons Service in Economics | Definition & Examples Consumerism: Lesson for Kids Consumerism Lesson Plan for Elementary School Goods & Services Lesson Plan Consumer Behavior Activities for High School Microeconomics Consumerism Lesson Plan Consumer Demand Definition, Theory & Factors Economic Mobility Definition, Types & Impact Say's Law in Economics | Theory, Criticisms & Examples Crude Materials: Definition, Categorization & Examples Alfred Weber's Model of Industrial Location | Overview & Examples Productive Efficiency | Definition & Examples Isoquant Curve Overview & Examples | What is an Isoquant? 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https://dec41.user.srcf.net/notes/II_L/logic_and_set_theory.pdf
Part II — Logic and Set Theory Based on lectures by I. B. Leader Notes taken by Dexter Chua Lent 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine. No specific prerequisites. Ordinals and cardinals Well-orderings and order-types. Examples of countable ordinals. Uncountable ordi-nals and Hartogs’ lemma. Induction and recursion for ordinals. Ordinal arithmetic. Cardinals; the hierarchy of alephs. Cardinal arithmetic. Posets and Zorn’s lemma Partially ordered sets; Hasse diagrams, chains, maximal elements. Lattices and Boolean algebras. Complete and chain-complete posets; fixed-point theorems. The axiom of choice and Zorn’s lemma. Applications of Zorn’s lemma in mathematics. The well-ordering principle. Propositional logic The propositional calculus. Semantic and syntactic entailment. The deduction and completeness theorems. Applications: compactness and decidability. Predicate logic The predicate calculus with equality. Examples of first-order languages and theories. Statement of the completeness theorem; sketch of proof. The compactness theorem and the Lowenheim-Skolem theorems. Limitations of first-order logic. Model theory. Set theory Set theory as a first-order theory; the axioms of ZF set theory. Transitive closures, epsilon-induction and epsilon-recursion. Well-founded relations. Mostowski’s collapsing theorem. The rank function and the von Neumann hierarchy. Consistency Problems of consistency and independence 1 Contents II Logic and Set Theory Contents 0 Introduction 3 1 Propositional calculus 4 1.1 Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Semantic entailment . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Syntactic implication . . . . . . . . . . . . . . . . . . . . . . . . . 7 2 Well-orderings and ordinals 13 2.1 Well-orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2 New well-orderings from old . . . . . . . . . . . . . . . . . . . . . 18 2.3 Ordinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.4 Successors and limits . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.5 Ordinal arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.6 Normal functions . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3 Posets and Zorn’s lemma 28 3.1 Partial orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.2 Zorn’s lemma and axiom of choice . . . . . . . . . . . . . . . . . 34 3.3 Bourbaki-Witt theorem . . . . . . . . . . . . . . . . . . . . . . . 36 4 Predicate logic 37 4.1 Language of predicate logic . . . . . . . . . . . . . . . . . . . . . 37 4.2 Semantic entailment . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.3 Syntactic implication . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.4 Peano Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.5 Completeness and categoricity . . . . . . . . . . . . . . . . . . . 48 5 Set theory 51 5.1 Axioms of set theory . . . . . . . . . . . . . . . . . . . . . . . . . 51 5.2 Properties of ZF . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 5.3 Picture of the universe . . . . . . . . . . . . . . . . . . . . . . . . 59 6 Cardinals 62 6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 6.2 Cardinal arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . 63 7 Incompleteness 66 Index 69 2 0 Introduction II Logic and Set Theory 0 Introduction Most people are familiar with the notion of “sets” (here “people” is defined to be mathematics students). However, most of the time, we only have an intuitive picture of what set theory should look like — there are sets, we can take intersections, unions, intersections and subsets. We can write down sets like {x : ϕ(x) is true}. Historically, mathematicians were content with this vague notion of sets. However, it turns out that our last statement wasn’t really correct. We cannot just arbitrarily write down sets we like. This is evidenced by the famous Russel’s paradox, where the set X is defined as X = {x : x ̸∈x}. Then we have X ∈X ⇔X ̸∈X, which is a contradiction. This lead to the formal study of set theory, where set theory is given a formal foundation based on some axioms of set theory. This is known as axiomatic set theory. This is similar to Euclid’s axioms of geometry, and, in some sense, the group axioms. Unfortunately, while axiomatic set theory appears to avoid paradoxes like Russel’s paradox, as G¨ odel proved in his incompleteness theorem, we cannot prove that our axioms are free of contradictions. Closely related to set theory is formal logic. Similarly, we want to put logic on a solid foundation. We want to formally define our everyday notions such as propositions, truth and proofs. The main result we will have is that a statement is true if and only if we can prove it. This assures that looking for proofs is a sensible way of showing that a statement is true. It is important to note that having studied formal logic does not mean that we should always reason with formal logic. In fact, this is impossible, as we ultimately need informal logic to reason about formal logic itself! Throughout the course, we will interleave topics from set theory and formal logic. This is necessary as we need tools from set theory to study formal logic, while we also want to define set theory within the framework of formal logic. One is not allowed to complain that this involves circular reasoning. As part of the course, we will also side-track to learn about well-orderings and partial orders, as these are very useful tools in the study of logic and set theory. Their importance will become evident as we learn more about them. 3 1 Propositional calculus II Logic and Set Theory 1 Propositional calculus Propositional calculus is the study of logical statements such p ⇒p and p ⇒ (q ⇒p). As opposed to predicate calculus, which will be studied in Chapter 4, the statements will not have quantifier symbols like ∀, ∃. When we say “p ⇒p is a correct”, there are two ways we can interpret this. We can interpret this as “no matter what truth value p takes, p ⇒p always has the truth value of “true”.” Alternatively, we can interpret this as “there is a proof that p ⇒p”. The first notion is concerned with truth, and does not care about whether we can prove things. The second notion, on the other hand, on talks about proofs. We do not, in any way, assert that p ⇒p is “true”. One of the main objectives of this chapter is to show that these two notions are consistent with each other. A statement is true (in the sense that it always has the truth value of “true”) if and only if we can prove it. It turns out that this equivalence has a few rather striking consequences. Before we start, it is important to understand that there is no “standard” logical system. What we present here is just one of the many possible ways of doing formal logic. In particular, do not expect anyone else to know exactly how your system works without first describing it. Fortunately, no one really writes proof with formal logic, and the important theorems we prove (completeness, compactness etc.) do not depend much on the exact implementation details of the systems. 1.1 Propositions We’ll start by defining propositions, which are the statements we will consider in propositional calculus. Definition (Propositions). Let P be a set of primitive propositions. These are a bunch of (meaningless) symbols (e.g. p, q, r), which are used as the basic building blocks of our more interesting propositions. These are usually interpreted to take a truth value. Usually, any symbol (composed of alphabets and subscripts) is in the set of primitive propositions. The set of propositions, written as L or L(P), is defined inductively by (i) If p ∈P, then p ∈L. (ii) ⊥∈L, where ⊥is read as “false” (also a meaningless symbol). (iii) If p, q ∈L, then (p ⇒q) ∈L. Example. If our set of primitive propositions is P = {p, q, r}, then p ⇒q, p ⇒⊥, ((p ⇒q) ⇒(p ⇒r)) are propositions. To define L formally, we let L0 = {⊥} ∪P Ln+1 = Ln ∪{(p ⇒q) : p, q ∈Ln}. Then we define L = L0 ∪L1 ∪L2 ∪· · · . 4 1 Propositional calculus II Logic and Set Theory In formal language terms, L is the set of finite strings of symbols from the alphabet ⊥⇒( ) p1 p2 · · · that satisfy some formal grammar rule (e.g. brackets have to match). Note here that officially, the only relation we have is ⇒. The familiar “not”, “and” and “or” do not exist. Instead, we define them to be abbreviations of certain expressions: Definition (Logical symbols). ¬p (“not p”) is an abbreviation for (p ⇒⊥) p ∧q (“p and q”) is an abbreviation for ¬(p ⇒(¬q)) p ∨q (“p or q”) is an abbreviation for (¬p) ⇒q The advantage of having just one symbol ⇒is that when we prove something about our theories, we only have to prove it for ⇒, instead of all ⇒, ¬, ∧and ∨ individually. 1.2 Semantic entailment The idea of semantic entailment is to assign truth values to propositions, where we declare each proposition to be “true” or “false”. This assignment is performed by a valuation. Definition (Valuation). A valuation on L is a function v : L →{0, 1} such that: – v(⊥) = 0, – v(p ⇒q) = ( 0 if v(p) = 1, v(q) = 0, 1 otherwise We interpret v(p) to be the truth value of p, with 0 denoting “false” and 1 denoting “true”. Note that we do not impose any restriction of v(p) when p is a primitive proposition. For those people who like homomorphisms, we can first give the set {0, 1} a binary operation ⇒by a ⇒b = ( 0 if a = 1, b = 0 1 otherwise as well as a constant ⊥= 0. Then a valuation can be defined as a homomorphism between L and {0, 1} that preserves ⊥and ⇒. It should be clear that a valuation is uniquely determined by its values on the primitive propositions, as the values on all other propositions follow from the definition of a valuation. In particular, we have Proposition. (i) If v and v′ are valuations with v(p) = v′(p) for all p ∈P, then v = v′. (ii) For any function w : P →{0, 1}, we can extend it to a valuation v such that v(p) = w(p) for all p ∈L. 5 1 Propositional calculus II Logic and Set Theory Proof. (i) Recall that L is defined inductively. We are given that v(p) = v′(p) on L0. Then for all p ∈L1, p must be in the form q ⇒r for q, r ∈L0. Then v(q ⇒r) = v(p ⇒q) since the value of v is uniquely determined by the definition. So for all p ∈L1, v(p) = v′(p). Continue inductively to show that v(p) = v′(p) for all p ∈Ln for any n. (ii) Set v to agree with w for all p ∈P, and set v(⊥) = 0. Then define v on Ln inductively according to the definition. Example. Suppose v is a valuation with v(p) = v(q) = 1, v(r) = 0. Then v((p ⇒q) ⇒r) = 0. Often, we are interested in propositions that are always true, such as p ⇒p. These are known as tautologies. Definition (Tautology). t is a tautology, written as | = t, if v(t) = 1 for all valuations v. To show that a statement is a tautology, we can use a truth table, where we simply list out all possible valuations and find the value of v(t). Example. (i) | = p ⇒(q ⇒p) “A true statement is implied by anything” v(p) v(q) v(q ⇒p) v(p ⇒(q ⇒p)) 1 1 1 1 1 0 1 1 0 1 0 1 0 0 1 1 (ii) | = (¬¬p) ⇒p. Recall that ¬¬p is defined as ((p ⇒⊥) ⇒⊥). v(p) v(p ⇒⊥) v((p ⇒⊥) ⇒⊥) v(((p ⇒⊥) ⇒⊥) ⇒p) 1 0 1 1 0 1 0 1 (iii) | = [p ⇒(q ⇒r)] ⇒[(p ⇒q) ⇒(p ⇒r)]. Instead of creating a truth table, which would be horribly long, we show this by reasoning: Suppose it is not a tautology. So there is a v such that v(p ⇒(q ⇒r)) = 1 and v((p ⇒q) ⇒(p ⇒r)) = 0. For the second equality to hold, we must have v(p ⇒q) = 1 and v(p ⇒r) = 0. So v(p) = 1, v(r) = 0, v(q) = 1. But then v(p ⇒(q ⇒r)) = 0. Sometimes, we don’t want to make statements as strong as “t is always true”. Instead, we might want to say “t is true whenever S is true”. This is known as semantic entailment. 6 1 Propositional calculus II Logic and Set Theory Definition (Semantic entailment). For S ⊆L, t ∈L, we say S entails t, S semantically implies t or S | = t if, for any v such that v(s) = 1 for all s ∈S, we have v(t) = 1. Here we are using the symbol | = again. This is not an attempt to confuse students. | = t is equivalent to the statement ∅| = t. Example. {p ⇒q, q ⇒r} | = (p ⇒r). We want to show that for any valuation v with v(p ⇒q) = v(q ⇒r) = 1, we have v(p ⇒r) = 1. We prove the contrapositive. If v(p ⇒r) = 0, then v(p) = 1 and v(r) = 0. If v(q) = 0, then v(p ⇒q) = 0. If v(q) = 1, then v(q ⇒r) = 0. So v(p ⇒r) = 0 only if one of v(p ⇒q) or v(q ⇒r) is zero. Note that {p} | = q and p ⇒q both intuitively mean “if p is true, then q is true”. However, these are very distinct notions. p ⇒q is a proposition within our theory. It is true (or false) in the sense that valuations take the value 0 (or 1). On the other hand, when we say {p} | = q, this is a statement in the meta-theory. It is true (or false) in the sense that we decided it is true (or false) based on some arguments and (informal) proofs, performed in the real world instead of inside propositional calculus. The same distinction should be made when we define syntactic implication later. Before we dive into syntactic implication, we will define a few convenient terms for later use. Definition (Truth and model). If v(t) = 1, then we say that t is true in v, or v is a model of t. For S ⊆L, a valuation v is a model of S if v(s) = 1 for all s ∈S. 1.3 Syntactic implication While semantic implication captures the idea of truthfulness, syntactic impli-cation captures the idea of proofs. We want to say S syntactically implies t if there we can prove t from S. To prove propositions, we need two things. Firstly, we need axioms, which are statements we can assume to be true in a proof. These are the basic building blocks from which we will prove our theorems. Other than axioms, we also need deduction rules. This allows as to make deduce statements from other statements. Our system of deduction composes of the following three axioms: 1. p ⇒(q ⇒p) 2. [p ⇒(q ⇒r)] ⇒[(p ⇒q) ⇒(p ⇒r)] 3. (¬¬p) ⇒p and the deduction rule of modus ponens: from p and p ⇒q, we can deduce q. At first sight, our axioms look a bit weird, especially the second one. We will later see that how this particular choice of axioms allows us to prove certain theorems more easily. This choice of axioms can also be motivated by combinatory logic, but we shall not go into details of these. 7 1 Propositional calculus II Logic and Set Theory Definition (Proof and syntactic entailment). For any S ⊆L, a proof of t from S is a finite sequence t1, t2, · · · tn of propositions, with tn = t, such that each ti is one of the following: (i) An axiom (ii) A member of S (iii) A proposition ti such that there exist j, k < i with tj being tk ⇒ti. If there is a proof of t from S, we say that S proves or syntactically entails t, written S ⊢t. If ∅⊢t, we say t is a theorem and write ⊢t. In a proof of t from S, t is the conclusion and S is the set of hypothesis or premises. Example. {p ⇒q, q ⇒r} ⊢p ⇒r We go for (p ⇒q) ⇒(p ⇒r) via Axiom 2. 1. [p ⇒(q ⇒r)] ⇒[(p ⇒q) ⇒(p ⇒r)] Axiom 2 2. q ⇒r Hypothesis 3. (q ⇒r) ⇒[p ⇒(q ⇒r)] Axiom 1 4. p ⇒(q ⇒r) MP on 2, 3 5. (p ⇒q) ⇒(p ⇒r) MP on 1, 4 6. p ⇒q Hypothesis 7. p ⇒r MP on 5, 6 Example. ⊢(p ⇒p) We go for [p ⇒(p ⇒p)] ⇒(p ⇒p). 1. [p ⇒((p ⇒p) ⇒p)] ⇒[(p ⇒(p ⇒p)) ⇒(p ⇒p)] Axiom 2 2. p ⇒((p ⇒p) ⇒p) Axiom 1 3. [p ⇒(p ⇒p)] ⇒(p ⇒p) MP on 1, 2 4. p ⇒(p ⇒p) Axiom 1 5. p ⇒p MP on 3, 4 This seems like a really tedious way to prove things. We now prove that the deduction theorem, which allows as to find proofs much more easily. Proposition (Deduction theorem). Let S ⊂L and p, q ∈L. Then we have S ⊢(p ⇒q) ⇔ S ∪{p} ⊢q. This says that ⊢behaves like the connective ⇒in the language. Proof. (⇒) Given a proof of p ⇒q from S, append the lines – p Hypothesis 8 1 Propositional calculus II Logic and Set Theory – q MP to obtain a proof of q from S ∪{p}. (⇐) Let t1, t2, · · · , tn = q be a proof of q from S ∪{p}. We’ll show that S ⊢p ⇒ti for all i. We consider different possibilities of ti: – ti is an axiom: Write down ◦ti ⇒(p ⇒ti) Axiom 1 ◦ti Axiom ◦p ⇒ti MP – ti ∈S: Write down ◦ti ⇒(p ⇒ti) Axiom 1 ◦ti Hypothesis ◦p ⇒ti MP – ti = p: Write down our proof of p ⇒p from our example above. – ti is obtained by MP: we have some j, k < i such that tk = (tj ⇒ti). We can assume that S ⊢(p ⇒tj) and S ⊢(p ⇒tk) by induction on i. Now we can write down ◦[p ⇒(tj ⇒ti)] ⇒[(p ⇒tj) ⇒(p ⇒ti)] Axiom 2 ◦p ⇒(tj ⇒ti) Known already ◦(p ⇒tj) ⇒(p ⇒ti) MP ◦p ⇒tj Known already ◦p ⇒ti MP to get S ⊢(p ⇒ti). This is the reason why we have this weird-looking Axiom 2 — it enables us to easily prove the deduction theorem. This theorem has a “converse”. Suppose we have a deduction system system that admits modus ponens, and the deduction theorem holds for this system. Then axioms (1) and (2) must hold in the system, since we can prove them using the deduction theorem and modus ponens. However, we are not able to deduce axiom (3) from just modus ponens and the deduction theorem. Example. We want to show {p ⇒q, q ⇒r} ⊢(p ⇒r). By the deduction theorem, it is enough to show that {p ⇒q, q ⇒r, p} ⊢r, which is trivial by applying MP twice. Now we have two notions: | = and ⊢. How are they related? We want to show that they are equal: if something is true, we can prove it; if we can prove something, it must be true. Aim. Show that S ⊢t if and only if S | = t. This is known as the completeness theorem, which is made up of two directions: 9 1 Propositional calculus II Logic and Set Theory (i) Soundness: If S ⊢t, then S | = t. “Our axioms aren’t absurd” (ii) Adequacy: If S | = t, S ⊢t. “Our axioms are strong enough to be able to deduce, from S, all semantic consequences of S.” We first prove the easy bit: Proposition (Soundness theorem). If S ⊢t, then S | = t. Proof. Given valuation v with v(s) = 1 for all s ∈S, we need to show that v(t) = 1. We will show that every line ti in the proof has v(ti) = 1. If ti is an axiom, then v(ti) = 1 since axioms are tautologies. If ti is a hypothesis, then by assumption v(s) = 1. If ti is obtained by modus ponens, say from tj ⇒ti, since v(tj) = 1 and v(tj ⇒ti) = 1, we must have v(ti) = 1. Note that soundness holds whenever our axioms are all tautologies. Even if we had silly axioms that are able to prove almost nothing, as long as they are all tautologies, it will be sound. Now we have to prove adequacy. It seems like a big and scary thing to prove. Given that a statement is true, we have to find a proof for it, but we all know that finding proofs is hard! We first prove a special case. To do so, we first define consistency. Definition (Consistent). S is inconsistent if S ⊢⊥. S is consistent if it is not inconsistent. The special case we will prove is the following: Theorem (Model existence theorem). If S | = ⊥, then S ⊢⊥. i.e. if S has no model, then S is inconsistent. Equivalently, if S is consistent, then S has a model. While we called this a “special case”, it is in fact all we need to know to prove adequacy. If we are given S | = t, then S ∪{¬t} | = ⊥. Hence using the model existence theorem, we know that S ∪{¬t} ⊢⊥. Hence by the deduction theorem, we know that S ⊢¬¬t. But ⊢(¬¬t) ⇒t by Axiom 3. So S ⊢t. As a result, some books call this the “completeness theorem” instead, because the rest of the completeness theorem follows trivially from this. The idea of the proof is that we’d like to define v : L →{0, 1} by p 7→ ( 1 if p ∈S 0 if p ̸∈S However, this is obviously going to fail, because we could have some p such that S ⊢p but p ̸∈S, i.e. S is not deductively closed. Yet this is not a serious problem — we take the deductive closure first, i.e. add all the statements that S can prove. But there is a more serious problem. There might be a p with S ̸⊢p and S ̸⊢¬p. This is the case if, say, p never appears in S. The idea here is to arbitrarily declare that p is true or false, and add p or ¬p to S. What we have to prove is that we can do so without making S consistent. We’ll prove this in the following lemma: Lemma. For consistent S ⊂L and p ∈L, at least one of S ∪{p} and S ∪{¬p} is consistent. 10 1 Propositional calculus II Logic and Set Theory Proof. Suppose instead that both S ∪{p} ⊢⊥and S ∪{¬p} ⊢⊥. Then by the deduction theorem, S ⊢p and S ⊢¬p. So S ⊢⊥, contradicting consistency of S. Now we can prove the completeness theorem. Here we’ll assume that the primitives P, and hence the language L is countable. This is a reasonable thing to assume, since we can only talk about finitely many primitives (we only have a finite life), so uncountably many primitives would be of no use. However, this is not a good enough excuse to not prove our things properly. To prove the whole completeness theorem, we will need to use Zorn’s lemma, which we will discuss in Chapter 3. For now, we will just prove the countable case. Proof. Assuming that L is countable, list L as {t1, t2, · · · }. Let S0 = S. Then at least one of S ∪{t1} and S ∪{¬t1} is consistent. Pick S1 to be the consistent one. Then let S2 = S1 ∪{t2} or S1 ∪{¬t2} such that S2 is consistent. Continue inductively. Set ¯ S = S0∪S1∪S2 · · · . Then p ∈¯ S or ¬p ∈¯ S for each p ∈L by construction. Also, we know that ¯ S is consistent. If we had ¯ S ⊢⊥, then since proofs are finite, there is some Sn that contains all assumptions used in the proof of ¯ S ⊢⊥. Hence Sn ⊢⊥, but we know that all Sn are consistent. Finally, we check that ¯ S is deductively closed: if ¯ S ⊢p, we must have p ∈¯ S. Otherwise, ¬p ∈¯ S. But this implies that ¯ S is inconsistent. Define v : L →{0, 1} by p 7→ ( 1 if p ∈¯ S 0 if not . All that is left to show is that this is indeed a valuation. First of all, we have v(⊥) = 0 as ⊥̸∈¯ S (since ¯ S is consistent). For p ⇒q, we check all possible cases. (i) If v(p) = 1, v(q) = 0, we have p ∈¯ S, q ̸∈¯ S. We want to show p ⇒q ̸∈¯ S. Suppose instead that p ⇒q ∈¯ S. Then ¯ S ⊢q by modus ponens. Hence q ∈¯ S since ¯ S is deductively closed. This is a contradiction. Hence we must have v(p ⇒q) = 0. (ii) If v(q) = 1, then q ∈¯ S. We want to show p ⇒q ∈¯ S. By our first axiom, we know that ⊢q ⇒(p ⇒q). So ¯ S ⊢p ⇒q. So p ⇒q ∈¯ S by deductive closure. Hence we have v(p ⇒q) = 1. (iii) If v(p) = 0, then p ̸∈¯ S. So ¬p ∈¯ S. We want to show p ⇒q ∈¯ S. – This is equivalent to showing ¬p ⊢p ⇒q. – By the deduction theorem, this is equivalent to proving {p, ¬p} ⊢q. – We know that {p, ¬p} ⊢⊥. So it is sufficient to show ⊥⊢q. – By axiom 3, this is equivalent to showing ⊥⊢¬¬q. – By the deduction theorem, this is again equivalent to showing ⊢⊥⇒ ¬¬q. – By definition of ¬, this is equivalent to showing ⊢⊥⇒(¬q ⇒⊥). 11 1 Propositional calculus II Logic and Set Theory But this is just an instance of the first axiom. So we know that ¯ S ⊢p ⇒q. So v(p ⇒q) = 1. Note that the last case is the first time we really use Axiom 3. By remark before the proof, we have Corollary (Adequacy theorem). Let S ⊂L, t ∈L. Then S | = t implies S ⊢t. Theorem (Completeness theorem). Le S ⊂L and t ∈L. Then S | = t if and only if S ⊢t. This theorem has two nice consequences. Corollary (Compactness theorem). Let S ⊂L and t ∈L with S | = t. Then there is some finite S′ ⊂S has S′ | = t. Proof. Trivial with | = replaced by ⊢, because proofs are finite. Sometimes when people say compactness theorem, they mean the special case where t = ⊥. This says that if every finite subset of S has a model, then S has a model. This result can be used to prove some rather unexpected result in other fields such as graph theory, but we will not go into details. Corollary (Decidability theorem). Let S ⊂L be a finite set and t ∈L. Then there exists an algorithm that determines, in finite and bounded time, whether or not S ⊢t. Proof. Trivial with ⊢replaced by | =, by making a truth table. This is a rather nice result, because we know that proofs are hard to find in general. However, this theorem only tells you whether a proof exists, without giving you the proof itself! 12 2 Well-orderings and ordinals II Logic and Set Theory 2 Well-orderings and ordinals In the coming two sections, we will study different orderings. The focus of this chapter is well-orders, while the focus of the next is partial orders. A well-order on a set S is a special type of total order where every non-empty subset of S has a least element. Among the many nice properties of well-orders, it is possible to do induction and recursion on well-orders. Our interest, however, does not lie in well-orders itself. Instead, we are interested in the “lengths” of well-orders. Officially, we call them them the order types of the well-orders. Each order type is known as an ordinal. There are many things we can do with ordinals. We can add and multiply them to form “longer” well-orders. While we will not make much use of them in this chapter, in later chapters, we will use ordinals to count “beyond infinity”, similar to how we count finite things using natural numbers. 2.1 Well-orderings We start with a few definitions. Definition ((Strict) total order). A (strict) total order or linear order is a pair (X, <), where X is a set and < is a relation on X that satisfies (i) x ̸< x for all x (irreflexivity) (ii) If x < y, y < z, then x < z (transitivity) (iii) x < y or x = y or y < x (trichotomy) We have the usual shorthands for total orders. For example, x > y means y < x and x ≤y means (x < y or x = y). It is also possible for a total order to be defined in terms of ≤instead of <. Definition ((Non-strict) total order). A (non-strict) total order is a pair (X, ≤), where X is a set and ≤is a relation on X that satisfies (i) x ≤x (reflexivity) (ii) x ≤y and y ≤z implies x ≤z (transitivity) (iii) x ≤y and y ≤x implies x = y (antisymmetry) (iv) x ≤y or y ≤x (trichotomy) Example. (i) N, Z, Q, R with usual the usual orders are total orders. (ii) On N+ (the positive integers), ‘x < y if x | y and x ̸= y’ is not trichotomous, and so not a total order. (iii) On P(S), define ‘x ≤y’ if x ⊆y. This is not a total order since it is not trichotomous (for |X| > 1). While there can be different total orders, the particular kind we are interested in is well-orders. 13 2 Well-orderings and ordinals II Logic and Set Theory Definition (Well order). A total order (X, <) is a well-ordering if every (non-empty) subset has a least element, i.e. (∀S ⊆X)[S ̸= ∅⇒(∃x ∈S)(∀y ∈S) y ≥x]. Example. (i) N with usual ordering is a well-order. (ii) Z, Q, R are not well-ordered because the whole set itself does not have a least element. (iii) {x ∈Q : x ≥0} is not well-ordered. For example, {x ∈X : x > 0} has no least element. (iv) In R, {1 −1/n : n = 2, 3, 4, · · · } is well-ordered because it is isomorphic to the naturals. (v) {1 −1/n : n = 2, 3, 4, · · · } ∪{1} is also well-ordered, If the subset is only 1, then 1 is the least element. Otherwise take the least element of the remaining set. (vi) Similarly, {1 −1/n : n = 2, 3, 4, · · · } ∪{2} is well-ordered. (vii) {1 −1/n : n = 2, 3, 4, · · · } ∪{2 −1/n : n = 2, 3, 4, · · · } is also well-ordered. This is a good example to keep in mind. There is another way to characterize total orders in terms of infinite decreasing sequences. Proposition. A total order is a well-ordering if and only if it has no infinite strictly decreasing sequence. Proof. If x1 > x2 > x3 > · · · , then {xi : i ∈N} has no least element. Conversely, if non-empty S ⊂X has no least element, then each x ∈S have x′ ∈S with x′ < x. Similarly, we can find some x′′ < x′ ad infinitum. So x > x′ > x′′ > x′′′ > · · · is an infinite decreasing sequence. Like all other axiomatic theories we study, we identify two total orders to be isomorphic if they are “the same up to renaming of elements”. Definition (Order isomorphism). Say the total orders X, Y are isomorphic if there exists a bijection f : X →Y that is order-preserving, i.e. x < y ⇒f(x) < f(y). Example. (i) N and {1 −1/n : n = 2, 3, 4, · · · } are isomorphic. (ii) {1−1/n : n = 2, 3, 4, · · · }∪{1} is isomorphic to {1−1/n : n = 2, 3, 4, · · · }∪ {2}. (iii) {1 −1/n : n = 2, 3, 4, · · · } and {1 −1/n : n = 2, 3, 4, · · · } ∪{1} are not isomorphic because the second has a greatest element but the first doesn’t. 14 2 Well-orderings and ordinals II Logic and Set Theory Recall from IA Numbers and Sets that in N, the well-ordering principle is equivalent to the principle induction. We proved that we can do induction simply by assuming that N is well-ordered. Using the same proof, we should be able to prove that we can do induction on any well-ordered set. Of course, a general well-ordered set does not have the concept of “+1”, so we won’t be able to formulate weak induction. Instead, our principle of induction is the form taken by strong induction. Proposition (Principle by induction). Let X be a well-ordered set. Suppose S ⊆X has the property: (∀x) (∀y) y < x ⇒y ∈S  ⇒x ∈S  , then S = X. In particular, if a property P(x) satisfies (∀x) (∀y) y < x ⇒P(y)  ⇒P(x)  , then P(x) for all x. Proof. Suppose S ̸= X. Let x be the least element of X \S. Then by minimality of x, for all y, y < x ⇒y ∈S. Hence x ∈S. Contradiction. Using proof by induction, we can prove the following property of well-orders: Proposition. Let X and Y be isomorphic well-orderings. Then there is a unique isomorphism between X and Y . This is something special to well-orders. This is not true for general total orderings. For example, x 7→x and x 7→x−13 are both isomorphisms Z →Z. It is also not true for, say, groups. For example, there are two possible isomorphisms from Z3 to itself. Proof. Let f and g be two isomorphisms X →Y . To show that f = g, it is enough, by induction, to show f(x) = g(x) given f(y) = g(y) for all y < x. Given a fixed x, let S = {f(y) : y < x}. We know that Y \ S is non-empty since f(x) ̸∈S. So let a be the least member of Y \ S. Then we must have f(x) = a. Otherwise, we will have a < f(x) by minimality of a, which implies that f −1(a) < x since f is order-preserving. However, by definition of S, this implies that a = f(f −1(a)) ∈S. This is a contradiction since a ∈Y \ S. By the induction hypothesis, for y < x, we have f(y) = g(y). So we have S = {g(y) : y < x} as well. Hence g(x) = min(Y \ S) = f(x). If we have an ordered set, we can decide to cut off the top of the set and keep the bottom part. What is left is an initial segment. Definition (Initial segment). A subset Y of a totally ordered X is an initial segment if x ∈Y, y < x ⇒y ∈Y, 15 2 Well-orderings and ordinals II Logic and Set Theory Y X Example. For any x ∈X, the set Ix = {y ∈X : y < x} is an initial segment. However, not every initial segment of X need to be in this form. For example, {x : x ≤3} ⊆R and {x : x < 0 or x2 < 2} ⊆Q are both initial segments not of this form. The next nice property of well-orders we have is that every proper initial segment is of this form. Proposition. Every initial segment Y of a well-ordered set X is of the form Ix = {y ∈X : y < x}. Proof. Take x = min X \ Y . Then for any y ∈Ix, we have y < x. So y ∈Y by definition of x. So Ix ⊆Y . On the other hand, if y ∈Y , then definitely y ̸= x. We also cannot have y > x since this implies x ∈Y . Hence we must have y < x. So y ∈Ix. Hence Y ⊆Ix. So Y = Ix. The next nice property we want to show is that in a well-ordering X, every subset S is isomorphic to an initial segment. Note that this is very false for a general total order. For example, in Z, no initial segment is isomorphic to the subset {1, 2, 3} since every initial segment of Z is either infinite or empty. Alternatively, in R, Q is not isomorphic to an initial segment. It is intuitively obvious how we can prove this. We simply send the minimum element of S to the minimum of X, and the continue recursively. However, how can we justify recursion? If we have the well-order { 1 2, 2 3, 3 4, · · · , 1}, then we will never reach 1 if we attempt to write down each step of the recursion, since we can only get there after infinitely many steps. We will need to define some sort of “infinite recursion” to justify our recursion on general well-orders. We first define the restriction of a function: Definition (Restriction of function). For f : A →B and C ⊆A, the restriction of f to C is f|C = {(x, f(x)) : x ∈C}. In this theorem (and the subsequent proof), we are treating functions as explicitly a set of ordered pairs (x, f(x)). We will perform set operations on functions and use unions to “stitch together” functions. Theorem (Definition by recursion). Let X be a well-ordered set and Y be any set. Then for any function G : P(X ×Y ) →Y , there exists a function f : X →Y such that f(x) = G(f|Ix) for all x. 16 2 Well-orderings and ordinals II Logic and Set Theory This is a rather weird definition. Intuitively, it means that G takes previous values of f(x) and returns the desired output. This means that in defining f at x, we are allowed to make use of values of f on Ix. For example, we define f(n) = nf(n −1) for the factorial function, with f(0) = 1. Proof. We might want to jump into the proof and define f(0) = G(∅), where 0 is the minimum element. Then we define f(1) = G(f(0)) etc. But doing so is simply recursion, which is the thing we want to prove that works! Instead, we use the following clever trick: We define an “h is an attempt” to mean h : I →Y for some initial segment I of X, and h(x) = G(h|Ix) for x ∈I. The idea is to show that for any x, there is an attempt h that is defined at x. Then take the value f(x) to be h(x). However, we must show this is well-defined first: Claim. If attempts h and h′ are defined at x, then h(x) = h′(x). By induction on x, it is enough to show that h(x) = h′(x) assuming h(y) = h′(y) for all y < x. But then h(x) = G(h|Ix) = G(h′|Ix) = h′(x). So done. Claim. For any x, there must exist an attempt h that is defined at x. Again, we may assume (by induction) that for each y < x, there exists an attempt hy defined at y. Then we put all these functions together, and take h′ = S y<x hy. This is defined for all y < x, and is well-defined since the hy never disagree. Finally, add to it (x, G(h′|Ix)). Then h = h′ ∪(x, G(h′|Ix)) is an attempt defined at x. Now define f : X →Y by f(x) = y if there exists an attempt h, defined at x, with h(x) = y. Claim. There is a unique such f. Suppose f and f ′ both work. Then if f(y) = f ′(y) for all y < x, then f(x) = f ′(x) by definition. So by induction, we know for all x, we have f ′(x) = f(x). With the tool of recursion, we can prove that every subset of a well-order. Lemma (Subset collapse). Let X be a well-ordering and let Y ⊆X. Then Y is isomorphic to an initial segment of X. Moreover, this initial segment is unique. Proof. For f : Y →X to be an order-preserving bijection with an initial segment of X, we need to map x to the smallest thing not yet mapped to, i.e. f(x) = min(X \ {f(y) : y < x}). To be able to take the minimum, we have to make sure the set is non-empty, i.e. {f(y) : y < x} ̸= X. We can show this by proving that f(z) < x for all z < x by induction, and hence x ̸∈{f(y) : y < x}. Then by the recursion theorem, this function exists and is unique. 17 2 Well-orderings and ordinals II Logic and Set Theory This implies that a well-ordered X can never be isomorphic to a proper initial segment of itself. This is since X is isomorphic to itself by the identity function, and uniqueness shows that it cannot be isomorphic to another initial segment. Using the idea of initial segments, we can define an order comparing different well-orders themselves. Notation. Write X ≤Y if X is isomorphic to an initial segment of Y . We write X < Y if X ≤Y but X is not isomorphic to Y , i.e. X is isomorphic to a proper initial segment of Y . Example. If X = N, Y = { 1 2, 2 3, 3 4, · · · } ∪{1}, then X ≤Y . We will show that this is a total order. Of course, we identify two well-orders as “equal” when they are isomorphic. Reflexivity and transitivity are straightforward. So we prove trichotomy and antisymmetry: Theorem. Let X, Y be well-orderings. Then X ≤Y or Y ≤X. Proof. We attempt to define f : X →Y by f(x) = min(Y \ {f(y) : y < x}). By the law of excluded middle, this function is either well-defined or not. If it is well-defined, then it is an isomorphism from X to an initial segment of Y . If it is not, then there is some x such that {f(y) : y < x} = Y and we cannot take the minimum. Then f is a bijection between Ix = {y : y < x} and Y . So f is an isomorphism between Y and an initial segment of X. Hence either X ≤Y or Y ≤X. Theorem. Let X, Y be well-orderings with X ≤Y and Y ≤X. Then X and Y are isomorphic. Proof. Since X ≤Y , there is an order-preserving function f : X →Y that bijects X with an initial segment of Y . Similarly, since Y ≤X, we get an analogous g : Y →X. Then g ◦f : X →X defines a bijection between X and an initial segment of X. Since there is no bijection between X and a proper initial segment of itself, the image of g ◦f must be X itself. Hence g ◦f is a bijection. Similarly, f ◦g is a bijection. Hence f and g are both bijections, and X and Y are isomorphic. 2.2 New well-orderings from old Given a well-ordering X, we want to create more well-orderings. We’ve previously shown that we can create a shorter one by taking an initial segment. In this section, we will explore two ways to make longer well-orderings. 18 2 Well-orderings and ordinals II Logic and Set Theory Add one element We can extend a well-ordering by exactly one element. This is known as the successor. Definition (Successor). Given X, choose some x ̸∈X and define a well-ordering on X ∪{x} by setting y < x for all y ∈X. This is the successor of X, written X+. We clearly have X < X+. Put some together More interestingly, we want to “stitch together” many well-orderings. However, we cannot just arbitrarily stitch well-orderings together. The well-orderings must satisfy certain nice conditions for this to be well-defined. Definition (Extension). For well-orderings (X, <X) and (Y, <Y ), we say Y extends X if X is a proper initial segment of Y and <X and <Y agree when defined. Note that we explicitly require X to be an initial segment of Y . X simply being a subset of Y will not work, for reasons that will become clear shortly. Definition (Nested family). We say well-orderings {Xi : i ∈I} form a nested family if for any i, j ∈I, either Xi extends Xj, or Xj extends Xi. Proposition. Let {Xi : i ∈I} be a nested set of well-orderings. Then there exists a well-ordering X with Xi ≤X for all i. Proof. Let X = S i∈I Xi with < defined on X as S i∈I <i (where <i is the ordering of Xi), i.e. we inherit the orders from the Xi’s. This is clearly a total ordering. Since {Xi : i ∈I} is a nested family, each Xi is an initial segment of X. To show that it is a well-ordering, let S ⊆X be a non-empty subset of X. Then S ∩Xi is non-empty for some i. Let x be the minimum element (in Xi) of S ∩Xi. Then also for any y ∈S, we must have x ≤y, as Xi is an initial segment of X. Note that if we didn’t require X to be an initial segment of Y when defining ’extension’, then the above proof will not work. For example, we can take the collection of all subsets Xn = {x ≥−n : x ∈Z}, and their union would be Z, which is not well-ordered. 2.3 Ordinals We have already shown that the collection of all well-orderings is a total order. But is it a well-ordering itself? To investigate this issue further, we first define ourselves a convenient way of talking about well-orderings. Definition (Ordinal). An ordinal is a well-ordered set, with two regarded as the same if they are isomorphic. We write ordinals as Greek letters α, β etc. 19 2 Well-orderings and ordinals II Logic and Set Theory We would want to define ordinals as equivalence classes of well-orders under isomorphism, but we cannot, because they do not form a set. We will provide a formal definition of ordinals later when we study set theory. Definition (Order type). If a well-ordering X has corresponding ordinal α, we say X has order type α, and write otp(X) = α. Notation. For each k ∈N, we write k for the order type of the (unique) well-ordering of size k. We write ω for the order type of N. Example. In R, {2, 3, 5, 6} has order type 4. { 1 2, 2 3, 3 4, · · · } has order type ω. Notation. For ordinals α, β, write α ≤β if X ≤Y for some X of order type α, Y of order type β. This does not depend on the choice of X and Y (since any two choices must be isomorphic). Proposition. Let α be an ordinal. Then the ordinals < α form a well-ordering of order type α. Notation. Write Iα = {β : β < α}. Proof. Let X have order type α. The well-orderings < X are precisely (up to isomorphism) the proper initial segments of X (by uniqueness of subset collapse). But these are the Ix for all x ∈X. So we can biject X with the well-orderings < X by x 7→Ix. Finally, we can prove that the ordinals are well-ordered. Proposition. Let S be a non-empty set of ordinals. Then S has a least element. Proof. Choose α ∈S. If it is minimal, done. If not, then S ∩Iα is non-empty. But Iα is well-ordered. So S ∩Iα has a least element, β. Then this is a minimal element of S. However, it would be wrong to say that the ordinals form a well-ordered set, for the very reason that they don’t form a set. Theorem (Burali-Forti paradox). The ordinals do not form a set. Proof. Suppose not. Let X be the set of ordinals. Then X is a well-ordering. Let its order-type be α. Then X is isomorphic to Iα, a proper initial subset of X. Contradiction. Recall that we could create new well-orderings from old via adding one element and taking unions. We can translate these into ordinal language. Given an ordinal α, suppose that X is the corresponding well-order. Then we define α+ to be the order type of X+. If we have a set {αi : i ∈I} of ordinals, we can stitch them together to form a new well-order. In particular, we apply “nested well-orders” to the initial segments {Iαi : i ∈I}. This produces an upper bound of the ordinals αi. Since the ordinals are well-ordered, we know that there is a least upper bound. We call this the supremum of the set {αi : i ∈I}, written sup{αi : i ∈I}. In fact, the upper bound created by nesting well-orders is the least upper bound. Example. {2, 4, 6, 8, · · · } has supremum ω. 20 2 Well-orderings and ordinals II Logic and Set Theory Now we have two ways of producing ordinals: +1 and supremum. We can generate a lot of ordinals now: 0 ω · 2 + 1 ω2 + 1 ω2 · 3 ωω+2 ε0 + 1 1 ω · 2 + 2 ω2 + 2 ω2 · 4 . . . . . . 2 ω · 2 + 3 ω2 + 3 ω2 · 5 ωω·2 ε0 · 2 . . . . . . . . . . . . . . . . . . ω ω · 3 ω2 + ω ω3 ωω2 ε2 0 ω + 1 ω · 4 . . . . . . . . . . . . ω + 2 ω · 5 ω2 + ω · 2 ωω ωωω εε0 0 . . . . . . . . . . . . . . . . . . ω + ω = ω · 2 ω · ω = ω2 ω2 · 2 ωω+1 ωω... = ε0 εε... 0 0 = ε1 Here we introduced a lot of different notations. For example, we wrote ω + 1 to mean ω+, and ω · 2 = sup{ω, ω + 1, ω + 2, · · · }. We will formally define these notations later. We have written a lot of ordinals above, some of which are really huge. However, all the ordinals above are countable. The operations we have done so far is adding one element and taking countable unions. So the results are all countable. So is there an uncountable ordinal? Theorem. There is an uncountable ordinal. Proof. This is easy by looking at the supremum of the set of all countable ordinals. However, this works only if the collection of countable ordinals is a set. Let A = {R ∈P(N × N) : R is a well-ordering of a subset of N}. So A ⊆ P(N × N). Then B = {order type of R : R ∈A} is the set of all countable ordinals. Let ω1 = sup B. Then ω1 is uncountable. Indeed, if ω1 were countable, then it would be the greatest countable ordinal, but ω1 + 1 is greater and is also countable. By definition, ω1 is the least uncountable ordinal, and everything in our previous big list of ordinals is less than ω1. There are two strange properties of ω1: (i) ω1 is an uncountable ordering, yet for every x ∈ω1, the set {y : y < x} is countable. (ii) Every sequence in ω1 is bounded, since its supremum is a countable union of countable sets, which is countable. In general, we have the following theorem: Theorem (Hartogs’ lemma). For any set X, there is an ordinal that does not inject into X. Proof. As before, with B = {α : α injects into X}. Notation. Write γ(X) for the least ordinal that does not inject into X. e.g. γ(ω) = ω1. 21 2 Well-orderings and ordinals II Logic and Set Theory 2.4 Successors and limits In general, we can divide ordinals into two categories. The criteria is as follows: Given an ordinal α, is there a greatest element of α? i.e. does Iα = {β : β < α} have a greatest element? If yes, say β is the greatest element. Then γ ∈Iα ⇔γ ≤β. So Iα = {β}∪Iβ. In other words, α = β+. Definition (Successor ordinal). An ordinal α is a successor ordinal if there is a greatest element β below it. Then α = β+. On the other hand, if no, then for any γ < α, there exists β < α such that β > γ. So α = sup{β : β < α}. Definition (Limit ordinal). An ordinal α is a limit if it has no greatest element below it. We usually write λ for limit ordinals. Example. 5 and ω+ are successors. ω and 0 are limits (0 is a limit because it has no element below it, let alone a greatest one!). 2.5 Ordinal arithmetic We want to define ordinals arithmetic such as + and ×, so that we can make formal sense out of our notations such as ω + ω in our huge list of ordinals. We first start with addition. Definition (Ordinal addition (inductive)). Define α + β by recursion on β (α is fixed): – α + 0 = α. – α + β+ = (α + β)+. – α + λ = sup{α + γ : γ < λ} for non-zero limit λ. Note that officially, we cannot do “recursion on the ordinals”, since the ordinals don’t form a set. So what we officially do is that we define α + γ on {γ : γ < β} recursively for each ordinal β. Then by uniqueness of recursions, we can show that this addition is well-defined. Example. ω + 1 = (ω + 0)+ = ω+. ω + 2 = (ω + 1)+ = ω++. 1 + ω = sup{1 + n : n ≤ω} = sup{1, 2, 3, · · · } = ω. It is very important to note that addition is not commutative! This asymmetry arises from our decision to perform recursion on β instead of α. On the other hand, addition is associative. Proposition. Addition is associative, i.e. (α + β) + γ = α + (β + γ). Proof. Since we define addition by recursion, it makes sense to prove this by induction. Since we recursed on the right-hand term in the definition, it only makes sense to induct on γ (and fix α + β). (i) If γ = 0, then α + (β + 0) = α + β = (α + β) + 0. 22 2 Well-orderings and ordinals II Logic and Set Theory (ii) If γ = δ+ is a successor, then α + (β + δ+) = α + (β + δ)+ = [α + (β + δ)]+ = [(α + β) + δ]+ = (α + β) + δ+ = (α + β) + γ. (iii) If γ is a limit ordinal, we have (α + β) + λ = sup{(α + β) + γ : γ < λ} = sup{α + (β + γ) : γ < λ} If we want to evaluate α + (β + λ), we have to first know whether β + λ is a successor or a limit. We now claim it is a limit: β + λ = sup{β + γ : γ < λ}. We show that this cannot have a greatest element: for any β + γ, since λ is a limit ordinal, we can find a γ′ such that γ < γ′ < λ. So β + γ′ > β + γ. So β + γ cannot be the greatest element. So α + (β + λ) = sup{α + δ : δ < β + λ}. We need to show that sup{α + δ : δ < β + λ} = sup{α + (β + γ) : γ < λ}. Note that the two sets are not equal. For example, if β = 3 and λ = ω, then the left contains α + 2 but the right does not. So we show that the left is both ≥and ≤the right. ≥: Each element of the right hand set is an element of the left. ≤: For δ < β + λ, we have δ < sup{β + γ : γ < λ}. So δ < β + γ for some γ < λ. Hence α + δ < α + (β + γ). Note that it is easy to prove that β < γ ⇒α + β < α + γ by induction on γ (which we implicitly assumed above). But it is not true if we add on the right: 1 < 2 but 1 + ω = 2 + ω. The definition we had above is called the inductive definition. There is an alternative definition of + based on actual well-orders. This is known as the synthetic definition. Intuitively, we first write out all the elements of α, then write out all the elements of β after it. The α + β is the order type of the combined mess. Definition (Ordinal addition (synthetic)). α + β is the order type of α ⊔β (α disjoint union β, e.g. α × {0} ∪β × {1}), with all α before all of β α + β = α β Example. ω + 1 = ω+. 1 + ω = ω. 23 2 Well-orderings and ordinals II Logic and Set Theory With this definition, associativity is trivial:. α + (β + γ) = α β γ = (α + β) + γ. Now that we have given two definitions, we must show that they are the same: Proposition. The inductive and synthetic definition of + coincide. Proof. Write + for inductive definition, and +′ for synthetic. We want to show that α + β = α +′ β. We induct on β. (i) α + 0 = α = α +′ 0. (ii) α + β+ = (α + β)+ = (α +′ β)+ = otp α β · = α +′ β+ (iii) α + λ = sup{α + γ : γ < λ} = sup{α +′ γ : γ < λ} = α +′ λ. This works because taking the supremum is the same as taking the union. α γ γ′ γ′′ · · · λ The synthetic definition is usually easier to work with, if possible. For example, it was very easy to show associativity using the synthetic definition. It is also easier to see why addition is not commutative. However, if we want to do induction, the inductive definition is usually easier. After addition, we can define multiplication. Again, we first give an inductive definition, and then a synthetic one. Definition (Ordinal multiplication (inductive)). We define α · β by induction on β by: (i) α · 0 = 0. (ii) α · (β+) = α · β + α. (iii) α · λ = sup{α · γ : γ < λ} for λ a non-zero limit. Example. – ω · 1 = ω · 0 + ω = 0 + ω = ω. – ω · 2 = ω · 1 + ω = ω + ω. – 2 · ω = sup{2 · n : n < ω} = ω. – ω · ω = sup{ω · n : n < ω} = sup{ω, ω2, ω3, · · · }. We also have a synthetic definition. Definition (Ordinal multiplication (synthetic)). β            α . . . α α Formally, α·β is the order type of α×β, with (x, y) < (x′, y′) if y < y′ or (y = y′ and x < x′). 24 2 Well-orderings and ordinals II Logic and Set Theory Example. ω · 2 = ω ω = ω + ω. Also 2 · ω = ω            · · . . . · · · · = ω We can check that the definitions coincide, prove associativity etc. similar to what we did for addition. We can define ordinal exponentiation, towers etc. similarly: Definition (Ordinal exponentiation (inductive)). αβ is defined as (i) α0 = 1 (ii) αβ+ = αβ · α (iii) αλ = sup{αγ : γ < λ}. Example. ω1 = ω0 · ω = 1 · ω = ω. ω2 = ω1 · ω = ω · ω. 2ω = sup{2n : n < ω} = ω. 2.6 Normal functions Note: These content were not lectured during the year. When we have ordinals, we would like to consider functions On →On. Since the ordinals are totally ordered, it would make sense to consider the order-preserving functions, i.e. the increasing ones. However, ordinals have an additional property — we could take suprema of ordinals. If we want our function to preserve this as well, we are lead to the following definition: Definition (Normal function). A function f : On →On is normal if (i) For any ordinal α, we have f(α) < f(α+). (ii) If λ is a non-zero limit ordinal, then f(λ) = sup{f(γ) : γ < λ}. Some replace the increasing condition by f(α) < f(α+). These are easily seen to be equivalent by transfinite induction. Example. By definition, we see that for each β > 1, the function α 7→βα is normal. We start by a few technical lemmas. Lemma. Let f be a normal function. Then f is strictly increasing. Proof. Let α be a fixed ordinal. We induct on all β > α that f(α) < f(β). If β = α+, then the result is obvious. If β = γ+ with γ ̸= α, then α < γ. So f(α) < f(γ) < f(γ+) = f(β) by induction. If β is a limit and is greater than α, then f(β) = sup{f(γ) : γ < β} ≥f(α+) > f(α), since α+ < β. So the result follows. 25 2 Well-orderings and ordinals II Logic and Set Theory Lemma. Let f be a normal function, and α an ordinal. Then f(α) ≥α. Proof. We prove by induction. It is trivial for zero. For successors, we have f(α+) > f(α) ≥α, so f(α+) ≥α+. For limits, we have f(λ) = sup{f(γ) : γ < λ} ≥sup{γ : γ < λ} = λ. The following is a convenient refinement of the continuity result: Lemma. If f is a normal function, then for any non-empty set {αi}i∈I, we have f(sup{αi : i ∈I}) = sup{f(αi) : i ∈I}. Proof. If {αi} has a maximal element, then the result is obvious, as f is increasing, and the supremum is a maximum. Otherwise, let α = sup{αi : i ∈I} Since the αi has no maximal element, we know α must be a limit ordinal. So we have f(α) = sup{f(β) : β < α}. So it suffices to prove that sup{f(β) : β < α} = sup{f(αi) : i ∈I}. Since all αi < α, we have sup{f(β) : β < α} ≥sup{f(αi) : i ∈I}. For the other direction, it suffices, by definition, to show that f(β) ≤sup{f(αγ) : i ∈I} for all β < α. Given such a β, since α is the supremum of the αi, we can find some particular αi such that β < αi. So f(β) < f(αi) ≤sup{f(αi) : i ∈I}. So we are done. Because of these results, some define normal functions to be functions that are strictly increasing and preserve all suprema. We now proceed to prove two important properties of normal functions (with easy proofs!): Lemma (Fixed-point lemma). Let f be a normal function. Then for each ordinal α, there is some β ≥α such that f(β) = β. Since the supremum of fixed points is also a fixed point (by normality), it follows that we can define a function g : On →On that enumerates the fixed points. Now this function itself is again normal, so it has fixed points as well. . . Proof. We thus define β = sup{f(α), f(f(α)), f(f(f(α))), · · · }. If the sequence eventually stops, then we have found a fixed point. Otherwise, β is a limit ordinal, and thus normality gives f(β) = sup{f(f(α)), f(f(f(α))), f(f(f(f(α)))), · · · } = β. So β is a fixed point, and β ≥f(α) ≥α. 26 2 Well-orderings and ordinals II Logic and Set Theory Lemma (Division algorithm for normal functions). Let f be a normal function. Then for all α, there is some maximal γ such that α ≥f(γ). Proof. Let γ = sup{β : f(β) ≤α}. Then we have f(γ) = sup{f(β) : f(b) ≤α} ≤α. This is clearly maximal. 27 3 Posets and Zorn’s lemma II Logic and Set Theory 3 Posets and Zorn’s lemma In this chapter, we study partial orders. While there are many examples of partial orders, the most important example is the power set P(X) for any set X, ordered under inclusion. We will also consider subsets of the power set. The two main theorems of this chapter are Knaster-Tarski fixed point theorem and Zorn’s lemma. We will use Zorn’s lemma to prove a lot of useful results in different fields, including the completeness theorem in propositional calculus. Finally, we will investigate the relationship between Zorn’s lemma and Axiom of Choice. 3.1 Partial orders Definition (Partial ordering (poset)). A partially ordered set or poset is a pair (X, ≤), where X is a set and ≤is a relation on X that satisfies (i) x ≤x for all x ∈X (reflexivity) (ii) x ≤y and y ≤z ⇒x ≤z (transitivity) (iii) x ≤y and y ≤x ⇒x = y (antisymmetry) We write x < y to mean x ≤y and x ̸= y. We can also define posets in terms of <: (i) x ̸< x for all x ∈X (irreflexive) (ii) x < y and y < z ⇒x < z (transitive) Example. (i) Any total order is (trivially) a partial order. (ii) N with “x ≤y” if x | y is a partial order. (iii) P(S) with ⊆for any set S is a partial order. (iv) Any subset of P(S) with inclusion is a partial order. (v) We can use a diagram a b c d e Where “above” means “greater”. So a ≤b ≤c, a ≤d ≤e, and what follows by transitivity. This is a Hasse diagram. 28 3 Posets and Zorn’s lemma II Logic and Set Theory Definition (Hasse diagram). A Hasse diagram for a poset X consists of a drawing of the points of X in the plane with an upwards line from x to y if y covers x: Definition (Cover). In a poset, y covers x if y > x and no z has y > z > x. Hasse diagrams can be useful — e.g. N, or useless, e.g. Q. (vi) The following example shows that we cannot assign “heights” or “ranks” to posets: a d b e c (vii) We can also have complicated structures: a b c d e (viii) Or the empty poset (let X be any set and nothing is less than anything else). While there are many examples of posets, all we care about are actually power sets and their subsets only. Often, we want to study subsets of posets. For example, we might want to know if a subset has a least element. All subsets are equal, but some subsets are more equal than others. A particular interesting class of subsets is a chain. Definition (Chain and antichain). In a poset, a subset S is a chain if it is totally ordered, i.e. for all x, y, x ≤y or y ≤x. An antichain is a subset in which no two things are related. Example. In (N, |), 1, 2, 4, 8, 16, · · · is a chain. In (v), {a, b, c} or {a, c} are chains. R is a chain in R. Definition (Upper bound and supremum). For S ⊂X, an upper bound for S is an x ∈X such that ∀y ∈S : x ≥y. x ∈X is a least upper bound, supremum or join of S, written x = sup S or x = W S, if x is an upper bound for S, and for all y ∈X, if y is an upper bound, then y ≥x. 29 3 Posets and Zorn’s lemma II Logic and Set Theory Example. (i) In R, {x : x < √ 2} has an upper bound 7, and has a supremum √ 2. (ii) In (v) above, consider {a, b}. Upper bounds are b and c. So sup = b. However, {b, d} has no upper bound! (iii) In (vii), {a, b} has upper bounds c, d, e, but has no least upper bound. Definition (Complete poset). A poset X is complete if every S ⊆X has a supremum. In particular, it has a greatest element (i.e. x such that ∀y : x ≥y), namely sup X, and least element (i.e. x such that ∀y : x ≤y), namely sup ∅. It is very important to remember that this definition does not require that the subset S is bounded above or non-empty. This is different from the definition of metric space completeness. Example. – R is not complete because R itself has no supremum. – [0, 1] is complete because every subset is bounded above, and so has a least upper bound. Also, ∅has a supremum of 0. – (0, 1) is not complete because (0, 1) has no upper bound. – P(S) for any S is always complete, because given any {Ai : i ∈A}, where each Ai ⊆S, S Ai is its supremum. Now we are going to derive fixed-point theorems for complete posets. We start with a few definitions: Definition (Fixed point). A fixed point of a function f : X →X is an x such that f(x) = x. Definition (Order-preserving function). For a poset X, f : X →X is order-preserving of x ≤y ⇒f(x) ≤f(y). Example. – On N, x 7→x + 1 is order-preserving – On Z, x 7→x −1 is order-preserving – On (0, 1), x 7→1+x 2 is order-preserving (this function halves the distance from x to 1). – On P(S), let some fixed i ∈S. Then A 7→A ∪{i} is order-preserving. Not every order-preserving f has a fixed point (e.g. first two above). However, we have Theorem (Knaster-Tarski fixed point theorem). Let X be a complete poset, and f : X →X be a order-preserving function. Then f has a fixed point. 30 3 Posets and Zorn’s lemma II Logic and Set Theory Proof. To show that f(x) = x, we need f(x) ≤x and f(x) ≥x. Let’s not be too greedy and just want half of it: Let E = {x : x ≤f(x)}. Let s = sup E. We claim that this is a fixed point, by showing f(s) ≤s and s ≤f(s). To show s ≤f(s), we use the fact that s is the least upper bound. So if we can show that f(s) is also an upper bound, then s ≤f(s). Now let x ∈E. So x ≤s. Therefore f(x) ≤f(s) by order-preservingness. Since x ≤f(x) (by definition of E) x ≤f(x) ≤f(s). So f(s) is an upper bound. To show f(s) ≤s, we simply have to show f(s) ∈E, since s is an upper bound. But we already know s ≤f(s). By order-preservingness, f(s) ≤f(f(s)). So f(s) ∈E by definition. While this proof looks rather straightforward, we need to first establish that s ≤f(s), then use this fact to show f(s) ≤s. If we decided to show f(s) ≤s first, then we would fail! The very typical application of Knaster-Tarski is the quick, magic proof of Cantor-Shr¨ oder-Bernstein theorem. Corollary (Cantor-Schr¨ oder-Bernstein theorem). Let A, B be sets. Let f : A → B and g : B →A be injections. Then there is a bijection h : A →B. Proof. We try to partition A into P and Q, and B into R and S, such that f(P) = R and g(S) = Q. Then we let h = f on R and g−1 on Q. Q P S R f g Since S = B \ R and Q = A \ P, so we want P = A \ g(B \ f(P)) Since the function P 7→A \ g(B \ f(P)) from P(A) to P(A) is order-preserving (and P(a) is complete), the result follows. The next result we have is Zorn’s lemma. The main focus of Zorn’s lemma is on maximal elements. Definition (Maximal element). In a poset X, x ∈X is maximal if no y ∈X has y > x. Caution! Under no circumstances confuse a maximal element with a maximum element, except under confusing circumstances! A maximum element is defined as an x such that all y ∈X satisfies y ≤x. These two notions are the same in totally ordered sets, but are very different in posets. Example. In the poset 31 3 Posets and Zorn’s lemma II Logic and Set Theory a b c d e c and e are maximal. Not every poset has a maximal element, e.g. N, Q, R. In each of these, not only are they incomplete. They have chains that are not bounded above. Theorem (Zorn’s lemma). Assuming Axiom of Choice, let X be a (non-empty) poset in which every chain has an upper bound. Then it has a maximal element. Note that “non-empty” is not a strictly necessary condition, because if X is an empty poset, then the empty chain has no upper bound. So the conditions can never be satisfied. The actual proof of Zorn’s lemma is rather simple, given what we’ve had so far. We “hunt” for the maximal element. We start with x0. If it is maximal, done. If not, we find a bigger x1. If x1 is maximal, done. Otherwise, keep go on. If we never meet a maximal element, then we have an infinite chain. This has an upper bound xω. If this is maximal, done. If not, find xω+1 > xω. Keep going on. We have not yet reached a contradiction. But suppose we never meet a maximal element. If X is countable, and we can reach xω1, then we have found uncountably many elements in a countable set, which is clearly nonsense! Since the ordinals can be arbitrarily large (Hartogs’ lemma), if we never reach a maximal element, then we can get find more elements that X has. Proof. Suppose not. So for each x ∈X, we have x′ ∈X with x′ > x. We denote the-element-larger-than-x by x′. We know that each chain C has an upper bound, say u(C). Let γ = γ(X), the ordinal-larger-than-X by Hartogs’ lemma. We pick x ∈X, and define xα for α < γ recursively by – x0 = x – xα+ = x′ α – xλ = u({xα : α < λ})′ for non-zero limit λ Of course, we have to show that {xα : α < λ} is a chain. This is trivial by induction. Then α 7→xα is an injection from γ →X. Contradiction. Note that we could as well have defined xλ = u({xα : α < λ}), and we can easily prove it is still an injection. However, we are lazy and put the “prime” just to save a few lines of proof. This proof was rather easy. However, this is only because we are given ordinals, definition by recursion, and Hartogs’ lemma. Without these tools, it is rather difficult to prove Zorn’s lemma. 32 3 Posets and Zorn’s lemma II Logic and Set Theory A typical application of Zorn’s lemma is: Does every vector space have a basis? Recall that a basis of V is a subset of V that is linearly independent (no finite linear combination = 0) and spanning (ie every x ∈V is a finite linear combination from it). Example. – Let V be the space of all real polynomials. A basis is {1, x, x2, x3, · · · }. – Let V be the space of all real sequences. Let ei be the sequence with all 0 except 1 in the ith place. However, {ei} is not a basis, since 1, 1, · · · cannot be written as a finite linear combination of them. In fact, there is no countable basis (easy exercise). It turns out that there is no “explicit” basis. – Take R as a vector space over Q. A basis here, if exists, is called a Hamel basis. Using Zorn’s lemma, we can prove that the answer is positive. Theorem. Every vector space V has a basis. Proof. We go for a maximal linearly independent subset. Let X be the set of all linearly independent subsets of V , ordered by inclusion. We want to find a maximal B ∈X. Then B is a basis. Otherwise, if B does not span V , choose x ̸∈span B. Then B ∪{x} is independent, contradicting maximality. So we have to find such a maximal B. By Zorn’s lemma, we simply have to show that every chain has an upper bound. Given a chain {Ai : i ∈I} in X, a reasonable guess is to try the union. Let A = S Ai. Then A ⊆Ai for all i, by definition. So it is enough to check that A ∈X, i.e. is linearly independent. Suppose not. Say λ1x1 + · · · + λnxn = 0 for some λ1 · · · λn scalars (not all 0). Suppose x1 ∈Ai1, · · · xn ∈Ain for some i1, · · · in ∈I. Then there is some Aim that contains all Aik, since they form a finite chain. So Aim contains all xi. This contradicts the independence of Aim. Hence by Zorn’s lemma, X has a maximal element. Done. Another application is the completeness theorem for propositional logic when P, the primitives, can be uncountable. Theorem (Model existence theorem (uncountable case)). Let S ⊆L(P) for any set of primitive propositions P. Then if S is consistent, S has a model. Proof. We need a consistent ¯ S ⊆S such that ∀t ∈L, t ∈¯ S or ¬t ∈¯ S. Then we have a valuation v(t) = ( 1 t ∈¯ S 0 t ̸∈¯ S , as in our original proof for the countable case. So we seek a maximal consistent ¯ S ⊇S. If ¯ S is maximal, then if t ̸∈¯ S, then we must have ¯ S ∪{t} inconsistent, i.e. ¯ S ∪{t} ⊢⊥. By deduction theorem, this means that ¯ S ⊢¬t. By maximality, we must have ¬t ∈¯ S. So either t or ¬t is in ¯ S. 33 3 Posets and Zorn’s lemma II Logic and Set Theory Now we show that there is such a maximal ¯ S. Let X = {T ⊆L : T is consistent , T ⊇S}. Then X ̸= ∅since S ∈X. We show that any non-empty chain has an upper bound. An obvious choice is, again the union. Let {Ti : i ∈I} be a non-empty chain. Let T = S Ti. Then T ⊇Ti for all i. So to show that T is an upper bound, we have to show T ∈X. Certainly, T ⊇S, as any Ti contains S (and the chain is non-empty). So we want to show T is consistent. Suppose T ⊢⊥. So we have t1, · · · , tn ∈T with {t1, · · · , tn} ⊢⊥, since proofs are finite. Then some Tk contains all ti since Ti are nested. So Tk is inconsistent. This is a contradiction. Therefore T must be consistent. Hence by Zorn’s lemma, there is a maximal element of X. This proof is basically the same proof that every vector space has a basis! In fact, most proofs involving Zorn’s lemma are similar. 3.2 Zorn’s lemma and axiom of choice Recall that in the proof of Zorn’s, we picked x, then picked x′, then picked x′′, ad infinitum. Here we are making arbitrary choices of x′. In particular, we have made infinitely many arbitrary choices. We did the same in IA Numbers and Sets, when proving a countable union of countable sets is countable, because we chose, for each An, a listing of An, and then count them diagonally. We needed to make a choice because each An has a lot of possible listings, and we have to pick exactly one. In terms of “rules for producing sets”, we are appealing to the axiom of choice, which states that you can pick an element of each Ai whenever {Ai : i ∈I} is a family of non-empty sets. Formally, Axiom (Axiom of choice). Given any family {Ai : i ∈I} of non-empty sets, there is a choice function f : i →S Ai such that f(i) ∈Ai. This is of a different character from the other set-building rules (e.g. unions and power sets exist). The difference is that the other rules are concrete. We know exactly what A ∪B is, and there is only one possible candidate for what A ∪B might be. “Union” uniquely specifies what it produces. However, the choice function is not. {Ai : i ∈I} can have many choice functions, and the axiom of choice does not give us a solid, explicit choice function. We say the axiom of choice is non-constructive. We are not saying that it’s wrong, but it’s weird. For this reason, it is often of interest to ask “Did I use AC?” and “Do I need AC?”. (It is important to note that the Axiom of Choice is needed only to make infinite choices. It is trivially true if |I| = 1, since A ̸= ∅by definition means ∃x ∈A. We can also do it for two sets. Similarly, for |I| finite, we can do it by induction. However, in general, AC is required to make infinite choices, i.e. it cannot be deduced from the other axioms of set theory) In the proof of Zorn’s we used Choice. However, do we need it? Is it possible to prove it without Choice? The answer is it is necessary, since we can deduce AC from Zorn’s. In other words, we can write down a proof of AC from Zorn’s, using only the other set-building rules. 34 3 Posets and Zorn’s lemma II Logic and Set Theory Theorem. Zorn’s Lemma ⇔Axiom of choice. As in the past uses of Zorn’s lemma, we have a big scary choice function to produce. We know that we can do it for small cases, such as when |I| = 1. So we start with small attempts and show that the maximal attempt is what we want. Proof. We have already proved that AC ⇒Zorn. We now proved the other way round. Given a family {Ai : i ∈I} of non-empty sets. We say a partial choice function is a function f : J →S i∈I Ai (for some J ⊆I) such that f(j) ∈A for all j ∈J. Let X = {(J, f) : f is a partial choice function with domain J}. We order by extension, i.e. (J, f) ≤(J′, f ′) iff J ⊆J′ and f ′ agrees with f when both are defined. Given a chain {(Jk, fk) : k ∈K}, we have an upper bound (S Jk, S fk), ie the function obtained by combining all functions in the chain. So by Zorn’s, it has a maximal element (J, f). Suppose J ̸= I. Then pick i ∈I \ J. Then pick x ∈Ai. Set J′ = J ∪{i} and f ′ = f ∪{(i, x)}. Then this is greater than (J, f). This contradicts the maximality of (J, f). So we must have J = I, i.e. f is a full choice function. We have shown that Zorn’s lemma is equivalent to the Axiom of Choice. There is a third statement that is also equivalent to both of these: Theorem (Well-ordering theorem). Axiom of choice ⇒every set X can be well-ordered. This might be very surprising at first for, say X = R, since there is no obvious way we can well-order R. However, it is much less surprising given Hartogs’ lemma, since Hartogs’ lemma says that there is a (well-ordered) ordinal even bigger than R. So well-ordering R shouldn’t be hard. Proof. The idea is to pick an element from X and call it the first; pick another element and call it the second, and continue transfinitely until we pick everything. For each A ⊆X with A ̸= X, we let yA be an element of X \ A. Here we are using Choice to pick out yA. Define xα recursively: Having defined xβ for all β < α, if {xβ : β < α} = X, then stop. Otherwise, set xα = y{xβ:β<α}, ie pick xα to be something not yet chosen. We must stop at some time. Otherwise, we have injected γ(X) (ie the ordinal larger than X) into X, which is a contradiction. So when stop, we have bijected X with an well-ordered set (i.e. Iα, where α is when you’ve stopped). Hence we have well-ordered X. Did we need AC? Yes, trivially. Theorem. Well-ordering theorem ⇒Axiom of Choice. Proof. Given non-empty sets {Ai : i ∈I}, well-order S Ai. Then define f(i) to be the least element of Ai. Our conclusion is: Axiom of Choice ⇔Zorn’s lemma ⇔Well-ordering theorem. 35 3 Posets and Zorn’s lemma II Logic and Set Theory Before we end, we need to do a small sanity check: we showed that these three statements are equivalents using a lot of ordinal theory. Our proofs above make sense only if we did not use AC when building our ordinal theory. Fortunately, we did not, apart from the remark that ω1 is not a countable supremum — which used the fact that a countable union of countable sets is countable. 3.3 Bourbaki-Witt theorem Finally, we’ll quickly present a second (non-examinable) fixed-point theorem. This time, we are concerned about chain-complete posets and inflationary functions. Definition (Chain-complete poset). We say a poset X is chain-complete if X ̸= ∅and every non-empty chain has a supremum. Example. – Every complete poset is chain-complete. – Any finite (non-empty) poset is chain complete, since every chain is finite and has a greatest element. – {A ⊆V : A is linearly independent} for any vector space V is chain-complete, as shown in the proof that every vector space has a basis. Definition (Inflationary function). A function f : X →X is inflationary if f(x) ≥x for all x. Theorem (Bourbaki-Witt theorem). If X is chain-complete and f : X →X is inflationary, then f has a fixed point. This is follows instantly from Zorn’s, since X has a maximal element x, and since f(x) ≥x, we must have f(x) = x. However, we can prove Bourbaki-Witt without choice. In the proof of Zorn’s, we had to “pick” and x1 > x0. Here, we can simply let x0 f 7− →x1 f 7− →x2 f 7− →x3 · · · Since each chain has a supremum instead of an upper bound, we also don’t need Choice to pick our favorite upper bound of each chain. Then we can do the same proof as Zorn’s to find a fixed point. We can view this as the “AC-free” part of Zorn’s. It can be used to prove Zorn’s lemma, but the proof is totally magic. 36 4 Predicate logic II Logic and Set Theory 4 Predicate logic In the first chapter, we studied propositional logic. However, it isn’t sufficient for most mathematics we do. In, say, group theory, we have a set of objects, operations and constants. For example, in group theory, we have the operations multiplication m : A2 →A, inverse i : A →A, and a constant e ∈A. For each of these, we assign a number known as the arity, which specifies how many inputs each operation takes. For example, multiplication has arity 2, inverse has arity 1 and e has arity 0 (we can view e as a function A0 →A, that takes no inputs and gives a single output). The study of these objects is known as predicate logic. Compared to proposi-tional logic, we have a much richer language, which includes all the operations and possibly relations. For example, with group theory, we have m, i, e in our language, as well as things like ∀, ⇒etc. Note that unlike propositional logic, different theories give rise to different languages. Instead of a valuation, now we have a structure, which is a solid object plus the operations and relations required. For example, a structure of group theory will be an actual concrete group with the group operations. Similar to what we did in propositional logic, we will take S | = t to mean “for any structure in which S is true, t is true”. For example, “Axioms of group theory” | = m(e, e) = e, i.e. in any set that satisfies the group axioms, m(e, e) = e. We also have S ⊢t meaning we can prove t from S. 4.1 Language of predicate logic We start with the definition of the language. This is substantially more compli-cated than what we’ve got in propositional logic. Definition (Language). Let Ω(function symbols) and Π (relation symbols) be disjoint sets, and α : Ω∪Π →N a function (’arity’). The language L = L(Ω, Π, α) is the set of formulae, defined as follows: – Variables: we have some variables x1, x2, · · · . Sometimes (i.e. always), we write x, y, z, · · · instead. – Terms: these are defined inductively by (i) Every variable is a term (ii) If f ∈Ω, α(f) = n, and t1, · · · , tn are terms, then ft1 · · · tn is a term. We often write f(t1, · · · , tn) instead. Example. In the language of groups Ω= {m, i, e}, Π = ∅, and α(m) = 2, α(i) = 1, α(e) = 0. Then e, x1, m(x1, x2), i(m(x1, x1)) are terms. – Atomic formulae: there are three sorts: (i) ⊥ (ii) (s = t) for any terms s, t. (iii) (ϕt1 · · · tn) for any ϕ ∈Π with α(ϕ) = n and t1, · · · , tn terms. Example. In the language of posets, Ω= ∅, Π = {≤} and α(≤) = 2. Then (x1 = x1), x1 ≤x2 (really means (≤x1x2)) are atomic formulae. 37 4 Predicate logic II Logic and Set Theory – Formulae: defined inductively by (i) Atomic formulae are formulae (ii) (p ⇒q) is a formula for any formulae p, q. (iii) (∀x)p is a formula for any formula p and variable x. Example. In the language of groups, e = e, x1 = e, m(e, e) = e, (∀x)m(x, i(x)) = e, (∀x)(m(x, x) = e ⇒(∃y)(m(y, y) = x)) are formu-lae. It is important to note that a formula is a string of meaningless symbol. It doesn’t make sense to ask whether it is true or false. In particular, the function and relation symbols are not assigned any meaning. The only thing the language cares is the arity of the symbol. Again, we have the usual abbreviations ¬p, p ∧q, p ∨q etc. Also, we have (∃x)p for ¬(∀x)(¬p). Definition (Closed term). A term is closed if it has no variables. Example. In the language of groups, e, m(e, e) are closed terms. However, m(x, i(x)) is not closed even though we think it is always e. Apart from the fact that it is by definition not closed (it has a variable x), we do not have the groups axioms stating that m(x, i(x)) = e. Definition (Free and bound variables). An occurrence of a variable x in a formula p is bound if it is inside brackets of a (∀x) quantifier. It is free otherwise. Example. In (∀x)(m(x, x) = e), x is a bound variable. In (∀y)(m(y, y) = x ⇒m(x, y) = m(y, x)), y is bound while x is free. We are technically allowed to have a formula with x both bound and free, but DO NOT DO IT. For example, m(x, x) = e ⇒(∀x)(∀y)(m(x, y) = m(y, x)) is a valid formula (first two x are free, while the others are bound). Definition (Sentence). A sentence is a formula with no free variables. Example. m(e, e) = e and (∀x)(m(x, x) = e) are sentences, while m(x, i(x)) = e is not. Definition (Substitution). For a formula p, a variable x and a term t, the substitution p[t/x] is obtained by replacing each free occurrence of x with t. Example. If p is the statement (∃y)(m(y, y) = x), then p[e/x] is (∃y)(m(y, y) = e. 4.2 Semantic entailment In propositional logic, we can’t say whether a proposition is true or false unless we have a valuation. What would be a “valuation” in the case of predicate logic? It is a set with the operations of the right arity. We call this a structure. For example, in the language of groups, a structure is a set G with the correct operations. Note that it does not have to satisfy the group axioms in order to qualify as a structure. 38 4 Predicate logic II Logic and Set Theory Definition (Structure). An L-structure is a non-empty set A with a function fA : An →A for each f ∈Ω, α(f) = n, and a relation ϕA ⊆An, for each ϕ ∈Π, α(ϕ) = n. Note that we explicitly forbid A from being empty. It is possible to formulate predicate logic that allows empty structures, but we will have to make many exceptions when defining our axioms, as we will see later. Since empty structures are mostly uninteresting (and don’t exist if there is at least one constant), it isn’t a huge problem if we ignore it. (There is a small caveat here — we are working with single-sorted logic here, so everything is of the same “type”. If we want to work with multi-sorted logic, where there can be things of different types, it would then be interesting to consider the case where some of the types could be empty). Example. In the language of posets A, a structure is a set A with a relation ≤A⊆A × A. In the language of groups, a structure is a set A with functions mA : A×A → A, iA : A →A and eA ∈A. Again, these need not be genuine posets/groups since we do not have the axioms yet. Now we want to define “p holds in A” for a sentence p ∈L and a L-structure A. For example, we want (∀x)(m(x, x) = e) to be true in A iff for each a ∈A, we have mA(a, a) = eA. So to translate p into something about A, you “add subscript A to each function-symbol and relation-symbol, insert ∈A after the quantifiers, and say it aloud”. We call this the interpretation of the sentence p. This is not great as a definition. So we define it formally, and then quickly forget about it. Definition (Interpretation). To define the interpretation pA ∈0, 1 for each sentence p and L-structure A, we define inductively: (i) Closed terms: define tA ∈A for each closed term t by (ft1, · · · , tn)A = fA(t1A, t2A · · · , tnA) for any f ∈Ω, α(f) = n, and closed terms t1, · · · , tn. Example. (m(m(e, e), e))A = mA(mA(eA, eA), eA). (ii) Atomic formulae: ⊥A = 0 (s = t)A = ( 1 sA = tA 0 sA ̸= tA (ϕt1 · · · tn)A = ( 1 (t1A, · · · , tnA) ∈ϕA 0 otherwise . 39 4 Predicate logic II Logic and Set Theory (iii) Sentences: (p ⇒q)A = ( 0 pA = 1, qA = 0 1 otherwise ((∀x)p)A = ( 1 p[¯ a/x] ¯ A for all a ∈A 0 otherwise where for any a ∈A, we define a new language L′ by adding a constant ¯ a and make A into an L′ structure ¯ A by setting ¯ a ¯ A = a. Now that we have formally defined truth, just forget about it! Note that we have only defined the interpretation only for sentences. We can also define it for functions with free variables. For any formula p with n free variables, we can define the interpretation as the set of all things that satisfy p. For example, if p is (∃y)(m(y, y) = a), then pA = {a ∈A : ∃b ∈A such that m(b, b) = a}. However, we are mostly interested in sentences, and don’t have to worry about these. Now we can define models and entailment as in propositional logic. Definition (Theory). A theory is a set of sentences. Definition (Model). If a sentence p has pA = 1, we say that p holds in A, or p is true in A, or A is a model of p. For a theory S, a model of S is a structure that is a model for each s ∈S. Definition (Semantic entailment). For a theory S and a sentence t, S entails t, written as S | = t, if every model of S is a model of t. “Whenever S is true, t is also true”. Definition (Tautology). t is a tautology, written | = t, if ∅| = t, i.e. it is true everywhere. Example. (∀x)(x = x) is a tautology. Example. (i) Groups: – The language L is Ω= (m, i, e) and Π = ∅, with arities 2, 1, 0 respectively. – Let T be {(∀x)(∀y)(∀z)m(x, m(y, z)) = m(m(x, y), z), (∀x)(m(x, e) = x ∧m(e, x) = x), (∀x)(m(x, i(x)) = e ∧m(i(x), x) = e)}. Then an L-structure A is a model for T iff A is a group. We say T axiomatizes the theory of groups/class of groups. Sometimes we call the members of T the axioms of T. Note that we could use a different language and theory to axiomatize group theory. For example, we can have Ω= (m, e) and change the last axiom to (∀x)(∃y)(m(x, y) = e ∧m(y, x) = e)}. 40 4 Predicate logic II Logic and Set Theory (ii) Fields: – The language L is Ω= (+, ×, −, 0, 1) and Π = ∅, with arities 2, 2, 1, 0, 0. – The theory T consists of: ◦Abelian group under + ◦× is commutative, associative, and distributes over + ◦¬(0 = 1) ◦(∀x)((¬(x = 0)) ⇒(∃y)(y × x = 1). Then an L-structure is a model of T iff it is a field. Then we have T | = ”inverses are unique”, i.e. T | = (∀x) ¬(x = 0)  ⇒(∀y)(∀z) (xy = 1 ∧xz = 1) ⇒(y = z)  (iii) Posets: – The language is Ω= ∅, and Π = {≤} with arity 2. – The theory T is {(∀x)(x ≤x), (∀x)(∀y)(∀z) (x ≤y) ∧(y ≤z) ⇒x ≤z  , (∀x)(∀y) x ≤y ∧y ≤z ⇒x = y  } Then T axiomatizes the theory of posets. (iv) Graphs: – The language L is Ω= ∅and Π = {a} with arity 2. This relation is “adjacent to”. So a(x, y) means there is an edge between x and y. – The theory is {(∀x)(¬a(x, x)), (∀x)(∀y)(a(x, y) ⇔a(y, x)} Predicate logic is also called “first-order logic”. It is “first-order” because our quantifiers range over elements of the structure only, and not subsets. It would be difficult (and in fact impossible) to axiomatize, say, a complete ordered field, since the definition requires says every bounded subset has a least upper bound. 4.3 Syntactic implication Again, to define syntactic implication, we need axioms and deduction rules. Definition (Axioms of predicate logic). The axioms of predicate logic consists of the 3 usual axioms, 2 to explain how = works, and 2 to explain how ∀works. They are 1. p ⇒(q ⇒p) for any formulae p, q. 2. [p ⇒(q ⇒r)] ⇒[(p ⇒q) ⇒(p ⇒r)] for any formulae p, q, r. 41 4 Predicate logic II Logic and Set Theory 3. (¬¬p ⇒p) for any formula p. 4. (∀x)(x = x) for any variable x. 5. (∀x)(∀y) (x = y) ⇒(p ⇒p[y/x])  for any variable x, y and formula p, with y not occurring bound in p. 6. [(∀x)p] ⇒p[t/x] for any formula p, variable x, term t with no free variable of t occurring bound in p. 7. [(∀x)(p ⇒q)] ⇒[p ⇒(∀x)q] for any formulae p, q with variable x not occurring free in p. The deduction rules are 1. Modus ponens: From p and p ⇒q, we can deduce q. 2. Generalization: From r, we can deduce (∀x)r, provided that no premise used in the proof so far had x as a free variable. Again, we can define proofs, theorems etc. Definition (Proof). A proof of p from S is a sequence of statements, in which each statement is either an axiom, a statement in S, or obtained via modus ponens or generalization. Definition (Syntactic implication). If there exists a proof a formula p form a set of formulae S, we write S ⊢p “S proves t”. Definition (Theorem). If S ⊢p, we say p is a theorem of S. (e.g. a theorem of group theory) Note that these definitions are exactly the same as those we had in propo-sitional logic. The only thing that changed is the set of axioms and deduction rules. Example. {x = y, x = z} ⊢y = z. We go for x = z giving y = z using Axiom 5. 1. (∀x)(∀y)((x = y) ⇒(x = z ⇒y = z)) Axiom 5 2. [(∀x)(∀y)((x = y) ⇒(x = z ⇒y = z))] ⇒(∀y)(x = y ⇒y = z) Axiom 6 3. (∀y)((x = y) ⇒x = z ⇒y = z) MP on 1, 2 4. [(∀y)((x = y) ⇒x = z ⇒y = z)] ⇒[(x = y) ⇒(x = z ⇒y = z) Axiom 6 5. (x = y) ⇒[(x = z) ⇒(y = z)] MP on 3, 4 6. x = y Premise 7. (x = z) ⇒(y = z) MP 6, 7 8. x = z Premise 9. y = z MP on 7, 8 42 4 Predicate logic II Logic and Set Theory Note that in the first 5 rows, we are merely doing tricks to get rid of the ∀ signs. We can now revisit why we forbid ∅from being a structure. If we allowed ∅, then (∀x)⊥holds in ∅. However, axioms 6 states that ((∀x)⊥) ⇒⊥. So we can deduce ⊥in the empty structure! To fix this, we will have to add some weird clauses to our axioms, or simply forbid the empty structure! Now we will prove the theorems we had for propositional logic. Proposition (Deduction theorem). Let S ⊆L, and p, q ∈L. Then S ∪{p} ⊢q if and only if S ⊢p ⇒q. Proof. The proof is exactly the same as the one for propositional logic, except in the ⇒case, we have to check Gen. Suppose we have lines – r – (∀x)r Gen and we have a proof of S ⊢p ⇒r (by induction). We want to seek a proof of p ⇒(∀x)r from S. We know that no premise used in the proof of r from S ∪{p} had x as a free variable, as required by the conditions of the use of Gen. Hence no premise used in the proof of p ⇒r from S had x as a free variable. Hence S ⊢(∀x)(p ⇒r). If x is not free in p, then we get S ⊢p ⇒(∀x)r by Axiom 7 (and MP). If x is free in p, then we did not use premise p in our proof r from S ∪{p} (by the conditions of the use of Gen). So S ⊢r, and hence S ⊢(∀x)r by Gen. So S ⊢p ⇒(∀x)r. Now we want to show S ⊢p iff S | = p. For example, a sentence that holds in all groups should be deducible from the axioms of group theory. Proposition (Soundness theorem). Let S be a set of sentences, p a sentence. Then S ⊢p implies S | = p. Proof. (non-examinable) We have a proof of p from S, and want to show that for every model of S, p holds. This is an easy induction on the lines of the proof, since our axioms are tautologies and our rules of deduction are sane. The hard part is proving S | = p ⇒S ⊢p. This is, by the deduction theorem, S ∪{¬p} | = ⊥⇒S ∪{¬p} ⊢⊥. This is equivalent to the contrapositive: S ∪{¬p} ̸⊢⊥⇒S ∪{¬p} ̸| = ⊥. 43 4 Predicate logic II Logic and Set Theory Theorem (Model existence lemma). Let S be a consistent set of sentences. Then S has a model. We need several ideas to prove the lemma: (i) We need to find a structure. Where can we start from? The only thing we have is the language. So we start form the language. Let A = set of all closed terms, with the obvious operations. For example, in the theory of fields, we have “1 + 1”, “0 + 1“ etc in the structure. Then (1 + 1) +A (0 + 1) = (1 + 1) + (0 + 1). (ii) However, we have a problem. In, say, the language of fields, and S our field axioms, our A has distinct elements “1 + 0”, “0 + 1”, “0 + 1 + 0” etc. However, S ⊢1 + 0 = 0 + 1 etc. So we can’t have them as distinct elements. The solution is to quotient out by the equivalence relation s ∼t if S ⊢(s = t), i.e. our structure is the set of equivalence classes. It is trivial check to check that the +, × operations are well-defined for the equivalence classes. (iii) We have the next problem: If S is ”field of characteristic 2 or 3“, i.e. S has a field axiom plus 1 + 1 = 0 ∨1 + 1 + 1 = 0. Then S ̸⊢1 + 1 = 0. Also S ̸⊢1 + 1 + 1 = 0. So [1 + 1] ̸= , and [1 + 1 + 1] ̸= . But then our S has neither characteristic 2 or 3. This is similar to the problem we had in the propositional logic case, where we didn’t know what to do with p3 if S only talks about p1 and p2. So we first extend S to a maximal consistent (or complete) ¯ S. (iv) Next problem: Let S = “fields with a square root of 2”, i.e. S is the field axioms plus (∃x)(xx = 1 + 1). However, there is no closed term t which is equivalent to √ 2. We say we lack witnesses to the statement (∃x)(xx = 1 + 1). So we add a witness. We add a constant c to the language, and add the axiom “cc = 1 + 1” to S. We do this for each such existential statement. (v) Now what? We have added new symbols to S, so our new S is no longer complete! Of course, we go back to (iii), and take the completion again. Then we have new existential statements to take care of, and we do (iv) again. Then we’re back to (iii) again! It won’t terminate! So we keep on going, and finally take the union of all stages. Proof. (non-examinable) Suppose we have a consistent S in the language L = L(Ω, Π). Extend S to a consistent S1 such that p ∈S1 or (¬p) ∈S for each sentence p ∈L (by applying Zorn’s lemma to get a maximal consistent S1). In particular, S1 is complete, meaning S1 ⊢p or S1 ⊢¬p for all p. Then for each sentence of the form (∃x)p in S1, add a new constant c to L and add p[c/x] to S1 — obtaining T1 in language L1 = L(Ω∪C1, Π). It is easy to check that T1 is consistent. Extend T1 to a complete theory S2 ⊆L1, and add witnesses to form T2 ⊆ L2 = L(Ω∪C1 ∪C2, Π). Continue inductively. Let ¯ S = S1 ∪S2 ∪· · · in language ¯ L = L1 ∪L2 ∪· · · (i.e. ¯ L = L(Ω∪C1 ∪ C2 ∪· · · , Π)). 44 4 Predicate logic II Logic and Set Theory Claim. ¯ S is consistent, complete, and has witnesses, i.e. if (∃x)p ∈¯ S, then p[t/x] ∈¯ S For some term t. It is consistent since if ¯ S ⊢⊥, then some Sn ⊢⊥since proofs are finite. But all Sn are consistent. So ¯ S is consistent. To show completeness, for sentence p ∈¯ L, we have p ∈Ln for some n, as p has only finitely many symbols. So Sn+1 ⊢p or Sn+1 ⊢¬p. Hence ¯ S ⊢p or ¯ S ⊢¬p. To show existence of witnesses, if (∃x)p ∈¯ S, then (∃x)p ∈Sn for some n. Hence (by construction of Tn), we have p[c/x] ∈Tn for some constant c. Now define an equivalence relation ∼on closed term of ¯ L by s ∼t if ¯ S ⊢(s = t). It is easy to check that this is indeed an equivalence relation. Let A be the set of equivalence classes. Define (i) fA([t1], · · · , [tn]) = [ft1, · · · , tn] for each formula f ∈Ω, α(f) = n. (ii) ϕA = {([t1], · · · , [tn]) : ¯ S ⊢ϕ(t1, · · · , tn)} for each relation ϕ ∈Π and α(ϕ) = n. It is easy to check that this is well-defined. Claim. For each sentence p, ¯ S ⊢p (i.e. p ∈¯ S) if and only if p holds in A, i.e. pA = 1. We prove this by an easy induction. – Atomic sentences: ◦⊥: ¯ S ̸⊢⊥, and ⊥A = 0. So good. ◦s = t: ¯ S ⊢s = t iff [s] = [t] (by definition) iff sA = tA (by definition of sA) iff (s = t)A. So done. ◦ϕt1, · · · , tn is the same. – Induction step: ◦p ⇒q: ¯ S ⊢(p ⇒q) iff ¯ S ⊢(¬p) or ¯ S ⊢q (justification: if ¯ S ̸⊢¬p and ¯ S ̸⊢q, then ¯ S ⊢p and ¯ S ⊢¬q by completeness, hence ¯ S ⊢¬(p ⇒q), contradiction). This is true iff pA = 0 or qA = 1 iff (p ⇒q)A = 1. ◦(∃x)p: ¯ S ⊢(∃x)p iff ¯ S ⊢p[t/x] for some closed term t. This is true since ¯ S has witnesses. Now this holds iff p[t/x]A = 1 for some closed term t (by induction). This is the same as saying (∃x)p holds in A, because A is the set of (equivalence classes of) closed terms. Here it is convenient to pretend ∃is the primitive symbol instead of ∀. Then we can define (∀x)p to be ¬(∃x)¬p, instead of the other way round. It is clear that the two approaches are equivalent, but using ∃as primitive makes the proof look clearer here. Hence A is a model of ¯ S. Hence it is also a model of S. So S has a model. Again, if L is countable (i.e. Ω, Π are countable), then Zorn’s Lemma is not needed. From the Model Existence lemma, we obtain: 45 4 Predicate logic II Logic and Set Theory Corollary (Adequacy theorem). Let S be a theory, and p a sentence. Then S | = p implies S ⊢p. Theorem (G¨ odel’s completeness theorem (for first order logic)). Let S be a theory, p a sentence. Then S ⊢p if and only if S | = p. Proof. (⇒) Soundness, (⇐) Adequacy. Corollary (Compactness theorem). Let S be a theory such that every finite subset of S has a model. Then so does S. Proof. Trivial if we replace “has a model” with “is consistent”, because proofs are finite. We can look at some applications of this: Can we axiomatize the theory of finite groups (in the language of groups)? i.e. is there a set of sentences T such that models of T are finite groups. Corollary. The theory of finite groups cannot be axiomatized (in the language of groups). It is extraordinary that we can prove this, as opposed to just “believing it should be true”. Proof. Suppose theory T has models all finite groups and nothing else. Let T ′ be T together with – (∃x1)(∃x2)(x1 ̸= x2) (intuitively, |G| ≥2) – (∃x1)(∃x2)(∃x3)(x1 ̸= x2 ̸= x3) (intuitively, |G| ≥3) – · · · Then T ′ has no model, since each model has to be simultaneously arbitrarily large and finite. But every finite subset of T ′ does have a model (e.g. Zn for some n). Contradiction. This proof looks rather simple, but it is not “easy” in any sense. We are using the full power of completeness (via compactness), and this is not easy to prove! Corollary. Let S be a theory with arbitrarily large models. Then S has an infinite model. “Finiteness is not a first-order property” Proof. Same as above. Similarly, we have Corollary (Upward L¨ owenheim-Skolem theorem). Let S be a theory with an infinite model. Then S has an uncountable model. Proof. Add constants {ci : i ∈I} to L for some uncountable I. Let T = S S{“ci ̸= cj” : i, j ∈I, i ̸= j}. Then any finite T ′ ⊆T has a model, since it can only mention finitely many of the Ci. So any infinite model of S will do. Hence by compactness, T has a model 46 4 Predicate logic II Logic and Set Theory Similarly, we have a model for S that does not inject into X, for any chosen set X. For example, we can add γ(X) constants, or P(X) constants. Example. There exists an infinite field (Q). So there exists an uncountable field (e.g. R). Also, there is a field that does not inject into P(P(R)), say, Theorem (Downward L¨ owenheim-Skolem theorem). Let L be a countable language (i.e. Ωand Π are countable). Then if S has a model, then it has a countable model. Proof. The model constructed in the proof of model existence theorem is count-able. Note that the proof of the model existence theorem is non-examinable, but the proof of this is examinable! So we are supposed to magically know that the model constructed in the proof is countable without knowing what the proof itself is! 4.4 Peano Arithmetic As an example, we will make the usual axioms for N into a first-order theory. We take Ω= {0, s, +, ×}, and Π = ∅. The arities are α(0) = 0, α(s) = 1, α(+) = α(×) = 2. The operation s is the successor operation, which should be thought of as “+1”. Our axioms are Definition (Peano’s axioms). The axioms of Peano’s arithmetic (PA) are (i) (∀x)¬(s(x) = 0). (ii) (∀x)(∀y)((s(x) = s(y)) ⇒(x = y)). (iii) (∀y1) · · · (∀yn)[p[0/x] ∧(∀x)(p ⇒p[s(x)/x])] ⇒(∀x)p. This is actually infinitely many axioms — one for each formula p, free variables y1, · · · , yn, x, i.e. it is an axiom scheme. (iv) (∀x)(x + 0 = x). (v) (∀x)(∀y)(x + s(y) = s(x + y)). (vi) (∀x)(x × 0 = 0). (vii) (∀x)(∀y)(x × s(y) = (x + y) + x). Note that our third axiom looks rather funny with all the (∀yi) in front. Our first guess at writing it would be [p[0/x] ∧(∀x)(p ⇒p[s(x)/x])] ⇒(∀x)p. However, this is in fact not sufficient. Suppose we want to prove that for all x and y, x + y = y + x. The natural thing to do would be to fix a y and induct on x (or the other way round). We want to be able to fix any y to do so. So we need a (∀y) in front of our induction axiom, so that we can prove it for all values of y all at once, instead of proving it once for y = 0, once for y = 1 , once for y = 1 + 1 etc. This is important, since we might have an uncountable model of PA, and we cannot name all y. When we actually think about it, we can just forget about the (∀yi)s. But just remember that formally we need them. 47 4 Predicate logic II Logic and Set Theory We know that PA has a model N that is infinite. So it has an uncountable model by Upward L¨ owenheim-Skolem. Then clearly this model is not isomorphic to N. However, we are also told that the axioms of arithmetic characterize N completely. Why is this so? This is since Axiom 3 is not full induction, but a “first-order” version. The proper induction axiom talks about subsets of N, i.e. (∀S ⊆N)((0 ∈S ∧x ∈S ⇒s(x) ∈S) ⇒S = N). However, there are uncountably many subsets of N, and countably many formulae p. So our Axiom 3 only talks about some subsets of N, not all. Now the important question is: is PA complete? G¨ odel’s incompleteness theorem says: no! There exists some p with PA ̸⊢p and PA ̸⊢¬p. Hence, there is a p that is true in N, but PA ̸⊢p. Note that this does not contradict G¨ odel’s completeness theorem. The completeness theorem tells us that if p holds in every model of PA (not just in N), then PA ⊢p. 4.5 Completeness and categoricity We now study some “completeness” property of theories. For convenience, we will assume that the language is countable. Definition (Complete theory). A theory T is complete if for all propositions p in the language, either T ⊢p or T ⊢¬p. Complete theories are rather hard to find. So for the moment, we will content ourselves by just looking at theories that are not complete. Example. The theory of groups is not complete, since the proposition (∀g)(∀h)(gh = hg) can neither be proven or disproven (there are abelian groups and non-abelian groups). Another “completeness” notion we might be interested in is categoricity. We will need to use the notion of a cardinal, which will be introduced later in the course. Roughly, a cardinal is a mathematical object that denotes the sizes of sets, and a set “has cardinality κ” if it bijects with κ. Definition (κ-categorical). Let κ be an infinite cardinal. Then a theory T is κ-categorical if there is a unique model of the theory of cardinality κ up to isomorphism. Here the notion of “isomorphism” is the obvious one — a homomorphism of models is a function between the structures that preserves everything, and an isomorphism is a homomorphism with an inverse (that is also a homomorphism). Example (Pure identity theory). The pure identity theory has an empty language and no axioms. Then this theory is κ-categorical for any κ, since if two models have the same cardinality, then by definition there is a bijection between them, and any such bijection is an isomorphism because there are no properties to be preserve. 48 4 Predicate logic II Logic and Set Theory One nice thing about categorical theories is that they are complete! Proposition. Let T be a theory that is κ categorical for some κ, and suppose T has no finite models. Then T is complete. Note that the requirement that T has no finite models is necessary. For example, pure identity theory is κ-categorical for all κ but is not complete, since the statement (∃x)(∃y)(¬(x = y)) is neither provable nor disprovable. Proof. Let p be a proposition. Suppose T ̸⊢p and T ̸⊢¬p. Then there are infinite models of T ∪{p} and T ∪{¬p} (since the models cannot be finite), and so by the L¨ owenhein–Skolem theorems, we can find such models of cardinality κ. But since one satisfies p and the other does not, they cannot be isomorphic. This contradicts κ-categoricity. We are now going to use this idea to prove the Ax-Grothendieck theorem Theorem (Ax-Grothendieck theorem). Let f : Cn →Cn be a complex polyno-mial. If f is injective, then it is in fact a bijection. We will use the following result from field theory without proof: Lemma. Any two uncountable algebraically closed fields with the same di-mension and same characteristic are isomorphic. In other words, the theory of algebraically closed fields of characteristic p (for p a prime or 0) is κ-categorical for all uncountable cardinals κ, and in particular complete. The rough idea (for the field theorists) is that an algebraically closed field is uniquely determined by its transcendence degree over the base field Q or Fp. In the following proof, we will also use some (very) elementary results about field theory that can be found in IID Galois Theory. Proof of Ax-Grothendieck. We will use compactness and completeness to show that we only have to prove this for fields of positive characteristic, and the result can be easily proven since we end up dealing with finite fields. Let ACF be the theory of algebraically closed fields. The language is the language of rings, and the axioms are the usual axioms of a field, plus the following axiom for each n > 0: (∀a0, a1, · · · , an−1)(∃x)(xn + an−1xn−1 + · · · + a1x + a0 = 0). Let ACF0 denote the theory of algebraically closed fields of characteristic 0, where we add the axiom 1 + 1 + · · · + 1 | {z } n times ̸= 0 (∗) for all n to ACFn. Let ACFp denote the theory of algebraically closed fields of characteristic p, where we add the axiom 1 + 1 + · · · + 1 | {z } p times = 0 to ACF. We now notice the following fact: if r is a proposition that is a theorem of ACFp for all p, then it is true of ACF0. Indeed, we know that ACF0 is complete. 49 4 Predicate logic II Logic and Set Theory So if r is not a theorem in ACF0, then ¬r is a theorem. But the proof is finite, so it can only use finitely many instances of (∗). So there is some large p where ¬r can be proven in ACFp, which is a contradiction. Now the statement “If f is a polynomial of degree d and f is injective, then f is surjective” can be expressed as a first-order statement. So we just have to prove it for all fields of characteristic p > 0. Moreover, by completeness, for each p, we only need to prove it for some algebraically complete field of characteristic p. Fix a prime p, and consider F = ¯ Fp, the algebraic closure of Fp. This is an algebraically closed field with the property that every element is algebraic over Fp, i.e. the field generated by any finite subset of elements is finite. Let f : F n →F n be a polynomial function involving coefficients a1, · · · , aK. Let b = (b1, · · · , bn) ∈F n be a point. Then F restricts to a function from the field ˜ F generated by {b1, · · · , bn, a1, · · · , aK} to itself. But ˜ F is finite, so any function f| ˜ F : ˜ F →˜ F that is injective must also be surjective. So b is in the image of f. So f is surjective. So done. We conclude the section by stating, without proof, a theorem by Morely: Theorem (Morley’s categoricity theorem). Let T be a theory with a countable language. If T is κ-categorical for some uncountable cardinal κ, then it is µ-categorical for all uncountable cardinals µ. Hence we often just say a theory is uncountably categorical when the theory is categorical for some (hence all) uncountable cardinals. 50 5 Set theory II Logic and Set Theory 5 Set theory Here we’ll axiomatize set theory as “just another first-order theory”, with signatures, structures etc. There are many possible formulations, but the most common one is Zermelo Fraenkel set theory (with the axiom of choice), which is what we will study. 5.1 Axioms of set theory Definition (Zermelo-Fraenkel set theory). Zermelo-Fraenkel set theory (ZF) has language Ω= ∅, Π = {∈}, with arity 2. Then a “universe of sets” will simply mean a model of these axioms, a pair (V, ∈), where V is a set and ∈is a binary relation on V in which the axioms are true (officially, we should write ∈V , but it’s so weird that we don’t do it). Each model (V, ∈) will (hopefully) contain a copy of all of maths, and so will look very complicated! ZF will have 2 axioms to get started with, 4 to build things, and 3 more weird ones one might not realize are needed. Axiom (Axiom of extension). “If two sets have the same elements, they are the same set”. (∀x)(∀y)((∀z)(z ∈x ⇔z ∈y) ⇒x = y). We could replace the ⇒with an ⇔, but the converse statement x = y ⇒ (z ∈x ⇔z ∈y) is an instance of a logical axiom, and we don’t have to explicitly state it. Axiom (Axiom of separation). “Can form subsets of sets”. More precisely, for any set x and a formula p, we can form {z ∈x : p(z)}. (∀t1) · · · (∀tn)(∀x)(∃y)(∀z)(z ∈y ⇔(z ∈x ∧p)). This is an axiom scheme, with one instance for each formula p with free variables t1, · · · , tn, z. Note again that we have those funny (∀ti). We do need them to form, e.g. {z ∈x : t ∈z}, where t is a parameter. This is sometimes known as Axiom of comprehension. Axiom (Axiom of empty set). “The empty-set exists” (∃x)(∀y)(y ̸∈x). We write ∅for the (unique, by extension) set with no members. This is an abbreviation: p(∅) means (∃x)(x has no members ∧p(x)). Similarly, we tend to write {z ∈x : p(z)} for the set given by separation. Axiom (Axiom of pair set). “Can form {x, y}”. (∀x)(∀y)(∃z)(∀t)(t ∈z ⇔(t = x ∨t = y)). We write {x, y} for this set. We write {x} for {x, x}. We can now define ordered pairs: 51 5 Set theory II Logic and Set Theory Definition (Ordered pair). An ordered pair (x, y) is {{x}, {x, y}}. We define “x is an ordered pair” to mean (∃y)(∃z)(x = (y, z)). We can show that (a, b) = (c, d) ⇔(a = c) ∧(b = d). Definition (Function). We define “f is a function” to mean (∀x)(x ∈f ⇒x is an ordered pair)∧ (∀x)(∀y)(∀z)[(x, y) ∈f ∧(x, z) ∈f] ⇒y = z. We define x = dom f to mean f is a function and (∀y)(y ∈x ⇔(∃z)((y, z) ∈f)). We define f : x →y to mean f is a function and dom f = x and (∀z)[(∃t)((t, z) ∈f) ⇒z ∈y]. Once we’ve defined them formally, let’s totally forget about the definition and move on with life. Axiom (Axiom of union). “We can form unions” Intuitively, we have a ∪b ∪c = {x : x ∈a or x ∈b or x ∈c}. but instead of a ∪b ∪c, we write S{a, b, c} so that we can express infinite unions as well. (∀x)(∃y)(∀z)(z ∈y ⇔(∃t)(t ∈x ∧z ∈t)). We tend to write S x for the set given above. We also write x ∪y for S{x, y}. Note that we can define intersection T x as a subset of y (for any y ∈x) by separation, so we don’t need an axiom for that. Axiom (Axiom of power set). “Can form power sets”. (∀x)(∃y)(∀z)(z ∈y ⇔z ⊆x), where z ⊆x means (∀t)(t ∈z ⇒t ∈x). We tend to write P(x) for the set generated above. We can now form x × y, as a subset of P(P(x ∪y)), because for t ∈x, s ∈y, we have (t, s) = {{t}, {t, s}} ∈P(P(x ∪y)). Similarly, we can form the set of all functions from x to y as a subset of P(x × y) (or P(P(P(x ∪y)))!). Now we’ve got the easy axioms. Time for the weird ones. Axiom of infinity From the axioms so far, we cannot prove that there is an infinite set! We start from the empty set, and all the operations above can only produce finite sets. So we need an axiom to say that the natural numbers exists. But how could we do so in finitely many words? So far, we do have infinitely many sets. For example, if we write x+ for x ∪{x}, it is easy to check that , ∅+, ∅++, · · · are all distinct. (Writing them out explicitly, we have ∅+ = {∅}, ∅++ = {∅, {∅}}, ∅+++ = {∅, {∅}, {∅, {∅}}}. We can also write 0 for ∅, 1 for ∅+, 2 for ∅++. So 0 = ∅, 1 = {0}, 2 = {0, 1}, · · · ) 52 5 Set theory II Logic and Set Theory Note that all of these are individually sets, and so V is definitely infinite. However, inside V , V is not a set, or else we can apply separation to obtain Russell’s paradox. So the collection of all 0, 1, · · · , need not be a set. Therefore we want an axiom that declares this is a set. So we want an axiom stating ∃x such that ∅∈x, ∅+ ∈x, ∅++ ∈x, · · · . But this is an infinite statement, and we need a smarter way of formulating this. Axiom (Axiom of infinity). “There is an infinite set”. (∃x)(∅∈x ∧(∀y)(y ∈x ⇒y+ ∈x)). We say any set that satisfies the above axiom is a successor set. A successor set can contain a lot of nonsense elements. How do we want to obtain N, without nonsense? We know that the intersection of successors is also a successor set. So there is a least successor set, i.e. the intersection of all successor sets. Call this ω. This will be our copy of N in V . So (∀x)(x ∈ω ⇔(∀y)(y is a successor set ⇒x ∈y)). Therefore, if we have (∀x)[(x ⊆ω ∧x is a successor set) ⇒x = ω], by definition of ω. By the definition of “is a successor set”, we can write this as (∀x)[(x ⊆ω ∧∅∈x ∧(∀y)(y ∈x ⇒y+ ∈x)) ⇒x = ω]. This is genuine induction (in V )! It is not just our weak first-order axiom in Peano’s axioms. Also, we have (∀x)(x ∈ω ⇒x+ ̸= ∅) and (∀x)(∀y)((x ∈ω ∧y ∈ω ∧x+ = y+) ⇒x = y) by ω-induction (i.e. induction on ω). Hence ω satisfies the axioms of natural numbers. We can now write, e.g. “x is finite” for (∃y)(∃f)(y ∈x ∧f bijects x with y). Similarly, we define “x is countable” to mean “x is finite or x bijects with ω”. That’s about it for this axiom. Time for the next one: Axiom of foundation “Sets are built out of simpler sets”. We want to ban weird things like x ∈x, or x ∈y ∧y ∈x, or similarly for longer chains. We also don’t want infinite descending chains like · · · x3 ∈x2 ∈x1 ∈x0. How can we capture the “wrongness” of these weird things all together? In the first case x ∈x, we see that {x} has no ∈-minimal element (we say y is ∈-minimal in x if (∀z ∈x)(z ̸∈y)). In the second case, {x, y} has no minimal element. In the last case, {x0, x1, x2, · · · } has no ∈-minimal element. 53 5 Set theory II Logic and Set Theory Axiom (Axiom of foundation). “Every (non-empty) set has an ∈-minimal member” (∀x)(x ̸= ∅⇒(∃y)(y ∈x ∧(∀z)(z ∈x ⇒z ̸∈y))). This is sometimes known as the Axiom of regularity. Note that most of the time, we don’t actually use the Axiom of Foundation. It’s here just so that our universe “looks nice”. Most results in set theory don’t rely on foundation. We will later show that this entails that all sets in V can be “built out of” the empty set, but that’s left for a later time. Axiom of replacement In ordinary mathematics, we often say things like “for each x ∈I, we have some Ai. Now take {Ai : i ∈I}”. For example, (after defining ordinals), we want to have the set {ω + i : i ∈ω}. How do we know that {Ai : i ∈I} is a set? How do we know that i 7→Ai is a function, i.e. that {(i, Ai) : i ∈I} is a set? It feels like it should be. We want an axiom that says something along the line of “the image of a set under a function is a set”. However, we do not know that the thing is a function yet. So we will have “the image of a set under something that looks like a function is a set”. To formally talk about “something that looks like a function”, we need a digression on classes: Digression on classes x 7→{x} for all x looks like a function, but isn’t it, since every function f has a (set) domain, defined as a suitable subset of S S f, but our function here has domain V . So what is this x 7→{x}? We call it a class. Definition (Class). Let (V, ∈) be an L-structure. A class is a collection C of points of V such that, for some formula p with free variable x (and maybe more funny parameters), we have x ∈C ⇔p holds. Intuitively, everything of the form {x ∈V : p(x)} is a class. Note that here we are abusing notation. When we say x ∈C, the symbol ∈ does not mean the membership relation in V . Inside the theory, we should view x ∈C as a shorthand of “p(x) holds”. Example. (i) V is a class, by letting p be “x = x”. (ii) For any t, {x ∈V : t ∈x} is a class, with p being t ∈x. Here t is a parameter. (iii) For any set y ∈V , y is a class — let p be “x ∈y”. 54 5 Set theory II Logic and Set Theory Definition (Proper class). We say C is a proper class if C is not a set (in V ), ie ¬(∃y)(∀x)(x ∈y ⇔p). Similarly, we can have Definition (Function-class). A function-class F is a collection of ordered pairs such that there is a formula p with free variables x, y (and maybe more) such that (x, y) ∈F ⇔p holds, and (x, y) ∈F ∧(x, z) ∈F ⇒y = z. Example. x 7→{x} is a function-class: take p to be y = {x}. Back to replacement How do we talk about function-classes? We cannot refer to classes inside V . Instead, we must refer to the first-order formula p. Axiom (Axiom of replacement). “The image of a set under a function-class is a set”. This is an axiom scheme, with an instance for each first-order formula p: (∀t1) · · · (∀tn) | {z } parameters [(∀x)(∀y)(∀z)((p ∧p[z/y]) ⇒y = z)] | {z } p defines a function-class ⇒[(∀x)(∃y)(z ∈y ⇔(∃t)(t ∈x ∧p[t/x, z/y]))] | {z } image of x under F is a set  . Example. For any set x, we can form {{t} : t ∈x} by p being “y = {x}”. However, this is a bad use of replacement, because we can already obtain that by power set (and separation). We will give a good example later. So that’s all the axioms of ZF. Note that we did not include the Axiom of choice! Axiom of choice Definition (ZFC). ZFC is the axioms ZF + AC, where AC is the axiom of choice, “every family of non-empty sets has a choice function”. (∀f)[(∀x)(x ∈dom f ⇒f(x) ̸= ∅) ⇒ (∃g)(dom g = dom f) ∧(∀x)(x ∈dom g ⇒g(x) ∈f(x))] Here we define a family of sets {Ai : i ∈I} to be a function f : I →V such that i 7→Ai. 5.2 Properties of ZF Now, what does V look like? We start with the following definition: Definition (Transitive set). A set x is transitive if every member of a member of x is a member of x: (∀y)((∃z)(y ∈z ∧z ∈x) ⇒y ∈x). This can be more concisely written as S x ⊆x, but is very confusing and impossible to understand! 55 5 Set theory II Logic and Set Theory This notion seems absurd — and it is. It is only of importance in the context of set theory and understanding what V looks like. It is an utterly useless notion in, say, group theory. Example. {∅, {∅}} is transitive. In general, for any n ∈ω, n is transitive. ω is also transitive. Lemma. Every x is contained in a transitive set. Note that this is a theorem of ZF, i.e. it officially means: let (V, ∈) be a model of ZF. Then in V , . Equivalently, ZF ⊢. We know that any intersection of transitive sets is transitive. So this lemma will tell us that x is contained in a least transitive set, called the transitive closure of x, or TC(x). Proof. We’d like to form “x ∪(S x) ∪(S S x) ∪(S S S x) ∪· · · ”. If this makes sense, then we are done, since the final product is clearly transitive. This will be a set by the union axiom applied to {x, S x, S S x, · · · }, which itself is a set by replacement applied to ω, for the function-class 0 7→x, 1 7→S x, 2 7→S S x etc. Of course we have to show that the above is a function class, i.e. can be expressed as a first order relation. We might want to write the sentence as: p(s, t) is (s = 0 ∧t = x) ∨(∃u)(∃v)(s = u + 1 ∧t = Sv ∧p(u, v)), but this is complete nonsense! We are defining p in terms of itself! The solution would be to use attempts, as we did previously for recursion. We define “f is an attempt” to mean “f is a function and dom f ∈ω and dom f ̸= ∅ and f(0) = x and (∀n)(n ∈ω∧n ∈dom f) ⇒f(n) = S f(n−1), i.e. f is defined for some natural numbers and meet our requirements when it is defined. Then it is easy to show that two attempts f and f ′ agree whenever both are defined. Also, ∀n ∈ω, there is an attempt f defined for n (both by ω-induction). Note that the definition of an attempt is a first-order formula. So our function class is p(s, t) is (∃f)(f is an attempt ∧y ∈dom f ∧f(y) = z). This is a good use of replacement, unlike our example above. We previously said that Foundation captures the notion “every set is built out of simpler sets”. What does that exactly mean? If this is true, we should be able to do induction on it: if p(y) for all y ∈x, then p(x). If this sounds weird, think about the natural numbers: everything is built from 0 using +1. So we can do induction on ω. Theorem (Principle of ∈-induction). For each formula p, with free variables t1, · · · , tn, x, (∀t1) · · · (∀tn) [(∀x)((∀y)(y ∈x ⇒p(y))) ⇒p(x)] ⇒(∀x)(p(x))  Note that officially, p(y) means p[y/x] and p(x) is simply x. Proof. Given t1, · · · , tn, suppose ¬(∀x)p(x). So we have some x with ¬p(x). Similar to how we proved regular induction on naturals from the well-ordering principle (in IA Numbers and Sets), we find a minimal x such that p(x) does not hold. 56 5 Set theory II Logic and Set Theory While foundation allows us to take the minimal element of a set, {y : ¬p(y)} need not be a set — e.g. if p(y) is y ̸= y. Instead, we pick a single x such that ¬p(x). Let u = TC({x}). Then {y ∈u : ¬p(y)} ̸= ∅, since x ∈u. So it has an ∈-minimal element, say y, by Foundation. Then each z ∈y has z ∈u since u is transitive. Hence p(z) by minimality of y. But this implies p(y). Contradiction. Note that here we used the transitive closure to reduce from reasoning about the whole scary V , to an actual set TC(x). This technique is rather useful. We have now used Foundation to prove ∈-induction. It turns out that the two are in fact equivalent. Proposition. ∈-induction ⇒Foundation. Proof. To deduce foundation from ∈-induction, the obvious p(x) — x has an ∈-minimal member, doesn’t work. Instead, consider p(x) given by (∀y) x ∈y ⇒y has an ∈-minimal member. If p(x) is true, we say x is regular. To show that (∀x)p(x), it is enough to show that: if every y ∈x is regular, then x is regular. Given any z with x ∈z, we want to show that z has an ∈-minimal member. If x is itself minimal in z, then done. Otherwise, then y ∈z for some y ∈x. But since y ∈x, y is regular. So z has a minimal element. Hence all x is regular. Since all non-empty sets contain at least one element (by definition), all sets have ∈-minimal member. This looked rather simple to prove. However, it is because we had the clever idea of regularity. If we didn’t we would be stuck for a long time! Now what about recursion? Can we define f(x) using f(y) for all y ∈x? Theorem (∈-recursion theorem). Let G be a function-class, everywhere defined. Then there is a function-class F such that F(x) = G(F|x) for all x. Moreover, F is unique (cf. definition of recursion on well-orderings). Note that F|x = {(y, F(y)) : y ∈x} is a set, by replacement. Proof. We first show existence. Again, we prove this with attempts. Define “f is an attempt” to mean “f is a function and dom f is transitive and (∀x)(x ∈ dom f ⇒f(x) = G(f|x))”. Then by simple ∈-induction, we have (∀x)(∀f ′)[(f an attempt defined at x∧ f ′ an attempt defined at x) ⇒f(x) = f ′(x)]. Also, (∀x)(∃f)(f an attempt defined at x), again by ∈-induction: suppose for each y ∈x, there exists an attempt defined at y. So there exists a unique attempt fy with domain TC({y}). Set f = S y∈x fy, and let f ′ = f S{(x, G(f|x)}. Then this is an attempt defined at x. So we take q(x, y) to be (∃f)(f is an attempt defined at x with f(x) = y). Uniqueness follows form ∈-induction. 57 5 Set theory II Logic and Set Theory Note that this is exactly the same as the proof of recursion on well-orderings. It is just that we have some extra mumbling about transitive closures. So we proved recursion and induction for ∈. What property of the relation-class ∈(with p(x, y) defined as x ∈y) did we use? We most importantly used the Axiom of foundation, which says that (i) p is well-founded: every set has a p-minimal element. (ii) p is local: ie, {x : p(x, y)} is a set for each y. We needed this to form the transitive closure. Definition (Well-founded relation). A relation-class R is well-founded if every set has a R-minimal element. Definition (Local relation). A relation-class R is local if {x : p(x, y)} is a set for each y. So we will have Proposition. p-induction and p-recursion are well-defined and valid for any p(x, y) that is well-founded and local. Proof. Same as above. Note that if r is a relation on a set a, then r is trivially local. So we just need r to be well-founded. In particular, our induction and recursion theorems for well-orderings are special cases of this. We have almost replicated everything we proved for well-orderings, except for subset collapse. We will now do that. This is motivated by the following question: can we model a given relation on a set by ∈? For example, let a = {b, c, d}, with b r c and c r d. Can we find a set {b′, c′, d′} such that b′ ∈c′ and c′ ∈d′? Yes. We can put b′ = ∅, c′ = {∅}, d′ = {{∅}}. Moreover, a′ = {b′, c′, d′} is transitive. Definition (Extensional relation). We say a relation r on set a is extensional if (∀x ∈a)(∀y ∈a)((∀z ∈a)(z r x ⇔z r y) ⇒x = y). i.e. it obeys the axiom of extension. Theorem (Mostowski collapse theorem). Let r be a relation on a set a that is well-founded and extensional. Then there exists a transitive b and a bijection f : a →b such that (∀x, y ∈a)(x r y ⇔f(x) ∈f(y)). Moreover, b and f are unique. Note that the two conditions “well-founded” and “extensional” are trivially necessary, since ∈is both well-founded and extensional. Proof. Existence: define f on a the obvious way — f(x) = {f(y) : y r x}. This is well-defined by r-recursion, and is a genuine function, not just of a function class by replacement — it is an image of a. Let b = {f(x) : x ∈a} (this is a set by replacement). We need to show that it is transitive and bijective. 58 5 Set theory II Logic and Set Theory By definition of f, b is transitive, and f is surjective as b is defined to be the image of f. So we have to show that f is injective. We’ll show that (∀x ∈a)(f(y) = f(x) ⇒y = x) for each x ∈a, by r-induction. Given y ∈a, with f(y) = f(x), we have {f(t) : t r y} = {f(s) : s r y} by definition of f. So {t : t r y} = {s : s r x} by the induction hypothesis. Hence x = y since r is extensional. So we have constructed such an b and f. Now we show it is unique: for any suitable f, f ′, we have f(x) = f ′(x) for all x ∈a by r-induction. Recall that we defined the ordinals to be the “equivalence class” of all well-orderings. But this is not a good definition since the ordinals won’t be sets. Hence we define them (formally) as follows: Definition (Ordinal). An ordinal is a transitive set, totally ordered by ∈. This is automatically well-ordered by ∈, by foundation. Example. ∅, {∅}, {∅, {∅}} are ordinals. Any n ∈ω as n = {0, 1, · · · , n −1}, as well as ω itself, are ordinals. Why is this a good definition? Mostowski says that each well-ordering is order-isomorphic to a unique ordinal (using our definition of ordinal above) — this is its order-type. So here, instead of saying that the ordinals is the equivalence class of well-orderings, we simply choose one representative of each equivalence class (given by Mostowski collapse), and call that the ordinal. For any ordinal α, Iα = {β : β < α} is a well-ordering of order-type α. Applying Mostowski collapse, we get α = {β : β < α}. So β < α iff β ∈α. So, for example, α+ = α ∪{α}, and sup{αk : k ∈I} = S{αi : i ∈I}, Set theorists are often write suprema as unions, but this is a totally unhelpful notation! 5.3 Picture of the universe What does the universe look like? We start with the empty set, and take the power set, repeatedly, transfinitely. Definition (von Neumann hierarchy). Define sets Vα for α ∈On (where On is the class of ordinals) by ∈-recursion: (i) V0 = ∅. (ii) Vα+1 = P(Vα). (iii) Vλ = S{Vγ : γ < λ} for λ a non-zero limit ordinal. 59 5 Set theory II Logic and Set Theory On V0 = ∅ V1 = {∅} V7 . . . . . . Vω Vω+1 . . . . . . Vω+ω . . . . . . Note that by definition, we have x ⊆Vα ⇔x ∈Vα+1. We would like every x to be in some Vα, and that is indeed true. We prove this though a series of lemmas: Lemma. Each Vα is transitive. Proof. Since we define Vα by recursion, it is sensible to prove this by induction: By induction on α: (i) Zero: V0 = ∅is transitive. (ii) Successors: If x is transitive, then so is P(x): given y ∈z ∈P(x), we want to show that y ∈P(x). Since y is in a member of P(x), i.e. a subset of x, we must have y ∈x. So y ⊆x since x is transitive. So y ∈P(x). (iii) Limits: Any union of transitive sets is transitive. Lemma. If α ≤β, then Vα ⊆Vβ. Proof. Fix α, and induct on β. (i) β = α: trivial (ii) Successors: Vβ+ ⊆Vβ since x ⊆P(x) for transitive x. So Vα ⊆Vβ ⇒Vα ⊆ Vβ+. (iii) Limits: Trivial by definition Finally we are ready for: Theorem. Every x belongs to some Vα. Intuitively, we want to say V = [ α∈On Vα, We’ll need some terminology to start with. If x ⊆Vα for some α, then there is a least α with x ⊆Vα. We call this the rank of x. For example, rank(∅) = 0, rank({∅}) = 1. Also rank(ω) = ω. In fact, rank(α) = α for all α ∈On. Note that we want the least α such that x ⊆Vα, not x ∈Vα. 60 5 Set theory II Logic and Set Theory Proof. We’ll show that (∀x)(∃α)(x ∈Vα) by ∈-induction on x. So we are allowed to assume that for each y ∈x, we have y ⊆Vα for some α. So y ⊆Vrank(y), or y ∈Vrank(y)+1. Let α = sup{(rank(y)+ : y ∈x}. Then y ∈Vα for every y ∈x. So x ⊆Vα. Our definition of rank is easy to get wrong — it is easy to be off by of 1. So the official definition is Definition (Rank). The rank of a set x is defined recursively by rank(x) = sup{(rank y)+ : y ∈x}. Then the initial definition we had is now a proposition. Proposition. rank(x) is the first α such that x ⊆Vα. 61 6 Cardinals II Logic and Set Theory 6 Cardinals In this chapter, we will look at the “sizes” of (infinite) sets (finite sets are boring!). We work in ZFC, since things become really weird without Choice. Since we will talk about bijections a lot, we will have the following notation: Notation. Write x ↔y for ∃f : f is a bijection from x to y. 6.1 Definitions We want to define card(x) (the cardinality, or size of x) in such a way that card(x) = card(y) ⇔x ↔y. We can’t define card(x) = {y : y ↔x} as it may not be a set. So we want to pick a “representative” of the sets that biject with x, like how we defined the ordinals. So why not use the ordinals? By Choice, we know that all x is well-orderable. So x ↔α for some α. So we define: Definition (Cardinality). The cardinality of a set x, written card(x), is the least ordinal α such that x ↔α. Then we have (trivially) card(x) = card(y) ⇔x ↔y. (What if we don’t have Choice, i.e. if we are in ZF? This will need a really clever trick, called the Scott trick. In our universe of ZF, there is a huge of blob of things that biject with x. We cannot take the whole blob (it won’t be a set), or pick one of them (requires Choice). So we “chop off” the blob at fixed, determined point, so we are left with a set. Define the essential rank of x to be the least rank of all y such that y ↔x. Then set card(x) = {y ∈Vessrank(x)+ : y ↔x}.) So what cardinals are there? Clearly we have 1, 2, 3, · · · . What else? Definition (Initial ordinal). We say an ordinal α is initial if (∀β < α)(¬β ↔α), i.e. it is the smallest ordinal of that cardinality. Then 0, 1, 2, 3, · · · , ω, ω1, γ(X) for any X are all initial. However, ωω is not initial, as it bijects with ω (both are countable). Can we find them all? Yes! Definition (Omega ordinals). We define ωα for each α ∈On by (i) ω0 = ω; (ii) ωα+1 = γ(ωα); (iii) ωλ = sup{ωα : α < λ} for non-zero limit λ. It is easy induction to show that each ωα is initial. We can also show that every initial δ (for δ ≥ω) is an ωα. We know that the ωα are unbounded since, say ωα ≥α for all α. So there is a least α with ωα ≥δ. If α is a successor, then let α = β+. Then ωβ < δ ≤ωα. But there is no initial ordinal between ωβ and ωα = γ(ωβ), since γ(X) is defined as the least ordinal that does not biject with X. So we must have δ = ωα. If α is a limit, then since ωα is defined as a supremum, by definition we cannot have δ < ωα, or else there is some β < α with δ < ωβ. So ωα = δ as well. 62 6 Cardinals II Logic and Set Theory Definition (Aleph number). Write ℵα (“aleph-α”) for card(ωα). Then from the argument above, we have Theorem. The ℵα are the cardinals of all infinite sets (or, in ZF, the cardinals of all infinite well-orderable sets). For example, card(ω) = ℵ0, card ω1 = ℵ1. We will use lower case letters to denote cardinalities and upper case for the sets with that cardinality. e.g. card(N) = n. Definition (Cardinal (in)equality). For cardinals n and m, write m ≤n if M injects into N, where card M = m, card N = n. This clearly does not depend on M and N. So m ≤n and n ≤m implies n = m by Schr¨ oder-Bernstein. Write m < n if m ≤n by m ̸= n. Example. card(P(ω)) > card(ω). This ≤is a partial order. Moreover, it is a total order (assuming AC). 6.2 Cardinal arithmetic Definition (Cardinal addition, multiplication and exponentiation). For cardinals m, n, write m+n for card(M ⊔N); mn for card(M ×N); and mn for card(M N), where M N = {f : f is a function N →M}. Note that this coincides with our usual definition of Xn for finite n. Example. R ↔P(ω) ↔2ω. So card(R) = card(Pω) = 2ℵ0. Similarly, define X i∈I mi = card G i∈I Mi ! . Example. How many sequences of reals are there? A real sequence is a function from ω →R. We have card(Rω) = (2ℵ0)ℵ0 = 2ℵ0×ℵ0 = 2ℵ0 = card(R) Note that we used facts like Proposition. (i) m + n = n + m since N ⊔M ↔N ⊔N with the obvious bijection. (ii) mn = nm using the obvious bijection (iii) (mn)p = mnp as (M N)P ↔M N×P since both objects take in a P and an N and returns an M. It is important to note that cardinal exponentiation is different from ordinal exponentiation. For example, ωω (ordinal exponentiation) is countable, but ℵℵ0 0 ≥2ℵ0 > ℵ0 (cardinal exponentiation). From IA Numbers and sets, we know that ℵ0ℵ0 = ℵ0. What about ℵ1ℵ1? Or ℵ3ℵ0? It turns out that cardinal sums and multiplications are utterly boring: 63 6 Cardinals II Logic and Set Theory Theorem. For every ordinal α, ℵαℵα = ℵα. This is the best we could ever ask for. What can be simpler? Proof. Since the Alephs are defined by induction, it makes sense to prove it by induction. In the following proof, there is a small part that doesn’t work nicely with α = 0. But α = 0 case (ie ℵ0ℵ0 = ℵ0) is already done. So assume α ̸= 0. Induct on α. We want ωα × ωα to biject with ωα, i.e. well-order ωα × ωα to an ordering of length ωα. Using the ordinal product clearly doesn’t work. The ordinal product counts the product in rows, so we have many copies of ωα. When we proved ℵ0ℵ0 = ℵ0, we counted them diagonally. But counting diagonally here doesn’t look very nice, since we will have to “jump over” infinities. Instead, we count in squares ωα ωα We set (x, y) < (x′, y′) if either max(x, y) < max(x′, y′) (this says that (x′, y′) is in a bigger square), or, (say max(x, y) = max(x′, y′) = β and y′ = β, y < β or x = x′ = β, y < y′ or y = y′ = β, x < x′) (nonsense used to order things in the same square — utterly unimportant). How do we show that this has order type ωα? We show that any initial segment has order type < ωα. For any proper initial segment I(x,y), we have I(x,y) ⊆β × β for some β < ωα, since ωα is a limit, with wlog β infinite. So β × β ↔β by induction hypothesis (their cardinality is less that ωα). So card(β × β) < card(ωα). Hence I(x,y) has order type < ωα. Thus the order type of our well-order is ≤ωα. So ωα × ωα injects into ωα. Since trivially ωα injects into ωα × ωα, we have ωα × ωα ↔ωα. So why did we say cardinal arithmetic is boring? We have Corollary. Let α ≤β. Then ℵα + ℵβ = ℵαℵβ = ℵβ. 64 6 Cardinals II Logic and Set Theory Proof. ℵβ ≤ℵα + ℵβ ≤ℵβ + ℵβ = 2ℵβ ≤ℵβ × ℵβ = ℵβ, So done Example. X ⊔X bijects with X, for infinite X (in ZFC). However, cardinal exponentiation is very hard. For example, is 2ℵ0 = ℵ1? This is the continuum hypothesis, and cannot be proved or disproved in ZFC! Even today, not all implications among values of ℵℵβ α are known, i.e. we don’t know whether they are true, false or independent! 65 7 Incompleteness II Logic and Set Theory 7 Incompleteness The big goal of this (non-examinable) chapter is to show that PA is incomplete, i.e. there is a sentence p such that PA ̸⊢p and PA ̸⊢¬p. The strategy is to find a p that is true in N, but PA ̸⊢p. Here we say “true” to mean “true in N”, and “provable” to mean “PA proves it”. The idea is to find a p that says “I am not provable”. More precisely, we want a p such that p is true if and only if p is not provable. We are then done: p must be true, since if p were false, then it is provable, i.e. PA ⊢p. So p holds in every model of PA, and in particular, p holds in N. Contradiction. So p is true. So p is not provable. We’ll have to “code” formulae, proofs etc. inside PA, i.e. as numbers. But this doesn’t seem possible — it seems like, in any format, “p is not provable” must be longer than p. So p cannot say “p is not provable“! So be prepared for some magic to come up in the middle for the proof! Definability We first start with some notions of definability. Definition (Definability). A subset S ⊆N is definable if there is a formula p with one free variable such that ∀m ∈N : m ∈S ⇔p(m) holds. Similarly, f : N →N is definable if there exists a formula p(x, y) such that ∀m, n ∈N : f(m) = n ⇔p(m, n) holds. Example. The set of primes is definable: p(x) is x ̸= 1 ∧(∀y)(∀z)(yz = x ⇒(y = 1 ∨z = 1)). We can say “m is prime” is definable. How about powers of 2? We don’t have exponentiation here. What can we do? We can take p(x) to be (∀y)((y is prime ∧y | x) ⇒y = 2), where 2 is a shorthand for s(s(0)), and y | x is a shorthand for (∃z)(yz = x). So this is also definable. The function m 7→m2 is also definable: take p(x, y) to be yy = x. Here we will assume: Fact. Any function given by an algorithm is definable. Proof will not be given here. See, eg, PTJ’s book for detailed proof. Example. m 7→2m is definable. 66 7 Incompleteness II Logic and Set Theory Coding Our language has 12 symbols: s, 0, ×, +, ⊥, ⇒, (, ), =, x,′ , ∀(where the variables are now called x, x′, x′′, x′′′. We assign values to them, say v(s) = 1, v(0) = 2, · · · , v(∀) = 12. To code a formula, we can take 2v(first symbol) · 3v(second symbol) · · · (nth prime)v(nth symbol). For example, (∀x)(x = x) is coded as 27312510781171310179238. Not every m codes a formula, e.g. 27312510 is translated to (∀x, which is clearly nonsense. Similarly, 2773 or 2100 can’t even be translated at all. However, “m codes a formula” is definable, as there is an algorithm that checks that. Write Sm for the formula coded by m (and set Sm to be “⊥” if m does not code a formula). Similarly, write c(p) for the code of p. Given a finite sequence p1, · · · pn for formulae, code it as 2c(p1)3c(p2) · · · (nth prime)c(pn). Alternatively, we can add a separator character to our 12 symbols and concatenate the sequence of formulae with the separator character. Now, “m codes an axiom (either logical or axiom of PA)” is definable, as there exists an algorithm to check it. For example, an instance of the first logical axiom can be validated by the following regex: ^([s0()=x’+×⊥⇒∀]+)⇒([s0()=x’+×⊥⇒∀]+ ⇒\1)$ (after translating to a sentence and verifying it is a valid logical formula) Also, ϕ(ℓ, m, n) =“Sn obtained from Sℓ, Sm via MP” is definable, and similarly for generalization. So Θ(m, n) = “n codes a proof of Sm” is definable. Thus Ψ(m) = “Sm is provable” is definable as Ψ(m) ⇔(∃n)Θ(m, n). So far, everything is rather boring. We all know that strings of symbols can be coded into numbers — that’s what computers do! 67 7 Incompleteness II Logic and Set Theory Clever bit Consider the statement χ(m) that states “m codes a formula Sm with one free variable, and Sm(m) is not provable.” This is clearly definable. Suppose this is defined by p. So χ(n) ⇔p[n/x] Suppose c(p) = N. Then χ(N) asserts “N codes a formula SN with one free variable and SN(N) is not provable.” But we already know that N codes the formula χ. So χ(N) asserts that χ(N) is not provable. So Theorem (G¨ odel’s incompleteness theorem). PA is incomplete. Maybe that’s because PA is rubbish. Could we add some clever axiom t (true in N) to PA, so that PA∪{t} is complete? No! Just run the same proof with “PA” replaced by “PA∪{t}” to show that PA∪{t} is incomplete. But we can certainly extend PA to a complete theory — let T = {p : p holds in N}. What will go wrong in our proof? It can only be because Theorem. “Truth is not definable” T = {p : p holds in N} is not definable. This officially means {m : m codes a member of T} is not a definable set. Next question: we proved that our clever statement p is true but not provable. Why doesn’t this formalize into PA? The answer is, in the proof, we also used the fact that PA has a model, N. By completeness, this means that we used the statement con(PA), i.e. the statement that PA is consistent, or (∀m)(m does not code a proof of ⊥). With this statement, the proof does formalize to PA. So PA∪{con(PA)} ⊢p. Hence Theorem. PA ̸⊢con(PA). The same proof works in ZF. So ZF is incomplete and ZF does not prove its own consistency. 68 Index II Logic and Set Theory Index ℵα, 63 ∈-induction, 56 ∈-recursion, 57 κ-categorical, 48 ωα, 62 adequacy theorem, 12, 46 antichain, 29 atomic formulae, 37 Ax-Grothendieck theorem, 49 axiom of choice, 34, 55 of empty set, 51 of extension, 51 of foundation, 53 of infinity, 52 of pair set, 51 of power set, 52 of replacement, 55 of separation, 51 of union, 52 bound variable, 38 Bourbaki-Witt theorem, 36 Burali-Forti paradox, 20 Cantor-Schr¨ oder-Bernstein, 31 cardinal addition, 63 cardinal exponentiation, 63 cardinal inequality, 63 cardinal multiplication, 63 cardinality, 62 chain, 29 chain-complete, 36 class, 54 closed term, 38 compactness theorem, 12, 46 complete poset, 30 complete theory, 48 completeness theorem, 12, 46 consistent, 10 decidability theorem, 12 deduction theorem, 8, 43 definability, 66 downward L¨ owenheim-Skolem theorem, 47 extension, 19 extension relation, 58 fixed point, 26, 30 fixed point lemma for normal functions, 26 formulae, 38 free variable, 38 function, 52 function class, 55 G¨ odel’s completeness theorem, 46 G¨ odel’s incompleteness theorem, 68 generalization, 42 Hartogs’ lemma, 21 inconsistent, 10 induction, 15 inflationary function, 36 initial ordinal, 62 initial segment, 15 interpretation, 39 L¨ owenheim-Skolem theorem, 46, 47 language, 37 limit ordinal, 22 local relation, 58 maximal, 31 model, 40 model existence theorem, 10, 33, 44 modus ponens, 7, 42 Morley’s categoricity theorem, 50 Mostowski collapse theorem, 58 nested family, 19 normal function, 25 fixed point lemma, 26 order isomorphism, 14 order type, 20 order-preserving function, 30 ordered pair, 52 ordinal, 19, 59 fixed point lemma, 26 ordinal addition, 22, 23 ordinal exponentiation, 25 69 Index II Logic and Set Theory ordinal multiplication, 24 PA, 47 partial order, 28 Peano’s axioms, 47 poset, 28 predicate logic, 37 proof, 8, 42 proper class, 55 propositional logic, 4 propositions, 4 pure identity theory, 48 rank, 61 recursion, 16 semantic entailment, 7, 40 sentence, 38 soundness theorem, 10, 43 structure, 39 subset collapse lemma, 17 substitution, 38 successor, 19, 47 successor ordinal, 22 supremum, 29 syntactic entailment, 8, 42 tautology, 6, 40 term, 37 theorem, 8, 42 theory, 40 complete, 48 total order, 13 transitive, 55 upper bound, 29 upward L¨ owenheim-Skolem theorem, 46 valuation, 5 variable, 37 von Neumann hierarchy, 59 well order, 14 well-founded relation, 58 well-ordering theorem, 35 Zermelo-Fraenkel set theory, 51 ZF, 51 ZFC, 55 Zorn’s lemma, 32 70
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https://mathoverflow.net/questions/314981/i-conjecture-inequalities-sum-k-1n-kx-le-fracn2x
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange I conjecture inequalities $\sum_{k=1}^{n}{kx}\le\frac{n}{2}x$ Ask Question Asked Modified 6 years, 10 months ago Viewed 2k times 13 $\begingroup$ I conjecture the following inequality: For $x > 1$, and $n$ a positive integer, $$\sum_{k=1}^{n}{kx}\le\dfrac{n}{2}x.$$ For $n=1$, the inequality becomes $${x}\le\dfrac{x}{2}\Longleftrightarrow {x}\le [x];$$ and, for $n = 2$, it becomes $${x}+{2x}\le x\Longleftrightarrow {2x}\le [x],$$ which is obvious since $[x]\ge 1>{2x}$. If $x\ge 2$, the inequality is obvious, since $$\dfrac{n}{2}x\ge n\ge\sum_{k=1}^{n}{kx}.$$ However, I cannot prove it for $1 < x < 2$. inequalities Share Improve this question edited Nov 10, 2018 at 2:33 LSpice 13.9k44 gold badges4747 silver badges7373 bronze badges asked Nov 10, 2018 at 1:20 math110math110 4,4982121 silver badges5151 bronze badges $\endgroup$ 12 5 $\begingroup$ Isn't MSE a right forum for such questions? $\endgroup$ user64494 – user64494 2018-11-10 03:41:26 +00:00 Commented Nov 10, 2018 at 3:41 2 $\begingroup$ I suppose that ${x}$ means fractional part and $[x]$ means integer part? $\endgroup$ Nate Eldredge – Nate Eldredge 2018-11-10 04:47:30 +00:00 Commented Nov 10, 2018 at 4:47 13 $\begingroup$ Look at the integer grid squares the line through the origin with slope $x$ visits: you get $n$ right triangles with heights ${kx}$ and bases $\leqslant1$. $\endgroup$ მამუკა ჯიბლაძე – მამუკა ჯიბლაძე 2018-11-10 05:09:27 +00:00 Commented Nov 10, 2018 at 5:09 3 $\begingroup$ I do not know the methods of its proof (I do not know the proof, and even am not sure that it is true), but the estimates of the remainder in the approximate formulae for the number of lattice points in certain regions (this specific question is about triangle with vertices $(0,0),(n,0),(n,xn)) attract a lot of interest. Aswell as the sums of periodic functions along arithmetic progressions. $\endgroup$ Fedor Petrov – Fedor Petrov 2018-11-10 14:14:53 +00:00 Commented Nov 10, 2018 at 14:14 4 $\begingroup$ From this paper I conclude that $\sum_{k=1}^n{kx}$ grows as $n/2+c\log n$ (the coefficient $c=1/256$), which is consistent with your upper bound of $nx/2$ for $x>1$. $\endgroup$ Carlo Beenakker – Carlo Beenakker 2018-11-10 18:28:46 +00:00 Commented Nov 10, 2018 at 18:28 | Show 7 more comments 3 Answers 3 Reset to default 28 $\begingroup$ As მამუკა ჯიბლაძე said, the proof is a picture though I had trouble reconstructing the picture of his and came up with a slightly different one. Let $0. Consider the rectangles with bases $[k-1,k]$ of heights $[ky]$ for $k=1,2,\dots,n$. They cover the triangle with the vertices $(0,0),(n,0),(n,ny)$ up to several triangles of height $\le 1$ with sum of bases $\le n$. This proves the inequality $$ \sum_{k=1}^n[ky]\ge \frac{n^2y}2-\frac n2\,. $$ Now, for $x=1+y\in(1,2)$, we have $$ \sum_{k=1}^n{kx}=\sum_{k=1}^n{ky}=\sum_{k=1}^n ky-\sum_{k=1}^n[ky] \ \le \frac{n(n+1)y}2-\frac{n^2y}2+\frac n2=\frac n2(1+y)=\frac n2x\,. $$ Share Improve this answer edited Nov 16, 2018 at 16:07 Martin Sleziak 4,78344 gold badges3838 silver badges4242 bronze badges answered Nov 10, 2018 at 19:51 fedjafedja 63.8k1111 gold badges164164 silver badges308308 bronze badges $\endgroup$ 13 $\begingroup$ Sorry I don't get the argument. You want the large triangle to contain the rectangles plus some small triangles whose areas sum up to $n/2$. What are your small triangles? $\endgroup$ abx – abx 2018-11-11 08:58:04 +00:00 Commented Nov 11, 2018 at 8:58 $\begingroup$ @abx. No, the large triangle does not contain the rectangles. They stick out a bit in places. The small triangles are what is in the large triangle but not in the rectangles. Have you tried to draw a picture? If you draw it, you'll see them immediately. $\endgroup$ fedja – fedja 2018-11-11 13:35:54 +00:00 Commented Nov 11, 2018 at 13:35 $\begingroup$ Ah, OK. I did draw a picture, but with the base of the rectangle with height $[ky]$ being wrongly $[k,k+1]$. Thanks! $\endgroup$ abx – abx 2018-11-11 13:45:12 +00:00 Commented Nov 11, 2018 at 13:45 2 $\begingroup$ "the picture of his": did you look up the name to determine that this user is male? Or is this an instance of the implicit gender biases we all carry, where we simply assume that people discussing mathematics are male? It's worth paying attention to such things, because implicit bias in mathematics really harms women who want to participate. $\endgroup$ Greg Martin – Greg Martin 2018-11-18 06:48:01 +00:00 Commented Nov 18, 2018 at 6:48 1 $\begingroup$ @MartinSleziak (and others). 5 seconds after I clicked on the CV on that website I've got the "Microsoft critical alert" scam message. May be a pure coincidence, of course, but I'd rather report it here :-) $\endgroup$ fedja – fedja 2018-11-18 16:58:12 +00:00 Commented Nov 18, 2018 at 16:58 | Show 8 more comments 10 +25 $\begingroup$ Here is a different, more algebraic proof. I don't know if there is any reason to prefer it to the visual proof. It is sufficient to prove the inequality if we take ${x}$ to round up instead of down, as that is a stronger inequality. If we raise $x$ by $\epsilon$, where $\epsilon$ is small enough that no fractional part wraps around, the left side increases by $n (n+1) \epsilon/2$ and the right side by $n \epsilon/2$. We can do this unless $xk$ is an integer for some $1 \leq k \leq n$. Because the increase on the left is greater than the increase on the right, we can reduce to the case where $xk$ is an integer for some $1 \leq k \leq n$, say $x=a/b$ with $1 \leq b \leq n$ and $a,b$ relatively prime. We have $$ \sum_{k=1}^n \left{ k \frac{a}{b} \right} = \sum_{k=1}^b \left{ k \frac{a}{b} \right} +\sum_{k=b+1}^n \left{ k \frac{a}{b} \right} $$ For the first term, we use the fact that $a$ is a permutation of residue classes mod $b$, so $$ \sum_{k=1}^b \left{ k \frac{a}{b} \right} = \sum_{k=1}^b \left{ \frac{k}{b} \right}= \sum_{k=1}^b \frac{k}{b} = \frac{b (b+1)}{2b} \leq \frac{b a}{2b} = \frac{bx}{2} $$ For the second term, we use periodicity and induction on $n$, so $$\sum_{k=b+1}^n \left{ k \frac{a}{b} \right} = \sum_{k=1}^{n-b} \left{ k \frac{a}{b} \right} \leq \frac{ (n-b) x}{2} $$ Summing, we get an upper bound of $\frac{nx}{2}$, as desired (and as necessary for the induction step). Share Improve this answer answered Nov 18, 2018 at 2:10 Will SawinWill Sawin 160k99 gold badges346346 silver badges606606 bronze badges $\endgroup$ Add a comment | 7 $\begingroup$ This is a proof smilar to Will Sawin's but with no induction. Set $y=x-1$. We need to prove that the average of ${y},\dots,{ny}$ is at most $x/2$. The numbers $y,2y,\dots,ny$ split into contiguous groups, each group with the same integer part. It suffices to show that the average of the fractional parts for each group is at most $x/2$. Those fractional parts form an arithmetical sequence, so their average is half the sum of the first and the last term. As the fractional part of the first term in the group is at most $y$, this average is bounded by $\frac{y+1}2=x/2$, as desired. Share Improve this answer edited Nov 19, 2018 at 9:20 answered Nov 19, 2018 at 7:21 Ilya BogdanovIlya Bogdanov 24.7k5656 silver badges9696 bronze badges $\endgroup$ 4 $\begingroup$ I don€™t understand this argument. Since $x>1$, all your groups have only a single element, right? $\endgroup$ Emil Jeřábek – Emil Jeřábek 2018-11-19 09:12:19 +00:00 Commented Nov 19, 2018 at 9:12 $\begingroup$ @EmilJerabek Sorry, I thought too much about $x-1$. Fixed. $\endgroup$ Ilya Bogdanov – Ilya Bogdanov 2018-11-19 09:21:49 +00:00 Commented Nov 19, 2018 at 9:21 $\begingroup$ Thank you, now I get it. This is a nice argument. $\endgroup$ Emil Jeřábek – Emil Jeřábek 2018-11-19 12:16:44 +00:00 Commented Nov 19, 2018 at 12:16 $\begingroup$ I think this is more similar to the picture proof, in that these groups correspond exactly to the triangles in the picture. But either way, slick! $\endgroup$ Will Sawin – Will Sawin 2018-11-20 03:05:03 +00:00 Commented Nov 20, 2018 at 3:05 Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions inequalities See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Related 3 Prove this conjecture inequality 5 Prove this conjecture inequality $x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k}\le \frac{1}{(k+2)^{k-1}}$ How to determine the maximums of certain cyclic sums? 8 Find the maximum of the value $c(n)$ (similar to Hardy's inequality) 4 Proving two inequalities involving the gamma and digamma functions 8 On the conjecture : $|m^2-n^3|>\frac{1}{5}\sqrt{m^2+n^3}$ Proving $\sum_{i=1}^{n}\sum_{j=1}^{n}\left{\frac{x_{i}}{x_{j}}\right}\le \frac{9}{14}n^2$？ 8 Prove $\frac{\sum_{j=1}^mj^{2n}}{\sum_{j=1}^m\big(j-\frac12\big)^{2n}}>\Big(\frac{m+\frac12}m\Big)^{2n+1}$ 7 How to prove $\sum_{i,j=1}^{n}\frac{a_{i}a_{j}}{1+|i-j|}\ge c\cdot\frac{\ln{n}}{n}$ Question feed
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https://thecontentauthority.com/blog/load-vs-weight
Load vs Weight: Meaning And Differences Skip to Content Home Grammar Capitalization Definitions Idioms Parts of Speech Word Lists Word Usage Blog About Write For Us Login Load vs Weight: Meaning And Differences Home » Grammar » Word Usage Are you confused about the difference between load and weight? Many people use these terms interchangeably, but they actually have distinct meanings. In this article, we’ll explore the differences between load and weight, and when to use each term. Let’s define our terms. Load refers to the amount of weight that is being carried or transported. Weight, on the other hand, is a measure of the force exerted on an object due to gravity. In other words, weight is the amount of force that is pushing down on an object. next stay CC Settings Off Arabic Chinese English French German Hindi Portuguese Spanish Font Color white Font Opacity 100%Font Size 100%Font Family Arial Text Shadow none Background Color black Background Opacity 50%Window Color black Window Opacity 0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25% 200%175%150%125%100%75%50% Arial Georgia Garamond Courier New Tahoma Times New Roman Trebuchet MS Verdana None Raised Depressed Uniform Drop Shadow White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% So, which term is the proper one to use? The answer is that it depends on the context. If you’re talking about the amount of weight that is being carried or transported, then load is the appropriate term. If you’re talking about the force that is pushing down on an object, then weight is the correct term. Now that we’ve defined our terms, let’s dive deeper into the differences between load and weight. Understanding these differences can help you use the correct term in the appropriate context, which can improve your communication and avoid confusion. Define Load Load refers to the external force or weight that is placed on a structure or object. It can be a physical weight, such as the weight of a person on a bridge, or a force, such as wind or water pressure on a building. In engineering, load is often measured in units of force, such as pounds or newtons. Loads can be classified into different types based on their characteristics. Some common types of loads include: Dead load: the weight of the structure or object itself Live load: the weight of people, furniture, or other movable objects Wind load: the force exerted by wind on a structure Seismic load: the force exerted by earthquakes on a structure Define Weight Weight is the measure of the force exerted on an object by gravity. It is a scalar quantity that is typically measured in units of mass, such as kilograms or pounds. The weight of an object can vary based on its location in the universe, as the force of gravity changes depending on the mass and distance of other objects. The weight of an object can be calculated using the formula: Weight = Mass x Gravity where gravity is the force of attraction between two objects, typically measured as 9.8 m/s^2 on Earth. It is important to note that weight is not the same as mass, which is a measure of the amount of matter in an object. Mass is a constant property of an object, while weight can vary based on the gravitational force acting on it. How To Properly Use The Words In A Sentence When it comes to discussing the amount of mass or force that an object exerts, the words “load” and “weight” are often used interchangeably. However, they have distinct meanings and should be used appropriately to avoid confusion and miscommunication. How To Use “Load” In A Sentence “Load” refers to the amount of weight or force that is applied to an object. It can be a static or dynamic force, and can be measured in units such as pounds, kilograms, or newtons. Here are some examples of how to use “load” in a sentence: The load on the bridge exceeded its weight capacity. The crane was able to lift the heavy load with ease. The electrical system experienced a power surge due to the increased load. It’s important to note that “load” can also refer to the amount of work that needs to be done, or the amount of data that needs to be processed. In these cases, it still refers to a measurable amount of force or weight: The server was unable to handle the load of incoming requests. The construction crew had a heavy load of materials to transport to the site. How To Use “Weight” In A Sentence “Weight” refers specifically to the amount of mass that an object has. It is a static measurement, and is typically expressed in units such as pounds or kilograms. Here are some examples of how to use “weight” in a sentence: The weight of the box was too much for me to carry alone. The recommended weight limit for this elevator is 1000 pounds. The athlete’s weight was within the acceptable range for their height and age. It’s important to note that “weight” can also be used figuratively to describe the importance or significance of something: The CEO’s opinion carries a lot of weight in the company. The judge’s decision will hold a lot of weight in the final outcome of the case. More Examples Of Load & Weight Used In Sentences In order to better understand the difference between load and weight, it is important to see how they are used in everyday language. Here are some examples: Examples Of Using Load In A Sentence The truck was carrying a heavy load of bricks. The crane was lifting a load of steel beams. The backpack felt heavy because it was carrying a load of books. The washing machine was overloaded with a heavy load of clothes. The cargo ship was carrying a load of containers from China. The forklift was moving a load of pallets from one warehouse to another. The airplane was carrying a heavy load of passengers and luggage. The train was transporting a load of coal to the power plant. The tractor was pulling a load of hay bales in the field. The delivery van was carrying a load of packages for delivery. Examples Of Using Weight In A Sentence The weight of the elephant was over 10,000 pounds. The weight of the dumbbell was too heavy for the beginner lifter. The weight of the car caused the bridge to collapse. The weight of the snow on the roof was causing it to sag. The weight of the anchor kept the boat from drifting away. The weight of the furniture made it difficult to move up the stairs. The weight of the backpack was causing the hiker to tire quickly. The weight of the barbell was too much for the weightlifter to lift. The weight of the package was too heavy for the postal worker to lift alone. The weight of the bookshelf caused it to break under the weight of the books. Common Mistakes To Avoid When it comes to understanding load and weight, people often make the mistake of using these terms interchangeably. However, there are some key differences between these two concepts that should be kept in mind. Here are some common mistakes to avoid: Mistake #1: Using Load And Weight Interchangeably Load and weight are not the same thing. Weight refers to the force exerted on an object due to gravity, while load refers to the amount of weight that a structure or object can support. The confusion between these two terms can lead to errors in calculations and can result in unsafe conditions. For example, if you are trying to determine the weight of a load that a bridge can support, you cannot simply use the weight of the load as the load itself. You need to take into account the capacity of the bridge to support the load, which is determined by factors such as the strength of the materials used and the design of the bridge. Mistake #2: Overloading Structures Another common mistake is overloading structures by exceeding their load capacity. This can happen when people assume that load and weight are interchangeable and do not take into account the load capacity of the structure or object they are using. For example, if you are using a ladder to reach a high place, you need to make sure that the ladder can support your weight and the weight of any tools or materials you are carrying. If you exceed the load capacity of the ladder, it may buckle or collapse, leading to serious injury or even death. Tips To Avoid These Mistakes To avoid these common mistakes, it is important to understand the difference between load and weight and to always take into account the load capacity of the structures and objects you are using. Here are some tips: Read the manufacturer’s instructions carefully to determine the load capacity of the object or structure you are using. Do not assume that load and weight are interchangeable. Use a load calculator or consult with an expert if you are unsure about the load capacity of a structure or object. Regularly inspect structures and objects for signs of wear and tear, as this can affect their load capacity. Context Matters When it comes to discussing load and weight, context is key. The choice between these two terms can depend on the situation in which they are used. While they may seem interchangeable, there are distinct differences between the two that can affect their usage. Load Vs Weight Load refers to the amount of weight that is being carried or supported by a structure or vehicle. It can also refer to the amount of work that is being performed by a machine or system. Weight, on the other hand, refers to the force exerted on an object due to gravity. It is a measure of the mass of an object and the acceleration due to gravity. Understanding the difference between load and weight is important in various contexts. For example, in the field of engineering, load is a critical factor in designing structures that can withstand the forces placed upon them. In this context, load refers to the amount of weight that a structure can safely support without collapsing or experiencing damage. On the other hand, weight is more commonly used in everyday situations to describe the heaviness of an object. For instance, when weighing luggage at an airport, weight is the appropriate term to use. It is also used in the medical field to assess a patient’s health, with doctors and nurses using scales to measure a patient’s weight. Examples Of Contexts Let’s take a closer look at some examples of how the choice between load and weight might change depending on the context: Construction: In construction, load is a crucial factor in determining the strength and stability of a building. Engineers need to consider the weight of the building materials, as well as the weight of the people and equipment that will be using the building. They need to ensure that the building can support the load without collapsing. Aerospace: In the aerospace industry, weight is a critical factor in designing aircraft. Engineers need to consider the weight of the aircraft, including the fuel, passengers, and cargo, in order to determine the amount of lift required to keep the aircraft in the air. Shipping: When it comes to shipping, both load and weight are important factors to consider. Load refers to the amount of cargo that can be safely carried by a ship, while weight is used to determine the cost of shipping the cargo. Exercise: In the fitness industry, weight is a common term used to describe the heaviness of weights used in strength training. Load, on the other hand, is used to describe the amount of weight being lifted by an individual. As you can see, the choice between load and weight can vary depending on the context in which they are used. It’s important to understand the difference between the two terms in order to use them correctly and effectively in different situations. Exceptions To The Rules While the rules for using load and weight are generally straightforward, there are some exceptions to keep in mind. Here are a few cases where the standard rules might not apply: 1. Fluids And Gases When dealing with fluids and gases, the terms load and weight can become a bit more complex. This is because these substances can change their shape and volume depending on the conditions they are in. In these cases, it is often more appropriate to use the term “mass” instead of weight. For example, a gallon of water has a mass of 3.78 kilograms, but its weight can vary depending on where it is located. On the surface of the Earth, it would weigh about 8.34 pounds, but on the moon, it would only weigh about 1.38 pounds due to the difference in gravity. 2. Moving Objects When dealing with moving objects, the rules for load and weight can also become more complicated. This is because the weight of an object can change depending on its velocity and direction of movement. In these cases, it is important to consider the object’s momentum and kinetic energy, as well as its weight. For example, a car traveling at 60 miles per hour has a much greater force and momentum than the same car traveling at 10 miles per hour. This means that the load on the car’s tires will be much greater at the higher speed, even though the weight of the car remains the same. 3. Non-standard Measurements In some cases, non-standard measurements might be used that require different rules for load and weight. For example, in the shipping industry, the term “deadweight tonnage” is often used to refer to the weight of cargo, fuel, and supplies on a ship. This measurement takes into account the weight of the cargo and any fuel or supplies needed for the journey, but does not include the weight of the ship itself. Another example might be in the construction industry, where the term “live load” is often used to refer to the weight of people, equipment, and materials that are temporarily on a building or structure. This is different from the “dead load,” which refers to the weight of the building or structure itself. Summary of Exceptions to the Rules| Case | Explanation | Example | --- | Fluids and Gases | Use “mass” instead of weight when dealing with substances that can change shape and volume. | A gallon of water has a mass of 3.78 kilograms, but its weight can vary depending on where it is located. | | Moving Objects | Consider the object’s momentum and kinetic energy, as well as its weight, when dealing with moving objects. | A car traveling at 60 miles per hour has a much greater force and momentum than the same car traveling at 10 miles per hour. | | Non-Standard Measurements | Some industries might use different terms or measurements that require different rules for load and weight. | The shipping industry uses “deadweight tonnage” to refer to the weight of cargo, fuel, and supplies on a ship. | Practice Exercises One of the best ways to improve your understanding and use of load and weight is to practice using them in sentences. Here are some practice exercises to help you do just that: Exercise 1: Fill In The Blank Choose the correct word (load or weight) to fill in the blank in each sentence. | Sentence | Answer | --- | | The _ of the package was too heavy for me to lift. | weight | | The truck was carrying a heavy _ of bricks. | load | | The _ on the bridge caused it to collapse. | load | | I can’t believe how much _ you can lift! | weight | Exercise 2: Sentence Writing Write a sentence using each of the following words: load weight overload underweight Example: The load of books in my backpack is too heavy for me to carry. Exercise 3: Multiple Choice Choose the correct word (load or weight) to complete each sentence. The ____ of the plane was too heavy for it to take off. The truck was carrying a heavy ____ of furniture. The ____ on the bridge caused it to collapse. I can’t believe how much ____ you can lift! Answers: weight load load weight By practicing these exercises, you will improve your understanding and use of load and weight in no time! Conclusion After exploring the differences between load and weight, it is clear that these two terms are often used interchangeably but have distinct meanings. Load refers to the amount of weight being carried or transported, while weight is the force exerted by an object due to gravity. It is important for writers and speakers to use these terms correctly in order to avoid confusion and accurately convey their intended meaning. In addition, understanding the nuances of grammar and language use can greatly improve communication and clarity in both personal and professional settings. Key Takeaways: Load and weight are often used interchangeably, but have distinct meanings. Load refers to the amount of weight being carried or transported, while weight is the force exerted by an object due to gravity. Using these terms correctly can improve communication and avoid confusion. Continuing to learn about grammar and language use can greatly enhance communication and clarity. By mastering the differences between load and weight, writers and speakers can improve their communication skills and avoid common mistakes. Remember to use these terms correctly and continue learning about grammar and language use for even greater success. Shawn Manaher Shawn Manaher is the founder and CEO of The Content Authority. He’s one part content manager, one part writing ninja organizer, and two parts leader of top content creators. You don’t even want to know what he calls pancakes. 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https://www.freemathhelp.com/forum/threads/repeating-or-terminating-decimal-number.117354/
Repeating or Terminating Decimal Number | Free Math Help Forum Home ForumsNew postsSearch forums What's newNew postsLatest activity Log inRegister What's newSearch Search [x] Search titles only By: SearchAdvanced search… New posts Search forums Menu Log in Register Install the app Install Forums Free Math Help Arithmetic Repeating or Terminating Decimal Number Thread starterharpazo Start dateAug 2, 2019 harpazo Full Member Joined Jan 31, 2013 Messages 891 Aug 2, 2019 #1 A rational number is defined as the quotient of two integers. When written as a decimal, the decimal will either repeat or terminate. By looking at the denominator of the rational number, there is a way to tell in advance whether its decimal representation will repeat or terminate. How is this done without getting too technical? H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Aug 2, 2019 #2 A fraction, reduced to lowest terms, will correspond to a terminating decimal if and only if the prime factorization of the denominator has only powers of 2 and 5. If there are any other factors it will be an infinitely repeating decimal. Reactions:Otis, harpazo, JeffM and 2 others Otis Elite Member Joined Apr 22, 2015 Messages 4,592 Aug 2, 2019 #3 harpazo said: … without getting too technical? Click to expand... If you'd like information about prime factorization, here is an explanation. The following serve as basic examples to go with Halls' explanation. The decimal form of 7/20 terminates; the prime factorization of 20 is 2 2×5 The decimal form of 7/30 repeats; the prime factorization of 30 is 2×3×5 ? Reactions:harpazo and topsquark harpazo Full Member Joined Jan 31, 2013 Messages 891 Aug 2, 2019 #4 Otis said: If you'd like information about prime factorization, here is an explanation. The following serve as basic examples to go with Halls' explanation. The decimal form of 7/20 terminates; the prime factorization of 20 is 2 2×5 The decimal form of 7/30 repeats; the prime factorization of 30 is 2×3×5 ? Click to expand... Nice explanation. Can you please provide at least two more examples? Last edited: Aug 2, 2019 Otis Elite Member Joined Apr 22, 2015 Messages 4,592 Aug 2, 2019 #5 harpazo said: Nice explanation. Can you please provide at least two more examples? Click to expand... ? If you understand prime factorizations, then you can create at least two more examples and post them. Use the explanation Halls provided in post #2. Otherwise, do you have a specific question? \; Reactions:topsquark harpazo Full Member Joined Jan 31, 2013 Messages 891 Aug 2, 2019 #6 Otis said: ? If you understand prime factorizations, then you can create at least two more examples and post them. Use the explanation Halls provided in post #2. Otherwise, do you have a specific question? \; Click to expand... This is it for tonight. More math tomorrow morning. Otis Elite Member Joined Apr 22, 2015 Messages 4,592 Aug 2, 2019 #7 harpazo said: … More math tomorrow morning. Click to expand... I'll wait to see whether that means you'll answer the question. \; Reactions:topsquark harpazo Full Member Joined Jan 31, 2013 Messages 891 Aug 3, 2019 #8 Otis said: I'll wait to see whether that means you'll answer the question. \; Click to expand... Tomorrow morning new questions will be posted from R.2, specific math questions, you know, with work shown. Otis Elite Member Joined Apr 22, 2015 Messages 4,592 Aug 3, 2019 #9 Do you have another question in this thread? Why are you asking for more examples? Reactions:topsquark topsquark Senior Member Joined Aug 27, 2012 Messages 2,370 Aug 3, 2019 #10 Probably because he wants you to do the work for him, and if you don't he'll move on to a new problem. That's been his pattern for a long while now. -Dan harpazo Full Member Joined Jan 31, 2013 Messages 891 Aug 3, 2019 #11 topsquark said: Probably because he wants you to do the work for him, and if you don't he'll move on to a new problem. That's been his pattern for a long while now. -Dan Click to expand... I requested two more samples for my personal math files. I not only read each reply. I also save what I think is important for me to know for further study. Nothing more, nothing less. topsquark Senior Member Joined Aug 27, 2012 Messages 2,370 Aug 3, 2019 #12 Well, then, go ahead and follow the advice in post #5. It'll help you get the idea in your mind more firmly. It's one thing to understanding a process but another to actually use that framework for yourself. It's working with the framework on new problems that actually instructs you. -Dan Reactions:mmm4444bot harpazo Full Member Joined Jan 31, 2013 Messages 891 Aug 3, 2019 #13 topsquark said: Well, then, go ahead and follow the advice in post #5. It'll help you get the idea in your mind more firmly. It's one thing to understanding a process but another to actually use that framework for yourself. It's working with the framework on new problems that actually instructs you. -Dan Click to expand... Thank you, Dan. Otis Elite Member Joined Apr 22, 2015 Messages 4,592 Aug 3, 2019 #14 harpazo said: I [want] more samples for my personal math files … Click to expand... As you have no further questions in this thread, you can create additional examples. \; Reactions:topsquark H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Aug 4, 2019 #15 I told you, in the first response to your question, that "A fraction, reduced to lowest terms, will correspond to a terminating decimal if and only if the prime factorization of the denominator has only powers of 2 and 5. If there are any other factors it will be an infinitely repeating decimal." Now, do you know what "prime numbers" are? Do you know how to "factor" a number? Do you know what "prime factorization" means? If you know those fairly basic things then you should be able to create your own examples. Try it and post a few here. Reactions:topsquark You must log in or register to reply here. 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https://www.studocu.com/en-us/document/syracuse-university/intermediate-microeconomics/slides-2/1778960
Slides #2 - CES production function: q = (k r + l r)g/r The Lagrangian expression for cost - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Slides #2 Notes from Slides Course Intermediate Microeconomics (ECN 301) 13 documents University Syracuse University Academic year:2016/2017 Uploaded by: Anonymous Student Syracuse University Recommended for you 2 ECN301 - Unit 1B - Economics notes with equations and definitions Intermediate Microeconomics Lecture notes 100% (3) 30 ECN 301 Entire Lecture Notes Intermediate Microeconomics Lecture notes 100% (1) Comments Please sign in or register to post comments. Report Document Students also viewed Hw1 - Notes from Slides HW2 - Notes from Slides HW3 - Notes from Slides HW5 - Notes from Slides Slides #4 Slides #1 Preview text CES production function: q = (k r + l r)g/r The Lagrangian expression for cost minimization of producing q0 is ℒ = vk + wl + l[q0 - (k r + l r)g/r] First-order conditions for a minimum ¶ ℒ/¶k = v - l(g/r)(kr + lr)(g-r)/r(r)kr-1 = 0 ¶ ℒ/¶l = w - l(g/r)(kr + lr)(g-r)/r(r)lr-1 = 0 ¶ ℒ/¶l = q0 - (k r + l r)g/r = 0 Total cost function –Shows that for any set of input costs and for any output level –The minimum cost incurred by the firm is C = C(v,w,q) –As output (q) increases, total costs Increase To produce one unit of output we need k1 units of capital –And l1 units of labor input C(q=1) = vk1 + wl 1 To produce m units of output –Assuming constant returns to scale C(q=m) = vmk1 + wml1 = m(vk1 + wl1) C(q=m) = m × C(q=1) Suppose that total costs start out as concave and then becomes convex as output increases –One possible explanation for this is that there is a third factor of production that is fixed as capital and labor usage expands –Total costs begin rising rapidly after diminishing returns set in Cost curves –Are drawn under the assumption that input prices and the level of technology are held constant –Any change in these factors will cause the cost curves to shift Homogeneity –Cost functions are all homogeneous of degree one in the input prices –A doubling of all input prices will not change the levels of inputs purchased Inflation will shift the cost curves up Nondecreasing in q, v, and w –Cost functions are derived from a costminimization process –Any decline in costs from an increase in one of the function’s arguments would lead to a contradiction Concave in input prices –Costs will be lower When a firm faces input prices that fluctuate around a given level Than when they remain constant at that level –The firm can adapt its input mix to take advantage of such fluctuations Some of these properties carry over to average and marginal costs –Homogeneity –Effects of v, w, and q are ambiguous With input prices w’ and v’ , total costs of producing q0 are C (v’, w’, q0). If the firm does not change its input mix, costs of producing q0 would follow the straight line CPSEUDO. With input substitution, actual costs C (v’, w, q0) will fall below this line, and hence the cost function is concave in w. Some of these properties carry over to average and marginal costs –Homogeneity –Effects of v, w, and q are ambiguous A change in the price of an input –Will cause the firm to alter its input mix –The change in k/l in response to a change in w/v, while holding q constant is Gives an alternative definition of the elasticity of substitution In the two-input case, s must be nonnegative Large values of s indicate that firms change their input mix significantly if input There are two types of short-run costs: –Short-run fixed costs are costs associated with fixed inputs (vk1) –Short-run variable costs are costs associated with variable inputs (wl) Short-run costs –Are not minimal costs for producing the various output levels –The firm does not have the flexibility of input choice –To vary its output in the short run, the firm must use nonoptimal input combinations –The RTS will not be equal to the ratio of input prices Because capital input is fixed at k, in the short run the firm cannot bring its RTS into equality with the ratio of input prices. Given the input prices, q0 should be produced with more labor and less capital than it will be in the short run, whereas q2 should be produced with more capital and less labor than it will be. The short-run average total cost (SAC) function is SAC = total costs/total output = SC/q The short-run marginal cost (SMC) function is SMC = change in SC/change in output = ¶SC/¶q By considering all possible levels of capital input, the long-run total cost curve (C ) can be traced. In (a), the underlying production function exhibits constant returns to scale: In the long run, although not in the short run, total costs are proportional to output. By considering all possible levels of capital input, the long-run total cost curve (C ) can be traced. In (b), the long-run total cost curve has a cubic shape, as do the short-run curves. Diminishing returns set in more sharply for the short-run curves, however, because of the assumed fixed level of capital input. This set of curves is derived from the total cost curves shown in Figure 10. The AC and MC curves have the usual U-shapes, as do the short-run curves. At q1, long-run average costs are minimized. The configuration of curves at this minimum point is important. At the minimum point of the AC curve: –The MC curve crosses the AC curve MC = AC at this point –The SAC curve is tangent to the AC curve SAC (for this level of k) is minimized at the same level of output as AC SMC intersects SAC also at this point AC = MC = SAC = SMC Slides #2 Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Slides #2 Course: Intermediate Microeconomics (ECN 301) 13 documents University: Syracuse University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save CES production function: q = (k r + l r)g/r • The Lagrangian expression for cost minimization of producing q 0 is = ℒ vk + wl + l[q 0 - (k r + l r)g/r] • First-order conditions for a minimum ¶ℒ/¶k = v - l(g/r)(k r + l r)(g-r)/r(r)k r-1 = 0 ¶ℒ/¶l = w - l(g/r)(k r + l r)(g-r)/r(r)l r-1 = 0 ¶ℒ/¶l = q 0 - (k r + l r)g/r = 0 Total cost function –Shows that for any set of input costs and for any output level –The minimum cost incurred by the firm is C = C(v,w,q) –As output (q) increases, total costs Increase To produce one unit of output we need – k 1 units of capital –And l 1 units of labor input C(q=1) = vk 1 + wl 1 • To produce m units of output –Assuming constant returns to scale C(q=m) = vmk 1 + wml 1 = m(vk 1 + wl 1) C(q=m) = m × C(q=1) Suppose that total costs start out as concave and then becomes convex as output increases –One possible explanation for this is that there is a third factor of production that is fixed as capital and labor usage expands –Total costs begin rising rapidly after diminishing returns set in Cost curves –Are drawn under the assumption that input prices and the level of technology are held constant –Any change in these factors will cause the cost curves to shift Homogeneity –Cost functions are all homogeneous of degree one in the input prices –A doubling of all input prices will not change the levels of inputs purchased – Inflation will shift the cost curves up Nondecreasing in q, v, and w –Cost functions are derived from a costmini mization process –Any decline in costs from an increase in one of the function’s arguments would lead to a contradiction Concave in input prices –Costs will be lower • When a firm faces input prices that fluctuate around a given level • Than when they remain constant at that level –The firm can adapt its input mix to take advantage of such fluctuations Some of these properties carry over to average and marginal costs –Homogeneity –Effects of v, w, and q are ambiguous With input prices w’ and v’ , total costs of producing q0 are C (v’, w’, q0). If the firm does not change its input mix, costs of producing q0 would follow the straight li ne CPSEUDO. With input substituti on, actual costs C (v’, w, q0) will fall below this line, and hence the cost function is concave in w. Some of these properties carry over to average and marginal costs –Homogeneity –Effects of v, w, and q are ambiguous A change in the price of an input –Will cause the firm to alter its input mix –The change in k/l in response to a change in w/v, while holding q constant is Gives an alternative definiti on of the elasticity of substituti on – In the two-input case, s must be nonnegative – Large values of s indicate that firms change their input mix significantl y if input prices change The increase in costs will be largely influenced by –The relative significance of the input in the production process –The ability of firms to substitute another input for the one that has risen in price Total costs are given by Ct(v,w,q) = qCt(v,w,1) = qC 0(v,w,1)/A(t) = C 0(v,w,q)/A(t) • Total costs –Decrease over time at the rate of technica l Change Contingent demand functions –For all of the firms inputs can be derived from the cost function • Shephard’s lemma –The contingent demand function for any input is given by the partial derivati ve of the total-cost function with respect to that input’s price Shepherd’s lemma – Is one result of the envelope theorem –The change in the optimal value • In a constrained optimization problem • With respect to one of the parameters • Can be found by differentiating the Lagrangian with respect to the changing Parameter In the short run –Economic actors have only limited flexibility in their actions • Assume –The capital input is held constant at k 1 –The firm is free to vary only its labor input • The production function becomes q = f(k 1,l) Short-run total cost for the firm is SC = vk 1 + wl Too long to read on your phone? Save to read later on your computer Save to a Studylist • There are two types of short-run costs: –Short-run fixed costs are costs associat ed with fixed inputs (vk 1) –Short-run variable costs are costs associated with variable inputs (wl) Short-run costs –Are not minimal costs for producing the various output levels –The firm does not have the flexibilit y of input choice –To vary its output in the short run, the firm must use nonoptimal input combinations –The RTS will not be equal to the ratio of input prices Because capital input is fixed at k, in the short run the firm cannot bring its RTS into equality with the ratio of input prices. Given the input price s, q0 should be produced with more labor and less capital than it will be in the short run, wherea s q2 should be produced with more capital and less labor than it will be. The short-run average total cost (SAC) function is SAC = total costs/total output = SC/q • The short-run marginal cost (SMC) function is SMC = change in SC/change in output = ¶SC/¶q By considering all possible levels of capital input, the long-run total cost curve (C ) can be traced. In (a), the underlying production function exhibit s constant returns to scale: In the long run, although not in the short run, total costs are proportional to output. By considering all possible levels of capital input, the long-run total cost curve (C ) can be traced. In (b), the long-run total cost curve has a cubic shape, as do the short-run curve s. Diminishing returns set in more sharply for the short-run curves, however, bec ause of the assumed fixed level of capital input. This set of curves is derived from the total cost curves shown in Figure 10.8. The AC a nd MC curves have the usual U-shapes, as do the short-run curves. At q1, long-run average costs are minimized. The configuration of curves at this minimum point is im portant. At the minimum point of the AC curve: –The MC curve crosses the AC curve • MC = AC at this point –The SAC curve is tangent to the AC curve • SAC (for this level of k) is minimized at the same level of output as AC • SMC intersects SAC also at this point AC = MC = SAC = SMC 1 out of 5 Share Download Download More from:Intermediate Microeconomics(ECN 301) More from: Intermediate MicroeconomicsECN 301Syracuse University 13 documents Go to course 2 ECN301 - Unit 1B - Economics notes with equations and definitions Intermediate Microeconomics Lecture notes 100% (3) 30 ECN 301 Entire Lecture Notes Intermediate Microeconomics Lecture notes 100% (1) 1 Unit 1C - Economic Lecture notes from unit 1C with equations and definitions Intermediate Microeconomics Lecture notes None 3 Unit 2A-2C - Economics notes with equations and definitions Intermediate Microeconomics Lecture notes None More from: Intermediate MicroeconomicsECN 301Syracuse University13 documents Go to course 2 ECN301 - Unit 1B - Economics notes with equations and definitions Intermediate Microeconomics 100% (3) 30 ECN 301 Entire Lecture Notes Intermediate Microeconomics 100% (1) 1 Unit 1C - Economic Lecture notes from unit 1C with equations and definitions Intermediate Microeconomics None 3 Unit 2A-2C - Economics notes with equations and definitions Intermediate Microeconomics None 2 HW5 - Notes from Slides Intermediate Microeconomics None 2 HW3 - Notes from Slides Intermediate Microeconomics None Recommended for you 2 ECN301 - Unit 1B - Economics notes with equations and definitions Intermediate Microeconomics Lecture notes 100% (3) 30 ECN 301 Entire Lecture Notes Intermediate Microeconomics Lecture notes 100% (1) 2 ECN301 - Unit 1B - Economics notes with equations and definitions Intermediate Microeconomics 100% (3) 30 ECN 301 Entire Lecture Notes Intermediate Microeconomics 100% (1) Students also viewed Hw1 - Notes from Slides HW2 - Notes from Slides HW3 - Notes from Slides HW5 - Notes from Slides Slides #4 Slides #1 Get homework AI help with the Studocu App Open the App English United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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https://www.youtube.com/watch?v=wy-mKM8aWoY
Stability of Geminal diol (Gem-diol). Chemistry : The Mystery of Molecules 9500 subscribers 60 likes Description 2252 views Posted: 14 May 2019 Geminaldiol, #Gemdiol, #dialcohol, #carbonyl, #Hydrogenbond, #Chloralhydrate, In this video, I shall discuss the stability of germinal diol. Actually, the equilibrium between the carbonyl group (C=O) and Gem-diol prefers to remain in the left side. Consequently, the percentage of the carbonyl group (C=O) in the equilibrium is more. However, changing the controlling factors like electron deficiency of the substituent attached with carbonyl group and further stabilization of the produced di alcohol (OH) by hydrogen bonding shifts the equilibrium towards the right side. I have discussed these phenomena through various examples which may be helpful for basic understanding from the molecular interaction point of view. Links for some related videos are given below. Please watch them for a better understanding. For "Video Lecture 5 (part II): Reactivity of borohydrides (MBH4) [mainly organic reactions]." please click on the link "Video Lecture: 5 (part I), Preparation, Bonding and Reactivity reactions) of borohydrides (MBH4)." For "Question Answer discussion about Luche reaction (for CSIR-NET, GATE, JAM)" please click on the link " For "Luche Reaction, selective reduction of ketone into secondary (2o) alcohol." please click on link " For "Wacker process for conversion of ethylene to acetaldehyde." please click on the link " For "Lindlar Catalyst, selective reduction of alkyne to cis-alkene." please click on the link " For "Stability of Geminal diol (Gem-diol)." please click on the link " 7 comments Transcript:
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https://pmc.ncbi.nlm.nih.gov/articles/PMC6129380/
Recent advances in the diagnosis and management of Gaucher disease - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice Expert Rev Endocrinol Metab . Author manuscript; available in PMC: 2019 Mar 12. Published in final edited form as: Expert Rev Endocrinol Metab. 2018 Mar 12;13(2):107–118. doi: 10.1080/17446651.2018.1445524 Search in PMC Search in PubMed View in NLM Catalog Add to search Recent advances in the diagnosis and management of Gaucher disease Sam E Gary Sam E Gary 1 Medical Genetics Branch, NHGRI, NIH, Bethesda, MD, USA Find articles by Sam E Gary 1,#, Emory Ryan Emory Ryan 1 Medical Genetics Branch, NHGRI, NIH, Bethesda, MD, USA Find articles by Emory Ryan 1,#, Alta M Steward Alta M Steward 1 Medical Genetics Branch, NHGRI, NIH, Bethesda, MD, USA Find articles by Alta M Steward 1, Ellen Sidransky Ellen Sidransky 1 Medical Genetics Branch, NHGRI, NIH, Bethesda, MD, USA Find articles by Ellen Sidransky 1 Author information Article notes Copyright and License information 1 Medical Genetics Branch, NHGRI, NIH, Bethesda, MD, USA ✉ CONTACT Ellen Sidransky sidranse@mail.nih.gov Medical Genetics Branch, National Human, Genome Research Institute, NIH, Building 35A, Room 1E623, Convent Drive, MSC 3708, Bethesda, MD, 20892-3708, USA, Contributed equally. Issue date 2018 Mar. PMC Copyright notice PMCID: PMC6129380 NIHMSID: NIHMS1500397 PMID: 30058864 The publisher's version of this article is available at Expert Rev Endocrinol Metab Abstract Introduction: Gaucher disease, the autosomal recessive deficiency of the lysosomal enzyme glucocerebrosidase, is associated with wide phenotypic diversity including non-neuronopathic, acute neuronopathic, and chronic neuronopathic forms. Overlap between types can render definitive diagnoses difficult. However, differentiating between the different phenotypes is essential due to the vast differences in clinical outcomes and response to therapy. Genotypic information is helpful, but cannot always be used to make clinical predictions. Current treatments for Gaucher disease, including enzyme replacement therapy and substrate reduction therapy, can reverse many of the non-neurological manifestations, but these therapies must be administered continually and are extremely costly. Areas covered: We reviewed the literature concerning the varied clinical presentations of Gaucher disease throughout the lifetime, along with treatment options, management goals, and current and future research challenges. A PubMed literature search was performed for relevant publications between 1991 to January 2018. Expert commentary: Interest and research in the field of Gaucher disease is rapidly expanding. However, significant barriers remain in our ability to predict phenotype, assess disease progression using objective biomarkers, and determine optimal treatment strategy on an individual basis. As the field grows, we anticipate identification of genetic modifiers, new biomarkers, and small-molecule chaperone therapies, which may improve patient quality of life. Keywords: Gaucher disease, glucocerebrosidase, genotype/phenotype correlation, enzyme replacement therapy, substrate reduction therapy, newborn screening, parkinsonism, chaperones, biomarkers 1. Introduction Gaucher disease (GD) is an autosomal recessive lysosomal storage disorder (LSD) caused by mutations in GBA1, resulting in a deficiency of the enzyme glucocerebrosidase (GCase; E. C3.2.1.45). GD is a rare pan-ethnic disease that affects approximately 1 in 50,000 to 1 in 100,000 people in the general population. Although it is a single-gene disorder, there is wide variation in the degree and severity of symptoms. The phenotypes encountered have historically been classified by type and severity of neurological involvement. Type 1 GD (GD1) is defined as the non-neuronopathic subclass and can present at any age. GD1 is far more common in individuals of Eastern and Central European (Ashkenazi) Jewish heritage, where the carrier frequency is as high as 1 in 16 . Acute neuronopathic GD (GD2) generally presents perinatally or within the first year of life and is characterized by rapid neurological decline. Chronic neuronopathic GD (GD3) often has an onset in early childhood, with a highly variable spectrum of associated neurological and non-neurological manifestations. The defining and most common feature of GD3 is slowing of the horizontal saccadic eye movements . While GD was first described almost 140 years ago, the past two decades have produced great advances in our appreciation of the phenotypic spectrum associated with this disorder, as well as our ability to treat patients. These new insights affect the diagnosis and treatment of patients at both ends of the age spectrum. This review aims to convey the varied clinical presentations along with the evolving diagnostic and treatment options for pediatric and adult patients with GD. 2. Pediatric diagnosis at different ages Establishing the initial diagnosis of GD can be challenging because of the highly variable phenotypic presentations and the overlap with other disorders. This is further complicated by an increasing number of asymptomatic children being diagnosed through genetic screenings. As treatment for GD can be costly, time consuming, and invasive, deciding when and how to begin treatment can be difficult. A clearer understanding of the nuances of different disease presentations at different ages (Box 1) can lead to better treatment and patient/parental counseling. Box 1. Presenting Features by Age. Newborn Congenital ichthyosis Organomegaly Failure to thrive Brain stem dysfunction- dysphagia, apnea, difficulty with secretions Hepatosplenomegaly Hematological abnormalities (anemia, thrombocytopenia) First year of life Failure to thrive Anemia/thrombocytopenia Brain stem dysfunction- progressive Saccadic gaze abnormalities Seizures Cardiac valvular stenosis Childhood Organomegaly Hematological abnormalities, including bleeding events Bone pain crisis Skeletal/bone involvement-avascular necrosis, osteopenia, pathologic fractures Saccadic gaze abnormalities Myoclonic epilepsy Adolescence Organomegaly Abnormal bleeding Delayed puberty Bone pain crises skeletal/bone involvement 2.1. Prenatal screening Although prenatal screening for lysosomal storage disorders (LSDs) is being performed more frequently in specific populations, there is no standard list of LSDs that are screened for across centers. While enzymatic assays can identify patients with different LSDs, typically these assays demonstrate limited reliability for identifying carriers. Population screening is now typically done by DNA analyses. Most often, GD is included in a panel of disorders that are targeted to the higher risk Ashkenazi Jewish population. In Israel, where such panels are frequently offered, more than half of the pediatric GD diagnoses since 2000 were a result of screening, rather than symptomatic presentation . In this population, screening for 4–8 common GBA1 mutations can identify up to 95% of mutant alleles, making such screens quite effective. In populations with different ethnic backgrounds, these panels identify a far lower percentage of disease-causing mutations. Therefore, sequencing of the entire GBA1 gene is recommended. This is not trivial, as the presence of a nearby highly homologous pseudogene complicates genotypic analyses, limiting wide-scale genomic sequencing on a population level. In affected families, where the GBA1 mutations are known, prenatal screening can easily be performed to detect specific mutations in carriers. Subsequent pregnancies can be screened via chorionic villi samples or amniotic fluid. If the mutations are not known, prenatal diagnosis can be performed by enzymatic assays in an experienced laboratory. 2.2. Newborn screening In the United States, several states have begun pilot programs that screen for selected LSDs. Currently, Missouri, New York, and Illinois have introduced programs which include GD among other LSDs. Several other states have a more limited approach or are still considering implementation of testing [4,5]. Newborn screening for disorders like GD is a highly debated topic, as such screens will also identify infants who may be asymptomatic for years or even a lifetime. In a disorder like GD, where many affected individuals never reach medical attention, the negative impact of knowing about this chronic condition, including parental anxiety and financial implications, can be substantial. Others argue that this impact is greatly overshadowed by the benefits of early symptom recognition and treatment, which may prevent numerous long-term sequelae . 2.3. Children identified at birth without symptoms Prenatal and newborn screenings, as well as early screening in families of children with GD, have led to the identification of a cohort of children with GD, who are asymptomatic. There have been three studies describing these children. The first, conducted in New York, followed 38 patients aged 1–18 years (mean age at last visit 6.9 ± 4.1 years) who were diagnosed with GD1 either prenatally or postnatally by molecular genetic testing. Only a minority had hematological (5%), bone (15%), or linear growth (19%) issues, and just 12% displayed splenomegaly. Disease severity, assessed by the pediatric Gaucher severity score, remained stable and within the mild disease range for most (95%). During the 1–18 years of follow-up, treatment was recommended for four of the children . A similar study, conducted in Israel, followed 40 children diagnosed through screening since 2000 . Only 4 of the 40 required therapy. A third study, reporting the results of newborn screening for LSDs in Illinois, described five infants with GD. None required treatment in the 15-month follow-up period . It is likely that asymptomatic children will be identified with greater frequency in the coming years. Longitudinal studies following these cohorts will be imperative. Currently, routine monitoring of asymptomatic individuals identified through screening is recommended, so treatment can be initiated quickly, should symptoms develop. 2.4. Symptomatic presentations in the prenatal and neonatal period Symptoms appearing prenatally or in the neonatal period are usually associated with GD2, though presentations differ. Hydrops fetalis, the abnormal accumulation of fluid in at least two body areas, has been described in cases of GD2 and may present at birth or perinatally . GD and other LSDs remain a small but significant cause of hydrops fetalis, with some variability in incidence based on population [9–12]. As such, GD is not usually considered until a couple experiences multiple spontaneous abortions or perinatal deaths. The etiology of approximately 20% of cases of hydrops fetalis remains unknown. Testing for GD and other LSDs may elucidate the underlying cause in some cases and inform prenatal counseling . Congenital ichthyosis, the collodion baby phenotype, can also be associated with GD2. The collodion phenotype presents clinically at birth as a thick and shiny membrane, resembling cellophane wrap, covering the skin. This is also associated with other autosomal recessive conditions and, in GD, is a hallmark of severe neuronopathic disease [13,14]. While dysmorphic features are not generally associated with GD, when present, they are most often associated with the collodion baby phenotype . Other infants with GD2 may present in the neonatal period with petechiae , blueberry muffin lesions, oozing, or retinal bleeding, secondary to thrombocytopenia. Many also have massive hepatosplenomegaly. In the newborn nursery, affected infants characteristically have stridor, poor suck, and difficulty handling secretions. GD2 is further characterized by rapid neurological decline, manifesting as brain stem dysfunction, which may include supranuclear gaze palsy, irritability, hypertonia, and hypokinesia. The progression of symptoms typically includes dysphagia, stridor, pyramidal signs, failure to thrive, cachexia, and convergent strabismus, with death occurring within the first months to years of life [14,16]. 2.5. Symptomatic presentations in the first year of life Both acute and chronic neuronopathic, as well non-neuronopathic GD, may manifest in middle-to-late infancy. An accurate assessment of the phenotype is essential, as the difference in prognosis is striking (Figure 1). Figure 1. Open in a new tab Photo of an 14-month-old child with hepatosplenomegaly and growth delay, recently diagnosed with Gaucher disease. Because he presented in the first year of life, his parents were initially told that he had type 2 Gaucher disease. While this is clearly not the case, it has been challenging to determine whether he has type 1 or type 3 Gaucher disease. He currently demonstrates no neurological features, yet his genotype is L444P/L444P. Picture with written parent consent. In classic GD2, infants may appear normal during the immediate newborn period but begin to develop symptoms as described in the above section regarding neonatal onset, with death generally ensuing in the first 18 months. In recent years, more aggressive management including tracheostomy, gastrostomy, and enzyme replacement therapy (ERT) has increased the lifespan. However, arrest or regression of developmental milestones, seizures, and myoclonus usually follow [14,16]. While the visceral symptoms can be improved with ERT, there is little impact on neurodegeneration, as ERT does not cross the blood–brain barrier (BBB). Patients with GD3 often present during the first year of life with massive organomegaly, anemia or thrombocytopenia, and a horizontal supranuclear gaze palsy. Most children meet developmental milestones, though their ability to crawl or walk can be impeded by their massive organomegaly. These children, in contrast to those with GD2, show remarkable improvement with early ERT initiation. Dramatic reversal of non-neurological manifestations, rapid catch-up of motor skills, and weight gain are frequently seen. In GD3 patients with neurodegeneration, progression, while not affected by any available treatment, is markedly slower than in GD2. Patients with GD1 can also present in the first year of life. Usually, they are first noted to have organomegaly or cytopenia, often following an unrelated viral illness. In recent years, it has become possible to establish the diagnosis of GD using dried blood spot analyses . This has enabled physicians to diagnose patients who do not have access to appropriate testing facilities in geographically remote regions. These dried blood spots can also be used to measure glucosylsphingosine levels, which show promise as a biomarker for disease severity and response to treatment. 2.6. Symptomatic presentations in childhood and adolescence Both GD1 and GD3 can present in the pediatric years, though GD1 is more common. A recent survey of 212 patients with GD1 in the United States found that over 50% of those who received medical attention were diagnosed in childhood or adolescence . In childhood and adolescence, presenting symptoms of GD1 are primarily hematological, leading to frequent misdiagnosis of GD1 as a hematological/oncological disorder . These symptoms may include anemia, thrombocytopenia, nosebleeds, excessive bleeding, and splenomegaly . Splenomegaly is present in about 95% of children diagnosed with GD and usually appears prior to other manifestations [6,20]. GD3 may present in childhood with splenomegaly, anemia, thrombocytopenia, bone crises, or kyphosis, along with neurological manifestations. In a rare subset of patients with GD3, severe cardiac valve involvement may develop. The diagnosis of GD3 is frequently made by a neuro-ophthalmologist when abnormal eye movements are noted . It is recommended that treatment begin early in symptomatic children with GD1 and GD3 to avoid irreversible bony and visceral damage as well as other long-term growth and development issues . Short stature or growth retardation is prevalent in patients with both GD1 and GD3 [6,23]. Prior to development of ERT, patients with severe phenotypes of GD often experienced a delay of puberty. When treated, these children had a normalized onset of puberty and a corrected growth curve, both in stature and lean muscle mass [3,24]. However, even treated patients may not reach a full expected height . Bone manifestations in GD, which typically appear in adolescence, are varied and can include osteonecrosis, bone pain crises, lytic lesions, osteoporosis (across the lifespan), pathological fractures, and the Erlenmeyer flask deformity. In one study, this characteristic deformity was present in 63% of patients who presented with GD prior to age 10 . Additionally, altered bone density is most pronounced during and after adolescence . Decreases in bone density during this time of bone maturation lead to a decrease in overall peak bone mass. Even in asymptomatic patients with normal bone marrow density, there may be underlying disruption of the bone’s trabecular architecture, causing a decrease in bone stability . While the basis of the bone pathology observed in GD is still only partially understood, it is estimated that up to 75–90% of patients diagnosed with GD will experience some bone findings throughout the course of their disease [26,28]. Treatment during adolescence, when needed, may ameliorate some of these affects, increasing bone health and stability later in life . 2.6.1. Neurological manifestations in childhood and adolescence Neurological manifestations are a hallmark of GD3 and can present in childhood, adolescence, or even adulthood . The most common manifestation is horizontal supranuclear gaze palsy, and in some individuals, this is the only neurological symptom . These eye movement abnormalities are characterized by a slowing, looping, or absence of the horizontal saccades. Vertical saccades may also be affected to a lesser degree [21,30]. Some individuals develop additional neurological manifestations including seizures, myoclonus, and progressive myoclonic epilepsy. Electroencephalography can reveal generalized background slowing or epileptiform discharges [21,31]. Abnormal brainstem auditory evoked potentials in addition to abnormal somatosensory evoked potentials have also been noted in some individuals [32,33]. The range of neurological manifestations is broad and the severity varies greatly. Some individuals exhibit intelligence quotients (IQs) well below average, with several learning and functional deficits, while others have IQs in the superior range and complete college and advanced degrees [21,31,34]. The performance IQ is typically lower than the verbal IQ, suggesting a visual-spatial deficit visuospatial deficit that may be related to eye movement or other motor issues. This disparity is inconsistent with what is seen in age-matched controls and persists across all ranges of the IQ scale [31,34]. The neurological manifestations appear resistant to ERT and can worsen with disease progression . However, it is important to note that decline is not seen in all individuals, and some improve over time . 3. Diagnosis and symptomatic presentations in adulthood Symptoms presenting in adulthood are often similar to those in childhood and include anemia, thrombocytopenia, hepatosplenomegaly, and bone manifestations. In general, patients who only develop symptoms in adulthood are considered to have milder GD, but there are exceptions. Women may present with an exacerbation of GD symptoms around menarche or initially notice heavier bleeding during their menstrual cycles . Pregnancy can exacerbate GD, and bleeding complications in pregnancy frequently lead to diagnosis in adulthood. Normal pregnancy outcomes are reported in both symptomatic treated and asymptomatic untreated patients with GD . Currently, the recommendations for pregnancy are that asymptomatic patients should not begin therapy unless necessary, and symptomatic-treated patients should continue ERT, as its safety has been established during pregnancy. Substrate reduction therapy (SRT) should be discontinued, as it could cause fetal harm . There are no indications that GD affects fertility . Other adult manifestations of GD may include pulmonary fibrosis and pulmonary hypertension. The prevalence of subclinical pulmonary hypertension is not well established, but is relatively rare. It may become life threatening, especially in patients who have undergone a splenectomy . ERT has been shown to decrease pulmonary hypertension but does not affect the fibrotic changes that have already occurred . Gallstones are also seen with an increased frequency in adults with GD . 3.1. Patients with malignancies During the pre-ERT era, autopsy studies demonstrated that cancers contributed frequently to premature death in patients with GD . Currently, multiple myeloma appears to be the most commonly linked malignancy, with a relative risk range of 37.5–51.5 in longitudinal studies [44–46]. Furthermore, a recent study demonstrated that clonal immunoglobulin in 17 of 20 patients with GD, as well as 6 of 6 gba−/− mice, with monoclonal gammopathy was reactive against glucoslysphingosine (LGL1) . Other hematological malignancies are implicated with GD at a relative risk as high as 12.7 [45,48,49]. Moreover, cases of rare malignancies such as ovarian dysgerminoma and neuroblastoma have been reported in patients with GD [50,51]. Many cancer therapeutic investigations have focused on the glycosphingolipid pathway, with particular interest in ceramide, one product of the hydrolysis of glucosylceramide by GCase, because of its proapoptotic and anticancer properties . Conversely, sphingosine-1-phosphate, another reactive biolipid that is a product of glycosphingolipid metabolism, has been shown to have promitogenic, antiapoptotic, and pro-angiogenic properties . In a mouse model of GD that develops myeloma, SRT administration with eliglustat tartrate reduced glucosylsphingosine levels and prevented the development of myelomas [47,54]. However, it is still unclear whether ERT and SRT impact the frequency of malignancies in patients with GD. 3.2. Parkinsonism and GD Interest in GBA1, and GD research in general, has increased markedly since the connection was established between GD and the more common disorder, Parkinson disease (PD) . A multicenter study of 5691 patients with PD and 4898 healthy controls demonstrated that patients with PD have an odds ratio of 5.43 for carrying a GBA1 mutation . Furthermore, heterozygous mutations in GBA1 are the most common known genetic risk factor for PD and associated Lewy body disorders. In one study, 25% of patients with GD had a first-degree relative who was a carrier for GD and developed parkinsonism . Many movement disorder centers now screen their patients for GBA1 mutations, and they have identified both patients and carriers of GD. GBA1-associated parkinsonian manifestations often resemble those of sporadic PD with good response to levadopa. Some studies show an earlier age of onset of parkinsonian symptoms, greater cognitive deficits, and a more rapid progression of motor impairment in GBA1-associated PD . In some individuals, manifestations can more closely reflect features of dementia with Lewy bodies (DLB) . Still, symptoms and progression can vary greatly. Studies investigating the molecular cause of GBA1-PD have focused on the inverse relationship between GCase levels and α-synuclein [59–61], the protein aggregated in Lewy bodies. Two predominating hypotheses have been proposed. The gain-of-function hypothesis states that α-synuclein accumulates and aggregates with misfolded GCase. However, null alleles in GBA1 are associated with an even higher risk of developing PD . Since the gain-of-function hypothesis depends on simultaneous mutant, misfolded GCase, and α-synuclein accumulation, the observation that null alleles are also associated with a higher risk of PD refutes this hypothesis. On the other hand, the loss-of-function hypothesis states that loss of GCase activity would result in substrate accumulation and change in α-synuclein homeostasis. However, the fact that most patients with GD do not develop PD suggests that low levels of GCase alone are insufficient to cause PD. 4. The challenges in predicting clinical phenotype Understanding the basis for the phenotypic variation in GD has proven to be challenging at all points throughout the lifetime. Neither the amount of residual enzymatic activity , nor the quantity of lipid stored , has correlated well with the patient phenotype. In the age of molecular diagnostics, there was considerable hope that the genotype might be used to predict the clinical course. To date, researchers have identified over 400 different mutations in GBA1 in patients with GD, but phenotypic implications of these mutations are not always clear. For example, siblings who share the same mutations can have different symptoms, responses to treatment, and disease courses. This has stimulated research to identify genetic modifiers that may impact Gaucher pheno-types. Among the modifier genes proposed for GBA1 are CLN8 , GRIN2b , and PGRN . Despite the lack of clear correlations, there are several mutations that are associated with neuronopathic and non-neuronopathic illness. The L444P allele has been encountered in all three types of GD, although homozygotes generally have GD3 and many have no progressive neurodegeneration . L444P homozygotes are common among all ethnicities, and pheno-types can range from highly successful college graduates to children with severe autism , indicating the contribution of genetic modifiers. R463C is another mutation frequently seen with GD3, often accompanied by an L444P, null, or recombinant mutation on the second allele. Among patients with GD3 and myoclonic epilepsy, mutations V394L, G377S, and N188S are frequently found. D409H homozygotes have a unique cardiac phenotype which can include cardiac valve abnormalities, cor-neal opacities, and, at times, hydrocephalus . Many genotypes are associated with GD2, including some rare mutations specific to individual families and others that are more common. An L444P allele might be seen in combination with a recombinant allele that includes L444P (L444P/rec), though true homozygosity for L444P is rarely associated with GD2. Homozygosity for a null allele, or a recombinant allele encompassing large segments of the GBA1 pseudogene, is associated with neonatal lethality . In the United States, Europe, and Israel, the majority of patients presenting after the second decade have at least one N370S allele, and the frequency of new patients who are N370S homozygotes increases as the population ages . N370S homozygosity is solely associated with non-neuronopathic GD although it is seen in patients with PD . The N370S allele is the most common mutation in the Ashkenazi Jewish population and is often associated with milder visceral disease, especially in the homozygous state. However, even among N370S homozygotes, symptoms vary, specifically the presence and severity of bone disease . Avascular necrosis, especially of the hip, and other bone infarcts, can have debilitating effect on joints, leading to a need for joint replacement [72,73]. While these genotype associations are helpful in some cases, there are still many instances where no clear phenotypic predictions can be made. Overlap in phenotype can occur throughout the entire lifetime and makes definitive diagnosis and prediction of disease progression difficult. During infancy and early in the disease course, saccadic eye movements and other diagnostic measures often cannot be assessed accurately. With an ambiguous genotype, it may be difficult to distinguish between a child with severe GD1 and a child with milder GD3 (Figure 1). Additionally, the distinction between GD2 and GD3 is not always clear. Goker-Alpan et al. described a series of children with an intermediate phenotype who were diagnosed in the first year of life and lived from 2 to 8 years. These children attained some developmental milestones, including walking and talking, but ultimately succumbed to neurological disease during the first decade of life . Distinguishing between disease subtypes is essential, as symptomatic infants with GD1 or GD3 should begin ERT as soon as possible. Studies completed by the International Gaucher Registry demonstrated that pediatric patients with GD3 show great improvement after treatment with ERT. Amelioration of most hematological and visceral symptoms is seen in under 12 months, with stabilization of all non-neurological parameters within 5 years . Because ERT does not affect the neurological symptoms or progression of the disease, treatment for GD2 is controversial, and management is generally supportive or palliative. As such, there is a great need for methods to distinguish between the different phenotypes during young ages. Skin biopsies performed for specific electron microscopy structural studies may be used to discriminate GD2 from GD3. In a series of 25 skin biopsies from individuals with GD, 20 babies (100%) with GD2 were found to have uniquely disrupted epidermal ultrastructure . However, this evaluation is not readily available. On the other end of the age spectrum, many questions have been raised regarding genotype–phenotype associations and the impact of genotype on the development of parkinsonism and dementia. Some studies suggest differential degrees of cognitive decline based upon the ‘severity’ of GBA1 mutations [62,77]. However, polymorphisms in GBA1 that do not cause GD, such as E326K and T369M, are common in patients with parkinsonism [78,79], as are mild mutations like N370S. Gary et al. suggest that because N370S is the most frequent GBA1 allele in patients with DLB [81–83], ‘severity’ of the genotype alone cannot account for the degree of cognitive decline . As GBA1 receives increased attention in the movement disorders community, it is likely that genetic modifiers will reveal the basis for some of this variability. Understanding the relationship of common mutations to different phenotypes presenting throughout the entire lifespan is important for patient counseling and treatment. New assays or biomarkers may provide additional information to inform phenotypic predictions. 5. Biomarkers of disease severity and therapeutic efficacy Objective measures that reflect disease severity and can be evaluated as therapeutic outcomes are necessary for assessing GD and developing a treatment strategy. This may be increasingly important as more patients are diagnosed pre-symptomatically through different screening programs. Chitotriosidase has been employed as a biomarker for over a decade, but some patients have an inherited deficiency of this enzyme [84,85]. Currently, LysoGb1 appears to be the most sensitive and predictive biomarker of GD symptoms such as thrombocytopenia and splenomegaly [86,87]. Other potential molecular biomarkers include low-density lipoprotein , osteoactivin , serum ferritin , progranulin , and CCL18 [85,92,93]. Imaging modalities such as magnetic resonance imaging [94,95], dual energy X-ray absorptiometry , Fourier transform infrared spectroscopy , and transient and shear-wave elastography have also been employed to study key features of GD. However, further studies are necessary to establish the efficacy of these imaging methodologies. To date, there is no gold standard biomarker that can confidently predict the key features of GD. While many potential biomarkers have been identified, each has inherent limitations; so, clinical parameters continue to define treatment goals and assess outcomes. 6. Therapeutic treatment goals for GD The European Working Group on GD separates treatment goals into short- and long-term categories (Table 1). This model includes fatigue, quality of life, and prevention of additional long-term sequelae not addressed in earlier guidelines. This more holistic view highlights a deeper understanding of the long-term impact of GD and establishes standard outcomes in treated patients across all therapeutic modalities. Table 1. Modified from Biegstraaten et al. . | | Short-term goals for management of Gaucher disease type 1 | Long-term goals for management of Gaucher disease type 1 | :---: | Anemia-related symptoms | Eliminated blood transfusion dependency | Maintain normal hemoglobin values | | | Increase hemoglobin levels within 12–24 months to normal values of age and sex | | | Bleeding tendency | Increase platelet counts sufficiently to prevent surgical, obstetrical, and spontaneous bleeding | Maintain platelet count of ≥100,000/mm 3 | | | In patients with splenectomy - normalization of platelet count by 1 year of treatment | | | | In patients with an intact spleen: achieve platelet count of ≥100,000/mm 3 by 3 years of treatment | | | Mobility | Lessen bone pain that is not related to irreversible bone disease within 1–2 years | Prevent bone complications: avascular necrosis, bone crises, bone infarcts, and pathological fractures | | | Decrease bone marrow involvement, as measured by a locally used scoring system (e.g. BMB score or DGS) in patients without severe irreversible bone disease at baseline | Prevent osteopenia and osteoporosis (i.e. maintain BMD T-scores (DEXA) of >−1) | | | Increase BMD by 2 years in adults for patients with a T-score below −2.5 at baseline | Prevent chronic use of analgesic medication for bone pain | | | Attain normal or ideal peak skeletal mass in children | Maintain normal mobility or, if impaired at diagnosis, improve mobility | | | Normalize growth such that the height of the patient is in line with target height, based upon population standards and parental height, within 2 years of treatment | | | Visceral complications | Avoid splenectomy, if at all possible | Maintain spleen volume of <2–8 times normal | | | Alleviate symptoms due to splenomegaly: abdominal distension, early satiety, new splenic infarction | Maintain normal or (near) normal liver volume | | | Eliminate hypersplenism | Prevent liver fibrosis, cirrhosis, and portal hypertension | | | Reduce spleen volume to <2–8 times normal (or in absence of volume measurement tools reduce spleen size) by year 1–2, depending on baseline spleen volume | | | | Reduce the liver volume to 1.0–1.5 times normal (or in absence of volume measurement tools aim for normal liver size) by year 1–2, depending on baseline liver volume | | | General well-being | Improve scores from baseline of a validated quality-of-life instrument within 2–3 years or less depending on disease burden | Maintain good quality of life as measured by a validated instrument | | | Reduce fatigue (not anemia related) as measured by a validated fatigue scoring system | Maintain normal participation in school and work activities | | | Improve or restore physical function for carrying out normal daily activities and fulfilling functional roles | Minimize psychosocial burdens of life-long treatment | | | | Achieve normal onset of puberty Normalize life expectancy | | Pulmonary complications Prevent or improve pulmonary disease, pulmonary hypertension, and hepatopulmonary syndrome | | Pregnancy and delivery Prevent GD-related complications during pregnancy and delivery | Open in a new tab BMB: bone marrow burden; BMD: bone marrow density; DGS: düsseldorf Gaucher score. 6.1. Therapeutic management of GD Historically, GD1 was treated with supportive measures such as splenectomy and orthopedic procedures. Today, new therapeutics have dramatically altered the natural history of the disease both in children and adults. Approved therapies include ERT and SRT (Box 2); other therapeutic strategies are currently in development. Box 2. Approved First-line Therapies for Gaucher Disease. Enzyme Replacement Therapy (imiglucerase, velaglucerase alfa, taliglucerase alfa) Recombinant enzyme Administered intravenously Few adverse events No demonstration of superiority between the three available ERTs Very costly, life-long therapy Not effective for GD2 Substrate Reduction Therapy (eliglustat tartrate and miglustat) Administered orally Approved for use in adults but not children Dosage and use depends on rate of CYP2D6 metabolism More frequent adverse events than ERT Similar cost to ERT, life-long therapy Not effective for GD2 6.1.1. ERT ERT has revolutionized the treatment of GD and has markedly improved the prognoses of patients with GD1 and GD3. The concept of treating GD with ERT was first introduced by Brady and was implemented clinically in 1991 as a US FDA-approved therapy for patients with GD1 and later, GD3 . Placenta-derived alglucersae (Ceredase™, Genzyme) was the first ERT employed to treat GD. The infused enzyme was sequentially deglycosylated to expose specific glycoforms in an attempt to better target ERT to affected macrophages . Alglucerase was replaced by human recombinant imiglucerase (Cerezyme™, Sanofi/Genzyme) in the mid-1990s and continues to be the most widely used ERT in clinics across the world. In 2009, a vesivirus 2117 infection of the bioreactors in which imiglucerase was manufactured led to a global shortage of imiglucerase and expedited the approval of two ‘new’ enzymes: gene-activated human recombinant velaglucerase alfa (VPRIV™, Shire) and plant cell-derived human recombinant taliglucerase alfa (Elelyso™ Protalix/Pfizer). Although these three ERTs are not biosimilars, there appears to be no difference in safety between them [102–104]. In terms of efficacy, when administered prior to irreversible skeletal manifestations, most patients do extremely well, with correction of disease parameters. While dosage varies depending on the country and health of the individual, the recommended starting doses are 60 U/kg every other week for children and 30–60 U/kg every other week for adults . Dose reduction is reserved for individuals who have met therapeutic goals and therefore have ‘stable GD1’ . Some reports have decreased the dosage to 30 U/kg/month and demonstrated positive visceral and hematological outcomes [107,108]. Doses below 15 U/kg every other week (30 U/kg/month), however, have failed to prevent painful bone crises in some patients [109,110]. Although ERT dramatically improves the systemic symptoms of GD, virtually eliminating splenectomy as a treatment, neurological manifestations are not impacted by available therapies. Despite this, ERT has improved the quality of life of those with GD in many ways. For example, after years of ERT, some patients with GD are often able to take ‘drug holidays,’ or temporary delays in enzyme administration of up to a few months, without redevelopment of symptoms . Early administration of ERT has also been shown to positively impact the growth of children [25,112]. Furthermore, ERT has been shown to decrease bleeding during pregnancy, delivery, and postpartum and improves overall outcomes of mothers who have suffered previous miscarriages . Substantial efforts have been made to improve clinical recognition of GD and to initiate early treatment. However, Mehta and colleagues recently reported that one in six patients with GD experience a delay in treatment initiation of 7 years or more, following their first physician consultation presenting with hallmark GD symptoms . To further compound the delay in treatment, there continues to be a disparity in ERT availability and affordability worldwide. This has led to children with GD1 who, despite the proven efficacy of the enzyme, are unable to afford or gain access to ERT and die due to GD-related complications . 6.1.2. SRT SRT is an alternate treatment strategy first proposed by Radin . While ERT aims to rectify the absence and/or inactivity of GCase, SRT decreases glucocerebroside (GlcCer) production, making residual GCase activity sufficient to cleave the remaining lysosomal GlcCer. SRT was predicted to be more appealing to patients because it is administered orally, unlike ERT. Additionally, it was predicted to cost less, act on tissues not affected by ERT, and avoid potential immune responses that may be associated with infused proteins . Miglustat (N-butyldeoxynojirimycin; Zavesca™, Actelion) was the first SRT approved by the FDA and EMA in 2002. Although miglustat improved several hallmark features of GD, including hemoglobin concentrations, platelet counts, hepatosplenomegaly, and bone density [116–119], several adverse effects were reported. These included diarrhea, weight loss, tremor, and peripheral neuropathy. As a result, 49 of 115 patients discontinued use of miglustat after experiencing such side effects [120–122]. A newly developed SRT, eliglustat tartrate (Cerdelga™, Sanofi/Genzyme), was approved in 2014. Eliglustat is administered to CYP2D6 extensive, intermediate, or poor metabolizers since it is eliminated via the CYP2D6 pathway. The recommended dosage of eliglustat is lower in poor metabolizers, and use is contraindicated in patients who are ultrarapid metabolizers due to difficulty obtaining reliable blood levels of the drug . Although eliglustat is currently used as a first-line therapeutic, it is necessary to continue to compare eliglustat’s performance against ERT using consistent parameters and nomenclature , as there has been a lack of clarity when comparing results from ERT and eliglustat trials . Concerns regarding lack of long-term safety data, in addition to high annual costs, have led many individuals to continue ERT, rather than switch to SRT. New forms of SRT engineered to cross the BBB are currently being developed and tested. 6.1.3. Pharmacological chaperones Though not currently approved for clinical use, pharmacological chaperones have been the subject of intense investigation due to their ability to pass through the BBB, potentially ameliorating the neuronopathic features of GD2 and GD3. Small molecule chaperones decrease the amount of GCase retained in the endoplasmic reticulum (ER) by facilitating proper GCase folding and translocation to the lysosome. This reduces physiological stress on the ER and ER-associated GCase degradation, increasing lysosomal levels of GCase. The first chaperone administered to patients with GD was afegostat tartrate (Plicera™ Amicus Therapeutics) . In a phase 2 randomized, open-label study to assess the safety, tolerability, and preliminary efficacy in treatment-naive adult patients with GD1, results were disappointing . Another chaperone, ambroxol, is a pH-dependent mixed inhibitor of GCase, which appears to facilitate translocation in cellular and mouse models; a human study is underway [127–129]. In a small, non-randomized study by Narita and colleagues , ambroxol combined with ERT was shown to cross the BBB, increase GCase activity in lymphocytes, and positively impact patients with myoclonic epilepsy. However, the efficacy of inhibitory active-site chaperones like ambroxol and afegostat tartrate may be limited because they must out-compete GlcCer at its active site to facilitate folding and translocation of GCase. Non-inhibitory chaperones, which bind to a site distant from the active site, offer some advantages. They do not need to outcompete GlcCer to act upon GCase, and binding to GCase can be ‘programmed’ to release the chaperone once GCase is delivered to the lysosome. This frees GCase to cleave its glycolipid substrates. Two such chaperones have been identified. In primary macrophages, differentiated from patient-derived monocytes and induced pluripotent stem cells (iPSCs), these chaperones increased GCase activity, decreased substrate accumulation, and restored chemotaxis . In iPSC-derived dopaminergic neurons, the same chaperones increased GCase activity, decreased substrate storage, and lowered levels of α-synuclein . These results suggest that non-inhibitory chaperones are promising drug candidates and might prove useful in patients with GBA1-associated parkinsonism, as long as there is residual GCase activity. However, development of new non-inhibitory chaperones has been slow due to lack of GCase-specific fluorescent-based substrates that can measure the GCase activity in the lysosomes . Ultimately, the optimal therapeutic strategy may include a combination of therapies. ERT may be useful to initiate the ‘debulking’ of diseased cells in patients with significant clinical disease burden. Subsequently, or in less-affected patients, SRT or chaperones may be able to prevent re-accumulation of the substrate. Combining ERT with chaperone therapy may lower the required dose and subsequent cost. 7. Conclusions The clinical features of GD have changed drastically due to early screening and the administration of ERT. Although there have been significant improvements to the disease prognosis since ERT was first approved over 25 years ago, there are still many unmet challenges (Box 3). There is an urgent need for less costly therapies, as well as treatments that ameliorate the neurological manifestations seen in patients with GD2 and GD3. Improved diagnostics are needed to facilitate earlier treatment, and reliable biomarkers are needed to assess the efficacy of new therapies. Next-generation sequencing and genotype/phenotype studies may identify potential biomarkers and uncover other factors contributing to the clinical variation encountered in this disorder. These studies may also improve our understanding of the link between GD and PD, as well as GD and multiple myeloma, and will likely inform our understanding of the pathogenesis of these diseases. Box 3. Unmet Needs in Gaucher Disease. Therapy that crosses the blood-brain-barrier Less costly therapies Identification of other factors that impact including disease severity Improved understanding of specific aspects Bone disease Chronic fatigue and inflammation Pulmonary hypertension Gaucheromas Parkinsonism Association with malignancies 7.1. Expert commentary The spectrum of phenotypes associated with GD continues to expand. The number of children without symptoms identified through screening programs is dramatically increasing, providing new challenges for clinicians in predicting outcomes. While genotypic assessments can be helpful, clearly, there are other factors impacting disease phenotype. Detecting symptomatic GD in very young patients is essential for disease management and genetic counseling. While early intervention provides great benefit in symptomatic children with GD1 and GD3, palliative care might be considered early on for those with GD2. Identifying the different factors that impact disease phenotype will likely provide new insights into disease pathogenesis. Today, there are several effective therapeutic options for patients with GD. Children can be treated with the various forms of ERT, and adults have the option of ERT or SRT. Both are life-long, extremely costly treatments and thus not available in all parts of the world. New therapies, and combinations of therapies, are in different stages of development and will require further evaluation. Other remaining challenges are treatments for neuronopathic GD, understanding inflammatory and parkinsonian phenotypes, and establishing the optimal time to intervene in specific patients. 7.2. Five-year view We anticipate new advances in our ability to diagnose and treat GD, as the association with parkinsonism is greatly increasing attention to this rare disorder. Newborn screening and next-generation whole-exome or whole-genome sequencing are with likely to identify more cases of GD, broadening the phenotypic spectrum even more. Genetic and epigenetic modifiers that impact phenotype are likely to be discovered. These factors may improve our ability to make phenotypic predictions and identify new biomarkers. Development of small molecule drugs that can cross the BBB and impact neuronopathic GD will continue. Combinations of therapeutic modalities for specific patients will be considered. An improved understanding of the role of GCase in PD pathogenesis could lead to new treatments for early parkinsonism. Key issues. GD manifests with vast phenotypic heterogeneity ranging from asymptomatic older adults to infants who succumb during the first days of life. Being able to distinguish the different phenotypes is imperative. While some genotypes are associated with specific Gaucher phenotypes, predictions are limited. Genetic modifiers likely play a role. Different forms of ERT and SRT have a significant effect on disease outcome in patients with GD1 and GD3. Current treatments do not significantly change the outcome of GD2. Not all identified patients require therapy, but treatment should be administered early in symptomatic children. Current therapies remain extremely expensive and are not available to all patients that need them. The benefit of combining therapies needs to be further evaluated. As infants are increasingly being identified at birth, close monitoring and longitudinal studies will be necessary to better guide the need and criteria for therapeutic interventions. Thus far, screening does not appear to be changing disease management. GD is being diagnosed in older adults with PD or multiple myeloma. While mutations in GBA1 are a risk factor for parkinsonism, most patients with GD do not develop PD. Current GD therapeutics do not impact parkinsonian manifestations, but new therapies are being considered. Acknowledgments Funding This work was supported by the Intramural Research Programs of the National Human Genome Research Institute and the National Institutes of Health Footnotes Declaration of interest The authors have no relevant affiliations or financial involvement with any organization or entity with a financial interest in or financial conflict with the subject matter or materials discussed in the manuscript. This includes employment, consultancies, honoraria, stock ownership or options, expert testimony, grants or patents received or pending, or royalties. Peer reviewers on this manuscript have no relevant financial or other relationships to disclose. 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https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-strtlines-2009-1.pdf
Equations of straight lines mc-TY-strtlines-2009-1 In this unit we find the equation of a straight line, when we are given some information about the line. The information could be the value of its gradient, together with the co-ordinates of a point on the line. Alternatively, the information might be the co-ordinates of two different points on the line. There are several different ways of expressing the final equation, and some are more general than others. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: • find the equation of a straight line, given its gradient and its intercept on the y-axis; • find the equation of a straight line, given its gradient and one point lying on it; • find the equation of a straight line given two points lying on it; • give the equation of a straight line in either of the forms y = mx + c or ax + by + c = 0. Contents 1. Introduction 2 2. The equation of a line through the origin with a given gradient 2 3. The y-intercept of a line 4 4. The equation of a straight line with a given gradient, passing through a given point 7 5. The equation of a straight line through two given points 8 6. The most general equation of a straight line 10 www.mathcentre.ac.uk 1 c ⃝mathcentre 2009 1. Introduction This unit is about the equations of straight lines. These equations can take various forms depending on the facts we know about the lines. So to start, suppose we have a straight line containing the points in the following list. x y 0 2 1 3 2 4 3 5 x y There are many more points on the line, but we have enough now to see a pattern. If we take any x value and add 2, we get the corresponding y value: 0 + 2 = 2, 1 + 2 = 3, 2 + 2 = 4, and so on. There is a fixed relationship between the x and y co-ordinates of any point on the line, and the equation y = x + 2 is always true for points on the line. We can label the line using this equation. 2. The equation of a line through the origin with a given gradient Suppose we have a line with equation y = x. Then for every point on the line, the y co-ordinate must be equal to the x co-ordinate. So the line will contain points in the following list. y = x: x y 0 0 1 1 2 2 3 3 x y y = x We can find the gradient of the line using the formula for gradients, m = y2 −y1 x2 −x1 , www.mathcentre.ac.uk 2 c ⃝mathcentre 2009 and substituting in the first two sets of values from the table. We get m = 1 −0 1 −0 = 1 so that the gradient of this line is 1. What about the equation y = 2x? This also represents a straight line, and for all the points on the line each y value is twice the corresponding x value. So the line will contain points in the following list. y = 2x: x y 0 0 1 2 2 4 x y y = 2x y = x If we calculate the gradient of the line y = 2x using the first two sets of values in the table, we obtain m = 2 −0 1 −0 = 2 so that the gradient of this line is 2. Now take the equation y = 3x. This also represents a straight line, and for all the points on the line each y value is three times the corresponding x value. So the line will contain points in the following list. y = 3x: x y 0 0 1 3 2 6 x y y = 3x y = 2x y = x If we calculate the gradient of the line y = 3x using the first two sets of values in the table, we obtain m = 3 −0 1 −0 = 3 so that the gradient of this line is 3. www.mathcentre.ac.uk 3 c ⃝mathcentre 2009 We can start to see a pattern here. All these lines have equations where y equals some number times x. And in each case the line passes through the origin, and the gradient of the line is given by the number multiplying x. So if we had a line with equation y = 13x then we would expect the gradient of the line to be 13. Similarly, if we had a line with equation y = −2x then the gradient would be −2. In general, therefore, the equation y = mx represents a straight line passing through the origin with gradient m. Key Point The equation of a straight line with gradient m passing through the origin is given by y = mx . 3. The y-intercept of a line Consider the straight line with equation y = 2x + 1. This equation is in a slightly different form from those we have seen earlier. To draw a sketch of the line, we must calculate some values. y = 2x + 1: x y 0 1 1 3 2 5 x y y = 2x + 1 Notice that when x = 0 the value of y is 1. So this line cuts the y-axis at y = 1. What about the line y = 2x + 4? Again we can calculate some values. www.mathcentre.ac.uk 4 c ⃝mathcentre 2009 y = 2x + 4: x y -1 2 0 4 1 6 x y y = 2x + 4 4 This line cuts the y-axis at y = 4. What about the line y = 2x −1? Again we can calculate some values. y = 2x −1: x y -1 -3 0 -1 1 1 x y y = 2x - 1 -1 This line cuts the y-axis at y = −1. The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis. Key Point The equation of a straight line with gradient m and intercept c on the y-axis is y = mx + c . www.mathcentre.ac.uk 5 c ⃝mathcentre 2009 We are sometimes given the equation of a straight line in a different form. Suppose we have the equation 3y −2x = 6. How can we show that this represents a straight line, and find its gradient and its intercept value on the y-axis? We can use algebraic rearrangement to obtain an equation in the form y = mx + c: 3y −2x = 6 , 3y = 2x + 6 , y = 2 3x + 2 . So now the equation is in its standard form, and we can see that the gradient is 2 3 and the intercept value on the y-axis is 2. We can also work backwards. Suppose we know that a line has a gradient of 1 5 and has a vertical intercept at y = 1. What would its equation be? To find the equation we just substitute the correct values into the general formula y = mx + c. Here, m is 1 5 and c is 1, so the equation is y = 1 5x + 1. If we want to remove the fraction, we can also give the equation in the form 5y = x + 5, or 5y −x −5 = 0. Exercises 1. Determine the gradient and y-intercept for each of the straight lines in the table below. Equation Gradient y-intercept y = 3x + 2 y = 5x −2 y = −2x + 4 y = 12x y = 1 2x −2 3 2y −10x = 8 x + y + 1 = 0 2. Find the equation of the lines described below (give the equation in the form y = mx + c): (a) gradient 5, y-intercept 3; (b) gradient −2, y-intercept −1; (c) gradient 3, passing through the origin; (d) gradient 1 3 passing through (0, 1); (e) gradient −3 4, y-intercept 1 2. www.mathcentre.ac.uk 6 c ⃝mathcentre 2009 4. The equation of a straight line with given gradient, passing through a given point Example Suppose that we want to find the equation of a line which has a gradient of 1 3 and passes through the point (1, 2). Here, whilst we know the gradient, we do not know the value of the y-intercept c. We start with the general equation of a straight line y = mx + c. We know the gradient is 1 3 and so we can substitute this value for m straightaway. This gives y = 1 3x + c. We now use the fact that the line passes through (1, 2). This means that when x = 1, y must be 2. Substituting these values we find 2 = 1 3(1) + c so that c = 2 −1 3 = 5 3 So the equation of the line is y = 1 3x + 5 3. We can work out a general formula for problems of this type by using the same method. We shall take a general line with gradient m, passing through the fixed point A(x1, y1). We start with the general equation of a straight line y = mx + c. We now use the fact that the line passes through A(x1, y1). This means that when x = x1, y must be y1. Substituting these values we find y1 = mx1 + c so that c = y1 −mx1 So the equation of the line is y = mx + y1 −mx1. We can write this in the alternative form y −y1 = m(x −x1) This then represents a straight line with gradient m, passing through the point (x1, y1). So this general form is useful if you know the gradient and one point on the line. Key Point The equation of a straight line with gradient m, passing through the point (x1, y1), is y −y1 = m(x −x1) . www.mathcentre.ac.uk 7 c ⃝mathcentre 2009 For example, suppose we know that a line has gradient −2 and passes through the point (−3, 2). We can use the formula y −y1 = m(x −x1) and substitute in the values straight away: y −2 = −2(x −(−3)) = −2(x + 3) = −2x −6 y = −2x −4 . Exercise 3. Find the equation of the lines described below (give the equation in the form y = mx + c): (a) gradient 3, passing through (1, 4); (b) gradient −2, passing through (2, 0); (c) gradient 2 5, passing through (5, −1); (d) gradient 0, passing (−1, 2); (e) gradient −1, passing through (1, −1). 5. The equation of a straight line through two given points What should we do if we want to find the equation of a straight line which passes through the two points (−1, 2) and (2, 4)? Here we don’t know the gradient of the line, so it seems as though we cannot use any of the formulæ we have found so far. But we do know two points on the line, and so we can use them to work out the gradient. We just use the formula m = (y2 −y1)/(x2 −x1). We get m = 4 −2 2 −(−1) = 2 3 . So the gradient of the line is 2 3. And we know two points on the line, so we can use one of them in the formula y −y1 = m(x −x1). If we take the point (2, 4) we get y −4 = 2 3(x −2) 3y −12 = 2x −4 3y = 2x + 8 y = 2 3x + 8 3 . As before, it will be useful to find a general formula that can be used for examples of this kind. So suppose the general line passes through two points A(x1, y1) and B(x2, y2). We shall let a general point on the line be P(x, y). x y P(x, y) A(x1, y1) B(x2, y2) www.mathcentre.ac.uk 8 c ⃝mathcentre 2009 Now we know that the gradient of AP must be the same as the gradient of AB, as all three points are on the same line. But the gradient of AP is mAP = y −y1 x −x1 , whereas the gradient of AB is mAB = y2 −y1 x2 −x1 . Then mAP = mAB, so we must have y −y1 x −x1 = y2 −y1 x2 −x1 . Now this formula is fairly complicated, but it is easier to remember if all the terms involving y are on one side, and all the terms involving x are on the other. If we manipulate the formula, we get first y −y1 = (x −x1) y2 −y1 x2 −x1 and then y −y1 y2 −y1 = x −x1 x2 −x1 . It might help you to remember this formula if you notice that the pattern on the left-hand side, involving y, is just the same as the pattern on the right-hand side, involving x. Key Point The equation of a straight line passing through the two points (x1, y1) and (x2, y2) is y −y1 y2 −y1 = x −x1 x2 −x1 . Now we can use this formula for an example. Suppose that we want to find the equation of the straight line which passes through the two points (1, −2) and (−3, 0). We just substitute into the formula, and rearrange. The various steps are y −(−2) 0 −(−2) = x −1 −3 −1 y + 2 2 = x −1 −4 y + 2 = x −1 −2 = −1 2(x −1) −2y −4 = x −1 −2y = x + 3 y = −1 2x −3 2 . www.mathcentre.ac.uk 9 c ⃝mathcentre 2009 So the line has gradient −1 2 and its intercept on the y-axis is −3 2. We can also rearrange the equation a little further to obtain 2y = −x −3, or 2y + x + 3 = 0. Exercise 4. Find the equation of the lines described below (give the equation in the form y = mx + c): (a) passing through (4, 6) and (8, 26), (b) passing through (1, 1) and (4, −8), (c) passing through (3, 4) and (5, 4), (d) passing through (0, 2) and (4, 0), (e) passing through (−2, 3) and (2, −5). 6. The most general equation of a straight line There is one more form of the equation for a straight line that is sometimes needed. This is the equation ax + by + c = 0 . We have written equations in this form for some of our examples. We can see some special cases of this equation by setting either a or b equal to zero. If a = 0 then we obtain lines with general equation by + c = 0, i.e. y = −c b. These lines are horizontal, so that they are parallel to the x-axis. If b = 0 then we obtain lines with general equation ax + c = 0, i.e. x = −c a. These lines are vertical, so that they are parallel to the y-axis. The equation of a vertical line cannot be written in the form y = mx+c. The equation ax+by +c = 0 is the most general equation for a straight line, and can be used where other forms of equation are not suitable. x y ax + c = 0 x y by + c = 0 Key Point The most general equation of a straight line is ax + by + c = 0 . If a = 0 then the line is horizontal, and if b = 0 then the line is vertical. www.mathcentre.ac.uk 10 c ⃝mathcentre 2009 Exercise 5. Find the equation of the lines described below (give the equation in the form ax + by + c = 0, where a, b and c are whole numbers and a > 0): (a) the line in Exercise 2 (b) (b) the line in Exercise 2 (e) (c) the line in Exercise 3 (c) (d) the line in Exercise 4 (b) (e) the line in Exercise 4 (d) (f) the line in Exercise 4 (e) (g) the line through (3, −2) and (3, 2) (h) the vertical line passing through the point 0, 2 3  . Answers 1. Equation Gradient y-intercept y = 3x + 2 3 2 y = 5x −2 5 -2 y = −2x + 4 -2 4 y = 12x 12 0 y = 1 2x −2 3 1 2 −2 3 2y −10x = 8 5 4 x + y + 1 = 0 −1 −1 2. (a) y = 5x + 3, (b) y = −2x −1, (c) y = 3x, (d) y = 1 3x + 1, (e) y = −3 4x + 1 2. 3. (a) y = 3x + 1, (b) y = −2x + 4, (c) y = 2 5x −3, (d) y = 2, (e) y = −x. 4. (a) y = 5x −14, (b) y = −3x + 4, (c) y = 4, (d) y = −1 2x + 2, (e) y = −2x −1. 5. (a) 2x + y + 1 = 0, (b) 3x + 4y −2 = 0, (c) 2x −5y −15 = 0, (d) 3x + y −4 = 0, (e) x + 2y −4 = 0, (f) 2x + y + 1 = 0, (g) x −3 = 0, (h) x = 0. www.mathcentre.ac.uk 11 c ⃝mathcentre 2009
948
https://flexbooks.ck12.org/cbook/ck-12-middle-school-math-concepts-grade-6/section/2.5/primary/lesson/area-of-squares-and-rectangles-msm6/
Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 2.5 Area of Squares and Rectangles Written by:Melissa Sanders | Jen Kershaw, M.ed Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Allen bought a new house and is planning to replace the carpet in two of the bedrooms. One bedroom is a shaped like a square and the other is rectangular. He needs to figure out how much carpet is needed to cover both bedroom floors. The square bedroom has side lengths of 9 feet. The rectangular bedroom has a length of 12 feet and a width of 8 feet. In order for Allen to figure out how much carpet he needs to buy for these two bedrooms, he needs to find the area of each bedroom. In this concept, you will learn how to find the area of squares and rectangles. Area The inside space of a figure is known as the area of the figure. Carpeting a floor, grass on the ground, or anything else covering the space inside of a figure are all examples of area. There is a formula for finding the area of different shapes. In this concept, this concept will show you how to find the area of both squares and rectangles. This is the formula used to find the area of a square: @$$\begin{align}A= s\cdot s\end{align}@$$ Remember, the dot is another symbol for multiplication. To figure out the area of the square, multiply one side times the other side. Since all sides in a square are equal, the two numbers will be the same. The unit of measurement is kept in the equation to help you remember what the correct label should be. Here is how to use the formula for finding the area of a square: @$$\begin{align}A=6 \ ft \cdot 6 \ ft\end{align}@$$ Now, think of the equation as multiplying two different things; the numbers and the units of measurement. @$$\begin{align}A &= 6 \cdot 6\ A &= ft \cdot ft\end{align}@$$ Solve both equations. @$$\begin{align}A &= 6 \times 6 = 36\ A &= ft \times ft = ft^2\end{align}@$$ Think back to what you know about exponents. When you multiply two of the same thing together, you can write it in exponential form. Another way to label area is to write square feet (sq. ft.). The answer is 36 sq. ft. or @$\begin{align}36 {ft}^{2}\end{align}@$ This is the formula used to find the area of a rectangle: @$$\begin{align}A= lw\end{align}@$$ To find the area of a rectangle, use the measurements for length and width instead of side length. These numbers may be different from each other because not all rectangles have the same side lengths like squares do. This rectangle measures a length of 5 meters and a width of 3 meters. Just like the area formula for a square, multiply those two numbers to find the area of the rectangle. Remember, two variables next to each other in an equation indicate multiplication, just like the dot symbol or traditional multiplication symbol. The symbols will be used interchangeably. To find the area of a rectangle, multiply length times width. @$$\begin{align}A &= 5m \cdot 3m\ A &= 5 \times 3\ A &= meters \times meters\\end{align}@$$ The formula shows 5 meters times 3 meters. Multiply the measurement part (5 @$\begin{align}\times\end{align}@$ 3), then multiply the units of measurement. The answer is 15 sq. m or @$\begin{align}15 \ m^2\end{align}@$. The label for area of a rectangle is similar to the label for area of a square from above. The label can be written in exponential form or written in abbreviation. Examples Example 1 Earlier, you were given a problem about Allen and his new house that he wants to re-carpet. The dimensions of Allen's two bedrooms are given in the beginning of the concept. His square bedroom has side lengths of 9 feet. The other room is a rectangle with dimensions of 12 feet by 8 feet. To begin solving this problem, find the area of each of the bedrooms. The square bedroom has a side length of 9 feet. @$$\begin{align}A &= s \cdot s\ A &= 9 \times 9 = 81 \ sq. \ feet\end{align}@$$ The square bedroom has an area of 81 square feet. The rectangular bedroom has a length of 12 feet and width of 8 feet. @$$\begin{align}A &= 12ft \cdot 8ft\ A &= 12 \times 8\ A &= feet \times feet\end{align}@$$ The rectangular bedroom has an area of 96 square feet. Allen wants the total area of his two bedrooms to know how much carpet to buy, so add the two areas together. @$$\begin{align}81 + 96 = 177 \ square \ feet\end{align}@$$ The answer is 177 square feet. Example 2 Find the area of a rectangle with a length of 7 ft. and a width of 4 ft. First, write out the equation @$$\begin{align}A&=lw \ A&=7ft \cdot 4ft\end{align}@$$ Next, multiply the two numbers. @$$\begin{align}A= 7 \cdot 4 = 28\end{align}@$$ Then, multiply the units of measurement. @$$\begin{align}A= ft \cdot ft = {ft}^{2}\end{align}@$$ Now, put your answer together to show the final solution. @$$\begin{align}A= 28 {ft}^{2}\end{align}@$$ The answer is 28 sq. ft. Once you do more examples you may find yourself skipping the step of multiplying the units of measurement. This may become natural to square the unit of measurement for any area problem. Example 3 Find the area of a square with side lengths of 10 inches. First, write your equation. @$$\begin{align}A&= s \cdot s \ A&= 10 in \cdot 10 in\end{align}@$$ Next, solve the equation by multiplying the two side lengths. @$$\begin{align}A= 10 \cdot 10 = 100\end{align}@$$ Then, write your final solution. Remember to write the squared unit of measurement after the numerical answer. @$$\begin{align}A= 100 {in}^{2}\end{align}@$$ The answer is 100 sq. in. Example 4 Find the area of a rectangle with a length of 6 feet and a width of 5 feet. First, write your equation. @$$\begin{align}A&= lw \ A&= 6ft \cdot 5ft\end{align}@$$ Next, solve the equation by multiplying the length times the width. @$$\begin{align}A= 6 \cdot 5 = 30\end{align}@$$ Then, write your final solution. Remember to write the squared unit of measurement after the numerical answer. @$$\begin{align}A= 30 {ft}^{2}\end{align}@$$ The answer is 30 sq. ft. Example 5 Find the area of a rectangle with a length of 9 meters and a width of 8 meters. First, write your equation. @$$\begin{align}A&= lw \ A&= 9m \cdot 8 m\end{align}@$$ Next, solve the equation by multiplying the length times the width. @$$\begin{align}A= 9 \cdot 8 = 72\end{align}@$$ Then, write your final solution. Remember to write the squared unit of measurement after the numerical answer. @$$\begin{align}A= 72 {m}^{2}\end{align}@$$ The answer is 72 sq. m Review Find the area of each of the following figures. Make sure you label your answer correctly. A square with a side length of 6 inches. A square with a side length of 4 inches. A square with a side length of 8 centimeters. A square with a side length of 12 centimeters. A square with a side length of 9 meters. A rectangle with a length of 6 inches and a width of 4 inches. A rectangle with a length of 9 meters and a width of 3 meters. A rectangle with a length of 4 meters and a width of 2 meters. A rectangle with a length of 17 feet and a width of 12 feet. A rectangle with a length of 22 feet and a width of 18 feet. A square with a side length of 13 feet. A square with a side length of 18 feet. A square with a side length of 21 feet. A rectangle with a length of 18 feet and a width of 13 feet. A rectangle with a length of 60 feet and a width of 27 feet. A rectangle with a length of 57 feet and a width of 22 feet. Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. Resources | Image | Reference | Attributions | --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? No Results Found Your search did not match anything in . Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)30/ 100 This lesson has been added to your library. |Searching in: | | | Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents
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https://education.ohio.gov/getattachment/Topics/Learning-in-Ohio/Mathematics/Ohio-s-Learning-Standards-in-Mathematics/SP-MATH-Standards.pdf.aspx?lang=en-US
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Table Of Contents What are the Factors of 495? How to Find Factors of 495? Finding Factors Using Multiplication Finding Factors Using Division Method Prime Factors and Prime Factorization Factor Tree Common Mistakes and How to Avoid Them in Factors of 495 Factors of 495 Examples FAQs on Factors of 495 Important Glossaries for Factor of 495 Explore More numbers Table Of Contents Table Of Contents Summarize this article: ChatGPT Perplexity Last updated on August 5, 2025 Factors of 495 Factors are the numbers that divide any given number evenly without remainder. In daily life, we use factors for tasks like sharing items equally, arranging things, etc. In this topic, we will learn about the factors of 495, how they are used in real life, and the tips to learn them quickly. What are the Factors of 495? The numbers that divide 495 evenly are known as factors of 495. A factor of 495 is a number that divides the number without remainder. The factors of 495 are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, and 495. Negative factors of 495: -1, -3, -5, -9, -11, -15, -33, -45, -55, -99, -165, and -495. Prime factors of 495: 3, 5, and 11. Prime factorization of 495: 3² × 5 × 11. The sum of factors of 495: 1 + 3 + 5 + 9 + 11 + 15 + 33 + 45 + 55 + 99 + 165 + 495 = 936 How to Find Factors of 495? Factors can be found using different methods. Mentioned below are some commonly used methods: Finding factors using multiplication Finding factors using division method Prime factors and Prime factorization Finding Factors Using Multiplication To find factors using multiplication, we need to identify the pairs of numbers that are multiplied to give 495. Identifying the numbers which are multiplied to get the number 495 is the multiplication method. Step 1: Multiply 495 by 1, 495 × 1 = 495. Step 2: Check for other numbers that give 495 after multiplying 3 × 165 = 495 5 × 99 = 495 9 × 55 = 495 11 × 45 = 495 15 × 33 = 495 Therefore, the positive factor pairs of 495 are: (1, 495), (3, 165), (5, 99), (9, 55), (11, 45), (15, 33). For every positive factor, there is a negative factor. Finding Factors Using Division Method Dividing the given numbers with whole numbers until the remainder becomes zero and listing out the numbers which result as whole numbers as factors. Factors can be calculated by following the simple division method - Step 1: Divide 495 by 1, 495 ÷ 1 = 495. Step 2: Continue dividing 495 by the numbers until the remainder becomes 0. 495 ÷ 1 = 495 495 ÷ 3 = 165 495 ÷ 5 = 99 495 ÷ 9 = 55 495 ÷ 11 = 45 495 ÷ 15 = 33 Therefore, the factors of 495 are: 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495. Prime Factors and Prime Factorization The factors can be found by dividing it with prime numbers. We can find the prime factors using the following methods: Using prime factorization Using factor tree Using Prime Factorization: In this process, prime factors of 495 divide the number to break it down in the multiplication form of prime factors till the remainder becomes 1. 495 ÷ 3 = 165 165 ÷ 3 = 55 55 ÷ 5 = 11 11 ÷ 11 = 1 The prime factors of 495 are 3, 5, and 11. The prime factorization of 495 is: 3² × 5 × 11. Factor Tree The factor tree is the graphical representation of breaking down any number into prime factors. The following steps show - Step 1: Firstly, 495 is divided by 3 to get 165. Step 2: Now divide 165 by 3 to get 55. Step 3: Then divide 55 by 5 to get 11. Step 4: Divide 11 by 11 to get 1. So, the prime factorization of 495 is: 3² × 5 × 11. Factor Pairs Two numbers that are multiplied to give a specific number are called factor pairs. Both positive and negative factors constitute factor pairs. Positive factor pairs of 495: (1, 495), (3, 165), (5, 99), (9, 55), (11, 45), (15, 33). Negative factor pairs of 495: (-1, -495), (-3, -165), (-5, -99), (-9, -55), (-11, -45), (-15, -33). Common Mistakes and How to Avoid Them in Factors of 495 Mistakes are common while finding factors. We can identify and correct those mistakes using the following common mistakes and the ways to avoid them. Mistakes are common while finding factors. We can identify and correct those mistakes using the following common mistakes and the ways to avoid them. Mistake 1 Forgetting the number itself and 1 is a factor Forgetting the number itself and 1 is a factor Children might forget to add the given number itself and 1 as a factor. The number itself and 1 are the factors for every number. Always remember to include 1 and the number itself. For example, in factors of 495, 1 and 495 are also factors. Children might forget to add the given number itself and 1 as a factor. The number itself and 1 are the factors for every number. Always remember to include 1 and the number itself. For example, in factors of 495, 1 and 495 are also factors. Mistake 2 Missing Negative Factors Missing Negative Factors We often mention only positive factors. There are also negative factors, always check whether you have mentioned negative factors. For example, the factors of 495 consist of -1, -3, -5, etc. We often mention only positive factors. There are also negative factors, always check whether you have mentioned negative factors. For example, the factors of 495 consist of -1, -3, -5, etc. Mistake 3 Including Fractions Including Fractions Children might sometimes add fractions as factors. Only whole numbers can be factors. For example, thinking 2.5 as a factor because 495/2.5 = 198, and it is not a whole number. Children might sometimes add fractions as factors. Only whole numbers can be factors. For example, thinking 2.5 as a factor because 495/2.5 = 198, and it is not a whole number. Mistake 4 Confusing Factors With Prime Numbers Confusing Factors With Prime Numbers Remember that factors can be any whole numbers, not only just primes. For example, thinking all the factors of 495 are prime numbers. For example, 3, 5, and 11 are the prime factors of 495, but it has other factors like 9, 45. Remember that factors can be any whole numbers, not only just primes. For example, thinking all the factors of 495 are prime numbers. For example, 3, 5, and 11 are the prime factors of 495, but it has other factors like 9, 45. Mistake 5 Mistake in Factorization Mistake in Factorization Children might skip the steps in factorization and end up writing wrong factors. For example, not breaking 495 as 3² × 5 × 11 and missing all the key factors. Children might skip the steps in factorization and end up writing wrong factors. For example, not breaking 495 as 3² × 5 × 11 and missing all the key factors. Factors of 495 Examples Problem 1 There are 11 students and 495 pencils. How will they distribute them equally? They will get 45 pencils each. They will get 45 pencils each. Explanation To distribute the pencils equally, we need to divide the total pencils by the number of students. 495/11 = 45 To distribute the pencils equally, we need to divide the total pencils by the number of students. 495/11 = 45 Problem 2 A plot is rectangular, the length of the plot is 15 meters and the total area is 495 square meters. Find the width. 33 meters. 33 meters. Explanation To find the width of the plot, we use the formula, Area = length × width 495 = 15 × width To find the value of width, we need to shift 15 to the left side. 495/15 = width Width = 33. To find the width of the plot, we use the formula, Area = length × width 495 = 15 × width To find the value of width, we need to shift 15 to the left side. 495/15 = width Width = 33. Problem 3 There are 9 boxes and 495 apples. How many apples will be in each box? Each box will have 55 apples. Each box will have 55 apples. Explanation To find the apples in each box, divide the total apples by the boxes. 495/9 = 55 To find the apples in each box, divide the total apples by the boxes. 495/9 = 55 Problem 4 In a school, there are 495 students and 5 buses. How many students are there in each bus? There are 99 students in each bus. There are 99 students in each bus. Explanation Dividing the students by the total buses, we will get the number of students in each bus. 495/5 = 99 Dividing the students by the total buses, we will get the number of students in each bus. 495/5 = 99 Problem 5 495 books need to be arranged in 3 shelves. How many books will go on each shelf? Each shelf has 165 books. Each shelf has 165 books. Explanation Divide total books by shelves. 495/3 = 165 Divide total books by shelves. 495/3 = 165 FAQs on Factors of 495 1.What are the factors of 495? 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495 are the factors of 495. 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495 are the factors of 495. 2.Mention the prime factors of 495. The prime factors of 495 are 3² × 5 × 11. The prime factors of 495 are 3² × 5 × 11. 3.Is 495 a multiple of 9? Yes, 495 is a multiple of 9. Yes, 495 is a multiple of 9. 4.Mention the factor pairs of 495? (1, 495), (3, 165), (5, 99), (9, 55), (11, 45), and (15, 33) are the factor pairs of 495. (1, 495), (3, 165), (5, 99), (9, 55), (11, 45), and (15, 33) are the factor pairs of 495. 5.What is the square of 495? The square of 495 is 245025. The square of 495 is 245025. 6.How can children in United States use numbers in everyday life to understand Factors of 495? Numbers appear everywhere—from counting money to measuring ingredients. Kids in United States see how Factors of 495 helps solve real problems, making numbers meaningful beyond the classroom. 7.What are some fun ways kids in United States can practice Factors of 495 with numbers? Games like board games, sports scoring, or even cooking help children in United States use numbers naturally. These activities make practicing Factors of 495 enjoyable and connected to their world. 8.What role do numbers and Factors of 495 play in helping children in United States develop problem-solving skills? Working with numbers through Factors of 495 sharpens reasoning and critical thinking, preparing kids in United States for challenges inside and outside the classroom. 9.How can families in United States create number-rich environments to improve Factors of 495 skills? Families can include counting chores, measuring recipes, or budgeting allowances, helping children connect numbers and Factors of 495 with everyday activities. Important Glossaries for Factor of 495 Factors: The numbers that divide the given number without leaving a remainder are called factors. For example, the factors of 495 are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, and 495. Prime factors: The factors which are prime numbers. For example, 3, 5, and 11 are prime factors of 495. Factor pairs: Two numbers in a pair that are multiplied to give the original number are called factor pairs. For example, the factor pairs of 495 are (1, 495), (3, 165), etc. Prime factorization: The expression of a number as the product of its prime factors. For example, the prime factorization of 495 is 3² × 5 × 11. Multiples: Numbers that can be divided by another number without leaving a remainder. For instance, 495 is a multiple of 9. Explore More numbers Important Math Links IconPrevious to Factors of 495 Important Math Links IconNext to Factors of 495 About BrightChamps in United States Hiralee Lalitkumar Makwana About the Author Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns. Fun Fact : She loves to read number jokes and games.
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CNS PATHOMA 5 | Demyelinating Disorders MBBS Lectures by Dr Zeenat 48000 subscribers 38 likes Description 3463 views Posted: 1 Oct 2021 1 comments Transcript: अदनान सामी को नेट प्रॉफिट इस बार उत्तर डिमाइज मीटिंग डिसऑर्डर्स ठीक है बजट के हमारे पास मांशु पॉपुलर होती है ग्रीन कि आपको पता एडिटिंग सीनियर केजी राइस आफ रिसचर्स हैव इन द नेशन के इंसुलेशन 9th एग्जाम्स बाय द हेल्प ऑफ ऑल उर एडवरटाइजिंग और सेंट्रल नर्वस सिस्टम से ज़्यादा रिव्युस एंड डिसाइड्स होते हैं वह अच्छी प्रोवाइड करते हैं - मिशन ठीक है और पेरीफेरल नर्व सिस्टम सोंग्स होते हैं वह प्रोवाइड करते हैं रिलेशन कितनी हॉट ठीक है अब यह क्या करते हैं मान्य नेशंस आपको पता है कि स्पीड एडिशन से और कंडक्शन जो निरोधक हो जाती है अब डिमांड रिलेटेड डिसऑर्डर्स में क्या होगा कि यह जो मारी होगी ठीक है डिस्ट्रेक्शन हो जाएगी और इसको डेंगू साइड खराब हो जाएंगे जिसकी वजह से क्या होगा कि उसने मेंशन की वजह से डिसऑर्डर जाएंगे अरे ठीक है ना प्रोडक्ट इन प्रॉपर नहीं होगी अब इसमें सबसे पहले हमारे पास शिव को डिस्टिक ऑफिस युधिष्ठिर ऑफिस के नाम से पता चल रहा है यह को मींस वाइट ठीक है तो इसमें क्या होता है कि ठीक है इसको मींस वाइट इन डिस्ट्रॉफी मतलब उसमें कोई प्रॉब्लम आ जाएगी मतलब जो हमारा वाइट मैटअप ड्रेन होता है इसको बैलेंस शीट कहते हैं वह व्हाइट मैटर की बनी होती है ना ठीक है तो उसमें प्रॉब्लम आ जाएगी तो लिक्विड डिस्ट्रॉफी जिसको हम कहते हैं अब यह कोड इसको ऑफिस वह डिफरेंट टाइप की होती है डिपेंडिंग अपऑन कि कौन सा एंजाइम इंवोल्वड है जिस घर में आप देखेंगे ठीक है यहां पर अगर हमारे पास क्या उत्तर नॉर्मली क्या होता है जो सेल्यूलाइट होते हैं ठीक है उनके ऊपर एक एनवायरनमेंट करता है एयरसेल स्टेज ठीक है और सल्फर टाइड टो कन्वर्ट कर देता एंटी-एजिंग इफेक्ट ओं सैलूलोइड यह एडजेक्टिव्स रिवरसाइड ठीक है और ग्लैड टो सेलिब्रेट साइड को जो है ग्लैड टो सेलिब्रेट साइडेड तो अंजाम होता है वह इसको जो एलेक्ट्रोंस में कंवर्ट कर देता है यह नार्मल हमारे मैकेनिज़्म होता है लाइफ में के अंदर हो रहा होता है ठीक है अब क्या होता है कि जब फॉर एग्जांपल हमारे पास क्या होगा ऐरसल एयरटेल की डेफिशियेंसी हो जाएगी यह लाइन की फिक्र उसकी वजह से क्या होगा कि गैलेक्टोज साइज जो 3 साल पर टाइट है वह कलेक्टरों से रिड्यूस आईटी कैन वे कन्वर्ट नहीं होगा जिसे रीसेंट अपडेट्स होंगे वह Bigg Boss रोड खंदारी कि मैं लेट हो जाएंगे यह लाइव जो हमारा है इसका यह हमारा जो लाइंस ऑन है ठीक है यह हमारे उनको डेंगू साइड का इलाज समय ठीक है तो आप खुद सोचेंगे जब उसके अंदर लेकर डेंडोर साइट के अंदर उनके लाइव्स उसके अंदर की इमिग्रेशन हो जाएगी सल्फर टाइट की तो इट इज द फॉर्म ऑफ लाइफ रोमांस स्टोरेज डिजर्टेड हॉलीवुड डेंजर फाइट अपना काम प्रॉपर नहीं कर पाएंगे और हमारी जो मिनिमाइजेशन होगी प्रॉपर ने लॉन्च की वेबसाइट हो जाएगी ठीक है और दूसरा हमारे पास होता है कि अगर गलत टो सेलिब्रेट थिस डे एस कि इन थे डिफिशिएंसी होती है तो उसको उसके अंदर क्या होगा कि क्लब लट्टू शरीर उस साइड जो का वह इसी में लेट हो जाएगा हमारे लाइक जोंग यह मैक्रोफेजेस के अंदर ठीक है तो इस मलिक को हम कहते हैं इसलिए हम कहते हैं कि रेड्डी ठीक है पहली बनी कौन सी मटक मटक लेकर डिस्ट्रोफी इसमें डाल सकते कि डेफिशियेंसी थी यहां पर क्या है यहां पर क्लिक टो सेलिब्रेट साइड की डिफरेंस है विदेश की डेफिशियेंसी है जिसकी वजह से क्या होलीडेज आधे जिसमें इलेक्ट्रोंस अरे उस साइड इफेक्ट मिनट हो रहा है इसके इलावा जो थर्ड होती है फोन डिटेल्स डिनो लिक्विड्स पे फ़िकर एड्रीनो लिक्विड ट्रोफी में क्या होता है कि हमारी जो यह देखिए एक हमारा प्रोक्सेस वह मैं ठीक है इसके रिसेप्टर होता है थिस आफ्टर होता है जिसके ऊपर एक बड़ी क्या होता है कि लोंग चेन फैटी एसिड होते हैं ठीक है अब नॉर्मली क्या होता है कि लौंग पार्टी के जो है उसके ऊपर कोई दूसरा इन्हें घी जो है वह एडिशन उन्हें होती है ठीक है लेकिन जिन लोगों के अंदर इसकी डिफिशिएंसी होती है जिन्हें एक प्रॉब्लम होती है डिफरेंट होता है उसमें क्या होता है यह कोई टाइम लॉन्चिंग के साथ ही मिल पाता है और क्या होता है कि उसकी वजह होती है इसको कहते हैं ठीक है तो यह हमारे पास ऐसे किसी के अंदर यह खराबी की वजह से इसकी वजह से बॉलीवुड वेबसाइट अपना प्रॉपर्टी कहानी करेंगे और अपनी करें कि काम यू नो दैट कुंठित को प्रॉपरली मैनेज करना था वह नहीं हो पाएगी उनको हम कहते हैं जिसमें को subscribe our Channel को पता होना चाहिए कि फिर यह इसका थोड़ा सा था कि उनकी प्रॉपर्ली यूटिलाइज्ड उनके पति तो पति है वह देंगे ठीक है और हमारे जो करेंगे और हमारे जो है इसलिए इस चैनल को सब्सक्राइब कर लें और लिक्विड थे वॉइस आफ डिफेक्ट्स डोमिनेंट ठीक है नेशनल हाईवे टॉप इंपॉर्टेंट मल्टीपल स्क्लेरोसिस ठीक होने में वृद्धि करना है मल्टीपल स्क्लेरोसिस क्या होता है इसमें और टो एलिमिनेट याचिका और टो रिन्यू डिस्क्रप्शन होती है स्माइलिंग एंड लिक्विड वेस्ट साइड अब इसमें भी हमारे खराब हो गए हैं अल्टीमेटली मालिश खराब हो जाएगी तो कंडक्टर प्रॉपर नहीं होगी बाद में याद रखने वाली यह है कि इसमें टो डिफिशिएंसी यहां पर क्या है यहां पर और यह 2030 के अंदर-अंदर और इफेक्टिव टू-डू लिस्ट अंदर प्रॉब्लम यह होती है कि कर दिया दो ठीक है उनके अंदर मोस्ट लिए देखी गई है ठीक है और वूमेन केंद्र देखी गई है और यंगस्टर्स में देखी गई ठीक है और उसके बाद में कि बिल्कुल मल्टीपल स्क्लेरोसिस की वजह से खराब हो गई और उसके न्यूरोलॉजिकल प्रॉब्लम को पता चल रहा है मल्टीपल स्क्लेरोसिस मतलब है कि जो लोग जिन लोगों ने उनके अंदर में मल्टीपल जगह-जगह पर हम क्लिक करते हैं तो अगर प्रॉब्लम होगी तो ऐसे स्कैनिंग स्पीच ऐसे लोगों को हम देखकर ऐसा लगेगा जैसे नल को हेल्प हुई है ठीक है लेकिन अगर हमें हिस्ट्री नहीं मिलती तो उसे अल्कोहल नहीं आती ठीक है सो हम उसकी डीडी में देखेंगे कि मल्टीपल स्क्लेरोसिस हो सकता है ठीक है इसके अलावा क्या होता है कि अगर हमारा मीडिया लांगीट्यूडिनल 65 ईयर्स इन वॉल्व होता है तो उसकी वजह से इंटरनेट लेयर ऑफ लोकल एरिया होता है अब इस फिगर में देखिए यह मीडिया लांगीट्यूडिनल वर्षीय प्रिंस दिखाया है इसमें तीन आसपास करिए और की जो मोटर प्रॉड्यूसर न ठीक है अब जब यह खराब होगा मीडिया लौंग रन पैसे जरा होगी तो उनके साइंस जाएंगे ठीक है जिसमें और थे लोकल एरिया मेनली जो है इंटरनल क्लॉथ शर्मा प्लेट या मतलब इसमें ऐड अगर बात की जाए तो आए इंवॉल्वमेंट होती है ठीक है इसके अलावा क्या है कि अगर जो ए सेलिब्रिटी व्हाइट मैटर इफेक्ट होता है कि जो प्रिवेंट क्लियर एरिया होता है वेंट्रीकल्स के इर्द-गिर्द वाला एरिया अगर वह इंवोल्वड है तो उसकी वजह से क्या होता है पर ऐसे हो जाती है यह लैटर लॉट्स आफ सेंसेशन हो जाती है प्लीज रिकॉर्डिंग वॉल्व होता है तो अब इसकी क्या होगा लोअर एक्सट्रीमिटीज और सेंसेशन बिजनेस होगी ठीक है और अगर ऑटोनॉमिक नर्वस सिस्टम इंवोल्वड है तो बाउल ब्लड सेक्सुअल डिस्फंक्शन हो जाएगा डायग्नोसिस हम इसके करते हैं हमारा इसे रोल नंबर पंक्चर से इस गांव में देखें कि आपको नजर आ रहा है आइए भी फैक्ट्री आफ्टर ऑप्टिक नर्व की वजह से इसके इलावा जो है अभी जो सिम्टम्स आ रहे हैं ठीक है और इसमें जो है वह सेंसेशन फ्लॉप हो गई है वह और होटलों में नंबर सिस्टम की जो फंक्शंस वह Bigg Boss हुए नजर आ रहे हैं ठीक है डायग्नोसिस मैंने आपको बताया कि है मैम आर ऐसे कर लेते हैं और नंबर पंक्चर से कर लेते हैं यह मैम आर है करवाते हैं मल्टीपल स्क्लेरोसिस के पेशंस का तो अब टेस्टी क्योंकि यह वाइट मेटल के डिमिनिशिंग हुई है ठीक है डिमोनिटाइजेशन जो है उसकी हमें इस वेबसाइट पर मिलेंगे लेकिन लेंगे जैसे कि मैंने आपको बताया कि लोग यह लोगों की डेट होती है तो उनके रूप बैन है उसके लिए हमें हार डिपोजिट मिलते हैं प्लेट फॉर्मेशन मिलती है इसमें देखिए नोटिस करें नार्मल ब्रेन को और जो हमारा है अफेक्टेड विद मेडिकल से क्रॉस इस के किस तरह एयरप्लेन फॉर्मेशन हुई है प्रीवेंटिव मेडिसीन में और जब हम इन पेशंस का नंबर पंक्चर करवाते हैं तो लिंफोसाइट्स मैं नजर आते हैं उसमें ठीक है और इम्यूनोग्लोबुलिंस में ज्यादा मिलती है क्योंकि और ट्रीटमेंट डिजाइन है इस वजह से और गुलैल IG बैंड मिलते हैं हाई रेजोल्यूशन इलेक्ट्रोफॉरेसिस में और मालिक मालिक बेसिक रोटी मिलती है विकास स्माइलिंग डे में जोड़िए cross-examination में हमें जो एग्री अपीयरिंग प्लेस मिलते हैं व्हाइट मैटर में सेम उसी तरफ और ट्रीटमेंट हम कैसे करते हैं इसका अपडेट स्टील मिलों और हमारे पर स्टेटमेंट दीजिए तो हमें मीडियम प्रश्न करवाएंगे स्टार्स देंगे ठीक है और long-term कमिटमेंट हमको देते हैं इंटरफेरॉन विद्या से जिससे जो प्रोग्रेशन और डिसीस कम हो जाती है इस दिन यादव क्यूट से लॉजिंग एंड फ्लाइट इस ठीक है यह क्या होता है आपने पढ़ा होगा कि ऐसे स्पीड होता यह मीज़ल्स वायरस की वजह से होता है अब मैं इस वायरस से अगर ज्यादा देर तक ख्याल से पता क्या होता है इसमें तो इसमें जो होता है कि इंफेक्शन होता ना ब्रीज़र्स वायरस का व हो जाता पेंसिल इन फैंसी में ठीक है बच्चा जो पैदा होगा ठीक है एक साल का उसके अंदर क्या एक साल केंद्र क्या होगा उसको मीटर्स वायरस अटैक कर जाएगा गर्म को वैक्सीन नहीं देते ठीक है अगर वह इस कदर भरा रहता है लेकिन काफी सालों बाद जब वह थोड़ा-थोड़ा से भरा होता है चाइल्डहुड में जाता है फिर जाकर जो है ना उसको अनिल रॉजर्स सिम्टम्स आना शुरू होते हैं ठीक है क्योंकि मींस वायरस वहां पर बचपन से इसके अंदर था वैक्सिनेशन उसकी नहीं हुई है मार उसको नहीं लगाया गया जिसकी वजह से क्या वाकई चाइल्ड में जाकर उसने प्रेजेंट किया ठीक है इससे पेशंस आते हैं हम मदद से पूछते हैं आपने इस कमीज वायरस के गैस वैक्सिनेशन करवाई थी यह नहीं करवाई थी ठीक है तो और नहीं करवाई होती तो इस वजह से क्या होता है कि परसिस्टेंट इंफेक्शन अगर हो प्रियंका जैसे यूनिट एनसीसी कैडेट्स में क्या और है कि जो हल्दी को डेंगू साइड से उसके अंदर वायरल इनफ्लुएंस नजर आ रहे हैं वहां पर मेल्स वायरस परसिस्टेंटली प्रेजेंट होता है और इसे प्रेस करता है इससे सारे रोल्स को डैमेज करता जाता है वह गुड एंड साइड्स को कि हमारे व्हाइट मैटर बना रहे होते हैं इसके अलावा ग्रे मैटर डूइंग रोमांस के अंदर दिए बार इनक्रीज होती है अल्टीमेटली क्या होता है कि डाट होता है पोजीशन ठीक है अगर इसको टूटना रोकता है नेक्स्ट यह प्रोग्रेसिव मल्टीफोकल यू को इन सिंपल उपाय थी तो इस टाइम का क्या मतलब होता है प्रोग्रेसिव मतलब यह आहिस्ता-आहिस्ता प्रेस करता है ऐसा खराब होता जाता है मल्टीफोकल मतलब के डिफरेंट जगह को यह इंवाइट करेगा लियू को मींस व्हाइट मैटर ऑफ द ब्रेन इन सब पहलुओं पर थी मतलब के डिजीज आफ ग्रीन है ठीक है तो यह क्या करता है कि ब्रेन के व्हाइट मैटर में मल्टीपल जगह पर क्या होता है कि यह जो है वह तय है ठीक है थिस वायरस की वजह से यह होता है जैसी वायरस की वजह से ठीक है जैसे वायरस है उसकी अगर वह पड़ा रहता है बॉडी के अंदर लेटेंट होता है और एकदम से पेशंट को ऐड हो जाता है लेकिन यह जाता है मतलब उसको मिलों से परेशन हो जाती है तो यह वायरस होता है एक्टिवेट हो जाता है और एक्टिवेट हो गए उसी तरह जैसे मिनरल्स वायरस ने उनको डाइऑक्साइड खराब करना शुरू कर दिया था खैर यह तो जैसी वायरस होता है यह भी क्या करता है यह कि पोलियो वायरस होते ना ही मेन पोलियो वायरस इसे फोल्ड करना होता है कि उसी तरह जवाहरात करना होता है हमारे व्हाइट मैटर को उसकी जैसे यूरोलॉजिकल साइन जाते हैं विजुअल और सुख जाता है विटनेस डिमेंशिया एंड नीड इनटू डेथ ठीक है सभी यादव सेंट्रल फोन टाइम मारिनो लाइसेस इसके नाम से मैं पता चल रहा है कि इसमें क्या होता है कि पॉर्न के अंदर डुबोकर डिमांड नेशन हो जाती है मतलब जो है यह होता है इस वजह से अगर प्रोग्रेसिव ली मतलब हम जो ए रैपिड जो है ना हम मतलब इलेक्ट्रोलाइट इलेक्ट्रोलाइट डिस्टरबेंसस अगर बहुत तेजी से ठीक है रैपिड एकदम से इलेक्ट्रोलाइट डिस्टरबेंस जाते हैं कि फॉर एग्जांपल हम जो ना हाइपो किसी को पेशंट को ना हाइपोनेट्रिमिया हुआ था हमने इसको एकदम से जुड़ना आईवी फ्लूइड दे ताकि उसका जो आफ यूनिटी में वह ठीक हो जाए तो फिर उनको जब हम ऐसे देते हैं ना तो उसकी इधर से क्या होता है कि जो है ना डिमांड 19th फोन हो जाती है यह है कि हर किसी में यह होगा जो लोग होते हैं - पीपल होते हैं कॉल एक होते हैं या जिनको पहले से को लीवर डिजीज कि उनके अंदर ऐसे होता है ठीक है और उसकी वजह से क्या होता है कि जब फोन जो है वह उसकी डोमिनेशन हो जाएगी तो अल्टीमेटली इश्यूड बाय लेटर पैरालिसिस हो जाता है इसको लॉक टेंशन फ्रॉम ठीक है मतलब के सारी बॉडी जो है वह लो कपूर जाती है मतलब कि सिर्फ पेशंट जो नाग क्लिक कर सकता है आंखें बाकी ना वह उठ कर सकता है ना वह बोल सकता है ना मतलब उसके सारे बॉडी जो है ना वह काम करना छोड़ देती है सिर्फ मां ही ब्लेंड कर सकते हैं तो इस वजह से हम कहते हैं लॉग इन सिंड्रोम ठीक है और जिस अगर जिंदा सेंट्रल प्वाइंट टाइम मालूम लाइक तुझे इस वाले बुद्धिमान रिलेटेड डिसऑर्डर्स इन विच वे डिसकस्ड अबाउट थे रिकॉर्डिस्ट ऑफिस मल्टीपल स्क्लेरोसिस तबीयत से क्रॉसिंग बैलेंस सेटेलाइट इस कॉर्ड टू मीटर्स एंड प्रोग्रेस मल्टीफोकल यू कैन सिंपली आर्थिक विश्वसनीय जैसी वायरस एंड विल डिसकस अबाउट सेंट्रल प्वाइंट एंड मार्जिनलाइज्ड सिचुएशन लौटे
952
https://www.ixuehai.cn/html/644.html
移动版 QQ登录 | 登录 | 注册 | 留言 | 加收藏 手机访问公众号网站建设商标申请中采网 | | | 当前位置： 首页 > 中学课本 > 中学数学 > 高二数学 > 正文 爱学海 > 高二数学 > 正文 【高中数学】第九章-立体几何 四、平面平行与平面垂直. 2021-10-05 iXueHai.cn 爱学海 字体 - 小 + 大 纠错指正 | | | 四、 平面平行与平面垂直.1. 空间两个平面的位置关系：相交、平行.2. 平面平行判定定理：如果一个平面内有两条相交直线都平行于另一个平面，哪么这两个平面平行.（“线面平行，面面平行”）推论：垂直于同一条直线的两个平面互相平行；平行于同一平面的两个平面平行.[注]：一平面间的任一直线平行于另一平面.3. 两个平面平行的性质定理：如果两个平面平行同时和第三个平面相交，那么它们交线平行.（“面面平行，线线平行”）4. 两个平面垂直性质判定一：两个平面所成的二面角是直二面角，则两个平面垂直.两个平面垂直性质判定二：如果一个平面与一条直线垂直，那么经过这条直线的平面垂直于这个平面.（“线面垂直，面面垂直”）注：如果两个二面角的平面对应平面互相垂直，则两个二面角没有什么关系.5. 两个平面垂直性质定理：如果两个平面垂直，那么在一个平面内垂直于它们交线的直线也垂直于另一个平面.推论：如果两个相交平面都垂直于第三平面，则它们交线垂直于第三平面. 简记为：成角比交线夹角一半大，且又比交线夹角补角一半长，一定有4条.成角比交线夹角一半大，又比交线夹角补角小，一定有2条.成角比交线夹角一半大，又与交线夹角相等，一定有3条或者2条.成角比交线夹角一半小，又与交线夹角一半小，一定有1条或者没有. | 上一篇：【高中数学】第九章-立体几何三、直线与平面平行、直线与平面垂直. 下一篇：【高中数学】第九章-立体几何 五、棱锥、棱柱 本栏目热门阅读 【高中数学】第十章-排列组合二项定理 四、排列、组合综合 【高中数学】第九章-立体几何 六. 空间向量. 【高中数学】第九章-立体几何二、空间直线 【高中数学】第九章-立体几何 五、棱锥、棱柱 【高中数学】第八章-圆锥曲线方程 四、圆锥曲线的统一定义 【高中数学】第十章-排列组合二项定理 二、排列 【高中数学】第十章-排列组合二项定理 三、组合 【高中数学】第九章-立体几何 四、平面平行与平面垂直. 【高中数学】第八章-圆锥曲线方程 二、双曲线方程 【高中数学】第九章-立体几何三、直线与平面平行、直线与平面垂.. 本栏目最新文章 【高中数学】第十三章-极 限考试内容和考试要求 【高中数学】第十章-排列组合二项定理 五、二项式定理 【高中数学】第十章-排列组合二项定理 四、排列、组合综合 【高中数学】第十章-排列组合二项定理 三、组合 【高中数学】第十章-排列组合二项定理 二、排列 【高中数学】第十章-排列组合二项定理 一、两个原理 【高中数学】第十章-排列组合二项定理的考试内容和考试要求 【高中数学】第九章-立体几何 六. 空间向量. 【高中数学】第九章-立体几何 五、棱锥、棱柱 【高中数学】第九章-立体几何 四、平面平行与平面垂直. 本栏目随机阅读 【高中数学】第十三章-极 限考试内容和考试要求 【高中数学】第十章-排列组合二项定理 五、二项式定理 【高中数学】第十章-排列组合二项定理 四、排列、组合综合 【高中数学】第十章-排列组合二项定理 三、组合 【高中数学】第十章-排列组合二项定理 二、排列 【高中数学】第十章-排列组合二项定理 一、两个原理 【高中数学】第十章-排列组合二项定理的考试内容和考试要求 【高中数学】第九章-立体几何 六. 空间向量. 【高中数学】第九章-立体几何 五、棱锥、棱柱 【高中数学】第九章-立体几何 四、平面平行与平面垂直. | | | | | 高三： 语文｜ 数学｜ 英语｜ 物理｜ 化学｜ 地理｜ 生物｜ 历史｜ 思想政治｜ 信息技术｜ 初三： 语文｜ 数学｜ 英语｜ 物理｜ 化学｜ 历史｜ 道德与法制 | | 高二： 语文｜ 数学｜ 英语｜ 物理｜ 化学｜ 地理｜ 生物｜ 历史｜ 思想政治｜ 信息技术｜ 初二： 语文｜ 数学｜ 英语｜ 物理｜ 地理｜ 生物｜ 历史｜ 道法 | | 高一： 语文｜ 数学｜ 英语｜ 物理｜ 化学｜ 地理｜ 生物｜ 历史｜ 思想政治｜ 信息技术｜ 初一： 语文｜ 数学｜ 英语｜ 物理｜ 地理｜ 生物｜ 历史｜ 道法 | | 小学： 语文｜ 数学｜ 英语｜ 道德与法制｜ 科学｜ 试题： 小学｜ 初中｜ 中考｜ 高中｜ 高考｜ 作文： 小学｜ 初中｜ 高中｜ 满分｜ 技巧｜素材｜英汉词典 | | 高校： 985/211｜ 各省市｜ 港澳台｜ 国外｜ 排名｜ 中考专辑｜ 高考专辑｜ 各省中学｜ 职业考试： 从业资格｜ 技术技能｜ 艺术等级｜ 公务员｜ 百年党史 | | 课外： 教育动态｜ 教育法规｜ 老师榜｜ 家长帮｜ 学生派｜ 安全说｜ 古诗古文｜ 世界名著｜ 近代文学｜ 寓言故事｜ 格言名句｜ 阅读技巧｜ 在线许愿祝福 | | | | | | | Copyright © 2019-2023 爱学海 ixuehai.cn 版权所有 | 关于爱学海 | 商标证书 | 投诉反馈 | 版权声明 | 公司简介 | 中采网 | 粤ICP备09029428号 首页 搜索 留言 我的
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https://www.chegg.com/homework-help/questions-and-answers/two-static-pressure-taps-elevation-different-locations-along-horizontal-pipe-containing-ai-q172512805
Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Two static pressure taps at the same elevation but at different locations along a horizontal pipe containing air are connected to a water U-tube manometer. An accurate measurement of the difference in static pressure at the two pressure taps is registered by:Group of answer choicesThe (density of water minus the density of air) times the acceleration due Variables and Formulas The difference in pre... Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
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https://byjus.com/physics/unit-of-resistance/
Resistance is a term that students come across often when they study about electric currents. While electric current has two factors namely conductance and resistance we will have a quick look at the latter here in this page. Talking about resistance, it is basically defined as the obstruction that a substance causes to the flow of electric current. Resistance is represented by the letter R and has the dimension length squared mass per time cubed electric current squared. Below we will look at the unit of resistance and other details. SI Unit of Resistance The SI unit of resistance is ohm, which is defined as a volt per ampere. The name has been kept in remembrance of German physicist George Simon Ohm. The unit is sometimes written out as a word or symbolised by the uppercase Greek letter omega. Ohm has the following formula; in which the following units also appear: ampere (A), siemens (S), volt (V), watt (W), second (s), metre (m), joule (J), farad (F), kilogram (kg), and coulomb (C). There is something called specific resistance or resistivity which is measured by the unit ohm-metre or ohm-m. Popular Resistance Units Some of the popular units of resistance include Abohm, megohm, statohm, preece, and planck-impedance. Check out the conversion below. | | | --- | | Conversion to ohm | | | | 1 x 10-9 ohm | | | 898,755,178,740 ohms | | Preece | 1 x 10-6 ohms | | Megaohm | 1000000 ohm | Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S – The Learning App” for loads of interactive, engaging Physics-related videos and an unlimited academic assist. Related articles: | | | --- | | Unit Of Current | Unit Of Heat | Test your Knowledge on unit of resistance Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Physics related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs
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https://www.cs.cmu.edu/~lblum/flac/Presentations/Szabo-Wexler_ApproximateSetMembership.pdf
1 Approximate Membership of Sets: A Survey Elias Szabo-Wexler szabowexler@cmu.edu Carnegie Mellon University Abstract—The task of representing a set so as to support membership queries is a common one in computer science. If space is no object, a complete dictionary may provide accurate QUERYing in good runtime. If space is at a premium and the size of the set uncertain, Bloom filters and hash compaction provide good approximations. If the size is completely unknown, a modified Bloom filter provides very good runtime, with minimal extra error. This survey briefly explores the applications of dictionaries, Bloom filters, and hash compaction techniques to the set membership problem, observing details of their imple-mentations and tradeoffs in their accuracy. We conclude with a recommendation for when to use each, respectively. Index Terms—Data structures, Bloom filters, Bloom-g filters, hash compaction, membership testing, on-line algorithms, dy-namic set approximation I. INTRODUCTION T HE question of determining membership shows up com-monly in algorithmic design, in topics as diverse as probabilistic verification , network protocol analysis , routing-table lookup, online traffic measurement, peer-to-peer systems, cooperative caching, firewall design, intrusion de-tection, bioinformatics, database QUERY processing, stream computing, and distributed storage systems . Informally, we are frequently concerned with answering questions of the form “Is element x a member of set S”? In the static variant, the size of the set S is specified ahead of time, while in the dynamic variant, the elements of the set are specified one by one, in an on-line fashion. Obviously, answering the question correctly all of the time with good runtime is optimal from an accuracy perspective, motivating the classical dictionary approach, however it comes with a trade-off in memory utilization. Namely, we note the memory required by a classical dictionary grows linearly in the size of the set. For such static problems as verification, where the phenomenon of state explosion yields sets of tremendous size, or the dynamic variant on large data sets, even linear memory usage may be too much . In the static variant, if we wish to optimize for the smallest memory footprint or fastest runtime, we are advised to employ a Bloom-g filter (defined below). If instead we wish to optimize for accuracy, then an exhaustive dictionary is the clear choice. Hash compaction provides a decent middle ground, when a tradeoff between the two is desired. In the dynamic variant, Qiao shows in that the Bloom-g filter may still perform well even in the dynamic setting, offering a good default. In this paper, we provide a collection of applications of the set membership problem. With these as motivation, we then explore the static and dynamic variants of the problem, surveying the three major algorithmic techniques and provid-ing a high level algorithmic description of each, along with pertinent characterization descriptors, including runtime and memory footprint. A. Motivation In Broder and Mitzenmacher’s seminal survey of ap-plications of Bloom filters, several problems based on the set approximation problem are discussed. All of these instances employ Bloom filters, but in each such case there is a tradeoff, and any of the three techniques enumerated above might suf-fice. Historically, UNIX spell checkers utilized Bloom filters . In the field of networking, the Summary Cache protocol proposed by utilized Bloom filters for web cache sharing. Bloom filters were also applied by several different authors to locate resources in a peer-to-peer network, by for resource routing in a resource discovery service, and by for implementing IP traceback, to trace the route of a packet through a network. Hash compaction has been applied to explicit state verifi-cation , as well as explicit state space exploration –, and model checking for verification of software and hardware . Together, these represent a broad host of applications which require the ability to perform set membership queries, both in static and dynamic contexts. The ability to manage the tradeoff between memory utilization and accuracy is invaluable, as modern applications deal with data sets of ever increasing size. We are well motivated to explore alternatives which approach optimality in accuracy and requisite memory as well. B. Notation In this paper, we discuss the set approximation problem, namely the task of checking if some element x is contained in the set S. We denote the universe from which elements are drawn by U, and assume that |S| ≪|U|, that is, that the universe is much larger than the set itself. In the static variant, the size of the set is paramaterized by n ∈N, where |S| = n. In the dynamic variant, the size of the set is unknown, but the elements of S are all drawn from some universe U. We paramaterize the problem by u ∈N, where u = |U| and 0 < ϵ < 1, where ϵ is the accuracy of a solution, which is the probability of a “false positive,” namely the probability that an element y / ∈S should be reported to be an element of S. We denote the probability of an event E occurring by Pr[E]. Similarly, we denote the expected value of a random variable X by E[X]. We use IX for the indicator variable for the event E. Finally, we use log for the base two logarithm. Therefore, 2 Fig. 1. x is hashed via h(x) to an index i in the table, for both insertion and QUERYing. In the exhaustive hash table dictionary, elements of S are stored directly in the table, so each block stored is the actual element itself, contributing an overall Θ(n) term in memory utilization. unless otherwise stated, log x = log2 x. We indicate scalars via normal typeface, e.g. in the numeral 5, or the variable x. We indicate vectors via bold typeface, as in the vector x. II. IMPLEMENTATIONS We begin by offering an overview of the differing algo-rithms, along with their associated memory and runtime costs. Our algorithms provide both INSERT and QUERY operations. A. Complete Dictionary A dictionary is a complete mapping or enumeration of elements in S. A common implementation is the hash table -elements in the set are stored in some distribution of m buckets or slots indexed via a hash function, h : U →[m], which is used both for storage and retrieval. The table is therefore an array of buckets, each of which may contain zero, one, or more elements in the set. More formally, a dictionary is defined by a hash function h, the number of buckets, m, and a collision resolution scheme. This can be visualized as in Figure 1, above. The hash table function INSERT(x) is defined by: 1) Calculate bucket index 0 ≤i = h(x) < m 2) Update slot/bucket at index i in table with x′, utilizing resolution scheme if required The hash table QUERY(x) is similarly defined as: 1) Calculate bucket index 0 ≤i = h(x) < m 2) Search slot/bucket at index i in the table for x′. If present, report x ∈S, else report x / ∈S From we have that the memory utilized by a dictionary representing a set of size n taken from the universe U of size u is bounded by log u n  = n log(u/n) + Θ(n) bits of space. Utilizing a hash table to implement a dictionary, we may perform QUERY in Θ(n) worst case, and both QUERY and INSERT in O(1) expected time , with ϵ = 0 (i.e. with no errors). However, the hidden constants may be large, since many accesses to disk may be made if the elements stored are large. In this case, despite the good expected runtime, the hard drive accesses may become a bottleneck. Since real machines are built with a hierarchical memory structure, and Fig. 2. x is hashed via H(x) to a new fixed-width representation, x′ = H(x) of length |x′| = k. Then, the table hashing function is applied to generate the index i = h(x′), and the resulting slot or bucket is utilized for insertion or QUERY. layered caches, these bottlenecks translate to real world slow downs. If they can be minimized, a minimal memory footprint is practically translated into a runtime boost. B. Hash Compaction Hash compaction, introduced by Wolper and Leroy in , attempts to build on the concept of a hash table by addressing the issue of large element size. The hash table stores every element in its entirety, resulting in a linear relationship be-tween the size of the table and the elements inside. For a large number of large elements, this is prohibitively expensive. To resolve the issue, we may use the hash compaction scheme to store a compressed version of each element in the table, commonly resulting in a ten-fold decrease in space utilization . Hash compaction uses a hashing function, H : U 7→ {0, 1}l, where l is the number of bits used for the compressed representation , to generate a compressed representation from each element. Thus, we store only the compressed version, forcing the table to have a smaller footprint. The revised hash table is diagrammed in Figure 2, above. The algorithms for INSERT and QUERY do not change overly much. The hash compacted INSERT(x) is: 1) Compute hash compacted element x′ = H(x) 2) Calculate bucket index 0 ≤i = h(x′) < m 3) Update slot/bucket at index i in table with x′, utilizing resolution scheme if required The hash compacted QUERY(x) is similarly defined as: 1) Compute hash compacted element x′ = H(x) 2) Calculate bucket index 0 ≤i = h(x) < m 3) Search slot/bucket at index i in the table for x′. If present, report x ∈S, else report x / ∈S The new parameter k ∈N offers a tradeoff between memory footprint and accuracy. Observe that if k = log n and the hashing function is uniformly random, then in expectation the hash compacted algorithm will behave exactly as an exhaustive dictionary would. However, if k < log n, then multiple elements must map to identical compacted elements (i.e. there is overlap). This means that the hash compacted version may occasionally erroneously report membership, even when the queried element is not in the set. The chance of this occurring is parameterized by 0 < ϵ < 1, i.e. the probability that an input will receive a false positive (though the choice of ϵ fixes 3 Fig. 3. Basic Bloom filter: x is hashed via H(x) to a vector of indices, I = H(x) of length |I| = m. If all bits at those indices are asserted, x is in the set, otherwise it is not. Note that there is a circled 0 at index 1, so in this example either QUERY(x) = x / ∈S, or x is being inserted for the first time. bounds on the choice of k). Observe that this is bounded by the probability that a given input hashes to the same representative vector. This is given by Pr[H(x1) = H(x2)]. Assuming that H represents k universal hash functions, this is bounded by 1 2k (where 2k represents the number of possible hashed vectors). Thus, ϵ ≤ 1 2k . The tradeoff is simple - as k grows, ϵ drops rapidly, as the memory utilizing slowly increases. An optimal tuning is one where k pushes ϵ low enough for a given application, and no is no larger, to be efficient with memory. We note that this tradeoff comes at no asymptotic change in runtime. The compression is done in constant time, and the table lookup/insertion remains asymptotically the same. Thus, the only difference between hash compaction and the exhaustive dictionary is the memory utilization, and the chance for false positives. In the next section, we explore another alternative to the exhaustive dictionary which is asymptotically faster. C. Bloom Filter The Bloom filter, introduced by Bloom in , presents an even lighter weight alternative than the method of hash com-paction. Rather than utilizing a table to store a representation of each element in S, we instead maintain a large array A, with |A| = m bits. Elements are hashed to form a fixed-width vector of indices, which in turn index bits in the array. An element is present in the set if and only if every bit indexed by its hash is asserted. More formally, an input x ∈U is hashed via a set of k hashing functions H(x) to produce an index vector, I, and we have that x ∈S if and only if for each i ∈I, Ai = 1. This suggests very natural algorithms for INSERT and QUERY. Namely, the Bloom filter INSERT(x) is defined by: 1) Compute index vector, I = H(x) 2) Set each Ai = 1, for all i ∈I It is matched by the Bloom filter QUERY(x), defined by: 1) Compute index vector, I = H(x) 2) Check if Ai = 0, for any i ∈I. If yes, report x / ∈S, otherwise report x ∈S Clearly there are significant savings in terms of memory and throughput. Namely, we need only store m bits to categorize the entire set, rather than the entire table. Further, assuming our hash functions can be computed in constant time, then an insertion or QUERY may be performed in constant time, i.e. O(k). Note that for a given set of size n, and a filter of size m, the optimal choice of k is forced (and by extension of ϵ), and that both are fixed constants. We proceed as in in our analysis. Consider the vector A after all the elements of S are inserted, and recall that |A| = m, |S| = n, and we have k hashing functions. Then the probability that any bit is a 0, assuming uniformly distributed and independent hashing functions, is the probability that an element of S does not hash to a given bit index i, taken over all of the n elements, and each of the k hashing functions. This is given by  1 −1 m kn ≈e−kn/m Letting p = e−kn/m, the probability of a false positive is then 1 −  1 −1 m kn!k ≈  1 −e−kn/mk = (1 −p)k Analysis then reveals that the false positive rate is minimized globally by a particular selection of k, namely k = ln 2 · m n In this case, the false positive rate is ϵ = 1 2 k = (0.6185) m n Of particular note here is that we may minimize the false positive rate only when we know ahead of time the size of the set S. When that is not known, it is impossible to minimize the error in this way. In the next section, we consider modifying the Bloom filter to make fewer memory accesses (i.e. lowering the hidden constant factor). D. Bloom-g Filters The Bloom-g filter, proposed by Qiao, is a variant on the conventional Bloom filter which bounds the number of words which must be fetched from memory for a QUERY or insertion, at no cost in additional error. A Bloom-g filter maps each member x to g words, instead of one, and spreads its k membership bits evenly in the g words . To understand why this arrangement may be superior, consider an instance of a Bloom filter, f, with m bits in the filter and k membership bits per element. If m ≫k, then in the worst case for a QUERY we may need to fetch k words from memory, or write k words to memory. This operation may present a bottleneck, since reading from and writing to hard disk are much slower than processing cached data. In the Bloom-g filter, we map x to g words, which must be fetched from memory, and then map x to its k membership bits within those words. In the worst case, we must fetch or write at most g words from memory, and we may set g ≪k. Since the optimal number of membership bits 4 Fig. 4. Bloom-g filter, with l words, g = 2, and k = 4: x is hashed via H(x) to a vector of g words. Then, x is hashed to a set of k indices, distributed evenly among the g words. If all bits at those indices are asserted, x is in the set, otherwise it is not. In this example, x is not a member of the set. k is a function of the load factor of the filter (i.e. more entries stored requires more membership bits to maintain a good error rate), the worst case number of memory reads or writes for a regular Bloom filter grows linearly as the filter expands. With the Bloom-g filter, it is bounded by a constant, representing a significant improvement . Qiao shows that the error for the Bloom-g filter is a bit more complex than for the generic Bloom filter. In the trivial case, the Bloom-k filter (i.e. the Bloom-g filter with one bit per word) behaves exactly as the Bloom filter. However, when g ̸= k, the error is more complex to analyze. We proceed with a brief analysis, as in . Consider a Bloom-g filter, with g words per element, a total of l words, each with w bits in them, and n members. Consider an arbitrary word D in the array, and let X be the number of times this word is selected as an encoded word during the filter setup. When an element selects a word to encode membership in, D has a 1 l probability of being selected. Let c be a constant in the range [0, gn]. Then, Pr[X = c] = gn c  1 l c  1 −1 l gn−c . We may utilize this fact to analyze the likelihood of a nonmember x′ being labeled a member, by analyzing the prob-ability that a single membership word reports x′ a member, and taking this probability over all membership words. The result is that the probability of a false positive, ϵ, is given by    gn X c=0    gn c  1 l c  1 −1 l gn−c  1 −  1 −1 w c k g   k g       g . Experimentally, we find that Bloom-g filters require fewer membership bits to match Bloom filters for the same error with a given load factor. For complete results, and a full derivation, see . However, we may also utilize Bloom-g filters in a dynamic environment. We observe that the number of memory accesses is independent of the number of membership bits, so even as we increase the count of membership bits, we may have constant lookup costs. In order to ensure good results, we may cap the the number of elements in the filter (i.e. bound the load factor). This would ensure that we may report accurately to within some threshold, for a given number of inputs. If we know in advance the order of magnitude we will be handling, we may choose decent values for the size of the filter and its load factor, to ensure good access behavior. If the input is unbounded (i.e. we have no idea), we may set it high to start, and expand the filter dynamically if we rise above our load threshold. In this case, we would have to throw out the existing filter and reinitialize it. If the input set is large enough, this would not be an issue, though we may lose some data this way. In the next section, we move away from the hash table representation, and briefly commend on some more sophisticated constructions. E. Advanced Techniques We have discussed complete dictionaries, hash compaction, basic Bloom filters, and Bloom-g filters. There are other, yet more sophisticated constructions, which are also worth considering, though beyond the scope of this survey. Naor proposes a sliding Bloom filter in , which builds a Bloom filter to approximate only the most recent m elements in an m-element window on an unbounded stream. The technique abandons elements seen in the distant past, but for applications which may desire only recently seen elements (e.g. a least recently used cache eviction protocol), this may actually be advantageous. In this case, Naor’s filter has O(1) QUERY and INSERT in the worst case with high probability, and space consumption of (1 + o(1))(n log 1 ϵ + n · max{log n m, log log 1 ϵ }) bits, with a window size of n elements, slackness parameter of m, and error of ϵ. Pagh proposes a dynamic dictionary based approximation construction in . He utilizes techniques to modify existing dynamic dictionary constructions for the approximation problem, and then utilizes a clever scheme of representing a subset of elements from the input to answer QUERY in constant time in the worst case with high probability, and INSERT in constant time in the worse case with high probability, utilizing (1 + o(1))n log( 1 ϵ ) + O(n log log n) bits of space. Cohen proposes a spectral Bloom filter extending Bloom filters to multisets in . The result is a generalized Bloom filter, with query time in Θ(log( 1 ϵ )). Chazelle proposes the Bloomier filter in , associating additional satellite data with elements in the filter. The result is a filter within a constant factor of optimal in space usage, when utilized on a static set. The lookup time is constant, assuming access to truly random hash functions. III. FIXED SIZE APPROXIMATION We briefly review those constructions best suited to approx-imating a set of a fixed size, i.e. a set S for which |S| = n is known in advance. We considered the complete hash-table backed dictionary for perfect set representation, which offers QUERY and INSERT in O(1) expected time, requires 5 space in Θ(n), and ensures ϵ = 0. We then considered the hash compaction hash-table for approximate set representation, which offers QUERY and INSERT in O(1) expected time, requires space in Θ(n) with a smaller constant factor than the complete hash-table, and has probability of false positive ϵ ≤ 1 2k . We concluded with the Bloom filter, which offers QUERY and INSERT in O(1) worst time, requires space in Θ(1) (though the constant may be very large), and has the minimized probability of false positive ϵ = (0.6185) m n . Clearly, when perfect accuracy is required, some form of complete dictionary must be utilized. In this case, we would use the hash-table backed dictionary. If perfect accuracy is not absolutely required, then we choose between the Bloom filter and the hash compaction hash-table. Here, we consider both accuracy and runtime. If runtime is most important, we are well advised to use the Bloom filter, as operations are performed in constant worst case time, rather than amortized constant time. If memory is of utmost importance, we are again advised to use the Bloom filter, as we may fix the memory utilization directly, by selecting the size of the representing bit vector. IV. DYNAMIC SIZE APPROXIMATION In the dynamic case where |S| is not known in advance, the set of constructions best suited is somewhat different. The complete hash-table remains a viable alternative, but suffers the same linear dependence on the size of the set, and polylogarithmic rather than constant QUERY and INSERT. We considered the Bloom−g filter, a modified variant of the Bloom filter. The Bloom-g filter offers QUERY and INSERT in constant time in the worst case, with a fixed number of memory accesses per QUERY. The Bloom-g filter may be adapted to the dynamic environment by imposing a limit on the load factor (i.e. ratio of storage bits to elements stored). If it grows too high, more bits may be allocated, and the filter reinitialized. In this manner, it is possible to bound the error, and maintain good runtime with a small footprint. V. CONCLUSION In this paper, we surveyed the problem of set approxima-tion. We considered hash-table complete dictionaries, hash compacted hash-table dictionaries, Bloom filters, and Bloom-g filters. We analyzed these data structures on three axes: mem-ory footprint, runtime, and accuracy. With these as our core metrics, we made recommendations for the best choice in each of the static and dynamic variants of the set approximation problem. Finally, we recommended some more sophisticated constructions which are theoretically interesting. While there is great design freedom, we are generally advised to utilize the hash-table complete dictionary to address any problem in which perfect accuracy is required. Otherwise, the Bloom-g filter provides very good performance for the static or dynamic variant. ACKNOWLEDGMENT The author would like to thank Asa Frank, for providing meaningful guidance in the preparation of this survey and for conversations about the topics, and Lenore Blum, for instructing the Formal Languages, Automata, and Computation course at Carnegie Mellon University and for providing the impetus to write this paper. REFERENCES P. Dillinger and P. Manolios, “Bloom filters in probabilistic verification,” in Formal Methods in Computer-Aided Design, ser. Lecture Notes in Computer Science, A. Hu and A. Martin, Eds. Springer Berlin Heidelberg, 2004, vol. 3312, pp. 367–381. [Online]. Available: 26 A. Broder and M. Mitzenmacher, “Network applica-tions of bloom filters: A survey,” Internet Mathematics, vol. 1, no. 4, pp. 485–509, 2004. [Online]. Available: Y. Qiao, T. Li, and S. Chen, “Fast bloom filters and their generalization,” Parallel and Distributed Systems, IEEE Transactions on, vol. 25, no. 1, pp. 93–103, Jan 2014. P. Wolper and D. Leroy, “Reliable hashing without collision detection,” in IN COMPUTER AIDED VERIFICATION. 5TH INTERNATIONAL CONFERENCE. Springer-Verlag, 1993, pp. 59–70. L. Fan, P. Cao, J. Almeida, and A. Z. Broder, “Summary cache: A scalable wide-area web cache sharing protocol,” IEEE/ACM Trans. Netw., vol. 8, no. 3, pp. 281–293, Jun. 2000. [Online]. Available: S. E. Czerwinski, B. Y. Zhao, T. D. Hodes, A. D. Joseph, and R. H. Katz, “An architecture for a secure service discovery service,” 1999, pp. 24–35. A. C. Snoeren, C. Partridge, L. A. Sanchez, C. E. Jones, F. Tchakountio, S. T. Kent, and W. T. Strayer, “Hash-based ip traceback,” SIGCOMM Comput. Commun. Rev., vol. 31, no. 4, pp. 3–14, Aug. 2001. [Online]. Available: U. Stern and D. L. Dill, “Combining state space caching and hash compaction,” in In Methoden des Entwurfs und der Verifikation digitaler Systeme, 4. GI/ITG/GME Workshop. Shaker Verlag, 1996, pp. 81–90. M. Westergaard, L. Kristensen, G. Brodal, and L. Arge, “The comback method extending hash compaction with backtracking,” in Petri Nets and Other Models of Concurrency ICATPN 2007, ser. Lecture Notes in Computer Science, J. Kleijn and A. Yakovlev, Eds. Springer Berlin Heidelberg, 2007, vol. 4546, pp. 445–464. [Online]. Available: 26 J. Barnat, J. Havlek, and P. Rokai, “Distributed {LTL} model checking with hash compaction,” Electronic Notes in Theoretical Computer Science, vol. 296, no. 0, pp. 79 – 93, 2013, proceedings the Sixth International Workshop on the Practical Application of Stochastic Modelling (PASM) and the Eleventh International Workshop on Parallel and Distributed Methods in Verification (PDMC). [Online]. Available: R. Pagh, G. Segev, and U. Wieder, “How to approximate a set without knowing its size in advance,” CoRR, vol. abs/1304.1188, 2013. T. H. Cormen, C. E. Leiserson, R. L. Rivest, and C. Stein, Introduction to Algorithms, 3rd ed. The MIT Press, 2009. U. Stern and D. L. Dill, “A new scheme for memory-efficient probabilis-tic verification,” in in IFIP TC6/WG6.1 Joint International Conference on Formal Description Techniques for Distributed Systems and Commu-nication Protocols, and Protocol Specification, Testing, and Verification, 1996, pp. 333–348. B. H. Bloom, “Space/time trade-offs in hash coding with allowable errors,” Commun. ACM, vol. 13, no. 7, pp. 422–426, Jul. 1970. [Online]. Available: M. Naor and E. Yogev, “Sliding bloom filters,” in Algorithms and Computation, ser. Lecture Notes in Computer Science, L. Cai, S.-W. Cheng, and T.-W. Lam, Eds. Springer Berlin Heidelberg, 2013, vol. 8283, pp. 513–523. [Online]. Available: 48 S. Cohen and Y. Matias, “Spectral bloom filters.” in SIGMOD Confer-ence, A. Y. Halevy, Z. G. Ives, and A. Doan, Eds. ACM, 2003, pp. 241–252. B. Chazelle, J. Kilian, R. Rubinfeld, and A. Tal, “The bloomier filter: An efficient data structure for static support lookup tables,” in Proceedings of the Fifteenth Annual ACM-SIAM Symposium on Discrete Algorithms, ser. SODA ’04. Philadelphia, PA, USA: Society for Industrial and Applied Mathematics, 2004, pp. 30–39. [Online]. Available:
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6.1: Everyday Stoichiometry - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 6: Introduction to Stoichiometry Madera Community College Introductory Chemistry v2 { } { "6.01:Everyday_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.02:_Mole_Ratios" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.03:_Mass-Mole_and_Mole-Mass_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.04:_Mass-Mass_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.05:_Enthalpy_Change_is_a_Measure_of_the_Heat_Evolved_or_Absorbed" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.06:_Stoichiometry_Applications" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "6.E:_Introduction_to_Stoichiometry(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:__Basics_of_Measurement" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Matter_and_Energy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Elements_and_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_The_Mole_Concept" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Chemical_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Introduction_to_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Electrons_and_Chemical_Bonding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Molecules_and_Attractive_Forces" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Solids_Liquids_and_Gases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Aqueous_Solutions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Acids_and_Bases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Chemistry_Applications-_more_stoichiometry_and_reaction_prediction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Radioactivity_and_Nuclear_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 02 Jun 2025 17:33:17 GMT 6.1: Everyday Stoichiometry 522091 522091 Jamie MacArthur { } Anonymous Anonymous User 2 false false [ "article:topic", "showtoc:no", "transcluded:yes", "authorname:anonymous", "source-chem-53789", "program:ck12", "source@ "source@ "license:mixed", "source-chem-367790" ] [ "article:topic", "showtoc:no", "transcluded:yes", "authorname:anonymous", "source-chem-53789", "program:ck12", "source@ "source@ "license:mixed", "source-chem-367790" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. 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Madera Community College Introductory Chemistry v2 5. 6: Introduction to Stoichiometry 6. 6.1: Everyday Stoichiometry Expand/collapse global location Madera Community College Introductory Chemistry v2 Front Matter 1: Basics of Measurement 2: Matter and Energy 3: Elements and Compounds 4: The Mole Concept 5: Chemical Reactions 6: Introduction to Stoichiometry 7: Electrons and Chemical Bonding 8: Molecules and Attractive Forces 9: Solids, Liquids, and Gases 10: Aqueous Solutions 11: Acids and Bases 12: Chemistry Applications- more stoichiometry and reaction prediction 13: Radioactivity and Nuclear Chemistry Back Matter 6.1: Everyday Stoichiometry Last updated Jun 2, 2025 Save as PDF 6: Introduction to Stoichiometry 6.2: Mole Ratios picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report View on CommonsDonate Page ID 522091 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. How much equipment do you need for an experiment? 2. Everyday Stoichiometry 1. Example 6.1.1: Ham Sandwich Stoichiometry 1. Solution 2. Step 1: List the known quantities and plan the problem. 3. Unknown 4. Step 2: Solve. Summary Review Figure 6.1.1 (Public Domain; User:Skatebiker/Wikipedia via Wikipedia) How much equipment do you need for an experiment? You are in charge of setting out the lab equipment for a chemistry experiment. If you have twenty students in the lab (and they will be working in teams of two) and the experiment calls for three beakers and two test tubes, how much glassware do you need to set out? Figuring this out involves a type of balanced equation and the sort of calculations that you would do for a chemical reaction. Everyday Stoichiometry You have learned about chemical equations and the techniques used in order to balance them. Chemists use balanced equations to allow them to manipulate chemical reactions in a quantitative manner. Before we look at a chemical reaction, let's consider an equation for the ideal ham sandwich. Figure 6.1.2: The ideal ham sandwich. (CK-12 Curriculum Materials license; CK-12 Foundation via CK-12 Foundation) Our ham sandwich is composed of 2 slices of ham (H), a slice of cheese (C), a slice of tomato (T), 5 pickles (P), and 2 slices of bread (B). The equation for our sandwich is: 2⁢H+C+T+5⁢P+2⁢B→H⁡A 2⁢CTP⁢A 5⁢B⁢A 2 Now let us suppose that you are having some friends over and need to make five ham sandwiches. How much of each sandwich ingredient do you need? You take the number of each ingredient required for one sandwich (its coefficient in the above equation) and multiply by five. Using ham and cheese as examples, and using a conversion factor, you can calculate: 5⁢H⁡A 2⁢CTP⁢A 5⁢B⁢A 2×2 H 1⁢H⁡A 2⁢CTP⁢A 5⁢B⁢A 2=10 H 5⁢H⁡A 2⁢CTP⁢A 5⁢B⁢A 2×1⁢C 1⁢H⁡A 2⁢CTP⁢A 5⁢B⁢A 2=5 C The conversion factors contain the coefficient of each specific ingredient as the numerator and the formula of one sandwich as the denominator. The result is what you would expect. In order to make five ham sandwiches, you need 10 slices of ham and 5 slices of cheese. This type of calculation demonstrates the use of stoichiometry. Stoichiometry is the calculation of the amount of substances in a chemical reaction from the balanced equation. The sample problem below is another stoichiometry problem involving ingredients of the ideal ham sandwich. Example 6.1.1: Ham Sandwich Stoichiometry Kim looks in the refrigerator and finds that she has 8 slices of ham. In order to make as many sandwiches as possible, how many pickles does she need? Use the equation above. Solution Step 1: List the known quantities and plan the problem. Have 8 ham slices (H) 2 H=5 P (conversion factor) Unknown How many pickles (P) needed? The coefficients for the two reactants (ingredients) are used to make a conversion factor between ham slices and pickles. Step 2: Solve. 8 H×5 P 2 H=20 P Since 5 pickles combine with 2 ham slices in each sandwich, 20 pickles are needed to fully combine with 8 ham slices. Step 3: Think about your result. The 8 ham slices will make 4 ham sandwiches. With 5 pickles per sandwich, the 20 pickles are used in the 4 sandwiches. Summary An example of everyday stoichiometry is given. Review A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equation to describe the relationship. Write a conversion factor to show the relationship between the number of ice cubes and the number of smoothies produced. How many strawberries would you need to make 12 smoothies? 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https://www.youtube.com/watch?v=pJ1S4TwIMq8
Which fraction is greater 3/2 or 4/3? MagnetsAndMotors (Dr. B's Other Channel) 88400 subscribers 4 likes Description 976 views Posted: 23 May 2024 To determine which is bigger, 3/2 or 4/3 (three-halves or four-thirds), there are two different techniques (both provide the same answer). Method 1: Finding a Common Denominator Find the least common multiple (LCM) of the denominators 2 and 3. The LCM is 6. Convert both fractions to equivalent fractions with a denominator of 6: 3/2 3/3 = 9/6 4/3 2/2 = 8/6 Now, compare the numerators: 9 (from 9/6) is larger than 8 (from 8/6) Therefore, 3/2 is larger than 4/3. Method 2: Converting to Decimals Divide the numerator (top number) by the denominator (bottom number) of each fraction: 3 / 2 = 1.5 4 / 3 = 1.3333... Compare the decimal values: 1.5 is larger than 1.3333... Therefore, 3/2 is larger than 4/3. Either way, we find that the fraction 3/2 (three-halves) is greater than the fraction 4/3 (four-thirds). 1 comments Transcript: so which one of these fractions three halfes or 4/3 is larger well they're both improper fractions because the numerator is larger than the denominator but that really doesn't matter that much here because if we could find a common denominator right now we have two and three but if that were the same we could just compare the numerators and we'd know whether three halfes is greater than less than or equal to 4/3 so let's do that let's multiply 2 3 2 3 that equals 6 down here let's multiply 3 2 cuz 3 2 that equals six but we can't just multiply the denominator by two we do need to multiply the numerator by two as well 2 over 2 is 1 so we're just multiplying by one we don't really change the value just the way it's written 4 2 is 8 86 that's an equivalent fraction to 4/3 they have the same value up here let's multiply by 3 so we have 3 over 3 3 3 is 9 and now you see we have the same denominator we're just going to compare the numerators and 9 that's greater than 8 so 96 is greater than 86 only by 1 six 9 - 8 is 1 so only by one six but it's still greater since these are equivalent here and these are equivalent that means the three halves that's going to be greater than 4/3 as well this is Dr B and thanks for watching
959
https://www.wcedeportal.co.za/eresource/87171
ePortal PHET Interactive Simulations: Radioactive Dating Game Topics Radiometric Dating Carbon Dating Half Life Radioactivity Description Learn about different types of radiometric dating, such as carbon dating. Understand how decay and half life work to enable radiometric dating. Play a game that tests your ability to match the percentage of the dating element that remains to the age of the object. Sample Learning Goals Explain the concept of half-life, including the random nature of it, in terms of single particles and larger samples. Describe the processes of decay, including how elements change and emit energy and/or particles Explain how radiometric dating works and why different elements are used for dating different objects. Identify that 1/2-life is the time for 1/2 of a radioactive substance to decay. Do you have an educational app, video, ebook, course or eResource? Contribute to the Western Cape Education Department's ePortal to make a difference. Home Contact us Terms of Use Privacy Policy Western Cape Government © 2025. All rights reserved.
960
https://en.wikipedia.org/wiki/Gas_turbine_engine_thrust
Jump to content Search Contents (Top) 1 Transferring thrust to the aircraft 2 Rotor thrust 3 Thrust calculation 4 Thrust augmentation 5 References Gas turbine engine thrust Español Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia The familiar study of jet aircraft treats jet thrust with a "black box" description which only looks at what goes into the jet engine, air and fuel, and what comes out, exhaust gas and an unbalanced force. This force, called thrust, is the sum of the momentum difference between entry and exit and any unbalanced pressure force between entry and exit, as explained in "Thrust calculation". As an example, an early turbojet, the Bristol Olympus Mk. 101, had a momentum thrust of 9300 lb. and a pressure thrust of 1800 lb. giving a total of 11,100 lb. Looking inside the "black box" shows that the thrust results from all the unbalanced momentum and pressure forces created within the engine itself. These forces, some forwards and some rearwards, are across all the internal parts, both stationary and rotating, such as ducts, compressors, etc., which are in the primary gas flow which flows through the engine from front to rear. The algebraic sum of all these forces is delivered to the airframe for propulsion. "Flight" gives examples of these internal forces for two early jet engines, the Rolls-Royce Avon Ra.14 and the de Havilland Goblin. Transferring thrust to the aircraft [edit] The engine thrust acts along the engine centreline. The aircraft "holds" the engine on the outer casing of the engine at some distance from the engine centreline (at the engine mounts). This arrangement causes the engine casing to bend (known as backbone bending) and the round rotor casings to distort (ovalization). Distortion of the engine structure has to be controlled with suitable mount locations to maintain acceptable rotor and seal clearances and prevent rubbing. A well-publicized example of excessive structural deformation occurred with the original Pratt & Whitney JT9D engine installation in the Boeing 747 aircraft. The engine mounting arrangement had to be revised with the addition of an extra thrust frame to reduce the casing deflections to an acceptable amount. Rotor thrust [edit] The rotor thrust on a thrust bearing is not related to the engine thrust. It may even change direction at some RPM. The bearing load is determined by bearing life considerations. Although the aerodynamic loads on the compressor and turbine blades contribute to the rotor thrust they are small compared to cavity loads inside the rotor which result from the secondary air system pressures and sealing diameters on discs, etc. To keep the load within the bearing specification seal diameters are chosen accordingly as, many years ago, on the backface of the impeller in the de Havilland Ghost engine. Sometimes an extra disc known as a balance piston has to be added inside the rotor. An early turbojet example with a balance piston was the Rolls-Royce Avon. Thrust calculation [edit] The net thrust (FN) of an engine is given by:: p16 | | | --- | | where: | | | ṁair | = the mass rate of air flow through the engine | | ṁfuel | = the mass rate of propellant flow entering the engine | | ve | = the effective exhaust velocity of the jet (the speed of the exhaust plume relative to the aircraft) | | v | = the velocity of the air intake = the true airspeed of the aircraft | | (ṁair + ṁfuel)ve | = the nozzle gross thrust (FG) | | ṁair v | = the ram drag of the intake air | Most types of jet engine have an air intake, which provides the bulk of the fluid exiting the exhaust. Conventional rocket engines, however, do not have an intake, so ṁair is zero. Therefore, rocket engines do not have ram drag and the gross thrust of the rocket engine nozzle is the net thrust of the engine. Consequently, the thrust characteristics of a rocket motor are different from that of an air breathing jet engine, and thrust is independent of velocity. If the velocity of the jet from a jet engine is equal to sonic velocity, the jet engine's nozzle is said to be choked. If the nozzle is choked, the pressure at the nozzle exit plane is greater than atmospheric pressure, and extra terms must be added to the above equation to account for the pressure thrust.[citation needed][dubious – discuss] However, ve is the effective exhaust velocity. If a turbojet engine has a purely convergent exhaust nozzle and the actual exhaust velocity reaches the speed of sound in air at the exhaust temperature and pressure, the exhaust gas cannot be further accelerated by the nozzle. In such a case, the exhaust gas retains a pressure which is higher than that of the ambient air. This is the source of 'pressure thrust'. The rate of flow of fuel entering the engine is often very small compared with the rate of flow of air. When the contribution of fuel to the nozzle gross thrust can be ignored, the net thrust is: The velocity of the jet (ve) must exceed the true airspeed of the aircraft (v) if there is to be a net forward thrust on the aircraft. The velocity (ve) can be calculated thermodynamically based on adiabatic expansion. Thrust augmentation [edit] Thrust augmentation has taken many forms, most commonly to supplement inadequate take-off thrust. Some early jet aircraft needed rocket assistance to take off from high altitude airfields or when the day temperature was high. A more recent aircraft, the Tupolev Tu-22 supersonic bomber, was fitted with four SPRD-63 boosters for take-off. Possibly the most extreme requirement needing rocket assistance, and which was short-lived, was zero-length launching. Almost as extreme, but very common, is catapult assistance from aircraft carriers. Rocket assistance has also been used during flight. The SEPR 841 booster engine was used on the Dassault Mirage for high altitude interception. Early aft-fan arrangements which added bypass airflow to a turbojet were known as thrust augmentors. The aft-fan fitted to the General Electric CJ805-3 turbojet augmented the take-off thrust from 11,650lb to 16,100lb. Water, or other coolant, injection into the compressor or combustion chamber and fuel injection into the jetpipe (afterburning/reheat) became standard ways to increase thrust, known as 'wet' thrust to differentiate with the no-augmentation 'dry' thrust. Coolant injection (pre-compressor cooling) has been used, together with afterburning, to increase thrust at supersonic speeds. The 'Skyburner' McDonnell Douglas F-4 Phantom II set a world speed record using water injection in front of the engine. At high Mach numbers afterburners supply progressively more of the engine thrust as the thrust from the turbomachine drops off towards zero at which speed the engine pressure ratio (epr) has fallen to 1.0 and all the engine thrust comes from the afterburner. The afterburner also has to make up for the pressure loss across the turbomachine which is a drag item at higher speeds where the epr will be less than 1.0. Thrust augmentation of existing afterburning engine installations for special short-duration tasks has been the subject of studies for launching small payloads into low earth orbits using aircraft such as McDonnell Douglas F-4 Phantom II, McDonnell Douglas F-15 Eagle, Dassault Rafale and Mikoyan MiG-31, and also for carrying experimental packages to high altitudes using a Lockheed SR-71. In the first case an increase in the existing maximum speed capability is required for orbital launches. In the second case an increase in thrust within the existing speed capability is required. Compressor inlet cooling is used in the first case. A compressor map shows that the airflow reduces with increasing compressor inlet temperature although the compressor is still running at maximum RPM (but reduced aerodynamic speed). Compressor inlet cooling increases the aerodynamic speed and flow and thrust. In the second case a small increase in the maximum mechanical speed and turbine temperature were allowed, together with nitrous oxide injection into the afterburner and simultaneous increase in afterburner fuel flow. References [edit] ^ "The Avro Type 698 Vulcan" David W. Fildes, Pen & Sword Aviation 2012, ISBN 978 1 84884 284 7, p.301, Gas Flow Diagram ^ The Aircraft Gas Turbine and its operation December 1982, P&W Oper. Instr. 200, United Technologies Pratt & Whitney ^ Jet Propulsion For Aerospace Applications" Second Edition 1964, Pitman Publishing Corp., Library of Congress Catalog card Number 64-18757, p.262 ^ "flight - flight pdf - pdf archive - 1957 - 1484 - Flight Archive". ^ "goblin - lb - flight - 1946 - 0353 - Flight Archive". ^ "1969 - 3201 - Flight Archive". ^ "Jet engine force frame". ^ "747 Creating the world's first jumbo jet and other adventures from a life in aviation" Joe Sutter, Smithsonian Books, ISBN 978-0-06-088241-9, p.185-188 ^ "de havilland - 1947 - 0202 - Flight Archive". ^ "rolls-royce avon - 1955 - 1778 - Flight Archive". ^ a b c Nicholas Cumpsty (2003). Jet Propulsion (2nd ed.). Cambridge University Press. ISBN 978-0-521-54144-2. ^ 16.Unified: Thermodynamics and Propulsion, Prof. Z. S. Spakovszky. Scroll down to "Performance of Turbojet Engines, Section 11.6.4. (Obtained from the website of the Massachusetts Institute of Technology) ^ "Tupolev Tu-22 Blinder" Sergey Burdin & Alan E Dawes 2006, Pen & Sword Aviation, ISBN 1 84415 241 3, p.130 ^ "atar - snecma - pressure ratio - 1960 - 0376 - Flight Archive". ^ Gas turbine aero-thermodynamics : with special reference to aircraft propulsion Sir Frank Whittle, Pergamon Press Ltd. 1981, ISBN 9780080267197. p.220 ^ "gas turbines - flight boosting - boosting gas - 1952 - 0092 - Flight Archive". ^ ^ "Flightdeck Friday: The YF4H-1 Phantom II – Operations Skyburner and Sageburner". ^ "Jet Propulsion For Aerospace Applications" Second Edition 1964, Hesse and Mumford, Pitman Publishing Corporation, Library of Congress Catalog Card Number 64-18757, p.375 ^ "F-12 Series Aircraft Propulsion System Performance and Development, David H. Campbell, J. Aircraft Vol.II, No.11, November 1974, p.672 ^ "Water injection pre-compressor cooling assist space access" Mehta, Huynh, Hagseth, The Aeronautical Journal, February 2015, Volume 19, Number 1212, p.147 ^ "Data" (PDF). ntrs.nasa.gov. June 1997. Retrieved from " Categories: Jet engines Aerodynamics Hidden categories: All articles with unsourced statements Articles with unsourced statements from July 2018 All accuracy disputes Articles with disputed statements from July 2018 Gas turbine engine thrust Add topic
961
https://www.combinatorics.org/ojs/index.php/eljc/article/download/v24i4p15/pdf/
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https://math.stackexchange.com/questions/3497638/why-is-the-dot-product-of-perpendicular-vectors-zero
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Why is the dot product of perpendicular vectors zero? Ask Question Asked Modified 2 years, 6 months ago Viewed 9k times 2 $\begingroup$ I've read that taking a dot product is just projecting one vector on the other, so a perpendicular vector will have no components in the other vectors direction. But shouldn't this leave the length unchanged so it has its original magnitude like multiplying it by 1? vectors Share asked Jan 4, 2020 at 20:49 user552217user552217 $\endgroup$ 2 3 $\begingroup$ "taking a dot product is just projecting one vector on the other" What? $\endgroup$ Angina Seng – Angina Seng 2020-01-04 20:50:41 +00:00 Commented Jan 4, 2020 at 20:50 1 $\begingroup$ When does the cosine rule reduce to the Pythagorean theorem? $\endgroup$ Karl – Karl 2020-01-04 21:33:54 +00:00 Commented Jan 4, 2020 at 21:33 Add a comment | 5 Answers 5 Reset to default 2 $\begingroup$ $$u.v=|u||v|\cos \theta $$ If $ \theta =\pi/2$ we have $\cos \theta =0$ Thus perpendicular vectors have zero dot product. Share edited Sep 19, 2021 at 18:14 Xander Henderson♦ 32.8k2525 gold badges7373 silver badges122122 bronze badges answered Jan 4, 2020 at 20:58 Mohammad Riazi-KermaniMohammad Riazi-Kermani 70.1k44 gold badges4444 silver badges9393 bronze badges $\endgroup$ Add a comment | 1 $\begingroup$ To clarify, the projection of $\vec u$ on $\vec v$ is the vector $$\left(\frac{\vec u\cdot \vec v}{\|\vec v\|^2}\right)\vec v = \left(\frac{\vec u\cdot\vec v}{\|\vec v\|}\right)\frac{\vec v}{\|\vec v\|}.$$ The dot product is a scalar quantity. But the length of the projection is always strictly less than the original length unless $\vec u$ is a scalar multiple of $\vec v$. Share answered Jan 4, 2020 at 21:16 Ted ShifrinTed Shifrin 127k77 gold badges113113 silver badges174174 bronze badges $\endgroup$ Add a comment | 1 $\begingroup$ I've read that taking a dot product is just projecting one vector on the other, so a perpendicular vector will have no components in the other vectors direction. Yes, that's the intuition. But shouldn't this leave the length unchanged Why? If you project the orthogonal vectors to each other, the length of the projected vector becomes zero. Also "by definition", two non-zero vectors are said to be orthogonal when (if and only if) their dot product is zero. This can also be demonstrated by Pythagorean theorem. Share edited Mar 3, 2023 at 22:23 answered Mar 2, 2023 at 16:55 aerinaerin 14911 silver badge77 bronze badges $\endgroup$ 2 $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ FD_bfa – FD_bfa 2023-03-02 22:59:20 +00:00 Commented Mar 2, 2023 at 22:59 $\begingroup$ Hi @FD_bfa, I added essential parts of the answer to the body. Thank you for the suggestion. $\endgroup$ aerin – aerin 2023-03-03 00:25:48 +00:00 Commented Mar 3, 2023 at 0:25 Add a comment | 0 $\begingroup$ The dot product is that way by definition, this particular definition gives the expected Euclidean Norm. A consistent dot product can be and is defined differently, for example in physics & differential geometry the metric tensor is solved for and ascribes a different inner product at every space-time coordinate, which is the means for modeling curved spaces. Share answered Jan 5, 2020 at 4:55 Ehab ShoubakiEhab Shoubaki 18466 bronze badges $\endgroup$ Add a comment | 0 $\begingroup$ Visually, nonzero $u, v \in \mathbb{R}^n$ are perpendicular iff the smaller (in $[0, \pi]$) angle between them is $\pi/2$. With $u \cdot v$ defined as $\sum_{i = 1}^n u_i v_i$, the the law of cosines implies that $$u \cdot v= \| u \| \|v\| \cos \theta $$ If $ \theta =\pi/2$, then $\cos \theta = 0$, whence $u \cdot v = 0$; perpendicular vectors have zero dot product. Conversely, if $u \cdot v = 0$, then multiplying by $\frac 1 {\| u \| \| v \|}$ gives $\cos \theta = 0$, so the angle between $u$ and $v$ must be $\theta = \arccos 0 = \pi/2$. In conclusion, two nonzero vectors are perpendicular iff their dot product is zero. If we assert that the zero vector is perpendicular to everything, then this equivalence applies to all vectors, so the geometric statement of "perpendicular" coincides with the algebraic statement of $u \cdot v = 0$. This motivates the definition that two vectors in any inner product space are "orthogonal" iff $\langle u, v \rangle = 0$; in the case of Euclidean $\mathbb{R}^n$, orthogonal vectors are perpendicular vectors. (This was originally an edit to Mohammad Riazi-Kermani's answer, but the edit was taken down.) Share answered Sep 19, 2021 at 19:03 jskattt797jskattt797 1,7911212 silver badges2929 bronze badges $\endgroup$ Add a comment | You must log in to answer this question. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked How to understand dot product is the angle's cosine? Related 2 Why is vectors A.B expressed in the form of $a_xb_x +a_yb_y +a_zb_z$ in the dot product? 3 Why is the dot product defined as a scalar and the cross product a normal vector? What's an intuitive explanation behind cross products being vectors, when dot products are not? 1 Perpendicular distance with dot product 2 Why is the vector dot product scaled? 2 What is the derivation of the dot product formula? 1 Why isn't the dot product of two identical vectors equal to the magnitude of that vector? 1 The dot product between basis vectors gives all the necessary information to know the dot product between two arbitrary vectors. 2 Meaning of dot product of two vectors Hot Network Questions Quantizing EM field by imposing canonical commutation relations Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? How to rsync a large file by comparing earlier versions on the sending end? How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? 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https://online.stat.psu.edu/stat414/lesson/24/24.3
Skip to main content Keyboard Shortcuts Help : F1 or ? Previous Page : ← + CTRL (Windows) : ← + ⌘ (Mac) Next Page : → + CTRL (Windows) : → + ⌘ (Mac) Search Site : CTRL + SHIFT + F (Windows) : ⌘ + ⇧ + F (Mac) Close Message : ESC 24.3 - Mean and Variance of Linear Combinations We are still working towards finding the theoretical mean and variance of the sample mean: If we re-write the formula for the sample mean just a bit: we can see more clearly that the sample mean is a linear combination of the random variables . That's why the title and subject of this page! That is, here on this page, we'll add a few a more tools to our toolbox, namely determining the mean and variance of a linear combination of random variables . Before presenting and proving the major theorem on this page, let's revisit again, by way of example, why we would expect the sample mean and sample variance to have a theoretical mean and variance. Example 24-2 Section A statistics instructor conducted a survey in her class. The instructor was interested in learning how many siblings, on average, the students at Penn State University have? She took a random sample of students, and asked each student how many siblings he/she has. The resulting data were: 0, 2, 1, 1. In an attempt to summarize the data she collected, the instructor calculated the sample mean and sample variance, getting: and The instructor realized though, that if she had asked a different sample of students how many siblings they have, she'd probably get different results. So, she took a different random sample of students. The resulting data were: 4, 1, 2, 1. Calculating the sample mean and variance once again, she determined: and Hmmm, the instructor thought that was quite a different result from the first sample, so she decided to take yet another sample of students. Doing so, the resulting data were: 5, 3, 2, 2. Calculating the sample mean and variance yet again, she determined: and That's enough of this! I think you can probably see where we are going with this example. It is very clear that the values of the sample mean and the sample variance depend on the selected random sample. That is, and are continuous random variables in their own right. Therefore, they themselves should each have a particular: probability distribution (called a "sampling distribution"), mean, and variance. We are still in the hunt for all three of these items. The next theorem will help move us closer towards finding the mean and variance of the sample mean . Theorem Suppose are independent random variables with means and variances . Then, the mean and variance of the linear combination , where are real constants are: and: respectively. Proof Let'sstart with the proof for the mean first: Now for the proof for the variance. Starting with the definition of the variance of , we have: Now, substituting what we know about and the mean of Y, we have: Because the summation signs have the same index ( to ), we can replace the two summation signs with one summation sign: And, we can factor out the constants : Now, let's rewrite the squared term as the product of two terms. In doing so, use an index of on the first summation sign, and an index of on the second summation sign: Now, let's pull the summation signs together: Then, by the linear operator property of expectation, we can distribute the expectation: Now, let's rewrite the variance of by evaluating each of the terms from to and to . In doing so, recognize that when , the expectation term is the variance of , and when , the expectation term is the covariance between and , which by the assumed independence, is 0: Simplifying then, we get: And, simplifying yet more using variance notation: Finally, we have: as was to be proved. Example 24-3 Section Let and be independent random variables. Suppose the mean and variance of are 2 and 4, respectively. Suppose, the mean and variance of are 3 and 5 respectively. What is the mean and variance of ? Solution The mean of the sum is: and the variance of the sum is: What is the mean and variance of ? Solution The mean of the difference is: and the variance of the difference is: That is, the variance of the difference in the two random variables is the same as the variance of the sum of the two random variables. What is the mean and variance of ? Solution The mean of the linear combination is: and the variance of the linear combination is:
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https://zhuanlan.zhihu.com/p/74981237
【复习专题】解析二元一次方程知识点及应用 - 知乎 关注推荐热榜专栏圈子 New付费咨询知学堂 ​ 直答 切换模式 登录/注册 【复习专题】解析二元一次方程知识点及应用 切换模式 【复习专题】解析二元一次方程知识点及应用 老刘说数学 老师 67 人赞同了该文章 二元一次方程是初中数学学习中的重要内容，上承一元一次方程，下接不等式组。之前的文章内容中我们先后讲述了有关一元一次方程和一元二次方程的内容，那么，我们接着来学习有关二元一次方程的知识点： 复习要求 1、认识二元一次方程（组）； 2、了解二元一次方程（组）的解以及求二元一次方程的正整数解； 3、解决有关二元一次方程（组）的实际应用。 二元一次方程的基本内容 1 01二元一次方程 （1）二元一次方程的概念 含有两个未知数，并且所含未知数的项的次数都是1的整式方程叫做二元一次方程。 二元一次方程的一般形式：ax+by+c=0（a≠0，b≠0）。 判定二元一次方程必须同时满足三个条件： ①方程两边的代数式都是整式——整式方程； ②含有两个未知数——“二元”； ③含有未知数的项的次数为 1——“一次”。 （2）二元一次方程的解 使二元一次方程左、右两边的值相等的两个未知数的值，叫做二元一次方程的解。一般情况下，一个二元一次方程有无数个解。 2二元一次方程组 （1）二元一次方程组的概念 由几个一次方程组成并且含有两个未知数的方程组，叫二元一次方程组。 注意：二元一次方程组不一定由两个二元一次方程合在一起：方程可以超过两个，有的方程可以只有一元（一元方程在这里也可看作另一未知数系数为 0 的二元方程）。 （2）二元一次方程组的解 二元一次方程组的解必须满足方程组中的每一个方程，同时它也必须是一个数对，而不能是一个数。 （3）二元一次方程组的解法 ●a.代入消元法 代入消元法是解二元一次方程组的基本方法之一。 通过等量代换，消去方程组中的一个未知数，使二元一次方程组转化为一元一次方程，从而求得一个未知数的值，然后再求出被消去未知数的值，从而确定原方程组的解的方法。 步骤： ①从方程组中选一个系数比较简单的方程，将这个方程中的一个未知数，例如 y，用另一个未知数如 x 的代数式表示出来，即写成 y = ax + b 的形式； ② y = ax + b 代入另一个方程中，消去 y ，得到一个关于 x 的一元一次方程； ③解这个一元一次方程，求出 x 的值； ④回代求解：把求得的 x 的值代入 y = ax + b 中求出 y 的值，从而得出方程组的解。 ●b.加减消元法 加减法是消元法的一种，也是解二元一次方程组的基本方法之一。加减法不仅在解二元一次方程组中适用，也是今后解其他方程（组）经常用到的方法。 步骤： ①变换系数：把一个方程或者两个方程的两边都乘以适当的数，使两个方程里的某一个未知数的系数互为相反数或相等； ②加减消元：把两个方程的两边分别相加或相减，消去一个未知数，得到一个一元一次方程； ③解这个一元一次方程，求得一个未知数的值； ④回代：将求出的未知数的值代入原方程组中，求出另一个未知数的值。 ●加减消元方法的选择： 1、一般选择系数绝对值最小的未知数消元； 2、当某一未知数的系数互为相反数时，用加法消元；当某一未知数的系数相等时，用减法消元； 3、某一未知数系数成倍数关系时，直接对一个方程变形，使其系数互为相反数或相等，再用加减消元求解； 4、当相同的未知数的系数都不相同时，找出某一个未知数的系数的最小公倍数，同时对两个方程进行变形，转化为系数的绝对值相同，再用加减消元求解。 二元一次方程的应用 数学来源于生活又服务于生活，我们把生活实际中的问题，用设未知数的方法用二元一次方程来刻画，就把实际问题，转化成了数学问题，这种解题就是数学中的建模思想，它能化难为易化抽象为具体，也是我们学习方程的重点。 列方程组与列一元一次方程基本类似，只不过列二元一次方程组解应用题时，应从题目中找出两个独立的相等关系，根据这两个相等关系列方程组求解。尤其是在七年级没学好一元一次方程的同学，需要及时有效的补缺。 1、列方程组解应用题的基本思想 列方程组解应用题是把“未知”转化为“已知”的重要方法，它的关键是把已知量和未知量联系起来，找出题目中的相等关系。 所列方程必须满足： (1) 方程两边表示的是同类量； (2) 同类量的单位要统一； (3) 方程两边的数值要相等。 2、二元一次方程组的应用步骤 （1）审题：弄清题意及题目中的数量关系 （2）设未知数：可直接设元，也可间接设元 （3）找等量关系：根据相关公式变量等，找出题目中的等量关系 （4）列方程组：根据题目中能表示全部含义的等量关系列出方程，并组成方程组 （5）解方程组：利用消元法等方法解所列的方程组 （6）检验：检验解的正确性，是否满足实际问题 （7）答话：回答题目问题 3常用的基本等量关系1、行程问题： (1)追击问题：追击问题是行程问题中很重要的一种，它的特点是同向而行。这类问题比较直观，画线段，用图便于理解与分析。 其等量关系式是:两者的行程差＝开始时两者相距的路程。 (2)相遇问题：相遇问题也是行程问题中很重要的一种，它的特点是相向而行。这类问题也比较直观，因而也画线段图帮助理解与分析。 这类问题的等量关系是：双方所走的路程之和＝总路程。 (3)航行问题： ①船在静水中的速度＋水速＝船的顺水速度； ②船在静水中的速度－水速＝船的逆水速度； ③顺水速度－逆水速度＝2×水速。 注意：飞机航行问题同样会出现顺风航行和逆风航行，解题方法与船顺水航行、逆水航行问题类似。 2、利润问题： (1)利润＝售价－成本(进价)； (2)利润＝成本（进价）×利润率； (3)标价＝成本(进价)×(1＋利润率)； (4)实际售价＝标价×打折率； 注意：“商品利润＝售价－成本”中的右边为正时，是盈利；为负时，就是亏损。打几折就是按标价的十分之几或百分之几十销售。 （例如八折就是按标价的十分之八即五分之四或者百分之八十） 3、储蓄问题： ■(1)基本概念 ①本金：顾客存入银行的钱叫做本金。 ②利息：银行付给顾客的酬金叫做利息。 ③本息和：本金与利息的和叫做本息和。 ④期数：存入银行的时间叫做期数。 ⑤利率：每个期数内的利息与本金的比叫做利率。 ⑥利息税：利息的税款叫做利息税。 ■(2)基本关系式 ①利息＝本金×利率×期数 ②本息和＝本金＋利息＝本金＋本金×利率×期数＝本金× (1＋利率×期数) ③利息税＝利息×利息税率＝本金×利率×期数×利息税率 ④税后利息＝利息× (1－利息税率) ⑤年利率＝月利率×12 注意：当题目中涉及免税利息时，需要明晰免税利息=利息 4、数字问题： 解决这类问题，首先要正确掌握自然数、奇数、偶数等有关概念、特征及其表示。如当n为整数时，奇数可表示为2n+1(或2n-1)，偶数可表示为2n等。 有关两位数的基本等量关系式为：两位数=十位数字10+个位数字 5、其他问题： （1）工程问题：工作效率×工作时间=工作量 （2）增长率问题：原量×(1＋增长率)＝增长后的量；原量×(1－减少率)＝减少后的量 （3）和差倍分问题：较大量＝较小量＋多余量，总量＝倍数×倍量 （4）几何问题：解决这类问题的基本关系式有关几何图形的性质、周长、面积等计算公式 （5）年龄问题：解决这类问题的关键是抓住两人年龄的增长数是相等，两人的年龄差是永远不会变的 老刘有话说 涉及二元一次方程需要注意以下要点： (1)解实际应用问题必须写“答”，而且在写答案前要根据应用题的实际意义，检查求得的结果是否合理，不符合题意的解应该舍去 (2)“设”、“答”两步，都要写清单位名称 (3)一般来说，设几个未知数就应该列出几个方程并组成方程组。 (4)列方程组解应用题应注意的问题： ①弄清各种题型中基本量之间的关系； ②审题时注意从文字，图表中获得有关信息； ③注意用方程组解应用题的过程中单位的书写，设未知数和写答案都要带单位，列方程组与解方程组时，不要带单位； ④正确书写速度单位，避免与路程单位混淆； ⑤在寻找等量关系时，应注意挖掘隐含的条件； ⑥列方程组解应用题一定要注意检验。 发布于 2019-07-24 06:12 数学 初中数学 方程 ​赞同 67​​6 条评论 ​分享 ​喜欢​收藏​申请转载 ​ 写下你的评论... 6 条评论 默认 最新 Chen 鼓掌👏（应用题使我抓狂🤬😤） 2019-08-08 ​回复​3 哄哄 谢谢 2022-05-11 ​回复​喜欢 仟莳 还以为是数论里面的唉 2021-07-11 ​回复​喜欢 逆风草 很详细 2021-02-06 ​回复​喜欢 软酱奶糖 想要学好数学不仅仅要在课上下功夫，还要在课外积累 2020-04-06 ​回复​喜欢 执首 诗酒 2020-02-03 ​回复​喜欢 关于作者 老刘说数学 老师 回答 0文章 269关注者 2,109 ​关注他​发私信 推荐阅读 数学知识篇15：二元一次方程组「二元一次方程」 ======================= 影子 发表于学习方法/... 一元二次方程知识点解析和题型汇总 ================ 老刘说数学1. 一元三次方程通用解 ============ 中学学过一元二次方程的求解公式，那有没有一元三次方程的求跟公式呢？ 目录1. 一元三次方程通用解1.1. 化归思想1.2. 一元三次方程1.2.1. 化归为缺省二次项的三次方程1.2.2. 归结为解二次方… Aemon 发表于数学通识 “解一元二次方程”常见三大错误，90%的学生中招！ ========================= 万唯中考 想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫 其他扫码方式：微信 下载知乎App 无障碍模式 验证码登录 密码登录 开通机构号 中国 +86 获取短信验证码 获取语音验证码 登录/注册 其他方式登录 未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》 扫码下载知乎 App 关闭二维码 打开知乎App 在「我的页」右上角打开扫一扫 其他扫码方式：微信 下载知乎App 无障碍模式 验证码登录 密码登录 开通机构号 中国 +86 获取短信验证码 获取语音验证码 登录/注册 其他方式登录 未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》 扫码下载知乎 App 关闭二维码
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https://mathworld.wolfram.com/HilbertsAxioms.html
TOPICS Hilbert's Axioms The 21 assumptions which underlie the geometry published in Hilbert's classic text Grundlagen der Geometrie. The eight incidence axioms concern collinearity and intersection and include the first of Euclid's postulates. The four ordering axioms concern the arrangement of points, the five congruence axioms concern geometric equivalence, and the three continuity axioms concern continuity. There is also a single parallel axiom equivalent to Euclid's parallel postulate. See also Congruence Axioms, Continuity Axioms, Incidence Axioms, Ordering Axioms, Parallel Postulate Explore with Wolfram|Alpha More things to try: axiom axioms (0.8333...)(0.1111...)/(0.22111111...) References Hilbert, D. The Foundations of Geometry, 2nd ed. Chicago, IL: Open Court, 1980.Iyanaga, S. and Kawada, Y. (Eds.). "Hilbert's System of Axioms." §163B in Encyclopedic Dictionary of Mathematics. Cambridge, MA: MIT Press, pp. 544-545, 1980. Referenced on Wolfram|Alpha Hilbert's Axioms Cite this as: Weisstein, Eric W. "Hilbert's Axioms." From MathWorld--A Wolfram Resource. Subject classifications
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https://en.wikipedia.org/wiki/Inhibitory_postsynaptic_potential
Jump to content Search Contents (Top) 1 Components 1.1 Types 1.2 Factors 2 Inhibitory neurotransmitters 3 Inhibitory receptors 3.1 Ionotropic receptors 3.2 Metabotropic receptors 4 Significance 5 Studies 6 See also 7 References Inhibitory postsynaptic potential العربية Català Deutsch Français 한국어 Polski Português Русский Svenska Türkçe Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Electrical signal inhibiting a neuron from firing An inhibitory postsynaptic potential (IPSP) is a kind of synaptic potential that makes a postsynaptic neuron less likely to generate an action potential. The opposite of an inhibitory postsynaptic potential is an excitatory postsynaptic potential (EPSP), which is a synaptic potential that makes a postsynaptic neuron more likely to generate an action potential. IPSPs can take place at all chemical synapses, which use the secretion of neurotransmitters to create cell-to-cell signalling. EPSPs and IPSPs compete with each other at numerous synapses of a neuron. This determines whether an action potential occurring at the presynaptic terminal produces an action potential at the postsynaptic membrane. Some common neurotransmitters involved in IPSPs are GABA and glycine. Inhibitory presynaptic neurons release neurotransmitters that then bind to the postsynaptic receptors; this induces a change in the permeability of the postsynaptic neuronal membrane to particular ions. An electric current that changes the postsynaptic membrane potential to create a more negative postsynaptic potential is generated, i.e. the postsynaptic membrane potential becomes more negative than the resting membrane potential, and this is called hyperpolarisation. To generate an action potential, the postsynaptic membrane must depolarize—the membrane potential must reach a voltage threshold more positive than the resting membrane potential. Therefore, hyperpolarisation of the postsynaptic membrane makes it less likely for depolarisation to sufficiently occur to generate an action potential in the postsynaptic neuron. Depolarization can also occur due to an IPSP if the reverse potential is between the resting threshold and the action potential threshold. Another way to look at inhibitory postsynaptic potentials is that they are also a chloride conductance change in the neuronal cell because it decreases the driving force. This is because, if the neurotransmitter released into the synaptic cleft causes an increase in the permeability of the postsynaptic membrane to chloride ions by binding to ligand-gated chloride ion channels and causing them to open, then chloride ions, which are in greater concentration in the synaptic cleft, diffuse into the postsynaptic neuron. As these are negatively charged ions, hyperpolarisation results, making it less likely for an action potential to be generated in the postsynaptic neuron. Microelectrodes can be used to measure postsynaptic potentials at either excitatory or inhibitory synapses. In general, a postsynaptic potential is dependent on the type and combination of receptor channel, reverse potential of the postsynaptic potential, action potential threshold voltage, ionic permeability of the ion channel, as well as the concentrations of the ions in and out of the cell; this determines if it is excitatory or inhibitory. IPSPs always tend to keep the membrane potential more negative than the action potential threshold and can be seen as a "transient hyperpolarization". IPSPs were first investigated in motorneurons by David P. C. Lloyd, John Eccles and Rodolfo Llinás in the 1950s and 1960s. Components [edit] Types [edit] This system IPSPs can be temporally summed with subthreshold or suprathreshold EPSPs to reduce the amplitude of the resultant postsynaptic potential. Equivalent EPSPs (positive) and IPSPs (negative) can cancel each other out when summed. The balance between EPSPs and IPSPs is very important in the integration of electrical information produced by inhibitory and excitatory synapses. Factors [edit] The size of the neuron can also affect the inhibitory postsynaptic potential. Simple temporal summation of postsynaptic potentials occurs in smaller neurons, whereas in larger neurons larger numbers of synapses and ionotropic receptors as well as a longer distance from the synapse to the soma enables the prolongation of interactions between neurons. Inhibitory neurotransmitters [edit] GABA is a very common neurotransmitter used in IPSPs in the adult mammalian brain and retina. Glycine molecules and their receptors work much in the same way in the spinal cord, brain, and retina. Inhibitory receptors [edit] There are two types of inhibitory receptors: Ionotropic receptors [edit] Ionotropic receptors (also known as ligand-gated ion channels) play an important role in inhibitory postsynaptic potentials. A neurotransmitter binds to the extracellular site and opens the ion channel that is made up of a membrane-spanning domain that allows ions to flow across the membrane inside the postsynaptic cell. This type of receptor produces very fast postsynaptic actions within a couple of milliseconds of the presynaptic terminal receiving an action potential. These channels influence the amplitude and time-course of postsynaptic potentials as a whole. Ionotropic GABA receptors (GABAA receptors) are pentamers most commonly composed of three different subunits (α, β, γ), although several other subunits (δ,ε, θ, π, ρ) and conformations exist. The open channels are selectively permeable to chloride or potassium ions (depending on the type of receptor) and allow these ions to pass through the membrane. If the electrochemical potential of the ion is more negative than that of the action potential threshold then the resultant conductance change that occurs due to the binding of GABA to its receptors keeps the postsynaptic potential more negative than the threshold and decreases the probability of the postsynaptic neuron completing an action potential. Ionotropic GABA receptors are used in binding for various drugs such as barbiturates (Phenobarbital, pentobarbital), steroids, and picrotoxin. Benzodiazepines (Valium) bind to the α and γ subunits of GABA receptors to improve GABAergic signaling. Alcohol also modulates ionotropic GABA receptors. Metabotropic receptors [edit] Metabotropic receptors are often G-protein-coupled receptors such as GABAB receptors. These do not use ion channels in their structure; instead they consist of an extracellular domain that binds to a neurotransmitter and an intracellular domain that binds to G-protein. This begins the activation of the G-protein, which then releases itself from the receptor and interacts with ion channels and other proteins to open or close ion channels through intracellular messengers. They produce slow postsynaptic responses (from milliseconds to minutes) and can be activated in conjunction with ionotropic receptors to create both fast and slow postsynaptic potentials at one particular synapse. Metabotropic GABA receptors, heterodimers of R1 and R2 subunits, use potassium channels instead of chloride. They can also block calcium ion channels to hyperpolarize postsynaptic cells. Significance [edit] There are many applications of inhibitory postsynaptic potentials to the real world. Drugs that affect the actions of the neurotransmitter can treat neurological and psychological disorders through different combinations of types of receptors, G-proteins, and ion channels in postsynaptic neurons. For example, studies researching opioid receptor-mediated receptor desensitizing and trafficking in the locus coeruleus of the brain are being performed. When a high concentration of agonist is applied for an extended amount of time (fifteen minutes or more), hyperpolarization peaks and then decreases. This is significant because it is a prelude to tolerance; the more opioids one needs for pain the greater the tolerance of the patient. These studies are important because it helps us to learn more about how we deal with pain and our responses to various substances that help treat pain. By studying our tolerance to pain, we can develop more efficient medications for pain treatment. In addition, research is being performed in the field of dopamine neurons in the ventral tegmental area, which deals with reward, and the substantia nigra, which is involved with movement and motivation. Metabotropic responses occur in dopamine neurons through the regulation of the excitability of cells. Opioids inhibit GABA release; this decreases the amount of inhibition and allows them to fire spontaneously. Morphine and opioids relate to inhibitory postsynaptic potentials because they induce disinhibition in dopamine neurons. IPSPs can also be used to study the input-output characteristics of an inhibitory forebrain synapse used to further study learned behavior—for example in a study of song learning in birds at the University of Washington. Poisson trains of unitary IPSPs were induced at a high frequency to reproduce postsynaptic spiking in the medial portion of the dorsalateral thalamic nucleus without any extra excitatory inputs. This shows an excess of thalamic GABAergic activation. This is important because spiking timing is needed for proper sound localization in the ascending auditory pathways. Songbirds use GABAergic calyceal synaptic terminals and a calcyx-like synapse such that each cell in the dorsalateral thalamic nucleus receives at most two axon terminals from the basal ganglia to create large postsynaptic currents. Inhibitory postsynaptic potentials are also used to study the basal ganglia of amphibians to see how motor function is modulated through its inhibitory outputs from the striatum to the tectum and tegmentum. Visually guided behaviors may be regulated through the inhibitory striato-tegmental pathway found in amphibians in a study performed at the Baylor College of Medicine and the Chinese Academy of Sciences. The basal ganglia in amphibians is very important in receiving visual, auditory, olfactory, and mechansensory inputs; the disinhibitory striato-protecto-tectal pathway is important in prey-catching behaviors of amphibians. When the ipsilateral striatum of an adult toad was electrically stimulated, inhibitory postsynaptic potentials were induced in binocular tegmental neurons, which affects the visual system of the toad. Studies [edit] Inhibitory postsynaptic potentials can be inhibited themselves through a signaling process called "depolarized-induced suppression of inhibition (DSI)" in CA1 pyramidal cells and cerebellar Purkinje cells. In a laboratory setting step depolarizations the soma have been used to create DSIs, but it can also be achieved through synaptically induced depolarization of the dendrites. DSIs can be blocked by ionotropic receptor calcium ion channel antagonists on the somata and proximal apical dendrites of CA1 pyramidal cells. Dendritic inhibitory postsynaptic potentials can be severely reduced by DSIs through direct depolarization. Along these lines, inhibitory postsynaptic potentials are useful in the signaling of the olfactory bulb to the olfactory cortex. EPSPs are amplified by persistent sodium ion conductance in external tufted cells. Low-voltage activated calcium ion conductance enhances even larger EPSPs. The hyperpolarization activated nonselective cation conductance decreases EPSP summation and duration and they also change inhibitory inputs into postsynaptic excitation. IPSPs come into the picture when the tufted cells membranes are depolarized and IPSPs then cause inhibition. At resting threshold IPSPs induce action potentials. GABA is responsible for much of the work of the IPSPs in the external tufted cells. Another interesting study of inhibitory postsynaptic potentials looks at neuronal theta rhythm oscillations that can be used to represent electrophysiological phenomena and various behaviors. Theta rhythms are found in the hippocampus and GABAergic synaptic inhibition helps to modulate them. They are dependent on IPSPs and started in either CA3 by muscarinic acetylcholine receptors and within C1 by the activation of group I metabotropic glutamate receptors. When interneurons are activated by metabotropic acetylcholine receptors in the CA1 region of rat hippocampal slices, a theta pattern of IPSPs in pyramidal cells occurs independent of the input. This research also studies DSIs, showing that DSIs interrupt metabotropic acetylcholine-initiated rhythm through the release of endocannabinoids. An endocannabinoid-dependent mechanism can disrupt theta IPSPs through action potentials delivered as a burst pattern or brief train. In addition, the activation of metabotropic glutamate receptors removes any theta IPSP activity through a G-protein, calcium ion–independent pathway. Inhibitory postsynaptic potentials have also been studied in the Purkinje cell through dendritic amplification. The study focused in on the propagation of IPSPs along dendrites and its dependency of ionotropic receptors by measuring the amplitude and time-course of the inhibitory postsynaptic potential. The results showed that both compound and unitary inhibitory postsynaptic potentials are amplified by dendritic calcium ion channels. The width of a somatic IPSP is independent of the distance between the soma and the synapse whereas the rise time increases with this distance. These IPSPs also regulate theta rhythms in pyramidal cells. On the other hand, inhibitory postsynaptic potentials are depolarizing and sometimes excitatory in immature mammalian spinal neurons because of high concentrations of intracellular chloride through ionotropic GABA or glycine chloride ion channels. These depolarizations activate voltage-dependent calcium channels. They later become hyperpolarizing as the mammal matures. To be specific, in rats, this maturation occurs during the perinatal period when brain stem projects reach the lumbar enlargement. Descending modulatory inputs are necessary for the developmental shift from depolarizing to hyperpolarizing inhibitory postsynaptic potentials. This was studied through complete spinal cord transections at birth of rats and recording IPSPs from lumbar motoneurons at the end of the first week after birth. Glutamate, an excitatory neurotransmitter, is usually associated with excitatory postsynaptic potentials in synaptic transmission. However, a study completed at the Vollum Institute at the Oregon Health Sciences University demonstrates that glutamate can also be used to induce inhibitory postsynaptic potentials in neurons. This study explains that metabotropic glutamate receptors feature activated G proteins in dopamine neurons that induce phosphoinositide hydrolysis. The resultant products bind to inositol triphosphate (IP3) receptors through calcium ion channels. The calcium comes from stores and activate potassium conductance, which causes a pure inhibition in the dopamine cells. The changing levels of synaptically released glutamate creates an excitation through the activation of ionotropic receptors, followed by the inhibition of metabotropic glutamate receptors. See also [edit] Nonspiking neurons Shunting inhibition References [edit] ^ a b c d e Purves et al. Neuroscience. 4th ed. Sunderland (MA): Sinauer Associates, Incorporated; 2008. ^ Thompson SM, Gähwiler BH (March 1989). "Activity-dependent disinhibition. I. Repetitive stimulation reduces IPSP driving force and conductance in the hippocampus in vitro". Journal of Neurophysiology. 61 (3): 501–11. doi:10.1152/jn.1989.61.3.501. PMID 2709096. ^ Levy M, Koeppen B, Stanton B (2005). Berne & Levy principles of physiology (4th ed.). Elsevier Mosby. ISBN 978-0-8089-2321-3. ^ Coombs JS, Eccles JC, Fatt P (November 1955). "The specific ionic conductances and the ionic movements across the motoneuronal membrane that produce the inhibitory post-synaptic potential". The Journal of Physiology. 130 (2): 326–74. doi:10.1113/jphysiol.1955.sp005412. PMC 1363415. PMID 13278905. ^ Llinas R, Terzuolo CA (March 1965). "Mechanisms of Supraspinal Actions Upon Spinal Cord Activities. Reticular Inhibitory Mechanisms Upon Flexor Motoneurons". Journal of Neurophysiology. 28 (2): 413–22. doi:10.1152/jn.1965.28.2.413. PMID 14283063. ^ Chavas J, Marty A (March 2003). "Coexistence of excitatory and inhibitory GABA synapses in the cerebellar interneuron network". The Journal of Neuroscience. 23 (6): 2019–31. doi:10.1523/JNEUROSCI.23-06-02019.2003. PMC 6742031. PMID 12657660. ^ a b Williams, JT, Vollum Institute of Oregon Health Sciences University, Interviewed by Saira Ahmed, November 11, 2008 ^ Person AL, Perkel DJ (April 2005). "Unitary IPSPs drive precise thalamic spiking in a circuit required for learning". Neuron. 46 (1): 129–40. doi:10.1016/j.neuron.2004.12.057. PMID 15820699. ^ Wu GY, Wang SR (December 2007). "Postsynaptic potentials and axonal projections of tegmental neurons responding to electrical stimulation of the toad striatum". Neuroscience Letters. 429 (2–3): 111–4. doi:10.1016/j.neulet.2007.09.071. PMC 2696233. PMID 17996369. ^ Morishita W, Alger BE (January 2001). "Direct depolarization and antidromic action potentials transiently suppress dendritic IPSPs in hippocampal CA1 pyramidal cells". Journal of Neurophysiology. 85 (1): 480–4. doi:10.1152/jn.2001.85.1.480. PMID 11152751. S2CID 17060042. ^ Solinas SM, Maex R, De Schutter E (March 2006). "Dendritic amplification of inhibitory postsynaptic potentials in a model Purkinje cell" (PDF). The European Journal of Neuroscience. 23 (5): 1207–18. doi:10.1111/j.1460-9568.2005.04564.x. PMID 16553783. S2CID 6139806. Archived from the original (PDF) on 2007-04-18. Retrieved 2019-09-22. ^ Liu S, Shipley MT (October 2008). "Intrinsic conductances actively shape excitatory and inhibitory postsynaptic responses in olfactory bulb external tufted cells". The Journal of Neuroscience. 28 (41): 10311–22. doi:10.1523/JNEUROSCI.2608-08.2008. PMC 2570621. PMID 18842890. ^ Reich CG, Karson MA, Karnup SV, Jones LM, Alger BE (December 2005). "Regulation of IPSP theta rhythm by muscarinic receptors and endocannabinoids in hippocampus" (PDF). Journal of Neurophysiology. 94 (6): 4290–9. doi:10.1152/jn.00480.2005. PMID 16093334. S2CID 10333266. Archived from the original (PDF) on 2019-02-27. ^ Brenowitz SD, Regehr WG (2003). "Calcium dependence of retrograde inhibition by endocannabinoids at synapses onto Purkinje cells". Journal of Neuroscience. 23 (15): 6373–6384. doi:10.1523/JNEUROSCI.23-15-06373.2003. PMC 6740543. PMID 12867523. ^ Jean-Xavier C, Pflieger JF, Liabeuf S, Vinay L (November 2006). "Inhibitory postsynaptic potentials in lumbar motoneurons remain depolarizing after neonatal spinal cord transection in the rat". Journal of Neurophysiology. 96 (5): 2274–81. CiteSeerX 10.1.1.326.1283. doi:10.1152/jn.00328.2006. PMID 16807348. ^ Fiorillo CD, Williams JT (July 1998). "Glutamate mediates an inhibitory postsynaptic potential in dopamine neurons". Nature. 394 (6688): 78–82. Bibcode:1998Natur.394...78F. doi:10.1038/27919. PMID 9665131. S2CID 4352019. | v t e Physiology of the nervous system | | Primarily CNS | Arousal + Wakefulness Intracranial pressure Lateralization of brain function Sleep Memory | | Primarily PNS | Reflex Sensation | | Both | | | | --- | | Evoked potential | Bereitschaftspotential P300 Auditory evoked potential Somatosensory evoked potentials Visual evoked potential | | Other short term | Neurotransmission Chronaxie Membrane potential Action potential Postsynaptic potential + Excitatory + Inhibitory | | Long term | Axoplasmic transport Neuroregeneration/Nerve regeneration Neuroplasticity/Synaptic plasticity + Long-term potentiation + Long-term depression | | Other | | | Retrieved from " Categories: Memory Neural synapse Graded potentials Hidden categories: Articles with short description Short description is different from Wikidata Inhibitory postsynaptic potential Add topic
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https://labdemos.physics.sunysb.edu/l.-geometrical-optics/l2.-plane-mirrors/optical_board_-_ray_diagram_-_plane_mirror
Skip Navigation Physics Lab Demo Sections A. General Materials & Mathematics B. Statics C. Kinematics & Dynamics D. Rotational Mechanics E. Gravitation & Astronomy F. Fluid Mechanics G. Vibrations & Mechanical Waves H. Sound I. Thermodynamics J. Electrostatics & Magnetostatics K. Electromagnetic Principles L. Geometrical Optics M. Wave Optics N. Spectra & Color O. Vision P. Modern Physics HomeL. Geometrical OpticsL2. Plane MirrorsL2-02. Optical Board - Ray Diagram - Plane Mirror L2-02. Optical Board - Ray Diagram - Plane Mirror Purpose To demonstrate how several light rays are used to locate the image in a plane mirror. Equipment Optical board with half-silvered and front surface object mirrors and large plane mirror. Suggestions A single converging lens before the object keeps the light ray narrow. Images Description A single central ray passes through a half-silvered mirror, with half deflected to act as the source at the tip of an object. By rotating the top mirror it can be shown that all rays starting at the object point (the rotating mirror) appear to come from the same point behind the plane mirror (the image point.) References Sue Allen, The Gaussian formula and the elusive fourth principal ray, AJP 60, 160-163 (1992). L. Geometrical Optics See pagesl1. light sources & light rays L1-04. Point Source - Flashlight Bulb L1-05. Persistence Of A Filament L1-12. Inverse Square Law - Overhead Projector And Two-Meter Stick L1-21. Pinhole Image - Ground Glass Screen L1-32. Visible Laser L1. Light Sources And Light Rays See pagesl2. plane mirrors L2-01. Optical Board - Plane Mirror L2-02. Optical Board - Ray Diagram - Plane Mirror L2-21. Optical Board - Half-Silvered Mirror L2-22. Infinity Mirror L2-41. Optical Board - Corner Reflector L2. Plane Mirrors See pagesl3. curved mirrors L3-01. Optical Board - Convex Spherical Mirror L3-02. Optical Board - Ray Diagram - Convex Mirror L3-03. Large_Convex_Mirror L3-11. Optical Board - Concave Spherical Mirror L3-12. Optical Board - Ray Diagram - Concave Mirror L3-13. Optical Board - Spherical Aberration In Concave Mirror L3-14. Large Concave Mirror L3-16. Focusing Of Heat Waves By Mirrors L3-19. Penny And Parabolic Mirrors L3-22. Pendulum And Spherical Mirror L3-23. Image On Screen Using Concave Mirror L3-24. Optical Board - Spherical And Parabolic Mirrors L3-31. Giant Mirror - Concave And Convex L3. Curved Mirrors See pagesl4. refraction L4-01. Optical Board - Rectangular Slab L4-02. Refraction - Beer Mug In Water L4-03. Refraction - Rod In Water L4-04. Refraction - Can In Water Tank L4-05. Refraction In Fish Tank - Portable L4-06. Refraction In Cloudy Water L4-21. Snell's Law - Image Distortion L4-23. Bending Of A Laser Beam In Sugar Solution L4-31. Disappearance Of Glass In Liquid L4. Refraction See pagesl5. total internal reflection L5-02. Total Internal Reflection In Long Tank L5-03. Fish In Tank - Total Internal Reflection L5-05. Image Inversion By Prism L5-11. Laser Waterfall L5-12. Plexiglass Spiral L5-13. Plexiglass Spiral With Laser L5. Total Internal Reflection See pagesl6. lenses L6-01. Optical Board - Converging Spherical Lens L6-02. Optical Board - Circular Slab L6-03. Optical Board - Hyperbolic Lens L6-04. Optical Board - Ray Diagram - Real Image Pos Lens L6-05. Optical Board - Ray Diagram - Virtual Image Pos Lens L6-08. Real Image Of Converging Lens - Light Bulb L6-09. Real Image Of Converging Lens L6-11. Optical Board - Depth Of Field L6-12. Magnifying Lens In Water L6-21. Optical Board - Diverging Spherical Lens L6-22. Optical Board - Ray Diagram - Virtual Image Neg Lens L6-32. Optical Board - Spherical Aberration In Lens L6-34. Optical Board - Chromatic Aberration In Lens L6-35. Chromatic Aberration - Point Source And 20 Cm Lens L6-37. Astigmatism L6-38. Coma L6-39. Distortion - Pincushion And Barrel With 2 Irises L6-40. Distortion - Pincushion And Barrel L6-41. Distorted Image - Pincushion And Barrel L6-42. Distortion And Correction L6-51. Optical Board - Fresnel Lens L6-52. Fresnel Lenses - Miscellaneous L6-53. Fresnel Lens Magnifier L6-55. Fresnel Flashlight L6-56. Fresnel Spotlight L6. Lenses See pagesl7. optical instruments L7-01. Light Meters L7-02. Focal Plane Shutter L7-03. Shutter Speed Of A Camera L7-06. Minicam With Wide-Angle, Telephoto And Macrozoom Lens L7-11. Optical Board - Astronomical Telescope L7-12. Optical Board - Telephoto Lens L7-13. Optical Board - Galilean Telescope L7-14. Optical Board - Reflecting Telescope L7-21. Magnifying Glass - TV L7-22. Microscope - TV L7-23. Astronomical Telescope - TV L7-24. Terrestrial Telescope - TV L7-25. Galilean Telescope - Tv L7-31. Principal Planes L7-32. Field Lens L7-33. Overhead Projector - Dissected L7-34. Slide Projector - Dissected L7-41. Real Image - Mirror And Lens L7-42. Virtual Image - Mirror And Lens L7. Optical Instruments L. Geometrical Optics See pagesl1. light sources & light rays L1-04. Point Source - Flashlight Bulb L1-05. Persistence Of A Filament L1-12. Inverse Square Law - Overhead Projector And Two-Meter Stick L1-21. Pinhole Image - Ground Glass Screen L1-32. Visible Laser L1. Light Sources And Light Rays See pagesl2. plane mirrors L2-01. Optical Board - Plane Mirror L2-02. Optical Board - Ray Diagram - Plane Mirror L2-21. Optical Board - Half-Silvered Mirror L2-22. Infinity Mirror L2-41. Optical Board - Corner Reflector L2. Plane Mirrors See pagesl3. curved mirrors L3-01. Optical Board - Convex Spherical Mirror L3-02. Optical Board - Ray Diagram - Convex Mirror L3-03. Large_Convex_Mirror L3-11. Optical Board - Concave Spherical Mirror L3-12. Optical Board - Ray Diagram - Concave Mirror L3-13. Optical Board - Spherical Aberration In Concave Mirror L3-14. Large Concave Mirror L3-16. Focusing Of Heat Waves By Mirrors L3-19. Penny And Parabolic Mirrors L3-22. Pendulum And Spherical Mirror L3-23. Image On Screen Using Concave Mirror L3-24. Optical Board - Spherical And Parabolic Mirrors L3-31. Giant Mirror - Concave And Convex L3. Curved Mirrors See pagesl4. refraction L4-01. Optical Board - Rectangular Slab L4-02. Refraction - Beer Mug In Water L4-03. Refraction - Rod In Water L4-04. Refraction - Can In Water Tank L4-05. Refraction In Fish Tank - Portable L4-06. Refraction In Cloudy Water L4-21. Snell's Law - Image Distortion L4-23. Bending Of A Laser Beam In Sugar Solution L4-31. Disappearance Of Glass In Liquid L4. Refraction See pagesl5. total internal reflection L5-02. Total Internal Reflection In Long Tank L5-03. Fish In Tank - Total Internal Reflection L5-05. Image Inversion By Prism L5-11. Laser Waterfall L5-12. Plexiglass Spiral L5-13. Plexiglass Spiral With Laser L5. Total Internal Reflection See pagesl6. lenses L6-01. Optical Board - Converging Spherical Lens L6-02. Optical Board - Circular Slab L6-03. Optical Board - Hyperbolic Lens L6-04. Optical Board - Ray Diagram - Real Image Pos Lens L6-05. Optical Board - Ray Diagram - Virtual Image Pos Lens L6-08. Real Image Of Converging Lens - Light Bulb L6-09. Real Image Of Converging Lens L6-11. Optical Board - Depth Of Field L6-12. Magnifying Lens In Water L6-21. Optical Board - Diverging Spherical Lens L6-22. Optical Board - Ray Diagram - Virtual Image Neg Lens L6-32. Optical Board - Spherical Aberration In Lens L6-34. Optical Board - Chromatic Aberration In Lens L6-35. Chromatic Aberration - Point Source And 20 Cm Lens L6-37. Astigmatism L6-38. Coma L6-39. Distortion - Pincushion And Barrel With 2 Irises L6-40. Distortion - Pincushion And Barrel L6-41. Distorted Image - Pincushion And Barrel L6-42. Distortion And Correction L6-51. Optical Board - Fresnel Lens L6-52. Fresnel Lenses - Miscellaneous L6-53. Fresnel Lens Magnifier L6-55. Fresnel Flashlight L6-56. Fresnel Spotlight L6. Lenses See pagesl7. optical instruments L7-01. Light Meters L7-02. Focal Plane Shutter L7-03. Shutter Speed Of A Camera L7-06. Minicam With Wide-Angle, Telephoto And Macrozoom Lens L7-11. Optical Board - Astronomical Telescope L7-12. Optical Board - Telephoto Lens L7-13. Optical Board - Galilean Telescope L7-14. Optical Board - Reflecting Telescope L7-21. Magnifying Glass - TV L7-22. Microscope - TV L7-23. Astronomical Telescope - TV L7-24. Terrestrial Telescope - TV L7-25. Galilean Telescope - Tv L7-31. Principal Planes L7-32. Field Lens L7-33. Overhead Projector - Dissected L7-34. Slide Projector - Dissected L7-41. Real Image - Mirror And Lens L7-42. Virtual Image - Mirror And Lens L7. Optical Instruments Physics Lab Demos Discrimination Sexual Misconduct Accessibility Barrier ©Admin Login2023Stony Brook University
968
https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus__An_Investigation_of_Functions_(Lippman_and_Rasmussen)/05%3A_Trigonometric_Functions_of_Angles/5.03%3A_Points_on_Circles_Using_Sine_and_Cosine
Search x Text Color Text Size Margin Size Font Type selected template will load here Error This action is not available. 5.3: Points on Circles Using Sine and Cosine ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\| #1 \|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\| #1 \|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\AA}{\unicode[.8,0]{x212B}}) ( \newcommand{\vectorA}{\vec{#1}} % arrow) ( \newcommand{\vectorAt}{\vec{\text{#1}}} % arrow) ( \newcommand{\vectorB}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vectorC}{\textbf{#1}} ) ( \newcommand{\vectorD}{\overrightarrow{#1}} ) ( \newcommand{\vectorDt}{\overrightarrow{\text{#1}}} ) ( \newcommand{\vectE}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} ) ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) While it is convenient to describe the location of a point on a circle using an angle or a distance along the circle, relating this information to the x and y coordinates and the circle equation we explored in Section 5.1 is an important application of trigonometry. A distress signal is sent from a sailboat during a storm, but the transmission is unclear and the rescue boat sitting at the marina cannot determine the sailboat’s location. Using high powered radar, they determine the distress signal is coming from a distance of 20 miles at an angle of 225 degrees from the marina. How many miles east/west and north/south of the rescue boat is the stranded sailboat? In a general sense, to investigate this, we begin by drawing a circle centered at the origin with radius (r), and marking the point on the circle indicated by some angle (\theta). This point has coordinates ((x), (y)). If we drop a line segment vertically down from this point to the x axis, we would form a right triangle inside of the circle. No matter which quadrant our angle (\theta) puts us in we can draw a triangle by dropping a perpendicular line segment to the (x) axis, keeping in mind that the values of (x) and (y) may be positive or negative, depending on the quadrant. Additionally, if the angle (\theta) puts us on an axis, we simply measure the radius as the (x) or (y) with the other value being 0, again ensuring we have appropriate signs on the coordinates based on the quadrant. Triangles obtained from different radii will all be similar triangles, meaning corresponding sides scale proportionally. While the lengths of the sides may change, as we saw in the last section, the ratios of the side lengths will always remain constant for any given angle. (\dfrac{y_{1} }{r_{1} } =\dfrac{y_{2} }{r_{2} }) (\dfrac{x_{1} }{r_{1} } =\dfrac{x_{2} }{r_{2} }) To be able to refer to these ratios more easily, we will give them names. Since the ratios depend on the angle, we will write them as functions of the angle (\theta). Note: sine and cosine For the point ((x), (y)) on a circle of radius (r) at an angle of (\theta), we can define two important functions as the ratios of the sides of the corresponding triangle: The sine function: (\sin (\theta )=\dfrac{y}{r}) The cosine function: (\cos (\theta )=\dfrac{x}{r}) In this chapter, we will explore these functions using both circles and right triangles. In the next chapter, we will take a closer look at the behavior and characteristics of the sine and cosine functions. Example (\PageIndex{1}) The point (3, 4) is on the circle of radius 5 at some angle (\theta). Find (\cos (\theta ))and (\sin (\theta )). Solution Knowing the radius of the circle and coordinates of the point, we can evaluate the cosine and sine functions as the ratio of the sides. [\cos (\theta )=\dfrac{x}{r} =\dfrac{3}{5} \sin (\theta )=\dfrac{y}{r} =\dfrac{4}{5}\nonumber] There are a few cosine and sine values which we can determine fairly easily because the corresponding point on the circle falls on the (x) or (y) axis. Example (\PageIndex{2}) Find (\cos (90{}^\circ )) and (\sin (90{}^\circ )). Solution On any circle, the terminal side of a 90 degree angle points straight up, so the coordinates of the corresponding point on the circle would be (0, r). Using our definitions of cosine and sine, [\cos (90{}^\circ )=\dfrac{x}{r} =\dfrac{0}{r} =0\nonumber] [\sin (90{}^\circ )=\dfrac{y}{r} =\dfrac{r}{r} =1\nonumber] Exercise (\PageIndex{1}) Find cosine and sine of the angle (\pi). [\cos (\pi )=-1 \sin (\pi )=0\nonumber] Notice that the definitions above can also be stated as: coordinates of the point on a circle at a given angle On a circle of radius (r) at an angle of (\theta), we can find the coordinates of the point ((x), (y)) Circles:Points on a Circle at that angle using [x=r\cos (\theta )] [y=r\sin (\theta )] On a unit circle, a circle with radius 1, (x=\cos (\theta )) and (y=\sin (\theta )). Utilizing the basic equation for a circle centered at the origin, (x^{2} +y^{2} =r^{2}), combined with the relationships above, we can establish a new identity. [x^{2} +y^{2} =r^{2}\nonumber] substituting the relations above, [(r\cos (\theta ))^{2} +(r\sin (\theta ))^{2} =r^{2}\nonumber] simplifying, [r^{2} (\cos (\theta ))^{2} +r^{2} (\sin (\theta ))^{2} =r^{2}\nonumber] dividing by (r^{2}) [(\cos (\theta ))^{2} +(\sin (\theta ))^{2} =1\nonumber] or using shorthand notation [\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\nonumber] Here (\cos ^{2} (\theta )) is a commonly used shorthand notation for ((\cos (\theta ))^{2}). Be aware that many calculators and computers do not understand the shorthand notation. In Section 5.1 we related the Pythagorean Theorem (a^{2} +b^{2} =c^{2}) to the basic equation of a circle (x^{2} +y^{2} =r^{2}), which we have now used to arrive at the Pythagorean Identity. pythagorean indentity The Pythagorean Identity. For any angle (\theta), [\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\nonumber] One use of this identity is that it helps us to find a cosine value of an angle if we know the sine value of that angle or vice versa. However, since the equation will yield two possible values, we will need to utilize additional knowledge of the angle to help us find the desired value. Example (\PageIndex{3}) If (\sin (\theta )=\dfrac{3}{7}) and (\theta) is in the second quadrant, find (\cos (\theta )). Solution Substituting the known value for sine into the Pythagorean identity, [\cos ^{2} (\theta )+\sin ^{2} (\theta )=1\nonumber] [\cos ^{2} (\theta )+\dfrac{9}{49} =1\nonumber] [\cos ^{2} (\theta )=\dfrac{40}{49}\nonumber] [\cos (\theta )=\pm \sqrt{\dfrac{40}{49} } =\pm \dfrac{\sqrt{40} }{7} =\pm \dfrac{2\sqrt{10} }{7}\nonumber] Since the angle is in the second quadrant, we know the (x) value of the point would be negative, so the cosine value should also be negative. Using this additional information, we can conclude that [\cos (\theta )=-\dfrac{2\sqrt{10} }{7}\nonumber] Values for Sine and Cosine At this point, you may have noticed that we haven’t found any cosine or sine values from angles not on an axis. To do this, we will need to utilize our knowledge of triangles. First, consider a point on a circle at an angle of 45 degrees, or (\dfrac{\pi }{4}). At this angle, the x and y coordinates of the corresponding point on the circle will be equal because 45 degrees divides the first quadrant in half. Since the (x) and (y) values will be the same, the sine and cosine values will also be equal. Utilizing the Pythagorean Identity, [\cos ^{2} \left(\dfrac{\pi }{4} \right)+\sin ^{2} \left(\dfrac{\pi }{4} \right)=1\nonumber] since the sine and cosine are equal, we can substitute sine with cosine [\cos ^{2} \left(\dfrac{\pi }{4} \right)+\cos ^{2} \left(\dfrac{\pi }{4} \right)=1\nonumber] add like terms [2\cos ^{2} \left(\dfrac{\pi }{4} \right)=1\nonumber] divide [\cos ^{2} \left(\dfrac{\pi }{4} \right)=\dfrac{1}{2}\nonumber] since the (x) value is positive, we’ll keep the positive root [\cos \left(\dfrac{\pi }{4} \right)=\sqrt{\dfrac{1}{2} }\nonumber] often this value is written with a rationalized denominator Remember, to rationalize the denominator we multiply by a term equivalent to 1 to get rid of the radical in the denominator. [\cos \left(\dfrac{\pi }{4} \right)=\sqrt{\dfrac{1}{2} } \sqrt{\dfrac{2}{2} } =\sqrt{\dfrac{2}{4} } =\dfrac{\sqrt{2} }{2}\nonumber] Since the sine and cosine are equal, (\sin \left(\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2} }{2}) as well. The ((x), (y)) coordinates for a point on a circle of radius 1 at an angle of 45 degrees are (\left(\dfrac{\sqrt{2} }{2} ,\dfrac{\sqrt{2} }{2} \right)). Example (\PageIndex{4}) Find the coordinates of the point on a circle of radius 6 at an angle of (\dfrac{\pi }{4}). Solution Using our new knowledge that (\sin \left(\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2} }{2}) and (\cos \left(\dfrac{\pi }{4} \right)=\dfrac{\sqrt{2} }{2}), along with our relationships that stated (x=r\cos (\theta )) and (y=r\sin (\theta )), we can find the coordinates of the point desired: [x=6\cos \left(\dfrac{\pi }{4} \right)=6\left(\dfrac{\sqrt{2} }{2} \right)=3\sqrt{2}\nonumber ] [y=6\sin \left(\dfrac{\pi }{4} \right)=6\left(\dfrac{\sqrt{2} }{2} \right)=3\sqrt{2}\nonumber] Exercise (\PageIndex{2}) Find the coordinates of the point on a circle of radius 3 at an angle of (90{}^\circ). [\begin{array}{l} {x=3\cos \left(\dfrac{\pi }{2} \right)=3\cdot 0=0} \ {y=3\sin \left(\dfrac{\pi }{2} \right)=3\cdot 1=3} \end{array}\nonumber] Next, we will find the cosine and sine at an angle of 30 degrees, or (\frac{\pi }{6}). To do this, we will first draw a triangle inside a circle with one side at an angle of 30 degrees, and another at an angle of -30 degrees. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be 60 degrees. Since all the angles are equal, the sides will all be equal as well. The vertical line has length (2y), and since the sides are all equal we can conclude that (2y = r), or (y=\dfrac{r}{2}). Using this, we can find the sine value: [\text{sin}(\dfrac{\pi}{6}) = \dfrac{y}{r} = \dfrac{r/2}{r} = \dfrac{r}{2} \cdot \dfrac{1}{r} = \dfrac{1}{2}\nonumber] Using the Pythagorean Identity, we can find the cosine value: [\cos ^{2} \left(\dfrac{\pi }{6} \right)+\sin ^{2} \left(\dfrac{\pi }{6} \right)=1\nonumber] [\cos ^{2} \left(\dfrac{\pi }{6} \right)+\left(\dfrac{1}{2} \right)^{2} =1\nonumber] [\cos ^{2} \left(\dfrac{\pi }{6} \right)=\dfrac{3}{4}\nonumber] since the (x) value is positive, we’ll keep the positive root [\cos \left(\dfrac{\pi }{6} \right)=\sqrt{\dfrac{3}{4} } =\dfrac{\sqrt{3} }{2}\nonumber] The ((x), (y)) coordinates for the point on a circle of radius 1 at an angle of 30 degrees are (\left(\dfrac{\sqrt{3} }{2} ,\dfrac{1}{2} \right)). By drawing a the triangle inside the unit circle with a 30 degree angle and reflecting it over the line (y = x), we can find the cosine and sine for 60 degrees, or (\dfrac{\pi }{3}), without any additional work. By this symmetry, we can see the coordinates of the point on the unit circle at an angle of 60 degrees will be (\left(\dfrac{1}{2} ,\dfrac{\sqrt{3} }{2} \right)), giving (\cos \left(\dfrac{\pi }{3} \right)=\dfrac{1}{2}) and (\sin \left(\dfrac{\pi }{3} \right)=\dfrac{\sqrt{3} }{2}) We have now found the cosine and sine values for all the commonly encountered angles in the first quadrant of the unit circle. Angle | (0) | (\dfrac{\pi }{6}), or 30(\mathrm{{}^\circ}) | (\dfrac{\pi }{4}), or 45(\mathrm{{}^\circ}) | (\dfrac{\pi }{3}), or 60(\mathrm{{}^\circ}) | (\dfrac{\pi }{2}), or 90(\mathrm{{}^\circ}) Cosine | 1 | (\dfrac{\sqrt{3} }{2}) | (\dfrac{\sqrt{2} }{2}) | (\dfrac{1}{2}) | 0 Sine | 0 | (\dfrac{1}{2}) | (\dfrac{\sqrt{2} }{2}) | (\dfrac{\sqrt{3} }{2}) | 1 For any given angle in the first quadrant, there will be an angle in another quadrant with the same sine value, and yet another angle in yet another quadrant with the same cosine value. Since the sine value is the (y) coordinate on the unit circle, the other angle with the same sine will share the same (y) value, but have the opposite (x) value. Likewise, the angle with the same cosine will share the same (x) value, but have the opposite (y) value. As shown here, angle (\alpha) has the same sine value as angle (\theta); the cosine values would be opposites. The angle (\beta) has the same cosine value as the angle (\theta); the sine values would be opposites. It is important to notice the relationship between the angles. If, from the angle, you measured the smallest angle to the horizontal axis, all would have the same measure in absolute value. We say that all these angles have a reference angle of (\theta). Definition: reference angle An angle’s reference angle is the size of the smallest angle to the horizontal axis. A reference angle is always an angle between 0 and 90 degrees, or 0 and (\dfrac{\pi }{2}) radians. Angles share the same cosine and sine values as their reference angles, except for signs (positive or negative) which can be determined from the quadrant of the angle. Example (\PageIndex{5}) Find the reference angle of 150 degrees. Use it to find (\cos (150{}^\circ )) and (\sin (150{}^\circ )). Solution 150 degrees is located in the second quadrant. It is 30 degrees short of the horizontal axis at 180 degrees, so the reference angle is 30 degrees. This tells us that 150 degrees has the same sine and cosine values as 30 degrees, except for sign. We know that (\sin (30{}^\circ )=\dfrac{1}{2}) and (\cos (30{}^\circ )=\dfrac{\sqrt{3} }{2}). Since 150 degrees is in the second quadrant, the (x) coordinate of the point on the circle would be negative, so the cosine value will be negative. The (y) coordinate is positive, so the sine value will be positive. [\sin (150{}^\circ )=\dfrac{1}{2}\text{ and }\cos (150{}^\circ )=-\dfrac{\sqrt{3} }{2}\nonumber] The ((x), (y)) coordinates for the point on a unit circle at an angle of (150{}^\circ) are (\left(\dfrac{-\sqrt{3} }{2} ,\dfrac{1}{2} \right)). Using symmetry and reference angles, we can fill in cosine and sine values at the rest of the special angles on the unit circle. Take time to learn the ((x), (y)) coordinates of all the major angles in the first quadrant! Example (\PageIndex{6}) Find the coordinates of the point on a circle of radius 12 at an angle of (\dfrac{7\pi }{6}). Solution Note that this angle is in the third quadrant, where both x and y are negative. Keeping this in mind can help you check your signs of the sine and cosine function. [x=12\cos \left(\dfrac{7\pi }{6} \right)=12\left(\dfrac{-\sqrt{3} }{2} \right)=-6\sqrt{3}\nonumber ] [y=12\sin \left(\dfrac{7\pi }{6} \right)=12\left(\dfrac{-1}{2} \right)=-6\nonumber] The coordinates of the point are ((-6\sqrt{3} ,-6)). Exercise (\PageIndex{3}) Find the coordinates of the point on a circle of radius 5 at an angle of (\dfrac{5\pi }{3}). [\left(5\cos \left(\dfrac{5\pi }{3} \right),5\sin \left(\dfrac{5\pi }{3} \right)\right)=\left(\dfrac{5}{2} ,\dfrac{-5\sqrt{3} }{2} \right)\nonumber] Example (\PageIndex{7}) We now have the tools to return to the sailboat question posed at the beginning of this section. Solution A distress signal is sent from a sailboat during a storm, but the transmission is unclear and the rescue boat sitting at the marina cannot determine the sailboat’s location. Using high powered radar, they determine the distress signal is coming from a point 20 miles away at an angle of 225 degrees from the marina. How many miles east/west and north/south of the rescue boat is the stranded sailboat? We can now answer the question by finding the coordinates of the point on a circle with a radius of 20 miles at an angle of 225 degrees. [x=20\cos \left(225{}^\circ \right)=20\left(\dfrac{-\sqrt{2} }{2} \right)\approx -14.142\text{ miles}\nonumber] [y=20\sin \left(225{}^\circ \right)=20\left(\dfrac{-\sqrt{2} }{2} \right)\approx -14.142\text{ miles}\nonumber] The sailboat is located 14.142 miles west and 14.142 miles south of the marina. The special values of sine and cosine in the first quadrant are very useful to know, since knowing them allows you to quickly evaluate the sine and cosine of very common angles without needing to look at a reference or use your calculator. However, scenarios do come up where we need to know the sine and cosine of other angles. To find the cosine and sine of any other angle, we turn to a computer or calculator. Be aware: most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When you evaluate “cos(30)” on your calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode. Most computer software with cosine and sine functions only operates in radian mode. Example (\PageIndex{8}) Evaluate the cosine of 20 degrees using a calculator or computer. Solution On a calculator that can be put in degree mode, you can evaluate this directly to be approximately 0.939693. On a computer or calculator without degree mode, you would first need to convert the angle to radians, or equivalently evaluate the expression [\cos \left(20 \cdot \dfrac{\pi }{180} \right)\nonumber] Important Topics of This Section This page titled 5.3: Points on Circles Using Sine and Cosine is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by David Lippman & Melonie Rasmussen (The OpenTextBookStore) via source content that was edited to the style and standards of the LibreTexts platform. Recommended articles The LibreTexts libraries are Powered by NICE CXone Expert and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. 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How to write one-step equations for word problems | 6th grade | Khan Academy Khan Academy 9090000 subscribers 227 likes Description 146063 views Posted: 11 Sep 2015 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Learn how to write basic equations to model real-world situations. Grade 6th on Khan Academy: By the 6th grade, you're becoming a sophisticated mathemagician. You'll be able to add, subtract, multiply, and divide any non-negative numbers (including decimals and fractions) that any grumpy ogre throws at you. Mind-blowing ideas like exponents (you saw these briefly in the 5th grade), ratios, percents, negative numbers, and variable expressions will start being in your comfort zone. Most importantly, the algebraic side of mathematics is a whole new kind of fun! And if that is not enough, we are going to continue with our understanding of ideas like the coordinate plane (from 5th grade) and area while beginning to derive meaning from data! (Content was selected for this grade level based on a typical curriculum in the United States.) About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy à ªs 6th grade channel: Subscribe to Khan Academy: 25 comments Transcript: [Voiceover] Anna wants to celebrate her birthday by eating pizza with her friends. For $42.50 total, they can buy p boxes of pizza. Each box of pizza costs $8.50. Select the equation that matches this situation. So before I even look at these, let's see if I can make sense of the sentence here. So for $42.50 total, and I'll just write 42.5, especially because in all these choices they didn't write 42.50, they just wrote 42.5 which is equivalent. So 42.50 that's the total amount they spent on pizza and if I wanted to figure out how many boxes of pizza they could buy, I could divide the total amount they spend, I could divide that by the price per box. That would give me the number of boxes. So this is the total, total dollars. This right over here is the dollar per box and then this would give me the number of boxes. # of boxes. Now other ways that I could think about it. I could say, well what's the total that they spend? So 42.50, but what's another way of thinking about the total they spend? Well you could have the amount they spend per box, times the number of boxes. So this is the total they spend and this another way of thinking about the total they spend, so these two things must be equal. So let's see, if I can see anything here that looks like this, well actually this first choice, this, is exactly, is exactly what I wrote over here. Let's see this choice right over here. P is equal to 8.5 x 42.5. Well we've already been able to write an equation that has explicitly, that has just a p on one side and so when you solve for just a p on one side, you get this thing over here, not this thing, so we could rule that out. Over here it looks kind of like this, except the p is on the wrong side. This has 8.5p is equal to 42.5, not 42.5p is equal to 8.5. If we try to get the p on the other side here, you could divide both sides by p, but then you would get p divided by p is one. You would get 42.5 is equal to 8.5/p which is not true. We have 8.5 times p is equal to 42.5, so this is, this is not going to be the case. One thing to realize, no matter what you come up with, if you came up with this first, or if you came up with this first, you can go between these two with some algebraic manipulations. So for example, to go from this blue one to what I wrote in red up here, you just divide both sides by 8.5. So you divide by 8.5 on the left, you divide 8.5 on the right. Obviously to keep the equal sign you have to do the same thing to the left and right, but now you would have 42.5/8.5 is equal to, is equal to p. Which is exactly what we have over there. Let's do one more of these. Good practice. Mr. Herman's class is selling candy for a school fundraiser. The class has a goal of raising $500 by selling c boxes of candy. For every box they sell, they make $2.75. Write an equation that the students could solve to figure out how many boxes of candy they need to sell. How many boxes of candy they need to sell. Well there's a couple of ways you could think about it. They have a goal of raising $500 and so they want to get a total of $500, and if each box is $2.75, divide the total by the amount they get each per box and then this is going to be equal to the number of boxes that they need to sell. So this we've done. This is an equation that the students could solve to figure out how many boxes of candy they need to sell. Another way you could think about it, it's 2.75 per box times c boxes. This is the total amount of money they will raise. Whoops, this is the amount. So this is the amount that they will raise and their goal is that, their goal is to raise $500. So they want this to be equal to $500. So this also could be an equation that the students could solve to figure out how many boxes of candy they need to sell.
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https://brainly.in/question/55557185
CaC₂ + H₂O-A(g) + B When gas A is passed through red hot iron tube at 873 K gives another compound 'C'. - Brainly.in Skip to main content Ask Question Log in Join for free For parents For teachers Honor code Textbook Solutions Brainly App LeeEunBi 26.02.2023 Chemistry Secondary School answered CaC₂ + H₂O-A(g) + B When gas A is passed through red hot iron tube at 873 K gives another compound 'C'. Compound A and C respectively are a)Ethene and benzene b)Propyne and 1,3,5-trimethyl benzene c) Ethyne and toluene d) Ethyne and benzene​ 2 See answers See what the community says and unlock a badge. 0:00 / 0:15 Read More Answer 1 person found it helpful rajputashwin682 rajputashwin682 Ace 135 answers 20.7K people helped Answer: the compound A is acetylene (ethyne) and the compound C is benzene. Explanation: The given chemical equation is: CaC₂ + H₂O → A(g) + B When gas A is passed through a red-hot iron tube at 873 K, it gives another compound 'C'. The reaction between calcium carbide (CaC₂) and water (H₂O) produces acetylene (C₂H₂) gas (A) and calcium hydroxide (Ca(OH)₂) (B). The chemical equation for this reaction is: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂ When gas A (C₂H₂) is passed through a red-hot iron tube at 873 K, it undergoes a thermal decomposition reaction and produces benzene (C₆H₆) (C). The chemical equation for this reaction is: C₂H₂ → C₆H₆ Therefore, the compound A is acetylene (ethyne) and the compound C is benzene. So, the correct option is d) Ethyne and benzene Explore all similar answers Thanks 1 rating answer section Answer rating 4.6 (5 votes) Find Chemistry textbook solutions? See all Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Preeti Gupta - All In One Chemistry 11 3080 solutions Selina - Concise Chemistry - Class 9 1071 solutions 1500 Selected Problems In Chemistry for JEE Main & Advanced 2928 solutions Lakhmir Singh, Manjit Kaur - Chemistry 10 1797 solutions Lakhmir Singh, Manjit Kaur - Chemistry 9 1137 solutions NCERT Class 11 Chemistry Part 1 431 solutions NEET Exam - Chemistry 360 solutions Chemistry 643 solutions Selina - Concise Chemistry - Class 8 487 solutions Selina - Chemistry - Class 7 394 solutions SEE ALL Advertisement Answer No one rated this answer yet — why not be the first? 😎 mestrya75 mestrya75 Virtuoso 178 answers 31.1K people helped Answer: The given chemical equation is: CaC₂ + H₂O → A(g) + B Passing gas A through a red hot iron tube at 873 K gives another compound C. The reaction of calcium carbide (CaC₂) with water (H₂O) produces acetylene (C₂H₂), which is gas A. The balanced chemical equation for the reaction is: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂ Passing acetylene (C₂H₂) through a red-hot iron tube at 873 K (600 °C) leads to the polymerization of acetylene and the formation of benzene (C₆H₆), which is compound C. The chemical equation for this reaction is: 2C₂H₂ → C₆H₆ Therefore, the answer is option d) Ethyne and benzene. Explore all similar answers Thanks 0 rating answer section Answer rating 0.0 (0 votes) Advertisement Still have questions? Find more answers Ask your question New questions in Chemistry write types of probital and it geometry​ Arrange the following in increasing order of their atomic radii Na+, Al+2 ,Mg+2, Ne ​ Find the architectural height of the world's tallest building, the Burj Khalifa, in meters (m). Quantum age begin : potential and challenge​ -16 निम्न में से प्रत्येक के 1 मोल में ऑक्सीजन के मोल परमाणुओं की गणना कीजिए (i) NaOH, (ii) H2SO4, (iii) K2 Cr2 O7, (iv) C6H12O6 उत्तर : (i) 1, (ii) PreviousNext Advertisement Advertisement Report ad Ask your question Free help with homework Why join Brainly? ask questions about your assignment get answers with explanations find similar questions I want a free account Company Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Community Brainly Community Brainly for Schools & Teachers Brainly for Parents Honor Code Community Guidelines Insights: The Brainly Blog Become a Volunteer Get the Brainly App ⬈(opens in a new tab)⬈(opens in a new tab) Brainly.in We're in the know (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Why are you reporting this ad? Please make a selection. Played audio Had inappropriate content Covered the page Other Additional Information Please help us by describing the ad. Only 500 characters are allowed. REPORT AD Thank you for letting us know. This ad has already been reported. Powered by ×
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Ch-10 Mechanical Properties Of Fluids | Class 11 Physics Reading | PDF Attached | NCERT Audio Books NCERT Audio Books 3640 subscribers 11 likes Description 531 views Posted: 24 Feb 2024 Hi Guys Explore the fascinating world of Mechanical Properties of Fluids in this comprehensive audio book with accompanying PDF. Delve into the concepts of viscosity, surface tension, and buoyancy without any distracting background music, allowing you to focus solely on the enriching content. Whether you're a student, researcher, or simply curious about fluid mechanics, this resource provides a valuable learning experience. Guys share audio books with your friends Join this channel to get access to perks: NCERT Audio Books 2.0 Class 11 physics audio book (without music)~ Class 11 physics audio book (with music)~ Class 11 chemistry audio book (with music)~ Class 11 biology audio book (with music)~ Class 12 physics audio book (with music)~ Class 12 chemistry audio book (with music)~ Class 12 biology audio book (with music)~ Thank You Guys❤️❤️ 5 comments Transcript: [Music] [Music] physics textbook of class 11th part two chapter 10 mechanical properties of fluid narrated by isnar aat Khan introduction in this chapter we shall study some common physical properties of liquid and gases liquid and gases can flow and are therefore called fluids it is this property that distinguishes liquids and gases from solids in a basic way fluids are everywhere around us Earth has an envelope of air and 2/3 of its surface is covered with water water is not only necessary for our existence every Maman body constitute mostly of water all the process occurring in living beings including plants are mediated by fluids thus understanding the behavior and properties of fluid is important how are fluids different from solids what is common in liquids and gases unlike a solid a fluid has no definite shape of its own solids and liquids have a fixed volume whereas the gas fills the entire volume of its container we have learned in the previous chapter that the volume of solids can be changed by stress the volume of solid liquid or gas depends on stress or pressure acting on it when we talk about fixed volume of solid or liquid we mean its volume under atmospheric pressure the difference between gases and solids or liquids is that for solids or liquids the change in volume due to the change of external pressure is rather small in other words solids and liquids have much lower compressibility as compared to gases sheer stress can change the shape of a solid keeping its volume fixed the keepy property of the fluid is that they offer very little resistance to sheare stress their shape changes by application of very small share stress the sharing stress of fluid is about million times smaller than that of solids pressure a sharp needle when pressed against our skin pierces it our skin however remains intact when when a blunt objects with a wider contact area say a back of a spoon is pressed against it with the same Force if an elephant were to step on man's chest his ribs would crack a circus performer across whose chest a large light but strong wooden plank is placed first is saved from this accident such everyday experience convinced us that both the force and its coverage area are important is smaller the area on which the force acts greater is the impact this concept is known as pressure when an object is submersed in a fluid at rest the fluid exerts a force on its surface this force is always normal to the object surface this is so because if there were a component of force parallel to the surface the object will also exert a force on the fluid parallel to it as a consequence of Newton's third law this force will cause the fluid to flow parallel to the surface since the fluid is at rest this cannot happen hence the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it this is shown in figure 10.1 a the normal force exerted by the fluid at a point may be measured an idealized form of one such pressure measuring device is shown in figure 10.1b it consists of an evacuated chamber with a spring that is calibrated to measure the force acting on the piston this device is placed at a point inside the fluid the invert force exerted by a fluid on the piston is balanced by the outward spring force and is thereby measured if f is the magnitude of this normal force on the Piston of area a then the average pressure p average is defined as the normal force acting per unit area B average is equals to F by a in principle the Piston area can be made arbitrarily small the pressure is then defined in a limiting sense as pressure is equals to limit Delta a approaching z f by a pressure is a scalar quantity we remind the reader that it is the component of the force normal to the area under consideration and not the vector force that appears in the numerator the SI unit of pressure is Newton per M squar it has has been named as Pascal in honor of the French scientist Blaze Pascal who carried out pionering is studied on fluid pressure a common unit of pressure is atmosphere that is the pressure exerted by the atmosphere at sea level another quantity that is indispensable in describing fluids is the density Row for a fluid of mass m or occupying volume v row is equal to M by V the dimensions of density are ml the^ minus 3 its SI unit is kilog per M Cube it is a positive scalar quantity a liquid is largely incompressible and its density is therefore nearly constant at all pressures gases on the other hand exhibit a large variation in densities with pressure the density of water at 4° C is 1 into 10^ 3 Kg per M Cube the relative density of a substance is the ratio of its density to the density of water at 4° C it is a dimensionless positive scalar quantity for example the relative density of aluminium is 2.7 its density is 2.7 into 10 ^ 3 kilogram per M Cube Pascal's Law the French scientist Blaze Pascal observed that the pressure in a fluid at rest is same at all points if they are at the same height this fact may be demonstrating in simple way figure 10.2 shows an El element in the interior of a fluid at rest this element ABC D EF is in the form of a right angled prism in principle this Prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore the effect of the gravity is the same at all these points but for cl clity we have enlarged this element the force on this element are those exerted by the rest of the fluid and they must be normal to the surface of the element discussed above thus the fluid exerts pressure PA a PB and PC on this element of area corresponding to the normal force F A FB and FC as shown in figure 10 .2 on the faces B EFC a DF C and A D E B denoted by a a a and a c respectively then F sin Theta is equal to FC FB cos Theta is equal to f a Ab sin Theta is equal to AC AB cos Theta is equals to A A thus FB by a is equal to F C by a c is equal to F A by A a pressure B is equals to pressure C is equals to pressure a hence pressure exerted is same in all directions in a fluid at rest it again remind us that like other types of a stress pressure is not a vector quantity no directions can be assigned to it the force against any area within or bounding a fluid at rest and under pressure is normal to the area regardless of orientation of the area now consider a fluid element in the form of a horizontal bar of uniform cross-section the bar is an equilibrium the horizontal force exerted at its two ends must be balanced or the pressure at the two ends should be equal this proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane suppose the pressure were not equal in the different parts of the fluid then there would be a flow as the fluid will have some net force acting on it hence in the absence of flow the pressure in the fluid must be same everywhere when is flow of air due to pressure differences variation of pressure with depth consider a fluid at rest in a container in figure 10.31 is at height Edge above a point two the pressure at points 1 and two are P1 and P2 respectively consider a cylinder element of a fluid having area of Base a and height hedge as the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical Force should balance the weight of the element the force acting in the vertical direction are due to the fluid pressure at the top P1 a acting downwards at the bottom p2a acting upward if mg is the weight of the fluid in the the cylinder we have P2 - P1 into a is equal to mg now if row is the mass density of the fluid we have the mass of fluid to be m is equal to Row V is equals to row h a so that P2 minus P1 is equal to row GH pressure difference depends on the vertical distance H between points 1 and two mass density of the fluid row and acceleration due to the gravity G if the point one under discussion is shifted to the top of the fluid say water which is open to the atmosphere P1 may be replaced by the atmospheric pressure PA and we replace P2 by P then p is equal to p a plus r g does the pressure p at depth below the surface of a liquid open to the atmosphere is greater than the atmospheric pressure by an amount R GH the excess of pressure p minus PA a at depth H is called the gauge pressure at that point the area of cylinder is not appearing in the expression of absolute pressure thus the height of the fluid column is important and not cross-sectional or the base area or the shape of the container the liquid pressure is the same at all points at the same horizontal level that is same depth the result is appreciated through the example of hydrostatic Paradox consider three vessels a b and c of different shapes they are connected at the bottom by a horizontal pipe on filling with water the level in the three vessels is same though they hold different amount of water this is so because water at the bottom has the same pressure below each section of the vessel atmospheric pressure and gauge pressure the pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere at sea level it is 1.013 into 10 to the power five pascals Italian scientist orista tesel deviced for the first time a method for measuring atmospheric pressure a long glass tube closed at one end filled with Mercury is inverted into a throw of mercury as shown in figure 10.5 a this device is known as mercury barometer the space above the Mercury column in the tube contains only mercury vapor whose pressure p is so small that it may be neglected the pressure inside the column at Point a must be equal to the pressure at point B which is at the same level pressure at B is equals to atmospheric pressure that is equal to PA a PA a is equals to row GH where row is the density of mercury and H is the height of mercury column in the tube in the experiment it is found that the Mercury column in the barometer has a height of about 76 CM at CA level equivalent to one atmosphere this can also be obtained using the value Row in equation 10.8 a common way of stating pressure is in the terms of CM or millim of Mercury Hg a pressure equivalent of 1 mm is called a t after tesly one t is equals to 233 pascals the millimeter of H and Tor are used in medicine and Physiology in meteorology a common unit is the bar and mli bar a bar is equal to 10 to the^ 5 pascals an open tube manometer is a useful instrument for measuring pressure differences it consists of a u tube containing a suitable liquid that is a low density liquid such as oil for measuring pressure difference and a high density liquid such as mercury for large pressure differences one end of the tube is open to the atmosphere and the other end is connected to the system whose pressure we want to measure the pressure p at a is equals to pressure at point B what we normally measure is the gauge pressure press which is p minus PA a given by equation 10.8 and is proportional to the manometer height Edge pressure is same at the same levels on both sides of the u tube containing a fluid for liquids the density varies very little over the wide ranges in pressure and temperature and we can treat it safely as a constant for our present purpose gas on the other hand exhibits large variations of density with changes in pressure and temperature unlike gases liquids are therefore largely treated as incompressible hydraulic machines let us now consider what happens when we change the pressure on a fluid contained in a vessel consider a horizontal cylinder with a piston and 3w itical tubes at different points the pressure in the horizontal cylinder is indicated by the height of the liquid column in the vertical tubes it is necessarily the same in all if we push the Piston the fluid level rises in all the tubes again reaching the same level in each of them this indicates that when the pressure on the cylinder was increased it was distributed uniform formally throughout we can say whenever external pressure is applied on any part of a fluid contained in a vessel it is transmitted undiminished and equally in all directions this is Pascal's Law of transmission of fluid pressure and has many applications in daily life a number of devices such as hydraulic lift hydraulic brakes are based on the vascal law in this devices fluids are used for transmitting pressure in a hydraulic lift as shown in figure 10.6 two Pistons are separated by the space filled with a liquid a piston of a small cross-sectional area A1 is used to exert a force fub1 directly on the liquid the pressure p is equal to fub1 by A1 is is transmitted throughout the liquid to the larger cylinder attached with the larger piston of area A2 which results in an upward force of P into A2 therefore the Piston is capable of supporting a large Force large weight of say a car or a truck placed on the platform FS2 is equal to P A2 is equal to fub1 A2 divided by A1 by changing the force at A1 the platform can be moved up or down thus the applied force has been increased by a factor of A2 by A1 and this factor is mechanical advantage of this device hydraulic brakes and Automobiles also work on the same principle when we apply a little force on the pedal with our foot the master piston moves inside the master cylinder the pressure caused is transmitted through the brake oil to act on the Piston of larger area the large Force acts on the piston and is pushed down expanding the brake shoes against the brake lining in this way a small force of the pedal produces a large retarding force on the wheel an important advantage of this system is that the pressure set up by by pressing p is transmitted equally to all the cylinders attached to the four wheels so that the breaking effort is equal on all the wheels Archimedes principle fluid appears to provide partial support to the object placed in it when a body is wholly or partially immersed in a fluid at rest the fluid exerts pressure on the surface of the body in in contact with the fluid the pressure is greater on the lower surface of the body than on the upper surface as pressure in a fluid increases with depth the resultant of all the forces in an upward Force called the bo Force suppose that a cylindrical body is immersed in the fluid the upward force on the bottom of the body is more than downward force on its top the fluid exerts a resultant upward force or Bo force on the body equals to P2 - P1 into a we have seen in equation 10.4 that P2 - P1 into a is equals to R GHA now h a is the volume of the solid and R H is the weight of an equivalent volume of the fluid P2 2 - P1 into a is equal to mg thus the upward force exerted is equal to the weight of the displaced fluid the result holds true irrespective of the shape of the object and here cylindrical object is considered only for convenience this is Archimedes principle for totally immersed objects the volume of the fluid displaced by the object is equal to its own volume if the density of the immersed object is more than that of fluid the object will sink as the weight of the body is more than the upward thrust if the density of the object is less than that of fluid it floats in the fluid partially submerged to calculate the volume submerged suppose the total volume of the object is v s and the part VP of it is submerged in the fluid then the upward Force which is the weight of the displaced fluid is equal to row f g VP which must be equal to the weight of the body row s g v s is equals to row f g VP or row s by row f is equals to VP by v s the apparent weight of the floating body is zero this principle can be summarized as the loss of weight of the body submerged partially or fully in a fluid is equal to the weight of the fluid displaced streamline flow so far we have studied the fluid at rest the study of fluids in motion is known as the fluid dynamics when a water tab is turned on slowly the water flow is a smooth initially but loses its smoothness when the speed of the outflow is increased in studying the motion of fluids we focus our attention on what is happening to various fluid particles at a particular point in the space at a particular time the flow of the fluid is set to Be steady if at the given point the velocity of each passing fluid particle remains constant in time this does not mean that the velocity at different points in the space is same the velocity of the particular particle may change as it moves from one point to another that is at some other point the particle may have a different velocity but every other particle which passes the second particle behaves exactly as the previous particle that has just passed that point each particle follows a smooth path and the paths of the particle do not cross each other the path taken by a fluid particle under a steady flow is a stream line it is defined as the curve whose tangent at any point is in the direction of the fluid velocity at that point consider the path of a particle as shown in figure 10.7 a the curves describe how a fluid particle moves with time the curve PQ is like a permanent map of the fluid flow indicating how the fluid streams no two streamlines can cross for if they do an oncoming fluid particle can go either one way or the other and the flow would be not steady hence in steady flow the map of the flow is stationary in time how do we draw closely spaced stream lines if we intend to show streamline of every flowing particle we would end up with a Continuum of lines consider the planes perpendicular to the direction of the fluid flow example at three points p r q in figure 10 . 7B the plain pieces are so chosen that their boundaries can be determined by the same set of the stream lines this means that the number of fluid crossing the surface as indicated at p r and Q is the same if the area of crosssections at these Point are AP a r and AQ the speeds of the fluid particles are VP v r and VQ then the mass of the fluid Delta MP Crossing AT AP in a small interval of time delta T is row P AP VP delta T similarly mass of the fluid Delta M flowing or crossing at a r in a small interval of time delta T is row r a r v r delta T and mass of the fluid Delta mq is row q a q VQ delta T Crossing at AQ the mass of the fluid flowing out equals the mass flowing in holds in all cases for the flow of incompressible fluids row P is equal to row R is equal to row Q the equation reduces to AP VP is equal to a r v r is equal to AQ VQ which is called the equation of continuity and it is a statement of conservation of mass in flow of incompressible fluids in general AE is equals to constant AOE gives the volume flux or the flow rate and remains constant throughout the pipe of flow thus at narrower portions where the streamlines are closely placed velocity increases and it's vice versa from figure 10.7 B it is clear that a is greater than AQ or VR is less than VQ the fluid is accelerated while passing from R to Q This is associated with the change in pressure in fluid in the horizontal pipe steady flow is achieved at low flow speeds Beyond a limiting value called the critical speed this flow loses steadiness and becomes turbulent one sees this when a fast flowing stream encounters Rock a small Foy warpool like regions called white water rapids are formed figure 10.8 displays stream lines for some typical flows for example figure 10.8 a describes a laminer flow where the velocities at different points in the fluid may have different magnitudes but the directions are parallel figure 10.8 B gives a sketch of turbulent flow bernal's principle fluid flow is a complex phenomenon but we can obtain some useful properties for a steady or streamline flows using the conservation of energy consider a fluent moving in a pipe of varing cross-sectional area let the pipe be at varing Heights as shown in figure 10.9 we now suppose that an incompressible fluid is flowing through the pipe in a steady flow its velocity must change as a consequence of equation of continuity a force is required to produce this acceleration which is caused by the fluid surrounding it the pressure must be different in different regions Bern's equation is a general expression that relates the pressure difference between two points in a pipe to both velocity changes kinetic energy change and elevation height changes potential energy change the sus physicist Daniel beri developed this relationship in 1738 consider the flow at two regions one and two consider the fluid initially lying between B and D in an infinite decimal time interval delta T the fluid would have moved suppose V1 is the speed at B and we2 at D then fluid initially at B has moved a distance W1 delta T to C W1 delta T is a small enough to assume constant cross-section along BC in the same interval delta T the fluid initially at D moves to e the distance equals to V2 delta T Pressure P1 and P2 acts as shown on the plain faces of areas A1 and A2 binding the two regions the work done on the fluid at Left End BC is W1 is equals to P1 A1 into V1 delta T is equals to P1 Delta V since the same volume Delta V passes through both the regions from equation of continuity the work done by the fluid at the other end D is equal to W2 is equal to P2 A2 into W2 delta T is equals to P2 Delta V or the work done on the fluid is minus P2 Delta V so the total work done on the fluid is W1 - W2 is equal to P1 - P2 into Delta V part of this work goes into changing the kinetic energy of the fluid and part goes into changing the gravitational potential energy if the density of the fluid is row and Delta m is equals to row A1 V1 delta T is equals to row Delta V is the mass passing through the pipe in time delta T then change in gravitational potential energy is Delta U is equal to r g Delta V into H2 - H1 the change in its kinetic energy is Delta K is equal to half row Delta V into V2 s minus W1 s we can imploy the work energy theorem to this volume of the fluid and this yields P1 - P2 into Delta V isal = to half R Delta V into V2 2 - V1 2 + r g Delta V into H2 - H1 now we divide the equation by Delta V and then we rearrange the above terms to obtain P1 + half row V1 S Plus row G1 is equals to P2 + half row V2 square + r g H2 this is Bern's equation since 1 and two refers to any two locations along the pipeline we may write the expression in general as p + half r² + r g h is equals to constant in other words the bernal's relation may be stated as follows as we move along a stream line the sum of the pressure p the kinetic energy per volume v² by 2 and the potential energy per unit volume r g remains a constant note that in applying the energy conservation principle there is an assumption that no energy is lost due to friction but in fact when fluids flow some energy get lost due to the internal friction this arises due to the fact that in a fluid flow the different layers of the fluid flow with different velocities these layers exert frictional forces on each other resulting in a loss of energy this property of the fluid is called the viscosity and is discussed in more detail in a later section the Lost kinetic energy of the fluid gets convert into the heat energy thus bernal's equation ideally applies to the fluids with zero viscosity or nonviscous fluids another restrictions on the application of Bern's theorem is that the fluids must be incompressible as the elastic energy of the fluid is also not taken into consideration in practice it has a large number of useful applications and can help explain a wide variety of phenomen for low viscosity in compressible fluids bernal's equation also does not hold for nonsteady or turbulent flows because in that situation velocity and pressure are constantly fluctuating in time speed of iflux tus slow the word iflux means the fluid outflow toris discovered that the speed of e flux from from an open tank is given by a formula identical to that of the freely falling body consider a tank containing a liquid density row with a small hole in its side at a height y1 From The Bottom the air above the liquid whose surface is at height Y2 is at pressure p from the equation of continuity equation 10.10 we have V1 A1 is equal to V2 A2 V2 is equals to A1 by A2 into V1 if the cross-sectional area of the tank A2 is much larger than that of the hole then we may take the fluid to be approximately at rest at the top that is V2 is equals to0 now applying the bernal's equation at Point 1 and 2 and noting that at the whole P1 is equals to PA a when p is much and much greater than PA and 2 G may be ignored the speed of e flux is determined by a container pressure such a situation occurs in rocket propulsion on the other hand if the tank is open to the atmosphere then p is = to PA a and V1 is equal to under < TK 2 GH this is the speed of a freely falling body equation 10.15 that is V1 is equals to under root 2gh is known as the tus Alise law when meter the V meter is a device to measure the flow speed of incompressible fluids it consists of a tube with a broad diameter and a small constrction at the middle is shown in figure 10.1 a manometer in the form of a u tube is also attached to it with one arm at the broad neck point of the tube and the other at the constrction as shown in figure 10.11 the manometer contains a liquid of density row M the speed W1 of the the fluid flowing through the tube at the broad neck area a is to be measured from the equation of continuity the speed at the construction becomes V2 is equals to Capital a by small a into V1 then using Bern's equation we get P1 minus P2 is equals to half row V1 s into bracket capital A upon small a² - 1 this pressure difference cause the fluid in the Ube connecting at the narrow neck to rise in comparison to the arm the difference in the height Edge measures the pressure difference the principle behind this meter has many applications the carburetor of automobiles has a ventury channel nozle through which the air flows with large speed the pressure is then lowered at the narrow neck and the bat roll or gasoline is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion filter pumps or aspirators bunson burners automizer sprayers used for perfumes or to spray spray insecticide work on the same principle blood flow and heart attack Bern's principle helps in explaining blood flow in artery the artery may get constricted due to the accumulation of plague on its inner walls in order to drive the blood through this constriction a greater demand is placed on the activity of the heart the speed of the flow of blood in this region is raised which lowers the pressure inside the artery May collapse due to the external pressure the heart exerts further pressure to open this artery and forces the blood through as the blood rushes through the opening the internal pressure once again drops due to the same reasons leading to a repeat collapse this may result in heart attack Dynamic lift Dynamic lift is the force that acts on a body such as airplane wing a hydrofoil or a spinning Ball by virtue of its motion through a fluid in many games such as Cricket tennis ball baseball or golf we notice that a spinning ball deviates from its parabolic trajectory as it moves through the air this deviation can be partly explained on the basis of Bern's theorem one Ball moving without spin figure 10.13 a shows the stream lines around the non-spinning ball moving relative to a fluid from the symmetry of a stream lines it is clear that the velocity of fluid or air above and below the ball at corresponding points is same resulting in zero pressure difference the air therefore exerts no upward or downward force on the ball two Ball moving with spin a ball which is spinning drags air along with it if the surface is rough more air will be dragged figure 10.13 B shows the streamlines of air for a ball which is moving and spinning at the same time the ball is moving forward and relative to it the air is moving backwards therefore the velocity of air above the ball relative to it is larger and below it is smaller the stream lines thus get crowded above and rarified below this difference in the velocities of the air results in the pressure difference between the lower and the upper faces and there is a net upward force on the ball this Dynamic lift due to spinning is called the Magnus effect aeroil or lift on aircraft Wing figure 10.13 C shows an aeroil which is a solid piece shaped to provide an upward Dynamic lift when it moves horizontally through the air the cross-section of the wings of an airoplane looks somewhat like the airall shown in figure 10.13 C which streamlines around it when the air foil moves Against the Wind the orientation of the Wings relative to the direction of the flow causes the streamlines to crowd together above the wing more than those below it the flow speed on the top is higher than below it there is an upward Force resulting in a dynamic lift of the Wings and this balance the weight of the plane viscosity most of the fluids are not ideal ones and offer some resistance to the motion this resistance of the fluid motion is like an internal friction analogous to the friction when a solid moves on the surface it is called viscosity this Force exist when there is a relative motion between layers of the liquid suppose we consider a fluid like oil enclosed between two glass plates shown in figure 10.15 a the bottom plate is fixed while the top plate is moved with a constant velocity V relative to the fixed plate if the oil is replaced by Honey a greater force is required to move the plate with the same velocity hence we say that honey is more riskless than oil the fluid in contact with the surface has the same velocity as that of the surface hence the layer of the liquid in contact with top surface moves with a velocity away and the layer of the liquid in contact with the fixed surface is stationary the velocity of layers increase uniformly from bottom zero velocity to the top layer velocity V for any layer of liquid its upper layer pulls it Forward while the lower lay layer pulls it backwards this results in force between the layers this type of flow is known as laminer the layers of the liquid slide over one another as the pages of book do when it is placed on a flat table and a horizontal force is applied to the top cover when a fluid is flowing in a pipe or a tube then velocity of the liquid layer along the axis of the tube tube is maximum and decreases gradually as we move towards the wall where it becomes zero the velocity on a cylindrical surface in a tube is constant on account of this motion a portion of liquid which at some instant has a shape ab b c d takes the shape a e FD after short interval of time delta T during this time interval the liquid has undergone a sheer strain of Delta X by L since the strain in a flowing fluid increases with the time continuously unlike a solid here the stress is found experimentally to depend on the rate of change of a strain or a strain rate that is Delta X by L delta T or W by L instead instead of a strain itself the coefficient of viscosity pronounced AA for the fluid is defined as the ratio of sharing stress to The Strain rate F by a ided w by L is equals to f l by VA the SI unit of viscosity is poison its other units are Newton Second per M squar or Pascal second the dimensions of viscosity are m l^ minus1 t ^ -1 generally thin liquids like water alcohol Etc are less viscous than the thick liquids like coal thar blood glycerin Etc the coefficients of viscosity for some common fluids are listed in the table 10.2 we point out two facts about blood and water that you may find interesting as table 10.2 indicates blood is thicker or more viscous than water further the relative viscosity ITA by ITA water of blood remains constant between 0° C and 37° C the viscosity of liquids decreases with the temperature while it increases in in the case of gases a stroke slow when a body falls through a fluid it drags a layer of the fluid in contact with it a relative motion between the different layers of the fluid is set as a result the body experience a retarding force falling of a rain drop and Swinging of a pendulum Bob are some common examples of such motion it is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion the other quantities on which the force F depends on viscosity EA of the fluid and the radius a of the sphere Sir George G Stokes an English scientist annunciated clearly the wiscus drag Force f as f is equals to 6 Pi ETA a this is known as The Stokes law we shall not drive the Stokes law this law is an interesting example of retarding force which is proportional to the velocity we can study its consequence on an object falling through a wiscus medium we consider a raindrop in air it accelerates initially due to gravity as the velocity increases the re starting Force also increases finally when the viscous Force plus Bo Force becomes equal to force due to gravity the net force becomes zero and so does the acceleration the sphere rain drop then descends with a constant velocity thus in equilibrium this terminal velocity is given by 6 Pi EA A1 is equals to 4 pi I by 3 into a cub into row minus Sigma G where row and the sigma are mass densities of a sphere and the fluid respectively we obtain VT is equals to 2 a² into r - Sigma into G / 9 R so the terminal velocity VT depends on the square of radius of a sphere and in versely on viscosity of the medium renals number when the rate of flow of a fluid is large the flow no longer remains laminer but becomes turbulent in a turbulent flow the viscosity of the fluid at any points in the space varies rapidly and randomly with time some circular motions called edes are also generated and stackler placed in the path of a fast moving fluid causes turbulence the smoke rising from a burning stack of wood Oceanic currents are turbulent twinkling of stars is the result of atmospheric turbulence the waves in the water and in the air left by cars aeroplanes and boards are also turbulence Osborne renolds observed that turbulent flow is less likely for viscous fluid flowing at low rates he defined a dimensionless number whose value is given on an approximate idea whether the flow would be turbulent this number is called the renold's number re re is equals to Row V ided EA where row is the density of the fluid flowing with a speed v d stands for the DI dimension of the pipe and EA is the viscosity of the fluid re is dimensionless number and therefore it remains same in any System of Units it is therefore found that the flow is streamline or L Miner for re less than th000 the flow is turbulent for re greater than 2,000 the flow becomes unsteady for re between 1,000 and 2,000 the critical value of re known as the critical venid number at which the turbulent sets is found to be the same for geometrically similar flows for example when oil and water with their different densities and viscosities flow in pipes for same shapes and size turbulence sets in at almost the same values of re using this fact a small scale laboratory model can be set up to study the character of the fluid flow they are useful in designing of ships submarines racing car and aeroplanes re can also be written as re is equals to r² by EA V by D is equals to row a v v² by EA a v by D is equals to inertial force or the force of viscosity thus ral's number represents the ratio of the inertial force force due to inertia that is mass of moving fluid due to inertia of obstacle in its path to wiscus force turbulence dissipates kinetic energy usually in the form of heat racing cars and planes are engineered to Precision in order to minimize turbulence the design of such vehicle involve experimentation and trial and error on the other hand turbulence like friction is sometimes desirable turbulence promotes mixing and increase the rate of transfer of mass momentum and energy the blades of a kitchen mixture induce turbulent flow and provide thick milkshakes as well as beat eggs into a uniform texture surface tension you must have noticed that oil and water do not mix water wets you and me but not Ducks Mercury does not wet glass but water sticks to it oil rise up a cotton wick in spite of gravity sap and water rise up to the top of the leaves of the tree hairs of a paintbrush do not cling together when dry and even when dipped in water but form a fine tip when taken out of it all of these and many more such experience are related with free surface of fluids as liquids have no definite shape but have a definite volume they acquire a free surface when put in a container these surfaces possess some additional energy this phenomena is known as surface tension and is concerned with only liquids as gases do not have free surface let us understand this phenomena surface energy a liquid stays together because of attraction between molecules consider a molecule well inside a liquid the inter molecular distance are such that it is attracted to all the surrounding molecules this attraction results in a negative potential energy of the molecule which depends on the number and distribution of molecules around the chosen one but the average potential energy of all the molecules is the same this is supported by the fact that to take a collection of such such molecules the liquid and to disperse them far away from each other in order to evaporate or vaporize the heat of evaporation required is quite large for water it is of the order of 40 K per mole let us consider a molecule near the surface only lower half side of it is surrounded by the liquid molecules there is some negative potential energy due to these but obviously it is less than that of molecules in bulk that is the one fully inside approximately is half of the later does the molecule on a liquid surface have some extra energy in comparison to the molecules in the interior a liquid does tends to have least surface area which external condition permits increasing surface area requires energy most surface phenomena can be understood in terms of this fact what is the energy required for having a molecule at the surface as mentioned above roughly it is the half of the energy required to remove it entirely from the liquid that is half the Heat of evaporation finally what is a Surface since a liquid consists of molecules moving about there cannot be perfectly sharp surface the density of the liquid molecules drops rapidly to Z around Z is equals to 0 as we move along the direction indicated figure 10.6 C in a distance of the order of a few molecular sides surface energy and surface tension as we have discussed that an extra energy is associated with the surface of liquids the creation of more surface is spreading of surface keeping other things like volume fixed requires additional energy to appreciate this consider a horizontal liquid film ending in the bar free to slide over parallel guide figure 10.17 suppose we move the bar by a small distance d as shown since the area of the surface increases the system now has more energy this means that some work has been done against the internal Force let this internal force be F the work done by the applied force is FD from conservation of energy this is stored as additional energy in the film if the surface energy of the film is s per unit area the extra area is 2 DL a film has two sides and the liquid in between so that there are two surface and extra energy is s into 2dl is equals to FD or S is equals to FD by 2dl is equals to F by 2L the quantity s is the magnitude of surface tension it is equal to the surface energy per unit area of the liquid interface and is also equal to the force per unit length exerted by the fluid on the movable bar so far we have talked about the surface of one liquid more generally we need to consider fluid surface in contact with other fluids or solid surfaces the surface energy in that case depends on the material on both sides of the surface for example if the molecules of the material attracted each other surface energy is reduced while if they repel each other the surface energy is increased thus more appropriately the surface energy is the energy of interface between two materials and depends on both of them we make the following observations from above one surface tension is a force per unit length or Surface energy per unit area acting in the plane of the interface between the plane of the liquid and any other substance it also is the extra energy that the molecules at the interface have as compared to the molecules in the interior two at any point on the interface besides the boundary we can draw a line and imagine equal and opposite surface tension forces s per unit length of the line acting perpendicular to the line in the plane of the interface the line is an equ equilibrium to be more specific imagine a line of atoms or molecules at the surface the atoms to the left pull the line towards them those to the right pull it towards them this line of atoms is in equilibrium under tension if the line really marks the end of the inter surface as in figure 10.16 A and B there is is only the force s per unit length acting inwards table 10.3 gives the surface tension of various liquids the value of surface tension depends on temperature like viscosity the surface tension of a liquid usually Falls with temperature a fluid will stick to a solid surface if the surface energy between fluid and the solid is smaller than the sum of the surface energies between solid air and fluid air now there is a cohesion between the solid surface and the liquid it can be directly measured experimentally and schematically shown in figure 10.18 a flat vertical glass plate below which a vessle of some liquid is kept from one arm of the balance the plate is balanced by weight on the other side with its horizontal Edge just over water the vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension weights are added till the plate just clears water suppose the additional weight required is W then from equation 10.25 four and the discussion given there the surface tension of the liquid air interface is s is equals to W by 2L is equal to mg by 2L where m is the extra mass and L is the length of the plate Edge the subscript La emphasizes the fact that the liquid air interface tension is involved angle of contact the surface of liquid near the plane of contact with another medium is in general curved the angle between the tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as the angle of contact it is denoted by Theta it is defined at interface of different pairs of liquids and solids the value of theta determines whether a liquid will spread on the surface of a solid or it will form droplets on it for example water forms droplets on a Lotus Leaf as shown in figure 10.9 a while it spreads on a clean plastic plate as shown in figure 10.19 bam we consider the three interfacial tensions at all three interfaces liquid air solid air and solid liquid denoted by SL SLA SSA SS respectively as given in figure 10.19 A and B at the line of contact the surface forces between the three media must be in equilibrium from figure 1019b the following relation is easily derived SL a COS Theta + SL is equal to SSA the angle of contact is an obtuse angle if SL is greater than SL SLA as in the case of water Leaf interface while it is an acute angle if s SL is less than SLA as in the case of water plastic interface when Theta is an obtuse angle then molecules of liquids are attracted strongly to the themselves and weakly to those of solids it cost a lot of energy to create solid liquid surface and liquid does not wet the solid this is what happens with water on vxi or oily surface and with Mercury on any surface on the other hand if the molecules of the liquid are strongly attracted to those of solids this will reduce SSL and therefore cos Theta may increase or Theta May decrease in this case Theta is an acute angle this is what happens for water on glass or on plastic or for kerosene oil on virtually anything it just spreads soap detergents and dying substances are wetting agents when they are added the angle of contact becomes small so that these May Pen pen iterate well and becomes effective waterproofing agents on the other hand are added to create a large angle of contact between water and fibers drops and Bubbles one consequence of surface tension is that free liquid drops and Bubbles are spherical if effects of gravity can be neglected you may have seen this especially clearly in small drops just formed in a highp speeed spray or Jet and in soap bubbles blown by most of us in childhood why are drops and Bubbles spherical what keeps so bubbles stable as we have been saying repeatedly a liquid air interface has energy so for a given volume the surface with minimum energy is the one with the least area the sphere has this property though it is out of the scope of this book but you can check that a sphere is better than at least a cube in this respect so if gravity and other forces example air resistance were ineffective liquid drops would be spherical another interesting consequence of surface tension is that the pressure inside a spherical drop is more than the pressure outside suppose a spherical drop of radius R is in equilibrium if its radius increased by delt R the extra surface energy is 8 pi r delt R into SLA if the drop is in equilibrium this energy cost is balanced by the energy gain due to the expansion under the pressure difference P inside minus P outside but when the inside of the bubble and outside the work done is W is equal = to P inside minus P outside into 4 Pi r² into delt R so that P inside minus P outside is equals to 2 s l a by R in general for a liquid gas interface the convex side has a higher pressure than the concave side for example an air bubble in a liquid would have higher pressure inside a bubble differs from a drop and a cavity in this it has two interfaces applying the above argument we have for a bubble P inside minus P outside is equals to 4 SL L A by R this is probably when you have to blow hard but not too hard to form a soap bubble a little extra air pressure is needed inside capillary rise one consequence of pressure difference across a curved liquid air interface is the well-known effect that the water rise up in a narrow tube in spite of gravity the word Capa means hair in Latin if the tube were hair thin the rise would be very large to see this consider a vertical capillary tube of circular cross-section radius a inserted and an open vessel of water the contact angle between the water and glass is AC Ute thus the surface of water in the capillary is concave this means there is a pressure difference between the two sides of the top surface this is given by P inside minus P outside is equal to 2 s by R is = to 2 s by a sec Theta thus the presence of the water inside the tube just at the miniscus air water interface is less than than the atmospheric pressure consider the two points A and B in figure 10.21 a they must be at the same pressure namely P outside plus h r g is equals to P inside is equals to PA a where row is the density of water and H is the height called the capillary rise using equation we have h r g is equals to pressure inside minus pressure outside is equal to 2 s cos Theta by a the discussion here the equation 10.28 and 10.29 make it clear that the cap rise due to the surface tension it is larger for a smaller a typically it is of the order of a few CM for fine capillaries for example if a is equals to 05 CM using the value of surface tension for water we find that H is equal to 2.98 CM notice that if the liquid meniscus is convex as for mercury that is if cos Theta is negative then it is clear that the liquid will be lower in the capillary detergents and surface tension we clean dirty clothes containing grease and oil stains is sticking to Cotton or other fabric by adding detergent or soap to the water soaking clothes in it and shaking let us understand this process better washing with water does not remove grease stains this is because water does not wet greasy dirt that is there is a very little area of contact between them if water could wet Grease the flow of water could carry some grease away something of this sort is achieved through detergents the molecules of detergent are hair pin shaped with one end attracted to the water and other to the molecules of Grease oil or wax thus tending to form water oil interface the result is shown in figure 10.22 as a sequence of figures in our language we would say that addition of detergent whose molecules attract at one end and say oil on the other reduces drastically the surface tension s water oil it may even become energetically favorable to form such interface that is Globes of dirt around the detergents and then by water this kind of process using surface active detergents or surfectants is important not only for cleaning but also in recovering oil mineral ORS Etc summary one the basic property of a fluid is that it can flow the fluid does not have any resistance to change of its shape thus the shape of a fluid is governed by shape of its container two a liquid is incompressible and has a free surface of its own a gas is compressible and it expands to occupy all the space available to it three if f is the normal force exerted by a fluid on an area a then the average pressure p average is defined as the ratio of force to area P average is equals to F by a fourth the unit of the pressure is the Pascal it is the same as Newton per M squar other common units of pressure r 1 atmosphere is equal to 1.01 into 10^ 5 pascals 1 bar is equal to 10 ^ 5 pascals 1 T is equal to 133 pascals 1 mm of EDG is equal to 1 T is equal to 133 pascals five pascals law states that pressure in a fluid at rest is same at all points which are at the same height a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the Ws of the containing vessel six the pressure in a fluid varies with depth HED according to the expression p is equal to P A + r GH where row is the density of fluid assumed uniform seventh the volume of an incompressible fluid passing at any point every second in a pipe of non-uniform cross-section is the same in the St flow VA is equals to constant the equation is due to mass conservation in incompressible fluid flow eth Bern's principle states that as we move along a streamline the sum of the pressure p and kinetic energy per unit volume and the potential energy per unit volume remains constant the equation is basically the conservation of energy applied to non-viscous fluid Motion in steady state there is no fluid which have zero viscosity so the above statement is true only approximately the viscosity is like friction and converts the kinetic energy to heat energy ninth though Shear strain in a fluid does not require shear stress when a sheer stress is applied to a fluid the motion is generated which causes a sheer strain growing with time the ratio of the shared stress to the time rate of sharing strain is known as coefficient of viscosity EA 10th The Stokes law states that a viscous drag Force F on a sphere of radius a moving with velocity v through a fluid of viscosity is f is equal to 6 Pi ETA a 11th the onset of turbulence in a fluid is determined by a dimensionless parameter called the ral's number given by re is equals to row mu D bya 12th surface tension is a force per unit length acting in a plane of interface between the liquid and the bounding surface it is the X extra energy that the molecules at the interface have as compared to the interior points to ponder pressure is a scalar quantity the definition of pressure as Force per unit area may give a false impression that pressure is a vector the force in the numerator of the definition is the component of the force normal to the area upon which it is impressed while describing fluids as a conceptual shf from the particle and rigid body mechanics is required we are concerned with properties that vary from point to point in a fluid one should not think of pressure of a fluid as being exerted only on a solid like walls of a container or piece of solid matter immersed in a fluid pressure exerts at all point in a fluid an element of a fluid is in equilibrium because the pressure exerted on the various faces are equal the expression of pressure is p is equal to P A + r GH holds true only if the fluid is incompressible practically speaking it holds for liquids which are largely incompressible and hence is a constant with height the gauge pressure is the difference of the actual pressure and the atmosphere pressure many pressure measuring devices measure the gauge pressure these include the tire pressure gauge and the blood pressure gauge is figma manometer a extrem line is a map of fluid flow in a steed flow two stream lines do not intersect as it means that the fluid particle will have two possible velocities at the point Bern's principle does not hold in the presence of viscous drag on the fluid the work done by by this dissipative viscous Force must be taken into account in this case and P2 will be lower than the value given by equation 10.2 as the temperature rises the atoms of the liquid becomes more mobile and the coefficient of viscosity ITA Falls in a gas the temperature rise increase the random motion of the atoms and ITA increases the critical r number for the onset of turbulence is in the range 1,000 to 10,000 depending on the geometry of the flow for most cases Rena's number is less than th000 signifies the minor flow Rena's number from 1,000 to 2,000 is UN steady flow and ral's number more than th000 implies turbulent flow surface tension arises due to the axis potential energy of the molecules on the surface in comparison to the potential energy in the interior such a surface energy is present at interface separating two substances at least one of which is a fluid it is not the property of a single fluid alone
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https://www2.math.uconn.edu/~hurley/math116/section4_docs/Handouts/elimit.pdf
The Number e as a Limit This document derives two descriptions of the number e, the base of the natural logarithm function, as limits: (8, 9) lim x→0(1 + x)1/x = e = lim n→∞ µ 1 + 1 n ¶n . These equations appear with those numbers in Section 7.4 (p. 442) and in Section 7.4 (p. 467) of Stewart’s text Calculus, 4th Ed., Brooks/Cole, 1999. Reversing the approach of the text, the following derivation first establishes (9) and then shows that (8) follows from that. Suppose that n > 0. (Very similar reasoning applies if n < 0.) From the figure, the area under the graph of y = 1/x between x = 1 and x = 1 + 1/n lies between the areas of the two rectangles above the interval [1, 1 + 1/n]. The larger rectangle has height 1, while the height of the smaller is n n+1. Since the base in each case is 1/n, the area relations translate into the inequalities 1 n · n n + 1 ≤ Z 1+1/n 1 1 x dx ≤1 n · 1 H ⇒ 1 n + 1 ≤ln µ 1 + 1 n ¶ ≤1 n . Mutliplying the last set of inequalities through by the positive number n gives n n + 1 ≤n ln µ 1 + 1 n ¶ = ln µ 1 + 1 n ¶n ≤1. Next, let n →+∞and use the fact that taking limits preserves the relation ≤: 1 ≤ lim n→+∞ln µ 1 + 1 n ¶n ≤1 H ⇒ lim n→+∞ln µ 1 + 1 n ¶n = 1. Since exp is continuous at x = 1, for u = ln(1 + 1/n)n we have eu →e1 = e as u →1). Hence, (9) as n →+∞, eln(1+1/n)n = µ 1 + 1 n ¶n →e. Next, let x = 1/n, so that n →+∞is equivalent to x →0+. Thus (9) is equivalent to lim x→0+(1 + x)1/x = e. What about the limit as x →0−? The similar reasoning referred to above shows that limn→−∞(1 + 1/n)n = e also holds. So letting x = 1/n leads to the left-hand limit also being e, which means that (8) holds.
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https://www.skillsyouneed.com/num/percentages.html
Shop Our Blog Contact Us NUMERACY SKILLS Introduction to Percentages % Search SkillsYouNeed: Numeracy Skills: A - Z List of Numeracy Skills How Good Are Your Numeracy Skills? Numeracy Quiz Money Management and Financial Skills Real-World Maths Numbers | An Introduction Special Numbers and Mathematical Concepts Systems of Measurement Common Mathematical Symbols and Terminology Apps to Help with Maths Arithmetic: Addition + Subtraction - Multiplication × Division ÷ Positive and Negative Numbers Ordering Mathematical Operations - BODMAS Mental Arithmetic – Basic Mental Maths Hacks Essentials of Numeracy: Fractions Decimals Ratio and Proportion Percentages % Percentage Calculators Percentage Change | Increase and Decrease Calculating with Time Estimation, Approximation and Rounding Geometry: Introduction to Geometry: Points, Lines and Planes Introduction to Cartesian Coordinate Systems Polar, Cylindrical and Spherical Coordinates Angles Properties of Polygons Simple Transformations of 2-Dimensional Shapes Circles and Curved Shapes Perimeter and Circumference Calculating Area Three-Dimensional Shapes Net Diagrams of 3D Shapes Calculating Volume Area, Surface Area and Volume Reference Sheet Data Analysis: Graphs and Charts Averages (Mean, Median & Mode) Simple Statistical Analysis Statistical Analysis: Types of Data Understanding Correlations Understanding Statistical Distributions Significance and Confidence Intervals Developing and Testing Hypotheses Multivariate Analysis More Advanced Mathematical Concepts: Introduction to Algebra Simultaneous and Quadratic Equations Introduction to Trigonometry Introduction to Probability Set Theory Subscribe to our FREE newsletter and start improving your life in just 5 minutes a day. Subscribe You'll get our 5 free 'One Minute Life Skills' and our weekly newsletter. We'll never share your email address and you can unsubscribe at any time. Introduction to Percentages % See also: Percentage Calculators The term ‘per cent’ means ‘out of a hundred’. In mathematics, percentages are used like fractions and decimals, as ways to describe parts of a whole. When you are using percentages, the whole is considered to be made up of a hundred equal parts. The symbol % is used to show that a number is a percentage, and less commonly the abbreviation ‘pct’ may be used. You will see percentages almost everywhere: in shops, on the internet, in advertisements and in the media. Being able to understand what percentages mean is a key skill that will potentially save you time and money and will also make you more employable. The Meaning of Percentages Percentage is a term from Latin, meaning ‘out of one hundred’. You can therefore consider each ‘whole’ as broken up into 100 equal parts, each one of which is a single percent. The box below shows this for a simple grid, but it works the same way for anything: children in a class, prices, pebbles on the beach, and so on. Visualising Percentages The grid below has 100 cells. Each cell is equal to 1% of the whole (the red cell is 1%). Two cells are equal to 2% (the green cells). Five cells are equal to 5% (the blue cells). Twenty five cells (purple cells) are equal to 25% of the whole or one quarter (¼). Fifty cells (yellow cells) are equal to 50% of the whole or half (½). | | | | | | | | | | | --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | How many unshaded (white) cells are there? What is the percentage of unshaded cells? Answer: There are two ways to work this out. Count the white cells. There are 17 of them. Out of 100 cells, 17% are therefore white. Add up the number of other cells, and take them from 100. There is one red cell, two green, five blue, 25 purple, and 50 yellow. That adds up to 83. 100−83 = 17. Again, out of 100 cells, 17 are white, or 17%. It is easy to work out the percentage when there are 100 individual ‘things’ making up the whole, as in the grid above. But what if there are more or less? The answer is that you convert the individual elements that make up the whole into a percentage. For example, if there had been 200 cells in the grid, each percentage (1%) would be two cells, and every cell would be half a percent. We use percentages to make calculations easier. It is much simpler to work with parts of 100 than thirds, twelfths and so on, especially because quite a lot of fractions do not have an exact (non-recurring) decimal equivalent. Importantly, this also makes it much easier to make comparisons between percentages (which all effectively have the common denominator of 100) than it is between fractions with different denominators. This is partly why so many countries use a metric system of measurement and decimal currency. Finding the Percentage The general rule for finding a given percentage of a given whole is: Work out the value of 1%, then multiply it by the percentage you need to find. This is easiest to understand with an example. Let’s suppose that you want to buy a new laptop computer. You have checked local suppliers and one company has offered to give you 20% off the list price of £500. How much will the laptop cost from that supplier? In this example, the whole is £500, or the cost of the laptop before the discount is applied. The percentage that you need to find is 20%, or the discount offered by the supplier. You are then going to take that off the full price to find out what the laptop will cost you. Start by working out the value of 1% One percent of £500 is £500 ÷ 100 = £5. 2. Multiply it by the percentage you are looking for Once you have worked out the value of 1%, you simply multiply it by the percentage you are looking for, in this case 20%. £5 × 20 = £100. You now know that the discount is worth £100. 3. Complete the calculation by adding or subtracting as necessary. The price of the laptop, including the discount, is £500−20%, or £500−£100 = £400. The easy way to work out 1% of any number 1% is the whole (whatever that may be) divided by 100. When we divide something by 100, we simply move the place values two columns to the right (or move the decimal point two places to the left). You can find out more about numbers and place values on our Numbers page, but here’s a quick recap: £500 is made up of 5 hundreds, zero tens and zero units. £500 also has zero pence (cents if you are working in dollars) so could be written as £500.00, with zero tenths or hundredths. | | | | | | | --- --- --- | | Hundreds | Tens | Units | Point | Tenths | Hundredths | | 5 | 0 | 0 | . | 0 | 0 | When we divide by 100, we move our number two columns to the right. 500 divided by 100 = 005, or 5. Leading zeros (zeros on the ‘outside left’ of a number, such as those in 005, 02, 00014) have no value, so we do not need to write them. You can also think of this as moving the decimal point two places to the left. | | | | | | | --- --- --- | | Hundreds | Tens | Units | Point | Tenths | Hundredths | | 0 | 0 | 5 | . | 0 | 0 | This rule applies to all numbers, so £327 divided by 100 is £3.27. This is the same as saying that £3.27 is 1% of £327. £1 divided by 100 = £0.01, or one pence. There are one hundred pence in a pound (and one hundred cents in a dollar). 1p is therefore 1% of £1. Once you have calculated 1% of the whole, you can then multiply your answer to the percentage you are looking for (see our page on multiplication for help). Mental Maths Hacks As your maths skills develop, you can begin to see other ways of arriving at the same answer. The laptop example above is quite straightforward and with practise, you can use your mental maths skills to think about this problem in a different way to make it easier. In this case, you are trying to find 20%, so instead of finding 1% and then multiplying it by 20, you can find 10% and then simply double it. We know that 10% is the same as 1/10th and we can divide a number by 10 by moving the decimal place one place to left (removing a zero from 500). Therefore 10% of £500 is £50 and 20% is £100. A useful mental maths hack is that percentages are reversible, so 16% of 25 is the same as 25% of 16. Invariably, one of those will be much easier to work out in our head…try it! Use our Percentage Calculators to quickly solve your percentage problems. Working with Percentages We calculated a 20% discount in the example above and then subtracted this from the whole to work out how much a new laptop would cost. As well as taking a percentage away, we can also add a percentage to a number. It works exactly the same way, but in the final step, you simply add instead of subtracting. For example: George is promoted and gets a 5% pay rise. George currently earns £24,000 a year, so how much will he earn after his pay rise? Work out 1% of the whole The whole in this example is George's current salary, £24,000. 1% of £24,000 is 24,000 ÷ 100 = £240. 2. Multiply that by the percentage you are looking for George is getting a 5% pay rise, so we need to know the value of 5%, or 5 times 1%. £240 × 5 = £1,200. 3. Complete the calculation by adding to the original amount George’s pay rise is £1,200 per year. His new salary will therefore be £24,000 + £1,200 = £25,200. Percentages over 100% It is possible to have percentages over 100%. This example is one: George’s new salary is actually 105% of his old one. However, his old salary is not 100% of his new one. Instead, it is just over 95%. When you are calculating percentages, the key is to check that you are working with the correct whole. In this case, the ‘whole’ is George’s old salary. Percentages as Decimals and Fractions One percent is one hundredth of a whole. It can therefore be written as both a decimal and a fraction. To write a percentage as a decimal, simply divide it by 100. For example, 50% becomes 0.5, 20% becomes 0.2, 1% becomes 0.01 and so on. We can calculate percentages using this knowledge. 50% is the same as a half, so 50% of 10 is 5, because five is half of 10 (10 ÷ 2). The decimal of 50% is 0.5. So another way of finding 50% of 10 is to say 10 × 0.5, or 10 halves. 20% of 50 is the same as saying 50 × 0.2, which equals 10. 17.5% of 380 = 380 × 0.175, which equals 66.5. George’s salary increase above was 5% of £24,000. £24,000 × 0.05 = £1,200. The conversion from decimal to percentage is simply the reverse calculation: multiply your decimal by 100. 0.5 = 50%0.875 = 87.5% To write a percentage as a fraction, put the percentage value over a denominator of 100, and divide it down into its lowest possible form. 50% = 50/100 = 5/10 = ½20% = 20/100 = 2/10 = 1/530% = 30/100 = 3/10 WARNING! It is possible to convert fractions to percentages by converting the denominator (the bottom number of the fraction) into 100. However, it is harder to convert fractions to percentages than percentages to fractions because not every fraction has an exact (non-recurring) decimal or percentage. If the denominator of your fraction does not divide a whole number of times into 100, then there will not be a simple conversion. For example, 1/3, 1/6 and 1/9 do not make ‘neat’ percentages (they are 33.33333%, 16.66666% and 11.11111%). Working out Percentages of a Whole So far we have looked at the basics of percentages, and how to add or subtract a percentage from a whole. Sometimes it is useful to be able to work out the percentages of a whole when you are given the numbers concerned. For example, let’s suppose that an organisation employs 9 managers, 12 administrators, 5 accountants, 3 human resource professionals, 7 cleaners and 4 catering staff. What percentage of each type of staff does it employ? Start by working out the whole. In this case, you do not know the ‘whole’, or the total number of staff in the organisation. The first step is therefore to add together the different types of staff. 9 managers + 12 administrators + 5 accountants + 3 HR professionals + 7 cleaners + 4 catering staff = 40 members of staff. 2. Work out the proportion (or fraction) of staff in each category. We know the number of staff in each category, but we need to convert that to a fraction of the whole, expressed as a decimal. The calculation we need to do is: Staff in Category ÷ Whole (See our division page for help with division sums or use a calculator) We can use managers as an example: 9 managers ÷ 40 = 0.225 In this case it can be helpful if, instead of thinking of the division symbol ‘÷’ as meaning ‘divided by’, we can substitute the words ‘out of’. We use this often in the context of test results, for example 8/10 or ‘8 out of 10’ correct answers. So we calculate the ‘number of managers out of the whole staff’. When we use words to describe the calculation, it can help it to make more sense. 3. Convert the fraction of the whole into a percentage 0.225 is the fraction of staff that are managers, expressed as a decimal. To convert this number to a percentage, we need to multiply it by 100. Multiplying by 100 is the same as dividing by a hundred except you move the numbers the other way on the place values scale. So 0.225 becomes 22.5. In other words, 22.5% of the organisation’s employees are managers. We then do the same two calculations for each other category. 12 administrators ÷ 40 = 0.3. 0.3 × 100 = 30%. 5 accountants ÷ 40 = 0.125. 0.125 × 100 = 12.5%. 3 HR professionals ÷ 40 = 0.075. 0.075 × 100 = 7.5%. 7 cleaners ÷ 40 = 0.175. 0.175 × 100 = 17.5%. 4 catering staff ÷ 40 = 0.1. 0.1 × 100 = 10%. TOP TIP! Check you have a total of 100% When you have finished calculating your percentages, it is a good idea to add them together to make sure that they equal 100%. If they don't, then check your calculations. In summary, we can say that the organisation is made up of: | | | | --- | Roles | Number of Staff | % of Staff | | Managers | 9 | 22.5% | | Administrators | 12 | 30% | | Accountants | 5 | 12.5% | | HR professionals | 3 | 7.5% | | Cleaners | 7 | 17.5% | | Catering staff | 4 | 10% | | Total | 40 | 100% | It can be useful to show percentage data representing a whole on a pie chart. You can quickly see the proportions of categories of staff in the example. For more on pie charts and other types of graphs and charts see our page: Graphs and Charts. Points to remember Percentages are a way to describe parts of a whole. They are a bit like decimals, except that the whole is always split into 100, instead of tenths, hundredths, thousandths and so on of a unit. Percentages are designed to make calculations easier. Further Reading from Skills You Need Proportion Part of The Skills You Need Guide to Numeracy This eBook covers proportion looking at numbers as parts of other numbers, as parts of a larger whole, or in relation to other numbers. The book covers fractions and decimals, ratio and percentages with worked examples for you to try and develop your skills. Whether you want to brush up on your basics, or help your children with their learning, this is the book for you. Continue to:Percentage Calculators Percentage Change - Increase and Decrease See also:Averages (Mean, Median and Mode) Estimation, Approximation and Rounding Introduction to Geometry TOP
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https://www.thoughtco.com/acids-and-bases-titration-curves-603656
Titration Curves of Acids and Bases Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Science› Chemistry› Titration Curves of Acids and Bases Print Nicola Tree/Digital Vision/Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. Learn about ourEditorial Process Updated on June 10, 2025 Close Key Takeaways Titration curves show how the pH of a solution changes as we add a strong base. Strong acids reach neutrality at pH 7 while weak acids become more basic after neutralization. Polyprotic acids, like sulfuric acid, have multiple spikes on their titration curves for each acidic hydrogen. Titration is a technique used in analytical chemistry to determine the concentration of an unknown acid or base. Titration involves the slow addition of one solution where the concentration is known to a known volume of another solution where the concentration is unknown until the reaction reaches the desired level. For acid/base titrations, a color change from a pH indicator is reached or a direct reading using apH meter. This information can be used to calculate the concentration of the unknown solution. If the pH of an acid solution is plotted against the amount of base added during a titration, the shape of the graph is called a titration curve. All acid titration curves follow the same basic shapes. In the beginning, the solution has a low pH and climbs as the strong base is added. As the solution nears the point where all of theH+are neutralized, the pH rises sharply and then levels out again as the solution becomes more basic as more OH-ions are added. Strong Acid Titration Curve Strong Acid Titration Curve. ThoughtCo / Todd Helmenstine The first curve shows a strong acid being titrated by a strong base. There is the initial slow rise in pH until the reaction nears the point where just enough base is added to neutralize all the initial acid. This point is called the equivalence point. For a strong acid/base reaction, this occurs at pH = 7. As the solution passes the equivalence point, the pH slows its increase where the solution approaches the pH of the titration solution. Weak Acids and Strong Bases Weak Acid Titration Curve. ThoughtCo / Todd Helmenstine A weak acid only partially dissociates from its salt. The pH will rise normally at first, but as it reaches a zone where the solution seems to be buffered, the slope levels out. After this zone, the pH rises sharply through its equivalence point and levels out again like the strong acid/strong base reaction. There are two main points to notice about this curve. The first is the half-equivalence point. This point occurs halfway through a buffered region where the pH barely changes for a lot of ​base added. The half-equivalence point is when just enough base is added for half of the acid to be converted to the conjugate base. When this happens, the concentration of H+ ions equals the K a value of the acid. Take this one step further, pH = pK a. The second point is the higher equivalence point. Once the acid has been neutralized, notice the point is above pH=7. When a weak acid is neutralized, the solution that remains is basic because of the acid's conjugate base remains in solution. Polyprotic Acids and Strong Bases Diprotic Acid Titration Curve. ThoughtCo / Todd Helmenstine The third graph results from acids that have more than one H+ ion to give up. These acids are called polyprotic acids. For example, sulfuric acid (H 2 SO 4) is a diprotic acid. It has two H+ ions it can give up. The first ion will break off in water by the dissociation H2SO4→ H++ HSO4- The second H+ comes from the dissociation of HSO 4- by HSO4-→ H++ SO42- This is essentially titrating two acids at once. The curve shows the same trend as a weak acid titration where the pH does not change for a while, spikes up and levels off again. The difference occurs when the second acid reaction is taking place. The same curve happens again where a slow change in pH is followed by a spike and leveling off. Each 'hump' has its own half-equivalence point. The first hump's point occurs when just enough base is added to the solution to convert half the H+ ions from the first dissociation to its conjugate base, or it's K a value. The second hump's half-equivalence point occurs at the point where half the secondary acid is converted to the secondary conjugate base or that acid's K a value. On many tables of K a for acids, these will be listed as K 1 and K 2. Other tables will list only the K a for each acid in the dissociation. This graph illustrates a diprotic acid. For an acid with more hydrogen ions to donate [e.g., citric acid (H 3 C 6 H 5 O 7) with 3 hydrogen ions] the graph will have a third hump with a half-equivalence point at pH = pK 3. Cite this Article Format mlaapachicago Your Citation Helmenstine, Todd. "Titration Curves of Acids and Bases." ThoughtCo, Jun. 10, 2025, thoughtco.com/acids-and-bases-titration-curves-603656.Helmenstine, Todd. (2025, June 10). Titration Curves of Acids and Bases. Retrieved from Helmenstine, Todd. "Titration Curves of Acids and Bases." ThoughtCo. (accessed September 29, 2025). copy citation Sponsored Stories Buy the Dip: Top 5 Dividend Stocks with Growth Potential Seeking Alpha Should You Buy the Dip? 5 AI Stocks With Long-Term Potential Seeking Alpha Worst Zip Codes for Car Insurance in South Carolina (Is Yours on the List?)ottoinsurance.com Buy the Dip: 5 AI Stocks With Strong Growth Potential SeekingAlpha.com Titration Basics Strong Acids and the World's Strongest Acid Acid-Base Titration Calculation Acids and Bases: Titration Example Problem Overview of High School Chemistry Topics What Is Titration? Chemistry 101 - Introduction & Index of Topics What Are Acids and Bases? 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https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-8
Published Time: 2016-01-28T15:19-05:00 NIST Guide to the SI, Chapter 8 | NIST Skip to main content An official website of the United States government Here’s how you know Here’s how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. 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When time relates to calendar cycles, the minute (min), hour (h), and day (d) might be necessary. For example, the kilometer per hour (km/h) is the usual unit for expressing vehicular speeds. Although there is no universally accepted symbol for the year, Ref.[4:ISO 80000-3]suggests the symbol a. The rotational frequency n of a rotating body is defined to be the number of revolutions it makes in a time interval divided by that time interval [4: ISO 80000-3]. The SI unit of this quantity is thus the reciprocal second (s-1). However, as pointed out in Ref. [4: ISO 80000-3], the designations "revolutions per second" (r/s) and "revolutions per minute" (r/min) are widely used as units for rotational frequency in specifications on rotating machinery. 8.2 Volume The SI unit of volume is the cubic meter (m 3) and may be used to express the volume of any substance, whether solid, liquid, or gas. The liter (L) is a special name for the cubic decimeter (dm 3), but the CGPM recommends that the liter not be used to give the results of high accuracy measurements of volumes [1, 2]. Also, it is not common practice to use the liter to express the volumes of solids nor to use multiples of the liter such as the kiloliter (kL) [see Sec.6.2.8, and also Table 6, footnote (b)]. 8.3 Weight In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame [4: ISO 80000-4]. Thus the SI unit of the quantity weight defined in this way is the newton (N). When the reference frame is a celestial object, Earth for example, the weight of a body is commonly called the local force of gravity on the body. Example: The local force of gravity on a copper sphere of mass 10 kg located on the surface of the Earth, which is its weight at that location, is approximately 98 N. Note: The local force of gravity on a body, that is, its weight, consists of the resultant of all the gravitational forces acting on the body and the local centrifugal force due to the rotation of the celestial object. The effect of atmospheric buoyancy is usually excluded, and thus the weight of a body is generally the local force of gravity on the body in vacuum. In commercial and everyday use, and especially in common parlance, weight is usually used as a synonym for mass. Thus the SI unit of the quantity weight used in this sense is the kilogram (kg) and the verb "to weigh" means "to determine the mass of" or "to have a mass of." Examples: the child's weight is 23 kg the briefcase weighs 6 kg Net wt. 227 g Inasmuch as NIST is a scientific and technical organization, the word "weight" used in the everyday sense (that is, to mean mass) should appear only occasionally in NIST publications; the word "mass" should be used instead. In any case, in order to avoid confusion, whenever the word "weight" is used, it should be made clear which meaning is intended. 8.4 Relative atomic mass and relative molecular mass The terms atomic weight and molecular weight are obsolete and thus should be avoided. They have been replaced by the equivalent but preferred terms relative atomic mass, symbol A r, and relative molecular mass, symbol M r, respectively [4:ISO 31-8], which better reflect their definitions. Similar to atomic weight and molecular weight, relative atomic mass and relative molecular mass are quantities of dimension one and are expressed simply as numbers. The definitions of these quantities are as follows [4:ISO 31-8]: Relative atomic mass (formerly atomic weight): ratio of the average mass per atom of an element to 1/12 of the mass of the atom of the nuclide 12 C. Relative molecular mass (formerly molecular weight): ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide 12 C. Examples: A r(Si) = 28.0855, M r(H 2) = 2.0159, A r(12 C) = 12 exactly Notes: It follows from these definitions that if X denotes a specified atom or nuclide and B a specified molecule or entity (or more generally, a specified substance), then Ar(X) = m(X) / [m(12C) / 12] and Mr(B) = m(B) / [m(12C) / 12], where m(X) is the mass of X, m(B) is the mass of B, and m(12C) is the mass of an atom of the nuclide 12C. It should also be recognized that m(12C) / 12 = u, the unified atomic mass unit, which is approximately equal to 1.66 3 10-27 kg [see Table 7, footnote (d)]. It follows from the examples and note 1 that the respective average masses of Si, H 2, and 12 C are m(Si) = A r(Si) u, m(H2) = M r(H 2) u, and m(12 C) = A r(12 C)u. In publications dealing with mass spectrometry, one often encounters statements such as "the mass-to-charge ratio is 15." What is usually meant in this case is that the ratio of the nucleon number (that is, mass number—see Sec. 10.4.2) of the ion to its number of charges is 15. Thus mass-to-charge ratio is a quantity of dimension one, even though it is commonly denoted by the symbol m / z. For example, the mass-to-charge ratio of the ion 12 C 7 1 H 7+ + is 91/2 = 45.5. 8.5 Temperature interval and temperature difference As discussed in Sec. 4.2.1.1, Celsius temperature (t) is defined in terms of thermodynamic temperature (T) by the equation t = T - T 0, where T 0 = 273.15 K by definition. This implies that the numerical value of a given temperature interval or temperature difference whose value is expressed in the unit degree Celsius (°C) is equal to the numerical value of the same interval or difference when its value is expressed in the unit kelvin (K); or in the notation of Sec.7.1, note 2, {Δ t }°C = {ΔT}K. Thus temperature intervals or temperature differences may be expressed in either the degree Celsius or the kelvin using the same numerical value. Example: The difference in temperature between the freezing point of gallium and the triple point of water is Δ t = 29.7546 °C = Δ T = 29.7546 K. 8.6 Amount of substance, concentration, molality, and the like The following section discusses amount of substance, and the subsequent nine sections, which are based on Ref. [4: ISO 31-8] and which are succinctly summarized in Table 12, discuss quantities that are quotients involving amount of substance, volume, or mass. In the table and its associated sections, symbols for substances are shown as subscripts, for example, x B, n B, b B. However, it is generally preferable to place symbols for substances and their states in parentheses immediately after the quantity symbol, for example n(H 2 SO 4). (For a detailed discussion of the use of the SI in physical chemistry, see the book cited in Ref.,note 3.) 8.6.1 Amount of substance Quantity symbol: n (also v).SI unit: mole (mol). Definition: See Sec. A.7. Notes: Amount of substance is one of the seven base quantities upon which the SI is founded (see Sec. 4.1 and Table 1). In general, n(x B) = n(B) / x, where x is a number. Thus, for example, if the amount of substance of H 2 SO 4 is 5 mol, the amount of substance of (1/3)H 2 SO 4 is 15 mol: n[(1/3)H 2 SO 4] = 3 n(H 2 SO 4). Example: The relative atomic mass of a fluorine atom is A r(F) = 18.9984. The relative molecular mass of a fluorine molecule may therefore be taken as M r(F 2) = 2 A r(F) = 37.9968. The molar mass of F 2 is then M(F 2) = 37.9968 × 10-3 kg/mol = 37.9968 g/mol (see Sec. 8.6.4). The amount of substance of, for example, 100 g of F 2 is then n(F 2) = 100 g / (37.9968 g/mol) = 2.63 mol. 8.6.2 Mole fraction of B; amount-of-substance fraction of B Quantity symbol: x B (also y B). SI unit: one (1) (amount-of-substance fraction is a quantity of dimension one). Definition: ratio of the amount of substance of B to the amount of substance of the mixture: x B = n B/n. Table 12. Summary description of nine quantities that are quotients involving amount of substance, volume, or mass | | | Quantity in numerator | --- | Amount of substance Symbol: n SI unit: mol | Volume Symbol: V SI unit: m 3 | Mass Symbol: m SI unit: kg | | Quantity in denominator | Amount of substance Symbol: n SI unit: mol | amount-of-substance fraction x B=n B n x B=n B n SI unit: mol/mol = 1 | molar volume V m=V n V m=V n SI unit: m 3/mol | molar mass M=m n M=m n SI unit: kg/mol | | Volume Symbol: V SI unit: m 3 | amount-of-substance concentration c B=n B V c B=n B V SI unit: mol/m 3 | volume fraction φ B=x B V∗m,B Σ x A V∗m,A φ B=x B V m,B∗Σ x A V m,A∗ SI unit: m 3/m 3 = 1 | mass density ρ=m V ρ=m V SI unit: kg/m 3 | | Mass Symbol: m SI unit: kg | molality b B=n B m A b B=n B m A SI unit: mol/kg | specific volume v=V m v=V m SI unit: m 3/kg | mass fraction w B=m B m w B=m B m SI unit: kg/kg = 1 | | | Adapted from Canadian Metric Practice Guide (see Ref., note 3; the book cited in Ref., note 5, may also be consulted). | Scroll Notes: This quantity is commonly called "mole fraction of B" but this Guide prefers the name "amount of- substance fraction of B," because it does not contain the name of the unit mole (compare kilogram fraction to mass fraction). For a mixture composed of substances A, B, C, . . . , n A + n B + n C + ... ≡∑A n A≡∑A n A A related quantity is amount-of-substance ratio of B (commonly called "mole ratio of solute B"), symbol rB. It is the ratio of the amount of substance of B to the amount of substance of the solvent substance: r B = n B/n S. For a single solute C in a solvent substance (a one-solute solution), r C = x C/(1 - x C). This follows from the relations n = n C + n S, x C = n C / n, and r C = n C / n S, where the solvent substance S can itself be a mixture. 8.6.3 Molar volume Quantity symbol: V m. SI unit: cubic meter per mole (m 3/mol). Definition: volume of a substance divided by its amount of substance: V m = V/n. Notes: The word "molar" means "divided by amount of substance." For a mixture, this term is often called "mean molar volume." The amagat should not be used to express molar volumes or reciprocal molar volumes. (One amagat is the molar volume V m of a real gas at p = 101 325 Pa and T = 273.15 K and is approximately equal to 22.4 × 10-3 m 3/mol. The name "amagat" is also given to 1/V m of a real gas at p = 101 325 Pa and T = 273.15 K and in this case is approximately equal to 44.6 mol/m 3.) solvent substance S can itself be a mixture. 8.6.4 Molar mass Quantity symbol:M. SI unit: kilogram per mole (kg/mol). Definition: mass of a substance divided by its amount of substance: M = m/n. Notes: For a mixture, this term is often called "mean molar mass." The molar mass of a substance B of definite chemical composition is given by M(B) = M r(B) × 10-3 kg/mol = M r(B) kg/kmol = M r g/mol, where M r(B) is the relative molecular mass of B (see Sec. 8.4). The molar mass of an atom or nuclide X is M(X) = A r(X) × 10-3 kg/mol = A r(X) kg/kmol = A r(X) g/mol, where A r(X) is the relative atomic mass of X (seeSec.8.4). 8.6.5 Concentration of B; amount-of-substance concentration of B Quantity symbol: c B. SI unit: mole per cubic meter (mol/m 3). Definition: amount of substance of B divided by the volume of the mixture: c B = n B/V. Notes: This Guide prefers the name "amount-of-substance concentration of B" for this quantity because it is unambiguous. However, in practice, it is often shortened to amount concentration of B, or even simply to concentration of B. Unfortunately, this last form can cause confusion because there are several different "concentrations," for example, mass concentration of B, ρB = m B/V; and molecular concentration of B, C B = N B/V, where N B is the number of molecules of B. The term normality and the symbol N should no longer be used because they are obsolete. One should avoid writing, for example, "a 0.5 N solution of H 2 SO 4" and write instead "a solution having an amount-of-substance concentration of c [(1/2)H 2 SO 4]) = 0.5 mol/dm 3" (or 0.5 kmol/m 3 or 0.5 mol/L since 1 mol/dm 3 = 1 kmol/m 3 = 1 mol/L). The term molarity and the symbol M should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of B and such units as mol/dm 3, kmol/m 3, or mol/L. (A solution of, for example, 0.1 mol/dm 3 was often called a 0.1 molar solution, denoted 0.1 M solution. The molarity of the solution was said to be 0.1 M.) 8.6.6 Volume fraction of B Quantity symbol: φ B. SI unit: one (1) (volume fraction is a quantity of dimension one). Definition: for a mixture of substances A, B, C, . . . , φ B=x B V∗m,B/∑x A V∗m,A φ B=x B V m,B∗/∑x A V m,A∗ where x A, x B, x C, . . . are the amount-of-substance fractions of A, B, C, . . ., Vm,A , Vm,B , Vm,C , . . . are the molar volumes of the pure substances A, B, C, . . . at the same temperature and pressure, and where the summation is over all the substances A, B, C, . . . so that Σx A = 1. 8.6.7 Mass density; density Quantity symbol: ρ. SI unit: kilogram per cubic meter (kg/m 3). Definition: mass of a substance divided by its volume: ρ = m / V. Notes: This Guide prefers the name "mass density" for this quantity because there are several different "densities," for example, number density of particles, n = N / V; and charge density, ρ=Q/V. Mass density is the reciprocal of specific volume (see Sec. 8.6.9): ρ = 1 / ν. 8.6.8 Molality of solute B Quantity symbol: b B (also m B). SI unit: mole per kilogram (mol/kg). Definition: amount of substance of solute B in a solution divided by the mass of the solvent: b B = n B / m A. Note: The term molal and the symbol m should no longer be used because they are obsolete. One should use instead the term molality of solute B and the unit mol/kg or an appropriate decimal multiple or submultiple of this unit. (A solution having, for example, a molality of 1 mol/kg was often called a 1 molal solution, written 1 m solution.) 8.6.9 Specific volume Quantity symbol: ν.SI unit: cubic meter per kilogram (m 3/kg). Definition: volume of a substance divided by its mass: ν = V / m. Note: Specific volume is the reciprocal of mass density (see Sec. 8.6.7): ν = 1 / ρ. 8.6.10 Mass fraction of B Quantity symbol: w B.SI unit: one (1) (mass fraction is a quantity of dimension one). Definition: mass of substance B divided by the mass of the mixture: w B B = m B / m. 8.7 Logarithmic quantities and units: level, neper, bel This section briefly introduces logarithmic quantities and units. It is based on Ref. [5: IEC 60027-3], which should be consulted for further details. Two of the most common logarithmic quantities are level-of a-field-quantity, symbol L F, and level-of-a-power-quantity, symbol L P; and two of the most common logarithmic units are the units in which the values of these quantities are expressed: the neper, symbol Np, or the bel, symbol B, and decimal multiples and submultiples of the neper and bel formed by attaching SI prefixes to them, such as the millineper, symbol mNp (1 mNp = 0.001 Np), and the decibel, symbol dB (1 dB = 0.1 B). Level-of-a-field-quantity is defined by the relation L F = ln(F/F 0), where F/F 0 is the ratio of two amplitudes of the same kind, F 0 being a reference amplitude. Level-of-a-power-quantity is defined by the relation L P = (1/2) ln(P/P 0), where P/P 0 is the ratio of two powers, P 0 being a reference power. (Note that if P/P 0 = (F/F 0)2, then L P = L F.) Similar names, symbols, and definitions apply to levels based on other quantities which are linear or quadratic functions of the amplitudes, respectively. In practice, the name of the field quantity forms the name of L F and the symbol F is replaced by the symbol of the field quantity. For example, if the field quantity in question is electric field strength, symbol E, the name of the quantity is "level-of-electric-field-strength" and it is defined by the relation L E = ln(E/E 0). The difference between two levels-of-a-field-quantity (called "field-level difference") having the same reference amplitude F 0 is Δ L F = L F1 - L F2 = ln(F 1/F 0) - ln(F 2/F 0) = ln(F 1/F 2), and is independent of F 0. This is also the case for the difference between two levels-of-a-power-quantity (called "power-level difference") having the same reference power P 0: Δ L P1 = L P2 = ln(P 1/P 0) - ln(P 2/P 0) = ln(P 1/P 2). It is clear from their definitions that both L F and L P are quantities of dimension one and thus have as their units the unit one, symbol 1. However, in this case, which recalls the case of plane angle and the radian (and solid angle and the steradian), it is convenient to give the unit one the special name "neper" or "bel" and to define these so-called dimensionless units as follows: One neper (1 Np) is the level-of-a-field-quantity when F/F 0 = e, that is, when ln(F/F 0) = 1. Equivalently, 1 Np is the level-of-a-power-quantity when P/P 0 = e 2, that is, when (1/2) ln(P/P 0) = 1. These definitions imply that the numerical value of L F when L F is expressed in the unit neper is {L F}Np = ln(F/F 0), and that the numerical value of L P when L P is expressed in the unit neper is {L P}Np = (1/2) ln(P/P 0); that is L F = ln(F/F 0) Np L P = (1/2) ln(P/P 0) Np. One bel (1 B) is the level-of-a-field-quantity when F/F 0=10−−√F/F 0=10 that is, when 2 lg(F/F 0) = 1 (note that lg x = log 10 x – see Sec. 10.1.2). Equivalently, 1 B is the level- of-a-power-quantity when P/P 0 = 10, that is, when lg(P/P 0) = 1. These definitions imply that the numerical value of LF when LF is expressed in the unit bel is {L F}B = 2 lg(F/F 0) and that the numerical value of LP when LP is expressed in the unit bel is {L P}B = lg(P/P 0); that is L F = 2 lg(F/F 0) B = 20 lg(F/F 0) dB L P = lg(P/P 0) B = 10 lg(P/P 0) dB. Since the value of L F (or L P) is independent of the unit used to express that value, one may equate LF in the above expressions to obtain ln(F/F 0) Np = 2 lg(F/F 0) B, which implies 1 B 1 d B=≈≈ln 10 2 N p e x a c t l y 1.151 293 N p 0.115 129 3 N p.1 B=ln⁡10 2 N p e x a c t l y≈1.151 293 N p 1 d B≈0.115 129 3 N p. When reporting values of L F and L P, one must always give the reference level. According to Ref. 5:IEC 60027-3, this may be done in one of two ways: L x (re x ref) or L x / x ref where x is the quantity symbol for the quantity whose level is being reported, for example, electric field strength E or sound pressure p, and x ref is the value of the reference quantity, for example, 1 μV/m for E 0, and 20 μPa for p 0. Thus L E (re 1 μV/m) = - 0.58 Np or L E/(1 μV/m) = - 0.58 Np means that the level of a certain electric field strength is 0.58 Np below the reference electric field strength E 0 = 1 μV/m. Similarly L p (re 20 μPa) = 25 dB or L p/(20 μPa) = 25 dB means that the level of a certain sound pressure is 25 dB above the reference pressure p 0 = 20 μPa. Notes: When such data are presented in a table or in a figure, the following condensed notation may be used instead: - 0.58 Np (1 μV/m); 25 dB (20 μPa). When the same reference level applies repeatedly in a given context, it may be omitted if its value is clearly stated initially and if its planned omission is pointed out. The rules of Ref. [5: IEC 60027-3] preclude, for example, the use of the symbol dBm to indicate a reference level of power of 1 mW. This restriction is based on the rule of Sec. 7.4, which does not permit attachments to unit symbols. 8.8 Viscosity The proper SI units for expressing values of viscosity η (also called dynamic viscosity) and values of kinematic viscosity ν are, respectively, the pascal second (Pa·s) and the meter squared per second (m 2/s) (and their decimal multiples and submultiples as appropriate). The CGS units commonly used to express values of these quantities, the poise (P) and the stoke (St), respectively [and their decimal submultiples the centipoise (cP) and the centistoke (cSt)], are not to be used; see Sec. 5.3.1 and Table 10, which gives the relations 1 P = 0.1 Pa·s and 1 St = 10-4 m 2/s. 8.9 Massic, volumic, areic, lineic Reference [4: ISO 31-0] has introduced the new adjectives "massic," "volumic," "areic," and "lineic" into the English language based on their French counterparts: "massique," "volumique," "surfacique," and "linéique." They are convenient and NIST authors may wish to use them. They are equivalent, respectively, to "specific," "density," "surface . . . density," and "linear . . . density," as explained below. (a) The adjective massic, or the adjective specific, is used to modify the name of a quantity to indicate the quotient of that quantity and its associated mass. Examples: massic volume or specific volume: ν = V / m massic entropy or specific entropy: s = S / m (b) The adjective volumic is used to modify the name of a quantity, or the term density is added to it, to indicate the quotient of that quantity and its associated volume. Examples: volumic mass or (mass) density: ρ = m / V volumic number or number density: n = N / V Note: Parentheses around a word means that the word is often omitted. (c) The adjective areic is used to modify the name of a quantity, or the terms surface . . . density are added to it, to indicate the quotient of that quantity (a scalar) and its associated surface area. Examples: areic mass or surface (mass) density: ρ A = m / A areic charge or surface charge density: σ = Q / A (d) The adjective lineic is used to modify the name of a quantity, or the terms linear . . . density are added to it, to indicate the quotient of that quantity and its associated length. Examples: lineic mass or linear (mass) density: ρ l = m / l lineic electric current or linear electric current density: A = I / b Metrology Created January 28, 2016, Updated August 18, 2025 ‹ NIST Guide to the SI, Chapter 7: Rules and Style Conventions for Expressing Values of Quantities NIST Guide to the SI, Chapter 9: Rules and Style Conventions for Spelling Unit Names › Was this page helpful? HEADQUARTERS 100 Bureau Drive Gaithersburg, MD 20899 301-975-2000 Webmaster | Contact Us | Our Other Offices X.comFacebookLinkedInInstagramYouTubeGiphyRSS FeedMailing List How are we doing? Feedback Site Privacy Accessibility Privacy Program Copyrights Vulnerability Disclosure No Fear Act Policy FOIA Environmental Policy Scientific Integrity Information Quality Standards Commerce.gov Science.gov USA.gov Vote.gov Back to top
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https://www.iso.org/standard/74965.html
ISO 13287:2019 - Personal protective equipment — Footwear — Test method for slip resistance Skip to main content Applications OBP English español français русский Menu Standards SectorsHealth IT & related technologies Management & services Security, safety & risk Transport Energy Diversity & inclusion Environmental sustainability Food & agriculture Materials Building & construction Engineering About ISO Insights & newsInsightsAll insights Healthcare Artificial intelligence Climate change Transport Cybersecurity Quality management Renewable energy Occupational health and safety NewsExpert talk Standards world Media kit Taking part Store Search Cart Reference number ISO 13287:2019 © ISO 2025 International Standard ISO 13287:2019 Personal protective equipment — Footwear — Test method for slip resistance Edition 3 2019-10 Read sample ISO 13287:2019 74965 ISO 13287:2019 Personal protective equipment — Footwear — Test method for slip resistance Published (Edition 3, 2019) ISO 13287:2019 ISO 13287:2019 74965 Language Format PDF + ePub Paper PDF + ePub Paper CHF 132 Add to cart Shipping costs not included Convert Swiss francs (CHF) to your currency Abstract This document specifies a method of test for the slip resistance of PPE footwear. It is not applicable to special purpose footwear containing spikes, metal studs or similar. Footwear claiming ?slip resistance' would be deemed an item of personal protective equipment. NOTE For product development purposes, sole units, outsoles or other soling components such as top pieces may be tested. General information Status :Published Publication date :2019-10 Stage : Close of review [90.60] Edition :3 Number of pages :23 Technical Committee: ISO/TC 94/SC 3 ICS: 13.340.50 RSSupdates Life cycle Previously Withdrawn ISO 13287:2012 Now Published ISO 13287:2019 A standard is reviewed every 5 years Stage: 90.60 (Under review) 00 Preliminary 10 Proposal 10.99 2017-11-08 New project approved 20 Preparatory) 30 Committee) 30.99 2018-05-23 CD approved for registration as DIS 40 Enquiry) 40.00 2018-05-25 DIS registered 40.20 2018-07-27 DIS ballot initiated: 12 weeks 40.60 2018-10-20 Close of voting 40.99 2019-05-24 Full report circulated: DIS approved for registration as FDIS 50 Approval) 50.00 2019-05-24 Final text received or FDIS registered for formal approval 50.20 2019-07-09 Proof sent to secretariat or FDIS ballot initiated: 8 weeks 50.60 2019-09-03 Close of voting. Proof returned by secretariat 60 Publication) 60.00 2019-09-04 International Standard under publication 60.60 2019-10-01 International Standard published 90 Review) 90.20 2024-10-15 International Standard under systematic review 90.60 2025-03-05 Close of review 90.20 International Standard under systematic review 90.60 Close of review 90.92 International Standard to be revised 90.93 International Standard confirmed 90.99 Withdrawal of International Standard proposed by TC or SC 95 Withdrawal) 95.99 Withdrawal of International Standard Amendments Provide additional content; available for purchase; not included in the text of the existing standard. Under development ISO 13287:2019/CD Amd 1 This standard contributes to the following Sustainable Development Goals 3 Good Health and Well-being8 Decent Work and Economic Growth Got a question? Check out our Help and Support Store Store ICS 13 13.340 13.340.50 ISO 13287:2019 Sitemap Standards Benefits Popular standards Conformity assessment SDGs Sectors Health IT & related technologies Management & services Security, safety & risk Transport Energy Environmental sustainability Materials About ISO What we do Structure Members Events Strategy Insights & news Insights All insights Healthcare Artificial intelligence Climate change Transport News Expert talk Standards world Media kit Taking part Who develops standards Deliverables Get involved Collaborating to accelerate effective climate action Resources Drafting standards Store Store Publications and products ISO name and logo Privacy Notice Copyright Cookie policy Media kit Jobs Help and support Making liveseasier,saferandbetter. Sign up for email updates © All Rights Reserved All ISO publications and materials are protected by copyright and are subject to the user’s acceptance of ISO’s conditions of copyright. 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https://www.youtube.com/watch?v=lJX8DxoPRfk
Pi bonds and sp2 hybridized orbitals | Structure and bonding | Organic chemistry | Khan Academy Khan Academy 9090000 subscribers 3637 likes Description 983314 views Posted: 24 Jul 2010 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Pi bonds and sp2 Hybridized Orbitals. Created by Sal Khan. Watch the next lesson: Missed the previous lesson? Organic Chemistry on Khan Academy: Carbon can form covalent bonds with itself and other elements to create a mind-boggling array of structures. In organic chemistry, we will learn about the reactions chemists use to synthesize crazy carbon based structures, as well as the analytical methods to characterize them. We will also think about how those reactions are occurring on a molecular level with reaction mechanisms. Simply put, organic chemistry is like building with molecular Legos. Let's make some beautiful organic molecules! About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Organic Chemistry channel: Subscribe to Khan Academy: 306 comments Transcript: In the last video, I touched on the idea of a sigma bond. And that was a bond-- well, let me draw two nucleuses and let me just draw one of the orbitals. Let's say this is an sp3 hybridized orbital, and that's on this atom and this is kind of this big lobe right there. And then this guy has an sp3 hybridized orbital as well. That's the small lobe, and then that's the big lobe like that. A sigma bond is one where there's an overlap kind of in the direction in which the lobes are pointed. And you might say, well, how can there be any other type of bond than that? Well, the other type of bond, so this right here-- let me make this clear. This right here is a sigma bond. And you say, well, what other kind of bond could there be where my two orbitals overlap kind of in the direction that they're pointing? And the other type of bond you could have, you can imagine if you have two p orbitals. So let me draw the nucleus of two atoms, and I'll just draw one of each of their p orbitals. So let's say that that's the nucleus and I'll just draw their p orbitals. So a p orbital is just that dumbbell shape. Let me draw them a little bit closer together. So a p orbital is that dumbbell shape. So let me draw this guy's-- one of his p orbitals. I want to draw it a little bit bigger than that, and you'll see why a second. So one of his p orbitals right there. It comes out like that. And then this guy over here also has a p orbital that is parallel to this p orbital, so it goes like that. Let me draw that other one a little bit straighter. It goes-- I want it to overlap more, so it goes like that. I think you get the idea. So here, our two p orbitals are parallel to each other. This, you can imagine, these are sp3 hybridized orbitals. They're pointing at each other. Here, they're parallel. p orbitals are parallel to each other, and you see that they overlap on this kind of top lobe here and in this bottom lobe here. And this is a pi bond. Let me make this clear. And this is one pi bond. So you could call it a pi, literally, with the Greek letter pi: pi bond. Sometimes you'll see this just written as pi bond. And it's called a pi bond because it's the Greek letter for essentially p, and we're dealing with p orbitals overlapping. Now sigma bonds, which are what form when you have a single bond, these are stronger than pi bonds; pi bonds come into play once you start forming double or triple bonds on top of a sigma bond. To kind of get a better visualization of how that might work, let's think about ethene. So it's molecular structure looks like this. So you have C double-bonded to C, and then each of those guys have two hydrogens. So let me draw what it would look like, or our best visual, or our best ability to kind of conceptualize what the orbitals around the carbon might look like. So first I'll draw the sp2 hybridized orbitals. So let me just make it very clear what's going on here. So when we were dealing with methane, which is literally just a carbon bonded to four hydrogens, and if I actually wanted to draw it in a way that it kind of looks a little three-dimensional with a tetrahedral structure, it might look like this. This hydrogen is pointing out a little bit. This hydrogen is kind of in the plane of the page, and then maybe that hydrogen is behind it, and then you have one hydrogen popping up. That's methane. And we saw that these were all sp3 hybridized orbitals around the carbon, and then they each formed sigma bonds with each of the hydrogens. We saw that in the last video. And when we drew its electron configuration, in order for this to happen, carbon's electron configuration when bonding in methane needed to look like this. It needed to look like 1s2. And then instead of having 2s2 and then 2p2, what you essentially have is-- let me try it this way, actually, even better. Let me write this better. In 1s, you had two electrons, and then instead of two s's, you had two electrons and on each of the p's, you had one, the s's and the p's all got mixed up and you had a 2sp3 hybridized orbital, another 2sp3 hybridized orbital, another 2sp3 hybridized orbital, and then another one, sp3. Normally, when carbon's sitting by itself, you would expect a 2s here, and then you'd have a 2p in the x-direction, a 2p in the y-direction, and then a 2p in the z-direction. But we saw in the last video, they all get mixed up and they all have a 25% s-character, and a 75% p-character when carbon bonds in methane and the electrons kind of separate out in that situation. When you're dealing with the carbons in ethene, remember, eth- is for two carbons and ene-, because we're dealing with an alkene. We have a double bond here. In this situation, the carbon's electron configuration when they bond in ethene looks more like this. So you have your 1s, and the 1s orbital is still completely full. It has two electrons in it. But then in your 2 shell, I'll just write-- let me do this in a different color. So in our 2 shell, I'll show you what I mean in a second. I'm not writing the s or p's so far on purpose, but we're going to have four electrons just like we had before. We're still forming four bonds. We're going to have these four unpaired electrons. We're still forming one, two, three, four bonds with each of the carbons, so they're going to be separated out. But in this situation, instead of all of them being a mixture, kind of one part s, three parts p, the s mixes with two of the p orbitals. So what you have is 2sp2 orbital. So you can imagine that the s orbital mixes with two of the p orbitals. So now it's one part s, two parts p. And then one of the p orbitals kind of stays by itself. And we need this p orbital to stay by itself because it is going to be what's responsible for the pi bond. And we're going to see that the pi bond does something very interesting to the molecule. It kind of makes it unrotatable around a bond axis. And you'll see what I mean in a second. So let me see if I can, in three dimensions, draw each of these carbons. So you have-- let me do it a different color. You have this carbon right there. So let's say that's the nucleus. I'll put a C there so you know which carbon we're dealing with. And then I'll draw-- you could assume that the 1s orbital, it's really small right around the carbon. And then you have these hybridized orbitals, The 2sp2 orbitals, and they're all going to be planar, kind of forming a triangle, or I guess maybe a peace sign on some level, but I'll try to draw it in three dimensions here. So you have one, this is kind of coming out a little bit. Then you have one that's going in a little bit. And then you have-- and they have another lobe a little bit on the other side, but I'm not going to draw them. It'll complicate it. They still have characteristics of p, so they'll have two lobes, but one is bigger than the other. And then you have one that's maybe going in this side. So you can imagine that this is kind of a Mercedes sign if you drew a circle around it, on its side. So that's this carbon right here. And, of course, it has its hydrogens. So you have this hydrogen there. And so this hydrogen might be sitting right here. It just has one electron in its 1s orbital. You have this hydrogen up here. It's sitting right over there. And now let's draw this carbon. This carbon will be sitting-- I'm drawing it pretty close together. This carbon will be sitting right there. He has his 1s orbital. They have the exact same electron configuration. He has his 1s orbital right around him, and then he has the exact same configuration. Either of these guys, we've so far only-- or in this first guy, I've only drawn these first three. I haven't drawn this unhybridized p orbital yet. So I'll do that in a second. But let me draw his bonds. So first of all, he has this, or you could imagine, that bond right there, which would be an sp2 hybridized bond. Let me do that in the same color. So he has this bond right here, which would be an sp2 hybridized bond, just like that. And notice, this is a sigma bond. They overlap in kind of the direction that they're pointing in. That's the best I could think about it. And then he's got these two hydrogens, so one-- he's got this guy in the back, and then there's one in the front. I'll draw it a little bigger so it's kind of pointing out at us, right? And then we have this hydrogen is sitting right over here. And these are also sigma bonds, just to be very clear about things. This is an s orbital overlapping with an sp2 orbital, but they're kind of overlapping in the direction that they're pointed, or kind of along the direction of each other, of the two atoms. This is a sigma bond, sigma bond, and then we have this hydrogen in the back, which is also going to form a sigma bond. So everything I've drawn so far is a sigma bond, so that, that. Maybe I don't want to make this picture too-- so I can just put sigma bond there, sigma bond there, sigma bond there, sigma, sigma. So far I've drawn this bond, this bond, this bond, this bond, and this bond, all of those sigma bonds. So, what happens to this last p orbital for each of these guys? Well, that's going to be kind of sticking out of the plane of the Mercedes sign, is the best way I can describe it. And let me see if I can do that in a color that I haven't done yet. Oh, maybe this purple color. So you can imagine a pure p orbital. So a pure p orbital, I'm going to need to draw it even bigger than that, actually. A pure p orbital, it normally wouldn't be that big relative to things, but I have to make them overlap. So it's a pure p orbital that's kind of going in, maybe you can imagine, the z-axis, that the other orbitals are kind of a Mercedes sign in the x, y plane. And now you have the z-axis going straight up and down, and those bottom two have to overlap so let me draw them bigger. So it looks like that and it looks like that. And they're going straight up and down. And notice, they are now overlapping. So this bond right here is this bond. I could've drawn them in either way, but it's that second bond. And so what's happening now to the structure? So let me make it very clear. This right here, that is a pi bond, and this right here is also-- it's the same pi bond. It's this guy right here. It's the second bond in the double bond. But what's happening here? Well, first of all, by itself it would be a weaker bond, but because we already have a sigma bond that's making these molecules come closer together, this pi bond will make them come even closer together. So this distance right here is closer than if we were to just have a single sigma bond there. Now, on top of that, the really interesting thing is, if we just had a sigma bond here, both of these molecules could kind of rotate around the bond axis. They would be able to rotate around the bond axis if you just had one sigma bond there. But since we have these pi bonds that are parallel to each other and they're kind of overlapping and they're kind of locked in to that configuration, you can no longer rotate. If one of these molecules rotates, the other one's going to rotate with it because these two guys are locked together. So what this pi bond does in the situation is it makes this carbon-carbon double bond-- it means that the double bonds are going to be rigid, that you can't have one molecule kind of flipping, swapping these two hydrogens, without the other one having to flip with it. So you wouldn't be able to kind of swap configurations of the hydrogens relative to the other side. That's what it causes. So, hopefully, that gives you a good understanding of the difference between sigma and pi bond. And if you're curious, when you're dealing with-- just to kind of make it clear, if we were dealing with ethyne, this is an example of ethene, but ethyne looks like this. You have a triple bond. And so you have each side pointing to one hydrogen. In this case, one of these, so the first bonds, you can imagine, so these bonds are all sigma bonds. They're actually sp hybridized. Your 2s orbital only mixes with one of the p's, so these are sp hybrid orbitals forming sigma bonds, so all of these right here. And then both of these-- let me do this in different color. Both of these are pi bonds. And if you had to imagine it, could imagine another pi bond kind of coming out of the page and another one here coming out of the page and into the page, out and into the page, and they, too, are overlapping, and you just have one hydrogen pointing out in each direction. Maybe I'll make another video on that. So, hopefully, you--
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https://www.youtube.com/watch?v=4V-zBkQCqWA
Finding Endpoint Given Midpoint Mario's Math Tutoring 454000 subscribers 5932 likes Description 503135 views Posted: 17 Oct 2016 Learn how to find the endpoint given the midpoint and only 1 of the other endpoints in this free math video tutorial by Mario's Math Tutoring. Timestamps: 00:00 Intro 0:10 Example 1 Given Endpoint A(1,-4), Midpoint M(3,2) Solve For Endpoint B(x,y) 0:26 Midpoint Formula 1:00 Writing 2 Equations to Solve for the Endpoint Coordinates Organized List of My Video Lessons to Help You Raise Your Scores & Pass Your Class. Videos Arranged by Math Subject as well as by Chapter/Topic. (Bookmark the Link Below) ➡️JOIN the channel as a CHANNEL MEMBER at the "ADDITIONAL VIDEOS" level to get access to my math video courses(Algebra 1, Algebra 2/College Algebra, Geometry, and PreCalculus), midterm & final exam reviews, ACT and SAT prep videos and more! (Over 390+ videos) 532 comments Transcript: Intro so here they're telling us that a is one negative for the midpoint is three two and we're trying to find the other end point B okay since we don't know what the coordinates of me are let's just Example 1 Given Endpoint A(1,-4), Midpoint M(3,2) Solve For Endpoint B(x,y) call this X comma Y okay so you're with me so far so the thing that we want to remember here is the midpoint formula is X 1 plus X 2 divided by 2 and y 1 plus y 2 divided by 2 so when you think of the midpoint you can think of an average Midpoint Formula like if you get an 80% on a test and 100% on a test and you add those together divided by 2 that's like your average that gives you a sense of the middle right so that's a 90% and we're doing the same thing here we're averaging the x coordinates of the endpoints so we add them up and divide by 2 and that gives us the x coordinate of the midpoint same thing with the Y's we add the Y's divided by 2 that gives us the y coordinate of midpoint so let's write two equations we've got 1 plus ax okay which we don't know divided by 2 should give us the x coordinate of the midpoint same thing with the Y's negative 4 plus y divided by 2 equals Writing 2 Equations to Solve for the Endpoint Coordinates the y coordinate the midpoint which is 2 now what I'm going to do here is I'm just going to clear the denominators I'm gonna multiply the left side by 2 so that these cancel and the right side by 2 same thing over here I'm going to multiply the left side by 2 and the right side by 2 so if we simplify we get 1 plus x equals 6 and negative 4 plus y equals 4 if I subtract 1 from both sides you can see that x equals 5 and if I add 4 to both sides because I'm doing the opposite operation right y equals 8 we can see that this end point here is going to be 5 comma 8 now if you're not convinced you can always test it out 5 plus 1 is 6 divided by 2 is 3 so that works out 8 plus negative 4 is 4 divided by 2 is 2 so you can you know verify after you get the solution real quick like that but the key is to use the midpoint formula and use the end point that you don't know put those as a variables x and y so I hope this helped you understand how to work with this type of problem better subscribe to my channel mario's math tutor and youtube channel and check out some other videos there and i look forward to helping in the future ones i'll talk to you soon
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https://www.rainbowresource.com/016808.html?srsltid=AfmBOopS_-qwlh9g4nc3E-KtgY1CtyfQJLHyseiRcXsQNYE2tfjnN8e6
The Art of Problem Solving: Introduction to Number Theory Set Press Option+1 for screen-reader mode, Option+0 to cancelAccessibility Screen-Reader Guide, Feedback, and Issue Reporting | New window The store will not work correctly in the case when cookies are disabled. Your company account is blocked and you cannot place orders. If you have questions, please contact your company administrator. FREE SHIPPING ON ORDER S OVER $50LEARN MORE Excludes Purchase Orders. U.S. addresses only. Other exclusions may apply. Shop Products Show Search Form Search Sign In My Wish Lists Create A Wish List Many of our customers plan their children’s curriculum by creating a list for each child. Go ahead, try it! My Cart My Cart 0 CloseYou have no items in your shopping cart. 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In Stock 1 Question, 3 Answersor Be the first to write a review Our Price $55.00 Qty -+ Add to Cart Add to Wish List Skip to the end of the images gallery Skip to the beginning of the images gallery Description Description Includes the Student Text and Solutions Manual for the Intro to Number Theory Level. Publisher's Description of The Art of Problem Solving: Introduction to Number Theory Set A thorough introduction for students in grades 7-10 to topics in number theory such as primes & composites, multiples & divisors, prime factorization and its uses, base numbers, modular arithmetic, divisibility rules, linear congruences, how to develop number sense, and more. Learn the fundamentals of number theory from former MATHCOUNTS, AHSME, and AIME perfect scorer Mathew Crawford. Topics covered in the book include primes & composites, multiples & divisors, prime factorization and its uses, base numbers, modular arithmetic, divisibility rules, linear congruences, how to develop number sense, and much more. The text is structured to inspire the reader to explore and develop new ideas. Each section starts with problems, so the student has a chance to solve them without help before proceeding. The text then includes motivated solutions to these problems, through which concepts and curriculum of number theory are taught. Important facts and powerful problem solving approaches are highlighted throughout the text. In addition to the instructional material, the book contains hundreds of problems. The solutions manual contains full solutions to nearly every problem, not just the answers. This book is ideal for students who have mastered basic algebra, such as solving linear equations. Middle school students preparing for MATHCOUNTS, high school students preparing for the AMC, and other students seeking to master the fundamentals of number theory will find this book an instrumental part of their mathematics libraries. Paperback. Text: 336 pages. Solutions: 144 pages. Show More Category Description for The Art of Problem Solving Introduction Series (Gr. 6-10) This is an outstanding math program for the math-gifted student. It is rigorous and oriented to the independent problem-solver. The texts are based on the premise that students learn math best by solving problems - lots of problems - and preferably difficult problems that they don't already know how to solve. Most sections, therefore, begin by presenting problems and letting students intuit solutions BEFORE explaining ways to solve them. Even if they find ways to answer the problems, they should read the rest of the section to see if their answer is correct and if theirs is the best or most efficient way to solve that type of problem. Textual instruction, then, is given in the context of these problems, explaining how to best approach and solve them. Throughout the text there are also special, blue-shaded boxes highlighting key concepts, important things to retain (like formulas), warnings for potential problem-solving pitfalls, side notes, and bogus solutions (these demonstrate misapplications). There are exercises at the end of most sections to see if the student can apply what's been learned. Review problems at the end of each chapter test understanding for that chapter. If a student has trouble with these, he should go back and re-read the chapter. Each chapter ends with a set of Challenge Problems that go beyond the learned material. Successful completion of these sets demonstrates a high degree of mastery. A unique feature in this series is the hints section at the back of the book. These are intended to give a little help to selected problems, usually the very difficult ones (marked with stars). In this way, students can get a little push in the right direction, but still have to figure out the solution for themselves. The solution manuals do contain complete solutions and explanations to all the exercises, review problems and challenge problems. It is best for students not to access these until they have made several attempts to solve the problems first. I particularly like one of the motivating boxes in the text that coaches, "If at first you don't know how to solve a problem, don't just stare at it. Experiment!". That pretty much sums up the philosophy of the course, encouraging children to take chances, become aggressive problem solvers, and attack problems with confidence. I wonder how far some children would go if they were encouraged this way instead of being spoon fed? Though this course is used in classroom settings, the texts are student-directed, making them perfect for the independent learner or homeschooler. Students should start the introductory sequence with the Prealgebrabook. Afterwards, begin the Introduction to Algebra. Students will be prepared for both the Introduction to Countingand Probability and Introduction to Number Theory courses after completing the first 11 chapters of Algebra. It won't matter whether they do these along with Algebra, put aside Algebra and complete the other two or finish Algebra first and then do them. All of them should be completed prior to the Introduction to Geometry book. If you are coming into this course from another curriculum, you will probably want to take a placement test to decide where to enter this program. Even if your student has finished Algebra 2 elsewhere, you will want to make sure that all of the material from this series has been covered before continuing on to the Intermediate series. Taken together, these constitute a complete curriculum for outstanding math students in grades 6-10 and one that prepares them for competitions such as MATHCOUNTS and the American Mathematics Competitions. The material is challenging and in-depth; this is not a course for the mathematically faint of heart. If your child loves math, is genuinely math-gifted, or is interested in participating in math competitions, you definitely need to give this one serious consideration. Show More Details Details More Information| Product Format: | Other | | Grades: | 6-10 | | Brand: | Art of Problem Solving | Videos Videos This product doesn't have a video Reviews 0 Reviews No Ratings Be the first to review this item Write a Review Q&A Product Q&A Have a question? Ask owners.Have a question about this? Ask people who own it. Start typing and see existing answers.Learn more Instant Answers Start typing and we'll see if it was already asked and answered. 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https://www.superteacherworksheets.com/multiplying-decimals.html
Log In Become a Member Membership Info Addition (Basic) Addition (Multi-Digit) Algebra & Pre-Algebra Comparing Numbers Daily Math Review (Math Buzz) Division (Basic) Division (Long Division) Hundreds Charts Multiplication (Basic) Multiplication (Multi-Digit) Order of Operations Place Value Skip Counting Telling Time Word Problems (Daily) More Math Worksheets Reading Comprehension Reading Comprehension Pre-K/K Reading Comprehension Grade 1 Reading Comprehension Grade 2 Reading Comprehension Grade 3 Reading Comprehension Grade 4 Reading Comprehension Grade 5 Reading Comprehension Grade 6 Reading Comprehension Reading & Writing Reading Worksheets Cause & Effect Daily ELA Review (ELA Buzz) Fact & Opinion Fix the Sentences Graphic Organizers Synonyms & Antonyms Writing Prompts Writing Story Pictures Writing Worksheets More ELA Worksheets Phonics Consonant Sounds Vowel Sounds Consonant Blends Consonant Digraphs Word Families More Phonics Worksheets Early Literacy Alphabet Build Sentences Sight Word Units Sight Words (Individual) More Early Literacy Grammar Nouns Verbs Adjectives Adverbs Pronouns Punctuation Syllables Subjects and Predicates More Grammar Worksheets Spelling Lists Spelling Grade 1 Spelling Grade 2 Spelling Grade 3 Spelling Grade 4 Spelling Grade 5 Spelling Grade 6 More Spelling Worksheets Chapter Books Bunnicula Charlotte's Web Magic Tree House #1 Boxcar Children More Literacy Units Science Animal (Vertebrate) Groups Butterfly Life Cycle Electricity Human Body Matter (Solid, Liquid, Gas) Simple Machines Space - Solar System Weather More Science Worksheets Social Studies 50 States Explorers Landforms Maps (Geography) Maps (Map Skills) More Social Studies Art & Music Colors Worksheets Coloring Pages Learn to Draw Music Worksheets More Art & Music Holidays Spring Summer Back to School Labor Day More Holiday Worksheets Puzzles & Brain Teasers Brain Teasers Logic: Addition Squares Mystery Graph Pictures Number Detective Lost in the USA More Thinking Puzzles Teacher Helpers Teaching Tools Award Certificates More Teacher Helpers Pre-K and Kindergarten Alphabet (ABCs) Numbers and Counting Shapes (Basic) More Kindergarten Worksheet Generator Word Search Generator Multiple Choice Generator Fill-in-the-Blanks Generator More Generator Tools S.T.W. Full Website Index Help Files Contact Multiplying Decimals Use these worksheets to practice multiplying decimal factors together. Multiplying Decimals By 1-Digit Numbers Multiplying Decimals by 1-Digit Numbers This worksheet has 10 vertical problems and 2 word problems that students can solve to practice multiplying decimals by single digit numbers. 4th through 7th Grades View PDF Word Problems: Multiplying Decimals This worksheet has word problems to solve multiplying decimals by single digit numbers. 4th through 6th Grades View PDF Shape Multiplication (Tenths) Find products by multiplying the numbers inside the shapes. 4th through 6th Grades View PDF Word Problems: Multiplying & Dividing Decimals by 1-Digit Numbers Solve each word problem by multiplying decimals to the tenths and hundredths place by single digit numbers. 4th through 6th Grades View PDF Multiplying Decimals FREE Multiply each decimal by a one-digit number. (example: 18.7 x 6) This worksheet requires students to rewrite each problem vertically to solve. 4th through 6th Grades View PDF Task Cards: Multiplying Decimals by 1-Digit Numbers Using this set of task cards, students will practice multiplying decimals to tenths and hundredths by single digit numbers. 4th through 6th Grades View PDF Task Cards: Multiplying Decimals by 2-Digit Numbers Multiply each two digit number by decimals to the tenths and hundredths place. 4th through 6th Grades View PDF Task Cards: Multiplying Decimals by 10, 100, and 1,000 Use these task cards to practice multiplying decimals by 10, 100 or 1,000. 5th and 6th Grades View PDF Find the Mistake: Multiplying Decimals by 1-Digit The problems on this page were not solved correctly. Find the mistakes and explain how solve the problems correctly. 4th through 6th Grades View PDF Multiplying Decimals By 2-Digit Numbers Multiplying Decimals by 2-Digit Numbers This page has ten vertical problems and one word problem. (example: 76.4 x 2.5) 5th through 7th Grades View PDF Word Problems: Multiplying Decimals by 2-Digit Numbers Students will solve decimal by 2-digit number multiplication word problems on this worksheet. 4th through 7th Grades View PDF Word Problems: Multiplying & Dividing Decimals by 2-Digit Numbers Decide whether you need to multiply or divide to solve these decimal word problems. 4th through 6th Grades View PDF Multiplying Decimals By 1 and 2-Digit Numbers Multiply decimals by 1 and 2-digit numbers. (examples: 3.4 x 4 and 1.35 x 5.5) Includes 8 vertical problems and 2 word problems. 5th and 6th Grades View PDF Shape Multiplication (Tenths & Hundredths) Multiply the decimals inside the shapes to find the product. 4th through 6th Grades View PDF Find the Mistake: Multiplying Decimals by 2-Digit Numbers These multiplication problems were not solved correctly. Find the mistakes, solve them correctly, and explain your thinking. 4th through 6th Grades View PDF 2-Digit Decimal Multiplication (Horizontal Problems) Multiply decimals by 2-Digit numbers. (example. 82 x 0.65) This worksheet requires students to rewrite problems vertically. 5th and 6th Grades View PDF Determine Weight on Other Planets Students will use decimal multiplication to calculate their weight on all eight planets in the solar system. 4th through 7th Grades View PDF Decimal Multiplication Patterns Decimal Multiplication Patterns Multiply each decimal by 10; 100; 1000; 10,000; etc. (examples: 3.4x100, 5.671x10,000) 5th and 6th Grades View PDF Shape Multiplication (Decimal Patterns) Find the products by multiplying decimals by 10, 100 and 1,000. 4th through 6th Grades View PDF Decimal Multiplication Patterns (Missing Numbers) Students will find the missing factor or product for each decimal multiplication problem. 4th through 6th Grades View PDF See Also... Dividing Decimals This page has lots of worksheets for teaching decimal division. Decimal Addition & Subtraction Add and subtract pairs of decimal numbers. Sample Worksheet Images My Account Become a Member Membership Information Site License Information Login Site Information Help / FAQ Full Site Index Privacy Policy Terms of Service Contact Us Useful Links What's New? Social Media Follow Us Not a Member? For complete access to thousands of printable lessons click the button or the link below. Become a Member © 2025 Super Teacher Worksheets
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https://askfilo.com/user-question-answers-smart-solutions/how-is-20-2-10-2-3134323335303135
Question asked by Filo student How is 20/√2=10/√2 Views: 5,641 students Updated on: Nov 21, 2024 Text SolutionText solutionverified iconVerified Concepts: Rationalizing the denominator, Simplifying fractions Explanation: To show that 2​20​=2​10​, we can simplify the left side. First, we can divide the numerator and denominator by 2. Step by Step Solution: Step 1 Start with the left side: 2​20​. Step 2 Divide the numerator by 2: 2​20÷2​=2​10​. Step 3 Now we see that both sides are equal: 2​20​=2​10​. Final Answer: Yes, 2​20​=2​10​ is true. Students who ask this question also asked Views: 5,307 Topic: Smart Solutions View solution Views: 5,711 Topic: Smart Solutions View solution Views: 5,064 Topic: Smart Solutions View solution Views: 5,810 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. | | | --- | | Question Text | How is 20/√2=10/√2 | | Updated On | Nov 21, 2024 | | Topic | All topics | | Subject | Smart Solutions | | Class | Class 9 | | Answer Type | Text solution:1 | Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
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https://www.khanacademy.org/science/physics/centripetal-force-and-gravitation/centripetal-acceleration-tutoria/v/race-cars-with-constant-speed-around-curve
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https://www.youtube.com/watch?v=549eLWIu0Xk
The Fundamental Counting Principle AlRichards314 20900 subscribers 2648 likes Description 245726 views Posted: 22 Nov 2011 This video shows how the Fundamental Counting Principle can be used to solve counting problems. This video was created for the Data Management (MDM4U) course in the province of Ontario, Canada. 231 comments Transcript: The Fundamental Counting Principle Example 1: The school cafeteria offers a choice of two main courses grilled cheese sandwiches or soup of the day and five desserts (jello, pudding, fruit cups, sundaes, or granola bars). How many in this video we'll take a look at how you use the fundamental counting principle in order to solve counting problems and in this first example we have a school cafeteria that offers uh a choice of two different main courses the person could either have a grilled cheese sandwich or the soup of the day and five possible desserts they could have Jello-O or a pudding or a fruit cup or a sundae or a granola bar and the question is how many different lunches could you have so I'm going to use some abbreviations to represent the main courses in desserts so we don't have to write out the words every time so the capital G will represent grilled cheese sandwich and the capital S soup of the day the desserts I'll use a lowercase letter J for Jell-O P for pudding f for fruit cup s for sunde and B for granola bar and so we could just list all the possibilities for example having a grilled cheese sandwich and a Jello-O would look like this having a grilled cheese and a pudding would look like this grilled cheese and a fruit cup grilled cheese and a sundae grilled cheese use and a granola bar and then we could do soup of the day with each of them and so those are all the lunches that start with soup of the day and so if you count them you notice that there are 10 so there are 10 possible lunches now the way the fundamental counting principle works is in order to create a lunch here there are two actions that have to be uh performed there's two ways to select the main course and for each of those two ways there's five ways to select the dessert that's why I've actually got this arranged in two rows of five so there's two ways to select the main course either grilled cheese or soup of the day and there's five ways to select the dessert so if the first action perform in two ways and for each of those the second action to complete the lunch can be performed in five ways then there should be two times five or 10 possible lunches so this is the fundamental accounting principle and I'll summarize it on the the next page now this can also be solved using a tree diagram and tree diagrams can either go across the page like this one is or you can start at a uh some point on the page and work down there's two different ways to draw them so this is the tree diagram this point right here represents the fact that you have two choices to make first of all for the uh the main course and so the main course could be this um Branch right here represents choosing a grilled cheese this Branch down here represents choosing a soup of the day so let's say we've chosen the grilled cheese there are five branches here because there's five possible selections for the dessert and so that's the Jello-O the pudding the fruit cup the sunde or the granola bar and even if you've selected soup of the day you still have of course the same choices for the dessert so this point right here would represent the person selected grilled cheese and a Jello and then this one here represent grilled cheese and a pudding so GP grilled cheese and a a fruit cup on down to the bottom the last one here would be selecting Su day and a granola bar which is the SB so these are the uh same thing same possibilities we listed here so the tree diagram is just a different way to organize it and the nice thing about a tree diagram is that it's a very organized way of doing it so you don't miss possibilities so that's the nice thing about a tree diagram so the fundamental counting The Fundamental Counting Principle: If an action can be done in m ways, and for each way, a second action can be done inn ways, then the two actions can be done in mn ways. The Fundamental Counting Principle can be extended to cover actions performed in more than two ways. principle says this if an action can be performed in M ways like for example selecting the uh lunch's main course in the previous page in two ways and for each way a second action can be done in N ways for example picking the dessert in five different ways then the two actions can be done in M n ways so the this is all part of the same thing like there's two parts here of selecting the first part and the second part so we would multiply the uh M by n to get how many ways this this whole thing can be accomplished now the fundamental accounting principle can be extended to cover actions that can be performed in more than two ways so for example let's say you purchasing uh an automobile perhaps there's a certain number of exterior colors times there's a certain number of interior colors times there's perhaps you know four different engines and uh three different Transmissions or whatever so you could actually figure out how many possible automobiles you could pick um for any particular model of course that way so it doesn't have to be just something can be done in two different ways in example two John is colorblind John is color blind. He has 5 different pairs of pants, 8 different shirts and the significance of that first sentence is simply that uh he could literally um make his outfits here uh to not particularly match so that's why the first sentence is there and we're told that he has five different pairs of pants eight different shirts nine pairs of socks we'll assume that the socks are all paired so there's not 18 socks he could choose uh so nine pairs of socks vary in color and then three pairs of shoes and we're the question is how many different outfits can John show up to school in if he has to wear a pair of pants a pair of socks a shirt and a pair of shoes and so so this is uh the fundamental accounting principle and there are four actions here there's five ways for him to select his pants and for each of those there's eight different ways to select his shirt so five time 8 and for each of that there's nine different pairs of socks he could put on and for each of those there's also three different pairs of shoes he could put on so we would multiply five by 8 by 9 by 3 and get 1,080 so there's 1,080 outfits he could possibly wear uh Assuming he's color blind and he's not going to make it look really nice he might have really really poor colors going together okay so there's 1 1080 possible outfits that John could wear example number three we have a A license plate is to have three letters followed by three numbers How many license plates are possible? license plate and license plates in this example are three letters followed by three numbers and the question is how many license plates are possible so there's no restrictions here we'll get into restrictions on the next page um actually sorry not the next page uh restrictions would be for example that uh all the letters have to be different or all the numbers have to be different that's an example of restriction there are no restrictions here so in uh there are 26 letters in the alphabet so this letter could be selected in 20 26 ways this one 26 ways this one 26 ways if there was a restriction where they had to be all different we could say this would be selected in 26 ways and since we've already used one particular letter then that would be 25 and then that would be 24 if they had to be different so there is no restriction here so they're all 26s there's um 10 different digits 0 to 9 for each of the numbers and the reason that we would multiply these together is because for each of these 26 ways to select that there's 26 ways to select this one and for each of those there's 26 ways to select that one and for each of those there's 10 ways to select that number Etc so 26 26 26 is 26 cubed times and this be 10 10 or 10 cubed so there's 17 mil 576,000 possible plates that could be made with no restrictions whatsoever and example number four says Specialty license plates have either 4 letters followed by 2 numbers or 5 letters with one number. How many specialty license plates are possible? specialy licensed plates have either four letters followed by two numbers or five letters follow with one number and the question is how many of these specialty license plates are possible so similar to what's in the last page there are four letters so it would be 26 26 26 26 so 26 the^ 4 times and there's two numbers following it so times 10 10 or 10 squ so there's a little over 45 million of those now so there could be four letters followed by two numbers or five letters followed with one number so the five letters followed with one number there's five letters so 26 the^ of 5times 10 the^ of 1 so there's 118 million 83,7 60 possible plates that have five letters and one number now these are two different categories of license plates or kinds of license plates so we would employ what some people call the sum Rule and add the 45 m697 600 to the 118 Million number and so altogether there's 164,50 11,360 specialty licensed plates that are possible we would add them because there's this many with four letters and two numbers and there's this many with five letters and one number so we would add those two together we would not multiply them CU these are are two different cases of how the license plates can be made in example five a president A president, treasurer and secretary are to be drawn from the people Amelia, Bruce, Carmen, and Deloris. This is the year that the president must be female. How Treasurer and secretary to be drawn from the people Amelia Bruce Carmen and Dolores and we'll use the capital letter at the beginning of each of their names to represent Amelia Bruce Carmen and Dolores uh this is the year that the president must be a female so in this particular School uh Bruce is O luck this year uh so they're only selecting uh female um presidents so they have a some kind of gender rule here so let's say that this selection here is for president and then this selection in the middle here for is for treasurer and uh the next one's the secretary so how many Executives can be chosen so there's only three possibilities for president because Bruce is not allowed so it's Amelia Carmen or Dolores for treasurer next um this point here would represent that Amelia has been chosen for president so she cannot also be Treasurer so we could have uh Bruce Carmen or Dolores for treasurer at this point here we have uh Carmen as president so Amelia Bruce or Dolores could be treasurer and this point here we've chosen Dolores for the president so there's possible Amelia Bruce or Carman for treasurer so at this point right here we've chosen Amelia for the president and Bruce for the treasurer so only two possibilities left for the secretary it could be Carmen or Dolores and we would just keep on going through for each one so for example right down here at the bottom uh We've selected Dolores for president Carmen for treasurer so there was only Amelia and Bruce available for secretary and so we would list all the possibilities so ABC is the top one Amilia president Bruce treasurer kman secretary the next one would be ABD which is Amilia for the president Bruce for the treasurer and Dolores for the secretary and so these are all the possibilities if you of course add them all up you get 18 possible Executives can be chosen to use the fundamental accounting principal it would look like this uh right here for a president there's 's three possibilities for president times there's three possibilities for treasurer because once one person's been chosen from the four there's three left for the treasurer times now once you selected two people there's only uh there's only two people left for the secretary so 3 3 2 also gives you 18 so there's two there's two different ways to solve that the uh the multiplication here of course is the uh fundamental accounting principle How many 3 digit even numbers can be created if a Repetition of digits is allowed in example six it says how many three different even three digigit even numbers can be created if and an a repetition of digits is allowed so that means you could have for example a number uh 334 because that the three has been repeated twice there or 555 using five all the time now if it's a three-digit number then you cannot have a zero here you can only have one to nine because you have a zero here it's not TR a three-digit number uh you could literally have any number from 0 to 9 for the 10's digit and of course if it's an even number you can only have 0 2 4 6 or eight on the end so there's nine ways to select the hundreds digit 0 to 9 is 10 there's 10 ways to select the 10's digit and there's only five ways to select the last number if it's going to be even you have to have a 0 2 4 6 or eight so we would multiply these because if there's nine ways to select this for each of the nine ways to select the hundreds digit there's 10 ways to select the 10's digit and for each of those there's five ways to select the uh last digit the ones digit so 9 10 5 is 450 so there are 450 thre digigit even numbers with repetition of digits allowed so that's using the an example of using the fundamental counting principle now let's say repetition of digits is not allowed so you cannot repeat a digit you're only allowed to use it once so again this has has to be 1 to 9 but of course it has to be different than the other numbers same with this uh this has to be some number from 0 to 9 but again different than the other two and on the end you can only have 0 2 4 6 or eight it has to be even now we're going to consider two possibilities so first of all let's say that there is a zero on the end here so that represents a digit not how many digits you could have here let's say there was a zero on the end here this number here could be anything from 1 to 9 then remember it has to be 0 to 9 but different than this so instead of there being 10 possibilities here there's only nine so the nine represents how many different numbers there could be here now this number already couldn't be zero okay so it has to be something from 1 to n so if that's a zero well this already couldn't be a zero uh but of course this digit has to be different than that so instead of being nine possible ways to to select the hundreds digit there are eight because again that digit has to be different than this one so 8 times 9 1 if you want uh would be 72 so there's 72 ways to make these numbers if it ends in a zero now if it ends in either uh 2 4 6 or eight so there's four possible numbers here we could do them separately but they're all the same if there's a 2 four 6 or eight here this number here has is one of nine numbers so for example uh if this ends in a two well this can't be a two it would have to be um zero 1 3 4 5 6 7 8 9 so there's nine possible ways to select this remember it just has to be different than the two the four the six or the eight and this number here now this number here has to be different than this one and this one so instead of there being nine possibilities here there are seven because again this number has to be different than the one in the 10's digit and the one in the end so we would multiply s by 9 we'd also multiply by four because there are four different digits you could have in the end here uh so that's what the four represents it could be a two the four the six or the eight so four 7 9 is 252 so we add these because this is the way the digit could start with uh or sorry end in a zero this is the way it could end in a two four 6 or eight so we would add those together and use the sum Rule and so we get 324 so our 324 three-digit even numbers with repetition of digits not allowed and that's the end of the lesson
984
https://pubmed.ncbi.nlm.nih.gov/15606501/
Coordinated expression of desmoglein 1 and desmocollin 1 regulates intercellular adhesion - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Coordinated expression of desmoglein 1 and desmocollin 1 regulates intercellular adhesion Spiro Getsios1,Evangeline V Amargo,Rachel L Dusek,Ken Ishii,Linda Sheu,Lisa M Godsel,Kathleen J Green Affiliations Expand Affiliation 1 Department of Pathology and Dermatology, Northwestern University Feinberg School of Medicine, Chicago, IL 60611, USA. PMID: 15606501 DOI: 10.1111/j.1432-0436.2004.07208008.x Item in Clipboard Coordinated expression of desmoglein 1 and desmocollin 1 regulates intercellular adhesion Spiro Getsios et al. Differentiation.2004 Oct. Show details Display options Display options Format Differentiation Actions Search in PubMed Search in NLM Catalog Add to Search . 2004 Oct;72(8):419-33. doi: 10.1111/j.1432-0436.2004.07208008.x. Authors Spiro Getsios1,Evangeline V Amargo,Rachel L Dusek,Ken Ishii,Linda Sheu,Lisa M Godsel,Kathleen J Green Affiliation 1 Department of Pathology and Dermatology, Northwestern University Feinberg School of Medicine, Chicago, IL 60611, USA. PMID: 15606501 DOI: 10.1111/j.1432-0436.2004.07208008.x Item in Clipboard Full text links Cite Display options Display options Format Abstract Desmoglein 1 (Dsg1) is a component of desmosomes present in the upper epidermis and can be targeted by autoimmune antibodies or bacterial toxins, resulting in skin blistering diseases. These defects in tissue integrity are believed to result from compromised desmosomal adhesion; yet, previous attempts to directly test the adhesive roles of desmosomal cadherins using normally non-adherent L cells have yielded mixed results. Here, two complementary approaches were used to better resolve the molecular determinants for Dsg1-mediated adhesion: (1) a tetracycline-inducible system was used to modulate the levels of Dsg1 expressed in L cell lines containing desmocollin 1 (Dsc1) and plakoglobin (PG) and (2) a retroviral gene delivery system was used to introduce Dsg1 into normal human epidermal keratinocytes (NHEK). By increasing Dsg1 expression relative to Dsc1 and PG, we were able to demonstrate that the ratio of Dsg1:Dsc1 is a critical determinant of desmosomal adhesion in fibroblasts. The distribution of Dsg1 was organized at areas of cell-cell contact in the multicellular aggregates that formed in these suspension cultures. Similarly, the introduction of Dsg1 into NHEKs was capable of increasing the aggregation of single cell suspensions and further enhanced the adhesive strength of intact epithelial sheets. Endogenous Dsc1 levels were also increased in NHEKs containing Dsg1, providing further support for the coordination of these two desmosomal cadherins in regulating adhesive structures. These Dsg1-mediated effects on intercellular adhesion were directly related to the presence of an intact extracellular domain as ETA, a toxin that specifically cleaves this desmosomal cadherin, inhibited adhesion in both fibroblasts and keratinocytes. Collectively, these observations demonstrate that Dsg1 promotes the formation of intercellular adhesion complexes and suggest that the relative level of Dsg and Dsc expressed at the cell surface regulates this adhesive process. PubMed Disclaimer Similar articles Analysis of desmosomal cadherin-adhesive function and stoichiometry of desmosomal cadherin-plakoglobin complexes.Kowalczyk AP, Borgwardt JE, Green KJ.Kowalczyk AP, et al.J Invest Dermatol. 1996 Sep;107(3):293-300. doi: 10.1111/1523-1747.ep12363000.J Invest Dermatol. 1996.PMID: 8751959 Review. Assembly of desmosomal cadherins into desmosomes is isoform dependent.Ishii K, Norvell SM, Bannon LJ, Amargo EV, Pascoe LT, Green KJ.Ishii K, et al.J Invest Dermatol. 2001 Jul;117(1):26-35. doi: 10.1046/j.0022-202x.2001.01400.x.J Invest Dermatol. 2001.PMID: 11442746 The expression of desmoglein isoforms in cultured human keratinocytes is regulated by calcium, serum, and protein kinase C.Denning MF, Guy SG, Ellerbroek SM, Norvell SM, Kowalczyk AP, Green KJ.Denning MF, et al.Exp Cell Res. 1998 Feb 25;239(1):50-9. doi: 10.1006/excr.1997.3890.Exp Cell Res. 1998.PMID: 9511724 Antisense expression of a desmocollin gene in MDCK cells alters desmosome plaque assembly but does not affect desmoglein expression.Roberts GA, Burdett ID, Pidsley SC, King IA, Magee AI, Buxton RS.Roberts GA, et al.Eur J Cell Biol. 1998 Jul;76(3):192-203. doi: 10.1016/S0171-9335(98)80034-4.Eur J Cell Biol. 1998.PMID: 9716266 The desmosome and the syndesmos: cell junctions in normal development and in malignancy.Franke WW, Koch PJ, Schäfer S, Heid HW, Troyanovsky SM, Moll I, Moll R.Franke WW, et al.Princess Takamatsu Symp. 1994;24:14-27.Princess Takamatsu Symp. 1994.PMID: 8983060 Review. See all similar articles Cited by The desmosome.Delva E, Tucker DK, Kowalczyk AP.Delva E, et al.Cold Spring Harb Perspect Biol. 2009 Aug;1(2):a002543. doi: 10.1101/cshperspect.a002543.Cold Spring Harb Perspect Biol. 2009.PMID: 20066089 Free PMC article.Review. Structural basis of adhesive binding by desmocollins and desmogleins.Harrison OJ, Brasch J, Lasso G, Katsamba PS, Ahlsen G, Honig B, Shapiro L.Harrison OJ, et al.Proc Natl Acad Sci U S A. 2016 Jun 28;113(26):7160-5. doi: 10.1073/pnas.1606272113. Epub 2016 Jun 13.Proc Natl Acad Sci U S A. 2016.PMID: 27298358 Free PMC article. Significance of anti-desmocollin autoantibodies in pemphigus.Ishii N.Ishii N.J Dermatol. 2023 Feb;50(2):132-139. doi: 10.1111/1346-8138.16660. Epub 2022 Dec 28.J Dermatol. 2023.PMID: 36578135 Free PMC article.Review. Host Invasion by Pathogenic Amoebae: Epithelial Disruption by Parasite Proteins.Betanzos A, Bañuelos C, Orozco E.Betanzos A, et al.Genes (Basel). 2019 Aug 14;10(8):618. doi: 10.3390/genes10080618.Genes (Basel). 2019.PMID: 31416298 Free PMC article.Review. Plakoglobin rescues adhesive defects induced by ectodomain truncation of the desmosomal cadherin desmoglein 1: implications for exfoliative toxin-mediated skin blistering.Simpson CL, Kojima S, Cooper-Whitehair V, Getsios S, Green KJ.Simpson CL, et al.Am J Pathol. 2010 Dec;177(6):2921-37. doi: 10.2353/ajpath.2010.100397. Epub 2010 Nov 12.Am J Pathol. 2010.PMID: 21075858 Free PMC article. See all "Cited by" articles Publication types Research Support, N.I.H., Extramural Actions Search in PubMed Search in MeSH Add to Search Research Support, Non-U.S. Gov't Actions Search in PubMed Search in MeSH Add to Search Research Support, U.S. Gov't, P.H.S. Actions Search in PubMed Search in MeSH Add to Search MeSH terms Animals Actions Search in PubMed Search in MeSH Add to Search Cadherins / analysis Actions Search in PubMed Search in MeSH Add to Search Cadherins / genetics Actions Search in PubMed Search in MeSH Add to Search Cadherins / metabolism Actions Search in PubMed Search in MeSH Add to Search Cell Adhesion / physiology Actions Search in PubMed Search in MeSH Add to Search Cytoskeletal Proteins / metabolism Actions Search in PubMed Search in MeSH Add to Search Desmocollins Actions Search in PubMed Search in MeSH Add to Search Desmoglein 1 Actions Search in PubMed Search in MeSH Add to Search Desmogleins Actions Search in PubMed Search in MeSH Add to Search Desmoplakins Actions Search in PubMed Search in MeSH Add to Search Desmosomes / physiology Actions Search in PubMed Search in MeSH Add to Search Fibroblasts / chemistry Actions Search in PubMed Search in MeSH Add to Search Fibroblasts / drug effects Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Keratinocytes / metabolism Actions Search in PubMed Search in MeSH Add to Search L Cells Actions Search in PubMed Search in MeSH Add to Search Membrane Glycoproteins / genetics Actions Search in PubMed Search in MeSH Add to Search Membrane Glycoproteins / metabolism Actions Search in PubMed Search in MeSH Add to Search Mice Actions Search in PubMed Search in MeSH Add to Search Tetracycline / pharmacology Actions Search in PubMed Search in MeSH Add to Search Up-Regulation Actions Search in PubMed Search in MeSH Add to Search gamma Catenin Actions Search in PubMed Search in MeSH Add to Search Substances Cadherins Actions Search in PubMed Search in MeSH Add to Search Cytoskeletal Proteins Actions Search in PubMed Search in MeSH Add to Search DSC1 protein, human Actions Search in PubMed Search in MeSH Add to Search DSG1 protein, human Actions Search in PubMed Search in MeSH Add to Search Desmocollins Actions Search in PubMed Search in MeSH Add to Search Desmoglein 1 Actions Search in PubMed Search in MeSH Add to Search Desmogleins Actions Search in PubMed Search in MeSH Add to Search Desmoplakins Actions Search in PubMed Search in MeSH Add to Search Dsg1a protein, mouse Actions Search in PubMed Search in MeSH Add to Search Membrane Glycoproteins Actions Search in PubMed Search in MeSH Add to Search gamma Catenin Actions Search in PubMed Search in MeSH Add to Search Tetracycline Actions Search in PubMed Search in MeSH Add to Search Related information Gene Gene (GeneRIF) Protein (RefSeq) PubChem Compound PubChem Compound (MeSH Keyword) PubChem Substance Grants and funding P01 DE12328/DE/NIDCR NIH HHS/United States R01 AR41836/AR/NIAMS NIH HHS/United States T32 CA09560/CA/NCI NIH HHS/United States LinkOut - more resources Full Text Sources Elsevier Science Full text links[x] Elsevier Science [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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985
https://openstax.org/books/university-physics-volume-1/pages/7-summary
Skip to ContentGo to accessibility pageKeyboard shortcuts menu University Physics Volume 1 Summary University Physics Volume 1Summary Search for key terms or text. Summary ## 7.1 Work The infinitesimal increment of work done by a force, acting over an infinitesimal displacement, is the dot product of the force and the displacement. The work done by a force, acting over a finite path, is the integral of the infinitesimal increments of work done along the path. The work done against a force is the negative of the work done by the force. The work done by a normal or frictional contact force must be determined in each particular case. The work done by the force of gravity, on an object near the surface of Earth, depends only on the weight of the object and the difference in height through which it moved. The work done by a spring force, acting from an initial position to a final position, depends only on the spring constant and the squares of those positions. ## 7.2 Kinetic Energy The kinetic energy of a particle is the product of one-half its mass and the square of its speed, for non-relativistic speeds. The kinetic energy of a system is the sum of the kinetic energies of all the particles in the system. Kinetic energy is relative to a frame of reference, is always positive, and is sometimes given special names for different types of motion. ## 7.3 Work-Energy Theorem Because the net force on a particle is equal to its mass times the derivative of its velocity, the integral for the net work done on the particle is equal to the change in the particle’s kinetic energy. This is the work-energy theorem. You can use the work-energy theorem to find certain properties of a system, without having to solve the differential equation for Newton’s second law. ## 7.4 Power Power is the rate of doing work; that is, the derivative of work with respect to time. Alternatively, the work done, during a time interval, is the integral of the power supplied over the time interval. The power delivered by a force, acting on a moving particle, is the dot product of the force and the particle’s velocity. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: William Moebs, Samuel J. Ling, Jeff Sanny Publisher/website: OpenStax Book title: University Physics Volume 1 Publication date: Sep 19, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
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https://www.rama.mahidol.ac.th/graded/sites/default/files/public/High_Diploma/handout_HD/handout_HD701/handout_HD64/Bleeding_Disorder.pdf
B L E E D I N G D I S O R D E R นงนุช สิระชัยนันท์ ภาควิชากุมารเวชศาสตร์ O U T L I N E S Basic knowledge Symptoms and signs Investigation Treatment B A S I C H E M O S T A S I S HEMOSTASIS TF-FVIIa Thrombin Fibrin FVIII FXI FV FIXa Platelet vWF GPIb/IX/V GPIIb/IIIa Platelet Platelet Fibrin Vascular Platelet Coagulation protein 2-Antiplasmin PAI-1 Plasminogen Activators Plasminogen Activation Plasmin Degradation Fibrin Degradation products Fibrin strands network --H E M O S TA S I S : F I B R I N O LY S I S D-dimer B L E E D I N G S Y M P T O M S Asymptomatic Hypermenorrhea Petechiae Epistaxis Ecchymosis and joint bleeding Severe bleeding CNS GI Bleeding after surgical procedure Primary Hemostasis Secondary Hemostasis HEMOSTASIS LEVELS OF CLOTTING FACTORS Deficiency Location gene (Chro.) Clinical features Hemostatic level Half life Afibrinogenemia 4 Severe 0.5 g/L 2-4 d Prothrombin 11 Severe 20-30 3-4 d Factor V 1 Severe 15-20 36 h Factor V and VIII 18 Moderate NA Factor VII 13 Severe 15-20 4-6 h Factor VIII X Severe 30-40 8-12 h Factor IX X Severe 25-30 12-24 h Factor X 13 Severe 15-20 40-60 h Factor XI 4 Moderate/mild 15-30 40-70 h Factor XII 5 No bleeding NA NA Factor XIII 6, 1 Severe Not known 11-14 d BLEEDING AND PLATELET COUNT • Usually occur when platelet count <100,000/cumm • Surgical bleeding platelet <50,000/cumm • Spontaneous bleeding platelet <20,000/cumm Slichter SJ Trans Med Review 2004;18:153-167 All bleed Excluded skin and epistaxis Gross bleeding H I S T O R Y O F S U S P E C T E D B L E E D I N G D I S O R D E R • Prolonged • Excessive • Inappropriate • Unusual site • Family history ISTH-BAT Adults and Pediatrics Q U E S T I O N 1 • Which of the following is the symptoms of platelet disorder? A. Muscle bleeding B. Hemarthrosis C. Large ecchymosis D. Epistaxis D I S O R D E R O F P L A T E L E T Quantityy Qualityy • Platelet count <150,000/ cumm. • Immune thrombocytopenic purpura <100,000/cumm Thrombocytopenia Platelet Dysfunction Platelet vWF GPIIb/IIIa Platelet C O A G U L AT I O N TF-FVIIa Thrombin Fibrin FVIII FXI FV FIXa Hemophilia A, B: Factor VIII, IX Liver disease: Multiple factors DIC: Multiple factors Drug: warfarin, heparin L A B O R AT O R Y T E S T I N H E M O S TA S I S M E A S U R E M E N T Measurement Laboratoryy Tube Procoagulant APTT, PT/INR, TT Platelet CBC + blood smear Platelet function PFA-100, platelet aggregation test (Bleeding time) Fibrinolysis Euglobulin lysis time Global hemostasis Thromboelastometryy Thrombin generation S P E C I M E N C O L L E C T I O N Good technique • Proper vein • Tourniquet time < 1 min • Needle size 20-22 G • Double syringe technique (discard 1-2 ml) Proper time - Test within 2 h in room temperature or within 4 h in ice water (2 -4oC) Good sample - Concentration and amount of sodium citrate 3.2% trisodium citrate ratio 9:1 or 4.5 ml. in 0.5 ml of sodium citrate (Hct 25-55%) - Volume of anticoagulant = 0.5x (100-Hct) or 55 (100-Hct) V (595- Hct) ปริมาณเลือดที่จะใส่ลงไปในหลอดเลือดที่มี anticoagulant อยู่ 0.5 มล = (60x4.5) / (100-Hct) Extrinsic pathway (monitored by PT) (tissue factor; thromboplastin) VIIa VII (III) Ca++ X Xa Va Ca++ Phospholipid Plus Prothrombin (II) Thrombin (IIa) Fibrinogen (I) Fibrin (Ia) Common pathwayy (monitored by aPTT,PT, TT) COAGULATION CASCADE Intrinsic pathway (monitored by aPTT) XII Hageman factor (collagen) XIIa XI XIa IX IXa X Ca++ VIIIa (IIa) Ca++ Phospholipid Plus FXIII Prothrombinase complex: Xa, Va, PL, ca 2+ Tenase complex: IXa, VIIIa, PL, Ca 2+ HK+PKKallikrein TT COAGULOGRAM Coagulogram Activator Sensitivityy Activated partial thromboplastin time Phospholipid <30-40% Prothrombin time Tissue thromboplastin <30-40% Thrombin time Thrombin Fibrinogen <100 or >400 mg/dL I N D I C AT I O N O F C O A G U L O G R A M • Screening hemostasis before major surgery • Monitoring anticoagulant: heparin, warfarin • Screening of bleeding disorder Test Heparin Warfarin Thrombin time I N Prothrombin time I I Partial thromboplastin time I Nb D = decreased; I = increased; N = no change aDecreased in 25% of cases; bIncreased with high drug dosage I N T E R P R E T A T I O N & I N V E S T I G A T I O N Prolonged APTT or PT Mixing APTT, PT with normal plasma 1:1 Uncorrected Lupus anticoagulant Corrected Incubate 2 h at 37c Uncorrected Inhibitor to clotting factor Corrected Factor deficiencyy R/O lipidemic, icteric, hemolyzed P L A T E L E T F U N C T I O N A N A LY S I S • Normal Col/Epi 154±33, Col/ADP 98±17 sec • Sensitive for diagnosis of vWD and platelet function defect when compared to bleeding time • Monitoring of ASA (prolonged Epi) • Limitation: prolonged CT in thrombocytopenia and anemia, shorten CT in high vWF , fibrinogen and RBC Membrane coated with agonists (Col, ADP EPI ) Citrated Whole Blood Aperture Agonist Concentration Pathway ADP 5-10 umoL ADP receptor and cyclooxygenase or G-protein pathway Epinephrine 2-10 umoL Epinephrine receptor and cyclooxygenase or G-protein pathway Collagen 5-10 ug/mL Membrane receptor and cyclooxygenase or G-protein pathway Thrombin 0.3 U/mL Protease activated factor (PAR) 1 and PAR 4 Ristocetin 1.0 mg/mL GPIb/IX integrin and aggregation with vWF Arachinodic acid 1 mM Cyclooxygenase pathway P L AT E L E T A G G R E G AT I O N T E S T C O M P L E T E B L O O D C O U N T A N D B L O O D S M E A R • Immature platelet fraction (IPF) -Young platelet has high RNA content -Marker of bone marrow function -Not affected by fragmented RBC or large platelet size • Normal <10% G LOBAL H E M OSTA S I S TE STS -Thromboelastometry P H A S E S O F C O A G U L AT I O N T H R O M B O E L A S T O M E T R Y Extem: tissue thromboplastin Intem: ellagic acid Fibtem: tissue thromboplastin and cytochalasin D Heptem: heparinase (inhibit heparin) Aptem: aprotinin (inhibit plasminogen) Clotting factor and anticoagulation protein Platelet and fibrinogen Platelet, and fibrinogen Platelet and fibrinogen Fibrinolysis, factor XIII T H R O M B O E L A S T O G R A P H Y Normal Hyperfibrinolysis Fibrinogen deficiency Platelet dysfunction L A B O R A T O R Y I N V E S T I G A T I O N Screening CBC, blood smear, PT, APTT, Bleeding time/PFA-100 Coagulogram PT, APTT abnormal APTT: FVIII, FIX, FXI PT:FVII PT and APTT: liver disease DIC BT/PFA-100+APTT BT/PFA-100, APTT normal or prolonged vWD BT/PFA-100 BT/PFA-100 prolonged: Platelet disorder Normal screening test : FXIII, fibrinolysis FXII kallikrein FXI FXIa FVIIIa PL, Ca2+ FlX FlXa Intrinsic pathway Tissue factor PL, Ca2+ FVIIa FVII Extrinsic pathway FX Prothrombin Thrombin Common pathway FXa FVIIIa PL, Ca2+ Fibrinogen Fibrin APTT PT TT Q U E S T I O N 2 • Which of the following screening test is abnormal in hemophilia? A. APTT B. PT C. TT D. APTT and PT E. APTT, PT and TT C O M M O N P L AT E L E T D I S E A S E S •Thrombocytopenia • Congenital • Acquired • Dysfunction E T I O L O G Y O F T H R O M B O C Y T O P E N I A • Destructive thrombocytopenia • Immune • Immune thrombocytopenia • Neonatal thrombocytopenia • Drug • Non-immune (platelet activation and consumption • Surgery or trauma • Thrombotic microangiopathy • Congenital heart disease • Kasabach-Merritt syndrome • IAHS • Hypersplenism Impaired production Hereditary disorders Small sized platelet Normal sized platelet Large sized platelet Acquired disorders Infection: CMV, EBV, HIV, parvovirus Nutritional deficiency Bone marrow failure, MDS or infiltration Drugs and radiation Neonatal hypoxia and placental insufficiency D R U G S I N D U C E D T H R O M B O C Y T O P E N I A • Usually occur about 1 week after initiation and recover 1-2 days after stopping medication • Mechanisms • Decrease production • Chemotherapyy • Thiazide diuretic • Accelerated platelet destruction • Drug dependent antibodyy • Heparin, quinine, penicillin, sulfonamide, NSAIDs, anticonvulsant, anti-rheumatic, antidiabetic diuretic, GPIIb/IIIa inhibitor, rifampicin and ranitidine A C Q U I R E D T H R O M B O C Y T O P E N I A DHF ITP KMS DIC Leukemia IAHS I M M U N E T H R O M B O C Y T O P E N I A • History of acute mucocutaneous bleeding • Normal physical examination Cooper N. BJH 2014;165:756-767. Immune mediated acquired thrombocytopenia platelet <100,000/uL Complete blood count with blood smear -Anemia from blood loss -Some large platelet, no abnormal cell E T I O L O G I E S O F P L A T E L E T D Y S F U N C T I O N Inherited Acquired Drug, herbals Liver and kidney disease Myelodysplastic syndrome Paraproteinemia Myeloproliferative disease APDE Hermanskyy Pudlak, Chidiak Higashi Gray platelet syndrome Scott syndrome Bernard Soulier Glanzmann thrombastenia Storage pool disease A N T I P L AT E L E T Drug Platelet effect Approximate duration of increased bleeding risk Aspirin Irreversible 5 days Ibuprofen Reversible 24 hours Naproxen Reversible Up to 4 days Thienopyridines (Clopidogrel) Irreversible 7 days Dipyridamole Reversible Minimal to no risk with procedures Long-acting dipyridamole / aspirin Reversible / irreversible 5 days Cilostazol Reversible Minimal to no risk with procedures Individual patient risks and circumstance should be considered. Based on minimal data; no time points between 2 hours and 4 days. A C Q U I R E D P L AT E L E T D Y S F U N C T I O N W I T H E O S I N O P H I L I A • Common age group 3-7 years • Laboratory • eosinophilia (>500) • 86%, pale stained and large platelet Parasite of allergen Immune complex Y Platelet dysfunction Platelet release Propose pathogenesis M I N I M A L P L AT E L E T A N D C O N D I T I O N • No risk of bleeding 10,000/uL • Sepsis, infants, DIC 20,000/uL • Minimal Procedure 50,000/uL • Internal, mucosal bleeding 50,000/uL • CNS, life threatening bleeding 75,000-100,000/uL Pisciotto PT, Transfusion 1995;53:498-502 C O M M O N C O A G U L AT I O N D I S O R D E R • Congenital • Most common vWD and Hemophilia • Acquired • Liver disease, DIC H E M O P H I L L I A Severe <1% Mild-Moderate ≥1-40% APTT Prolonged B L O O D C O M P O N E N T A N D F A C T O R C O N C E N T R A T E F O R T R E A T M E N T O F H E M O P H I L I A ส่วนประกอบของเลือด ขนาดที่ใช้ เพิ่ม FFP, FDP 10 มล./กก. ปัจจัยการแข็งตัวของเลือด 10-15% ยกเว้นแฟคเตอร์ IX 7-10% Cryoprecipitate 0.2 ยูนิต/กก. แฟคเตอร์ VIII 15-20%, fibrinogen 80-100 มก./ดล. von Willebrand factor, factor XIII Cryo-removed plasma 10 ยูนิต/กก. เพิ่มทุกแฟคเตอร์ 10-15% ยกเว้นแฟคเตอร์ IX 7-10% และไม่เพิ่มแฟคเตอร์ V, VIII fibrinogen Factor VIII concentrate 1 ยูนิต/กก. เพิ่ม factor VIII:C 2% Lyophilized cryoprecipitate 1 ยูนิต/กก. เพิ่ม factor VIII:C 2% Factor IX concentrate หรือ PCC 1 ยูนิต/กก. เพิ่ม factor IX:C 1% cryoprecipitate 1 ยูนิต มี FVIII:C 40-60 ยูนิต ถ้าเป็น cryoprecipitate ที่มี FVIII:C 80-100 ยูนิต/ถุง การให้ 0.1 ยูนิต/กก. จะเพิ่ม FVIII:C ได้ 15-20% R E C O M M E N D E D P L A S M A F A C T O R L E V E L A N D D U R A T I O N ชนิดของอาการเลือดออก ระดับปัจจัยการแข็งตัวของเลือด (%) จุดเริ่มต้น ระดับต ่าสุดที่ยอมรับได้ 1. เลือดออกที่กล้ามเนื้อ การเย็บแผลหัตถการทางทันตกรรม 20-30 -2. เลือดออกในกล้ามเนื้อขนาดใหญ่ (ยกเว้น iliopsoas) เลือดออกในข้อ ปัสสาวะเป็นเลือด แผลฉีดลึก 40-60 20-30 (นาน 3-7 วัน) 3. ผ่าตัดขนาดเล็กถึงปานกลาง เช่น ผ่าตัดไส้ติ่งอักเสบ เลือดออกในสมอง, ทางเดินอาหาร, ล าคอ, อวัยวะส าคัญ และ iliopsoas) 80-100 40-50 (1-2 สัปดาห์ หรือจนแผลหายยกเว้น iliopsoas ให้ จนกว่า hematoma หายไปและ เลือดออกในสมองให้ เพื่อป้องกันอย่างน้อย 3-6 เดือน) 4. ผ่าตัดขนาดใหญ่ เช่น ผ่าตัดข้อหรือผ่าตัดสมอง 80-100 40-50 (นาน 1-2 สัปดาห์ หรือจนแผลหาย ยกเว้นเลือดออกใน สมองให้เพื่อป้องกันอย่างน้อย 3-6 เดือน) T R E A T M E N T O F H E M O P H I L I A ผู้ป่วยนอก หรือ หอสังเกตอาการ 150,000 บาท ไม่เกิน 2 ครั้ง ต่อ เดือน ผู้ป่วยในภาวะฉุกเฉิน 300,000 บาท ต่อ ครั้ง “Early treatment to primary prophylaxis” ลงทะเบียนในสถานพยาบาลที่จดทะเบียนดูแลผู้ป ่วย VO N W I L L E B R A N D D I S E A S E Platelet vWF GPIb/IX/V GPIIb/IIIa Platelet Von Willebrand factor functions -Platelet adhesion to endothelium and Attachment between platelet -Carrier protein of factor VIII Classification and mode of inheritance -Type 1 Quantitative defect: AD -Type 2 Qualitative defect: AD except 2N -Type 3 Severe deficiency: AR LABORATORY DIAGNOSIS Specific Test (consult hematologist) • vWF:Ag vWF multimer • Ristocetin cofactor activityy Genetic studyy • Collagen binding assay Platelet aggregation induced by ristocetin Screening test • Normal plt count and smear except type 2B • Normal or prolonged bleeding time, PFA-100 • Normal or prolonged APTT Diagnosis • Type 1: VWD <30% (if no bleeding symptom) 30-50% (if bleeding symptoms) • Type 2: VWF:Rco/VWF:Ag <0.6 • Type 3: VWF: Ag <10% Nochols WL et al Haemophilia ;14:171-232. T R E A T M E N T O F V O N W I L L E B R A N D D I S E A S E • DDAVP 0.3 mcg/kg/dose in 0.9% NSS 15 mL intravenous in 15-30 mins in VWD type 1, type 2A • Tranxenamic acid 10 mg/kg/dose IV, 15-20 mg/kg/dose oral q 4-6 h • Factor VIII conc. (vWF /FVIII): Alphanate, Immunate Mannuccio PM Engl J Med 2004;351:683-94. S U P P O R T I V E T R E AT M E N T • Antifibrinolytic agent • Tranexamic acid • 10 mg/kg/dose intravenouslyy • 15-20 mg/kg/dose orallyy • 5% solution mouth rinse • Desmopressin • DDAVP 0.3 mcg/kg D I S S E M I N A T E D I N T R A V A S C U L A R C O A G U L A T I O N Acquired syndrome, characterized by the systemic intravascular activation arising from different causes, can cause damage to the microvasculature result in organ dysfunction P A T H O G E N E S I S O F D I S S E M I N A T E D I N T R A V A S C U L A R C O A G U L A T I O N Mononuclear cell Endothelium Activated plt Inflammatory cell TF Thrombin Fibrin Cytokines Inhibit fibrinolysis Impaired anticoagulant function Neutrophil Intravascular clot C O N D I T I O N S A S S O C I A T E D W I T H D I S S E M I N A T E D I N T R A V A S C U L A R C O A G U L A T I O N Sepsis and severe infection Trauma Organ destruction e.g. pancreatitis Malignancy Solid tumors Leukemia Obstetric Amniotic fluid embolism Placental abruption Pre-eclampsia Vascular abnormalities Large hemangioma Vascular aneurysm Severe liver failure Toxic and immunological insults Recreational drugs ABO transfusion incompatibility Transplant rejection D I A G N O S I S I S T H S C O R I N G S Y S T E M Laboratoryy Score Platelet count >100 <100 <50 0 1 2 Level of fibrin markers No increase Increased but <5x upper limit of normal Increased but ≥5x upper limit of normal 0 2 3 Prolonged prothrombin time <3 sec ≥3 sec but <6 sec ≥6 sec 0 1 2 Fibrinogen >1.0 g/L ≤1.0 g/L 0 1 Score ≥ 5 Overt DIC Levi M Blood 2018;131:845-854 O U T C O M E O F D I C A N D S C O R I N G S Y S T E M ( J M H L W ) Wada H et al Thromb. Hemost. 1995;74:848-52 I complete remission II partial remission III unchanged IV exacerbation V dead Leukemia Non-leukemia H E M O S TA S I S I N L I V E R D I S E A S E M O D E L O F H E M O S T A S I S I N L I V E R D I S E A S E • Impaired synthetic of coagulation factors and anticoagulant • Hyperfibrinolysis vs hypofibrinolysis • Decrease vitamin K absorption • Platelet dysfunction vs hyperactivityy L A B O R AT O R Y T E S T I N G Loss of Reserve Laboratory test Procoagulant INR, APTT, fibrinogen Natural anticoagulant PC, PS, AT Fibrinolysis D-dimer Platelet Platelet count Global coagulation assayy Thromboelastometryy, thrombin generation Endothelial vWF Magnusson M et al. Arch Dis Child 2016;101:854–859 W A R F A R I N A N D B L E E D I N G M A N A G E M E N T กรณี อาการเลือดออก การรักษา มี ไม่มี INR < 5  หยุดยา 1 วัน ติดตามค่า INR เมื่อได้ค่าที่ต้องการ ปรับยาลดลง INR > 5 - <10  หยุดยา 1-2 วัน ติดตามค่า INR เมื่อได้ค่าที่ต้องการ ปรับยาลดลง INR > 10  หยุดยา warfarin ให้วิตามิน เค 2-5 มก.ทางปาก INR จะลดลง ภายใน 24-48 ชั่วโมง เมื่อค่าอยู่ในระดับที่ ต้องการ ปรับยาลดลง INR เท่าไรก็ ตาม  วิตามิน เค1-10 มก. ทางหลอดเลือด พิจารณาให้ FFP หรือ prothrombin complex concentrate ขึ้นกับความรุนแรงของอาการ เลือดออก Common anticoagulant related bleeding T R E A T M E N T O F H E P A R I N R E L A T E D B L E E D I N G ระยะเวลาตั้งแต่ได้รับยา heparin ครั้ง สุดท้าย (นาที) ขนาดยา protamine ต่อ 100 ยูนิต ของ heparin (มก.) <30 1 30-60 0.5-0.75 60-120 0.375-0.5 >120 0.25-0.375 M A S S I V E B L O O D L O S S • Hemorrhage is a leading cause of preventable traumatic death for patients of all ages, accounting for 20%-40% of all early trauma-related mortality Nosanov L. The American Journal of Surgery 2013; 206:655-660. M A S S I V E T R A N S F U S I O N • Replacement blood components >1 blood volume in 24 hr • Replacement of 50% BV in 3 h - Premature infant: 90-100 ml/kg - Term infant-3 months: 80-90 ml/kg - Children older than 3 months: 70 ml/kg - Very obese children: 65 ml/kg - Adult: 60 ml/kg • Adults >10 U of PRBCs in 24 hr C A U S E O F M A S S I V E B L E E D I N G Trauma Surgeryy Uncontrolled bleeding Uncontrolled bleeding Massive Blood loss Coagulopathyy Platelet disorder Anticoagulant, antiplatelet H O W T O E VA L U AT E ? Classification of hemorrhage Parameter I II III IV Blood loss (ml) <750 750-1500 1500-2000 >2000 Blood loss (%) <15% 15-30% 30-40% >40% Pulse rate (beats/min) <100 >100 >120 >140 Blood pressure Normal Decreased Decreased Decreased Respiratory rate (breaths/min) 14-20 20-30 30-40 >35 Urine output (ml/hr) >30 20-30 5-15 Negligible CNS symptoms Normal Anxious Confused Lethargic Modified from Committee on Trauma . CNS = central nervous systemVital signsUrineConsciousHct Gross JB et al Anesthesiology 1983 Trigger volume 40 ml/kg (older children) C O A G U L O P A T H Y A N D B L O O D C O M P O N E N T T R A N S F U S I O N Coagulopathy Acidosis Bleeding Severe trauma Tissue hypoxia Hypothermia Colloid and Crystalloid infusion Massive RBC transfusion Dilution of coagulation factors and platelet ↓37oC↓Clotting factor •Hypotension •Platelet dysfunction •Enzyme defect Hypothemia Coagulopathy Acidosis Lethal trauma triad COMPLICATION M A N A G E M E N T O F M A S S I V E B L E E D I N G • Stop bleeding • Identified site of bleeding • Surgical or interventional management • Restoration of coagulation Crystalloids Colloids RBC FFP/PCC Fibrinogen Platelet 0 0.5 1.0 1.5 2.0 Blood volume replacement 2.5 Hct = 21-24% Fig = 1.0 g litre-1 Plt < 50x109 litre-1 PT, aPTT >1.5 x normal FLUID & BLOOD COMPONENT INVESTIGATION • Complete blood count • Coagulogram (APTT, PT and TT) • Fibrinogen, euglobulin clot lysis time • Thromboelastometryy • Liver function and renal function test • Urine examination • Arterial blood gas • Serum lactate, electrolyte Ca, ionized Ca and albumin T H R O M B O E L A S T O M E T R Y A N D G U I D A N C E O F T R A N S F U S I O N • A systematic review or 9 randomized clinical trials, mostly cardiac surgery in 776 adults • Benefit: reduced bleeding, reduction of patients receiving transfusion as a combined FFP and platelet • No difference in mortality event rate Wikkelsoe AJ et al Acta Anaesthesiol Scand 2011;55:1174-1189 R E P L A C E M E N T T H E R A P Y Coagulation Therapy Dose Hct <24% RBC 10 ml/kg (10%) 1 U adult (3%) APTT, PT >1.5x FFP 20 ml/kg then 10 ml/kg q 6-12 hr Fibrinogen <150 mg/dL CPP 0.2-0.4 u/kg q 12-24 hr Platelet <50,000/mcL Platelet 0.4 u/kg increase 40,000-80,000/mcL or LPPC 8 U A D D I T I O N A L T R E A T M E N T Condition Management Hemophilia A Factor VIII conc 50 U/Kg Hemophilia B Factor IX conc 50 U/Kg Factor VII deficiency Factor VII 15-30 mcg/Kg/dose Hyperfibrinolysis Tranexamic acid 10 mg/kg/dose Warfarin induced bleeding 4-PCC 25-50 U/Kg/dose Immune thrombocytopenia IVIg 0.8-1 g/Kg, Methylprednisolone 30 mg/Kg/dayy R E P L A C E M E N T T H E R A P Y: A D U LT • Ratio of PRC: FFP: Platelet (and/or) 1:1:1 is now recommendation especially in adult Zehtabchi S. Academic Emergency Medicine 2009; 16:371-378. S Y S T E M A T I C R E V I E W R E P L A C E M E N T T H E R A P Y : C H I L D R E N • Important message • Early use of thromboelastometryy • Ratio of transfusion (FFP:Plt:RBC) 1:1:1 or 1:1:2 did not change the outcomes but increase FFP and plt transfusion in lower ratio • Ratio 1:1:1 is required when replacement is > 40 mL/Kg • Tranexamic acid in severe trauma could reduce transfusion requirement Maw G Pediatr Emergency Care 2018;34:594-8 R E C O M B I N A N T A C T I V AT E D F V I I • Fail conventional therapyy • rFVIIa should be used when • Fibrinogen concentration >100 mg/dL • Platelet count >50,000/mcL • PH >7.2 • Dose of rFVIIa • 200 follow by 100 mcg/kg within 1-3 hr • 40-150 mcg/kg can stop bleeding in 75% of patients FVIIa +TF FXa Thrombin Persistent massive bleeding rFVIIa 100 mcg/kg (dose แรก) เลือดหยุดไหล ภาวะเลือดออก ลดลงเป็นอย่างมาก Dose ต่อๆ ไป เว้นช่วง 1-4 ชม. (ปกติใช้ประมาณ 1-4 doses) rFVIIIa 100 mcg/kg (dose ที่ 2) เว้นช่วง 30 นาที จาก dose แรก. พยายามแก้ไข • Hct >24% • Plt >50,000-10,000/uL. • Fibrinogen 100 mg/dL. • T 37oC • pH >7.2 ภาวะเลือดออก ลดลงบ้าง rFVIIa PROTOCOL C O N C E P T O F M A S S I V E T R A N S F U S I O N G U I D E L I N E Massive blood loss Crystalloid loading PRC gr O low titer PRC: FFP (common ratio 1:1) PRC: FFP: platelet (common ration1:1:1) PRC: FFP: platelet: CPP Lab. test S U M M A R Y • Hemostasis consists of • Vascular, platelet, coagulation protein • Symptoms of suspected bleeding disorder • Prolonged bleeding after injuryy • Excessive bleeding • Multisystem Common hemostasis defect • Platelet • Thrombocytopenia • Dysfunction • Coagulation • Congenital • Acquired Platelet 0.2 U/Kg 20,000-40,000/uL rFVIIa 100 mcg/kg FIX 1 U/kg increase 1% FVIII 1 U/kg increase2% FFP 10 cc/Kg 10% CPP 0.2 U/Kg Fib 80-100 mg/dL THANK YOU
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https://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics)
Effective mass (solid-state physics) - Wikipedia Jump to content Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search Appearance Donate Create account Log in Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Simple case: parabolic, isotropic dispersion relation 2 Intermediate case: parabolic, anisotropic dispersion relation 3 General case Toggle General case subsection - 3.1 Inertial effective mass tensor - 3.2 Cyclotron effective mass - 3.3 Density of states effective masses (lightly doped semiconductors) - 4 Determination Toggle Determination subsection - 4.1 Experimental - 4.2 Theoretical - 5 Significance - 6 See also - 7 Footnotes - 8 References - 9 External links Toggle the table of contents Effective mass (solid-state physics) 22 languages বাংলা Català Čeština Deutsch Español Euskara فارسی Français 한국어 हिन्दी Italiano עברית Қазақша 日本語 Polski Русский Slovenščina Svenska Türkçe Українська Tiếng Việt 中文 Edit links Article Talk English Read Edit View history Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. From Wikipedia, the free encyclopedia Sorry to interrupt, but our fundraiser won't last long. This Tuesday, we ask you to join the 2% of readers who give. If everyone reading this right now gave just $2.75, we'd hit our goal quickly. $2.75 is all we ask. Give $2.75 Maybe later Close September 16: Knowledge is human. We're sorry we've asked you a few times recently, but it's Tuesday, September 16—please don't wait until tomorrow to help. We're happy you consult Wikipedia often. If just 2% of our most loyal readers gave $2.75 today, we'd reach our goal quickly. Most readers donate because Wikipedia is useful, others because they realize knowledge needs humans. If you agree, please give. Any contribution helps, whether it's $2.75 one time or monthly. 25 years ago Wikipedia was a dream. A dream built piece by piece by people, not machines. 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By submitting, you are agreeing to our donor privacy policy and to sharing your information with the Wikimedia Foundation and its service providers in the U.S. and elsewhere. The Wikimedia Foundation is a recognized public welfare institution (ANBI). If you make a recurring donation, you will be debited by the Wikimedia Foundation until you notify us to stop. We’ll send you an email which will include a link to easy cancellation instructions. Back Almost done: Please, make it monthly. Monthly support is the best way to ensure that Wikipedia keeps thriving. No thanks! I'll make a one-time donation of Yes, I'll donate each month Yes, I'll donate monthly, but for a different amount Back Thank you for your support! Enter your monthly donation amount Please select an amount (minimum $1) We cannot accept donations greater than 25000 USD through our website. Please contact our major gifts staff at benefactors@wikimedia.org. Donate monthly Maybe later I already donated Close How would you like to be reminded? We can send you an email or text message reminder to donate later. Send me an email Send me a text message Send me an email reminder We’ll gladly send you an email reminder and get out of your way so you can get back to reading. Email address Submit Please enter a valid email address i.e. name@domain.com We never sell your information. By submitting, you are agreeing to our donor privacy policy. The Wikimedia Foundation is a nonprofit, tax-exempt organization. We never sell your information. By submitting, you are agreeing to our donor privacy policy and to sharing your information with the Wikimedia Foundation and its service providers in the U.S. and elsewhere. The Wikimedia Foundation is a recognized public welfare institution (ANBI). We never sell your information. By submitting, you are agreeing to our donor privacy policy and to sharing your information with the Wikimedia Foundation and its service providers in the USA and elsewhere. Donations to the Wikimedia Foundation are likely not tax-deductible outside the USA. Send me a reminder We’ll gladly send you a reminder and get out of your way so you can get back to reading. Please enter your mobile phone number and check the opt in checkbox to receive a text message reminder. Mobile phone number I would like to receive text messages such as donation reminders and appeals from Wikimedia at the number I have provided. By participating, you consent to receive recurring updates through automated text messages from Wikimedia to the phone number you provide. Message frequency varies. For text messages, Msg&Data rates may apply. Text STOP to cancel or HELP for help. Terms of Service and Privacy Policy. Submit Please enter a valid phone number e.g. (201) 555-0123 Please check the box to consent to receive messages. Thank you! We will send you a reminder. 🎉 Thank you for donating recently! 🎉 Your support means the world to us. We'll hide banners in this browser for the rest of our campaign. Close Mass of a particle when interacting with other particles For negative mass in theoretical physics, see Negative mass. In solid state physics, a particle's effective mass (often denoted m ∗ {\textstyle m^{}} ) is the mass that it seems to have when responding to forces, or the mass that it seems to have when interacting with other identical particles in a thermal distribution. One of the results from the band theory of solids is that the movement of particles in a periodic potential, over long distances larger than the lattice spacing, can be very different from their motion in a vacuum. The effective mass is a quantity that is used to simplify band structures by modeling the behavior of a free particle with that mass. For some purposes and some materials, the effective mass can be considered to be a simple constant of a material. In general, however, the value of effective mass depends on the purpose for which it is used, and can vary depending on a number of factors. For electrons or electron holes in a solid, the effective mass is usually stated as a factor multiplying the rest mass of an electron, me (9.11 × 10−31 kg). This factor is usually in the range 0.01 to 10, but can be lower or higher—for example, reaching 1,000 in exotic heavy fermion materials, or anywhere from zero to infinity (depending on definition) in graphene. As it simplifies the more general band theory, the electronic effective mass can be seen as an important basic parameter that influences measurable properties of a solid, including everything from the efficiency of a solar cell to the speed of an integrated circuit. Simple case: parabolic, isotropic dispersion relation [edit] At the highest energies of the valence band in many semiconductors (Ge, Si, GaAs, ...), and the lowest energies of the conduction band in some semiconductors (GaAs, ...), the band structure E(k) can be locally approximated as : E ( k ) = E 0 + ℏ 2 k 2 2 m ∗ {\displaystyle E(\mathbf {k} )=E\_{0}+{\frac {\hbar ^{2}\mathbf {k} ^{2}}{2m^{\}}}} {\displaystyle E(\mathbf {k} )=E_{0}+{\frac {\hbar ^{2}\mathbf {k} ^{2}}{2m^{}}}} where E(k) is the energy of an electron at wavevector k in that band, E0 is a constant giving the edge of energy of that band, and m is a constant (the effective mass). It can be shown that the electrons placed in these bands behave as free electrons except with a different mass, as long as their energy stays within the range of validity of the approximation above. As a result, the electron mass in models such as the Drude model must be replaced with the effective mass. One remarkable property is that the effective mass can become negative, when the band curves downwards away from a maximum. As a result of the negative mass, the electrons respond to electric and magnetic forces by gaining velocity in the opposite direction compared to normal; even though these electrons have negative charge, they move in trajectories as if they had positive charge (and positive mass). This explains the existence of valence-band holes, the positive-charge, positive-mass quasiparticles that can be found in semiconductors. In any case, if the band structure has the simple parabolic form described above, then the value of effective mass is unambiguous. Unfortunately, this parabolic form is not valid for describing most materials. In such complex materials there is no single definition of "effective mass" but instead multiple definitions, each suited to a particular purpose. The rest of the article describes these effective masses in detail. Intermediate case: parabolic, anisotropic dispersion relation [edit] Constant energy ellipsoids in silicon near the six conduction band minima. For each valley (band minimum), the effective masses are mℓ = 0.92me ("longitudinal"; along one axis) and mt = 0.19me ("transverse"; along two axes). In some important semiconductors (notably, silicon) the lowest energies of the conduction band are not symmetrical, as the constant-energy surfaces are now ellipsoids, rather than the spheres in the isotropic case. Each conduction band minimum can be approximated only by : E ( k ) = E 0 + ℏ 2 2 m x ∗ ( k x − k 0 , x ) 2 + ℏ 2 2 m y ∗ ( k y − k 0 , y ) 2 + ℏ 2 2 m z ∗ ( k z − k 0 , z ) 2 {\displaystyle E\left(\mathbf {k} \right)=E\_{0}+{\frac {\hbar ^{2}}{2m\_{x}^{\}}}\left(k\_{x}-k\_{0,x}\right)^{2}+{\frac {\hbar ^{2}}{2m\_{y}^{\}}}\left(k\_{y}-k\_{0,y}\right)^{2}+{\frac {\hbar ^{2}}{2m\_{z}^{\}}}\left(k\_{z}-k\_{0,z}\right)^{2}} {\displaystyle E\left(\mathbf {k} \right)=E_{0}+{\frac {\hbar ^{2}}{2m_{x}^{}}}\left(k_{x}-k_{0,x}\right)^{2}+{\frac {\hbar ^{2}}{2m_{y}^{}}}\left(k_{y}-k_{0,y}\right)^{2}+{\frac {\hbar ^{2}}{2m_{z}^{}}}\left(k_{z}-k_{0,z}\right)^{2}} where x, y, and z axes are aligned to the principal axes of the ellipsoids, and m x, m y and m z are the inertial effective masses along these different axes. The offsets k0,x, k0,y, and k0,z reflect that the conduction band minimum is no longer centered at zero wavevector. (These effective masses correspond to the principal components of the inertial effective mass tensor, described later.) In this case, the electron motion is no longer directly comparable to a free electron; the speed of an electron will depend on its direction, and it will accelerate to a different degree depending on the direction of the force. Still, in crystals such as silicon the overall properties such as conductivity appear to be isotropic. This is because there are multiple valleys (conduction-band minima), each with effective masses rearranged along different axes. The valleys collectively act together to give an isotropic conductivity. It is possible to average the different axes' effective masses together in some way, to regain the free electron picture. However, the averaging method turns out to depend on the purpose: For calculation of the total density of states and the total carrier density, via the geometric mean combined with a degeneracy factor g which counts the number of valleys (in silicon g = 6): : m density ∗ = g 2 m x m y m z 3 {\displaystyle m\_{\text{density}}^{\}={\sqrt[{3}]{g^{2}m\_{x}m\_{y}m\_{z}}}} ![{\displaystyle m_{\text{density}}^{}={\sqrt[{3}]{g^{2}m_{x}m_{y}m_{z}}}}]( (This effective mass corresponds to the density of states effective mass, described later.) For the per-valley density of states and per-valley carrier density, the degeneracy factor is left out. - For the purposes of calculating conductivity as in the Drude model, via the harmonic mean : m conductivity ∗ = 3 [ 1 m x ∗ + 1 m y ∗ + 1 m z ∗ ] − 1 {\displaystyle m\_{\text{conductivity}}^{\}=3\left[{\frac {1}{m\_{x}^{\}}}+{\frac {1}{m\_{y}^{\}}}+{\frac {1}{m\_{z}^{\}}}\right]^{-1}} ![{\displaystyle m_{\text{conductivity}}^{}=3\left[{\frac {1}{m_{x}^{}}}+{\frac {1}{m_{y}^{}}}+{\frac {1}{m_{z}^{}}}\right]^{-1}}]( Since the Drude law also depends on scattering time, which varies greatly, this effective mass is rarely used; conductivity is instead usually expressed in terms of carrier density and an empirically measured parameter, carrier mobility. General case [edit] In general the dispersion relation cannot be approximated as parabolic, and in such cases the effective mass should be precisely defined if it is to be used at all. Here a commonly stated definition of effective mass is the inertial effective mass tensor defined below; however, in general it is a matrix-valued function of the wavevector, and even more complex than the band structure. Other effective masses are more relevant to directly measurable phenomena. Inertial effective mass tensor [edit] A classical particle under the influence of a force accelerates according to Newton's second law, a = m−1F, or alternatively, the momentum changes according to ⁠d/dt⁠p = F. This intuitive principle appears identically in semiclassical approximations derived from band structure when interband transitions can be ignored for sufficiently weak external fields. The force gives a rate of change in crystal momentum pcrystal: : F = d ⁡ p crystal d ⁡ t = ℏ d ⁡ k d ⁡ t , {\displaystyle \mathbf {F} ={\frac {\operatorname {d} \mathbf {p} \_{\text{crystal}}}{\operatorname {d} t}}=\hbar {\frac {\operatorname {d} \mathbf {k} }{\operatorname {d} t}},} {\displaystyle \mathbf {F} ={\frac {\operatorname {d} \mathbf {p} _{\text{crystal}}}{\operatorname {d} t}}=\hbar {\frac {\operatorname {d} \mathbf {k} }{\operatorname {d} t}},} where ħ = h/2π is the reduced Planck constant. Acceleration for a wave-like particle becomes the rate of change in group velocity: : a = d d ⁡ t v g = d d ⁡ t ( ∇ k ω ( k ) ) = ∇ k d ⁡ ω ( k ) d ⁡ t = ∇ k ( d ⁡ k d ⁡ t ⋅ ∇ k ω ( k ) ) , {\displaystyle \mathbf {a} ={\frac {\operatorname {d} }{\operatorname {d} t}}\,\mathbf {v} \_{\text{g}}={\frac {\operatorname {d} }{\operatorname {d} t}}\left(\nabla \_{k}\,\omega \left(\mathbf {k} \right)\right)=\nabla \_{k}{\frac {\operatorname {d} \omega \left(\mathbf {k} \right)}{\operatorname {d} t}}=\nabla \_{k}\left({\frac {\operatorname {d} \mathbf {k} }{\operatorname {d} t}}\cdot \nabla \_{k}\,\omega (\mathbf {k} )\right),} {\displaystyle \mathbf {a} ={\frac {\operatorname {d} }{\operatorname {d} t}}\,\mathbf {v} _{\text{g}}={\frac {\operatorname {d} }{\operatorname {d} t}}\left(\nabla _{k}\,\omega \left(\mathbf {k} \right)\right)=\nabla _{k}{\frac {\operatorname {d} \omega \left(\mathbf {k} \right)}{\operatorname {d} t}}=\nabla _{k}\left({\frac {\operatorname {d} \mathbf {k} }{\operatorname {d} t}}\cdot \nabla _{k}\,\omega (\mathbf {k} )\right),} where ∇k is the del operator in reciprocal space. The last step follows from using the chain rule for a total derivative for a quantity with indirect dependencies, because the direct result of the force is the change in k(t) given above, which indirectly results in a change in E(k)=ħω(k). Combining these two equations yields : a = ∇ k ( F ℏ ⋅ ∇ k E ( k ) ℏ ) = 1 ℏ 2 ( ∇ k ( ∇ k E ( k ) ) ) ⋅ F = M inert − 1 ⋅ F {\displaystyle \mathbf {a} =\nabla \_{k}\left({\frac {\mathbf {F} }{\hbar }}\cdot \nabla \_{k}\,{\frac {E(\mathbf {k} )}{\hbar }}\right)={\frac {1}{\hbar ^{2}}}\left(\nabla \_{k}\left(\nabla \_{k}\,E(\mathbf {k} )\right)\right)\cdot \mathbf {F} =M\_{\text{inert}}^{-1}\cdot \mathbf {F} } {\displaystyle \mathbf {a} =\nabla _{k}\left({\frac {\mathbf {F} }{\hbar }}\cdot \nabla _{k}\,{\frac {E(\mathbf {k} )}{\hbar }}\right)={\frac {1}{\hbar ^{2}}}\left(\nabla _{k}\left(\nabla _{k}\,E(\mathbf {k} )\right)\right)\cdot \mathbf {F} =M_{\text{inert}}^{-1}\cdot \mathbf {F} } using the dot product rule with a uniform force (∇kF=0). ∇ k ( ∇ k E ( k ) ) {\displaystyle \nabla _{k}\left(\nabla _{k}\,E(\mathbf {k} )\right)} is the Hessian matrix of E(k) in reciprocal space. We see that the equivalent of the Newtonian reciprocal inertial mass for a free particle defined by a = m−1F has become a tensor quantity : M inert − 1 = 1 ℏ 2 ∇ k ( ∇ k E ( k ) ) . {\displaystyle M\_{\text{inert}}^{-1}={\frac {1}{\hbar ^{2}}}\nabla \_{k}\left(\nabla \_{k}\,E(\mathbf {k} )\right).} {\displaystyle M_{\text{inert}}^{-1}={\frac {1}{\hbar ^{2}}}\nabla _{k}\left(\nabla _{k}\,E(\mathbf {k} )\right).} whose elements are : [ M inert − 1 ] i j = 1 ℏ 2 [ ∇ k ( ∇ k E ( k ) ) ] i j = 1 ℏ 2 ∂ 2 E ∂ k i ∂ k j . {\displaystyle \left[M\_{\text{inert}}^{-1}\right]\_{ij}={\frac {1}{\hbar ^{2}}}\left[\nabla \_{k}\left(\nabla \_{k}\,E(\mathbf {k} )\right)\right]\_{ij}={\frac {1}{\hbar ^{2}}}{\frac {\partial ^{2}E}{\partial k\_{i}\partial k\_{j}}}\,.} ![{\displaystyle \left[M_{\text{inert}}^{-1}\right]_{ij}={\frac {1}{\hbar ^{2}}}\left[\nabla _{k}\left(\nabla _{k}\,E(\mathbf {k} )\right)\right]_{ij}={\frac {1}{\hbar ^{2}}}{\frac {\partial ^{2}E}{\partial k_{i}\partial k_{j}}}\,.}]( This tensor allows the acceleration and force to be in different directions, and for the magnitude of the acceleration to depend on the direction of the force. For parabolic bands, the off-diagonal elements of Minert−1 are zero, and the diagonal elements are constants For isotropic bands the diagonal elements must all be equal and the off-diagonal elements must all be equal. For parabolic isotropic bands, Minert−1 = ⁠1/m⁠I, where m is a scalar effective mass and I is the identity. In general, the elements of Minert−1 are functions of k. The inverse, Minert = (Minert−1)−1, is known as the effective mass tensor. Note that it is not always possible to invert Minert−1 For bands with linear dispersion E ∝ k {\displaystyle E\propto k} such as with photons or electrons in graphene, the group velocity is fixed, i.e. electrons travelling with parallel with k to the force direction F cannot be accelerated and the diagonal elements of Minert−1 are obviously zero. However, electrons travelling with a component perpendicular to the force can be accelerated in the direction of the force, and the off-diagonal elements of Minert−1 are non-zero. In fact the off-diagonal elements scale inversely with k, i.e. they diverge (become infinite) for small k. This is why the electrons in graphene are sometimes said to have infinite mass (due to the zeros on the diagonal of Minert−1) and sometimes said to be massless (due to the divergence on the off-diagonals). Cyclotron effective mass [edit] Classically, a charged particle in a magnetic field moves in a helix along the magnetic field axis. The period T of its motion depends on its mass m and charge e, : T = | 2 π m e B | {\displaystyle T=\left\vert {\frac {2\pi m}{eB}}\right\vert } {\displaystyle T=\left\vert {\frac {2\pi m}{eB}}\right\vert } where B is the magnetic flux density. For particles in asymmetrical band structures, the particle no longer moves exactly in a helix, however its motion transverse to the magnetic field still moves in a closed loop (not necessarily a circle). Moreover, the time to complete one of these loops still varies inversely with magnetic field, and so it is possible to define a cyclotron effective mass from the measured period, using the above equation. The semiclassical motion of the particle can be described by a closed loop in k-space. Throughout this loop, the particle maintains a constant energy, as well as a constant momentum along the magnetic field axis. By defining A to be the k-space area enclosed by this loop (this area depends on the energy E, the direction of the magnetic field, and the on-axis wavevector kB), then it can be shown that the cyclotron effective mass depends on the band structure via the derivative of this area in energy: : m ∗ ( E , B ^ , k B ^ ) = ℏ 2 2 π ⋅ ∂ ∂ E A ( E , B ^ , k B ^ ) {\displaystyle m^{\}\left(E,{\hat {B}},k\_{\hat {B}}\right)={\frac {\hbar ^{2}}{2\pi }}\cdot {\frac {\partial }{\partial E}}A\left(E,{\hat {B}},k\_{\hat {B}}\right)} {\displaystyle m^{}\left(E,{\hat {B}},k_{\hat {B}}\right)={\frac {\hbar ^{2}}{2\pi }}\cdot {\frac {\partial }{\partial E}}A\left(E,{\hat {B}},k_{\hat {B}}\right)} Typically, experiments that measure cyclotron motion (cyclotron resonance, De Haas–Van Alphen effect, etc.) are restricted to only probe motion for energies near the Fermi level. In two-dimensional electron gases, the cyclotron effective mass is defined only for one magnetic field direction (perpendicular) and the out-of-plane wavevector drops out. The cyclotron effective mass therefore is only a function of energy, and it turns out to be exactly related to the density of states at that energy via the relation g ( E ) = g v m ∗ π ℏ 2 {\displaystyle \scriptstyle g(E)\;=\;{\frac {g_{v}m^{}}{\pi \hbar ^{2}}}} , where gv is the valley degeneracy. Such a simple relationship does not apply in three-dimensional materials. Density of states effective masses (lightly doped semiconductors) [edit] Density of states effective mass in various semiconductors | Group | Material | Electron | Hole | --- --- | | IV | Si (4 K) | 1.06 | 0.59 | | Si (300 K) | 1.09 | 1.15 | | Ge | 0.55 | 0.37 | | III–V | GaAs | 0.067 | 0.45 | | InSb | 0.013 | 0.6 | | II–VI | ZnO | 0.29 | 1.21 | | ZnSe | 0.17 | 1.44 | In semiconductors with low levels of doping, the electron concentration in the conduction band is in general given by : n e = N C exp ⁡ ( − E C − E F k T ) {\displaystyle n\_{\text{e}}=N\_{\text{C}}\exp \left(-{\frac {E\_{\text{C}}-E\_{\text{F}}}{kT}}\right)} {\displaystyle n_{\text{e}}=N_{\text{C}}\exp \left(-{\frac {E_{\text{C}}-E_{\text{F}}}{kT}}\right)} where EF is the Fermi level, EC is the minimum energy of the conduction band, and NC is a concentration coefficient that depends on temperature. The above relationship for ne can be shown to apply for any conduction band shape (including non-parabolic, asymmetric bands), provided the doping is weak (EC − EF ≫ kT); this is a consequence of Fermi–Dirac statistics limiting towards Maxwell–Boltzmann statistics. The concept of effective mass is useful to model the temperature dependence of NC, thereby allowing the above relationship to be used over a range of temperatures. In an idealized three-dimensional material with a parabolic band, the concentration coefficient is given by : N C = 2 ( 2 π m e ∗ k T h 2 ) 3 2 {\displaystyle \quad N\_{\text{C}}=2\left({\frac {2\pi m\_{\text{e}}^{\}kT}{h^{2}}}\right)^{\frac {3}{2}}} {\displaystyle \quad N_{\text{C}}=2\left({\frac {2\pi m_{\text{e}}^{}kT}{h^{2}}}\right)^{\frac {3}{2}}} In semiconductors with non-simple band structures, this relationship is used to define an effective mass, known as the density of states effective mass of electrons. The name "density of states effective mass" is used since the above expression for NC is derived via the density of states for a parabolic band. In practice, the effective mass extracted in this way is not quite constant in temperature (NC does not exactly vary as T3/2). In silicon, for example, this effective mass varies by a few percent between absolute zero and room temperature because the band structure itself slightly changes in shape. These band structure distortions are a result of changes in electron–phonon interaction energies, with the lattice's thermal expansion playing a minor role. Similarly, the number of holes in the valence band, and the density of states effective mass of holes are defined by: : n h = N V exp ⁡ ( − E F − E V k T ) , N V = 2 ( 2 π m h ∗ k T h 2 ) 3 2 {\displaystyle n\_{\text{h}}=N\_{\text{V}}\exp \left(-{\frac {E\_{\text{F}}-E\_{\text{V}}}{kT}}\right),\quad N\_{\text{V}}=2\left({\frac {2\pi m\_{\text{h}}^{\}kT}{h^{2}}}\right)^{\frac {3}{2}}} {\displaystyle n_{\text{h}}=N_{\text{V}}\exp \left(-{\frac {E_{\text{F}}-E_{\text{V}}}{kT}}\right),\quad N_{\text{V}}=2\left({\frac {2\pi m_{\text{h}}^{}kT}{h^{2}}}\right)^{\frac {3}{2}}} where EV is the maximum energy of the valence band. Practically, this effective mass tends to vary greatly between absolute zero and room temperature in many materials (e.g., a factor of two in silicon), as there are multiple valence bands with distinct and significantly non-parabolic character, all peaking near the same energy. Determination [edit] Experimental [edit] Traditionally effective masses were measured using cyclotron resonance, a method in which microwave absorption of a semiconductor immersed in a magnetic field goes through a sharp peak when the microwave frequency equals the cyclotron frequency f c = e B 2 π m ∗ {\displaystyle \scriptstyle f_{c}\;=\;{\frac {eB}{2\pi m^{}}}} . In recent years effective masses have more commonly been determined through measurement of band structures using techniques such as angle-resolved photoemission spectroscopy (ARPES) or, most directly, the de Haas–van Alphen effect. Effective masses can also be estimated using the coefficient γ of the linear term in the low-temperature electronic specific heat at constant volume c v {\displaystyle \scriptstyle c_{v}} . The specific heat depends on the effective mass through the density of states at the Fermi level and as such is a measure of degeneracy as well as band curvature. Very large estimates of carrier mass from specific heat measurements have given rise to the concept of heavy fermion materials. Since carrier mobility depends on the ratio of carrier collision lifetime τ {\displaystyle \tau } to effective mass, masses can in principle be determined from transport measurements, but this method is not practical since carrier collision probabilities are typically not known a priori. The optical Hall effect is an emerging technique for measuring the free charge carrier density, effective mass and mobility parameters in semiconductors. The optical Hall effect measures the analogue of the quasi-static electric-field-induced electrical Hall effect at optical frequencies in conductive and complex layered materials. The optical Hall effect also permits characterization of the anisotropy (tensor character) of the effective mass and mobility parameters. Theoretical [edit] A variety of theoretical methods including density functional theory, k·p perturbation theory, and others are used to supplement and support the various experimental measurements described in the previous section, including interpreting, fitting, and extrapolating these measurements. Some of these theoretical methods can also be used for ab initio predictions of effective mass in the absence of any experimental data, for example to study materials that have not yet been created in the laboratory. Significance [edit] The effective mass is used in transport calculations, such as transport of electrons under the influence of fields or carrier gradients, but it also is used to calculate the carrier density and density of states in semiconductors. These masses are related but, as explained in the previous sections, are not the same because the weightings of various directions and wavevectors are different. These differences are important, for example in thermoelectric materials, where high conductivity, generally associated with light mass, is desired at the same time as high Seebeck coefficient, generally associated with heavy mass. Methods for assessing the electronic structures of different materials in this context have been developed. Certain group III–V compounds such as gallium arsenide (GaAs) and indium antimonide (InSb) have far smaller effective masses than tetrahedral group IV materials like silicon and germanium. In the simplest Drude picture of electronic transport, the maximum obtainable charge carrier velocity is inversely proportional to the effective mass: v → = ‖ μ ‖ ⋅ E → {\textstyle {\vec {v}}\;=\;\left\Vert \mu \right\Vert \cdot {\vec {E}}} , where ‖ μ ‖ = e τ / ‖ m ∗ ‖ {\textstyle \left\Vert \mu \right\Vert \;=\;{e\tau }/{\left\Vert m^{}\right\Vert }} with e {\textstyle e} being the electronic charge. The ultimate speed of integrated circuits depends on the carrier velocity, so the low effective mass is the fundamental reason that GaAs and its derivatives are used instead of Si in high-bandwidth applications like cellular telephony. In April 2017, researchers at Washington State University claimed to have created a fluid with negative effective mass inside a Bose–Einstein condensate, by engineering the dispersion relation. See also [edit] Models of solids and crystals: Tight-binding model Free electron model Nearly free electron model Footnotes [edit] ^ Kittel, Introduction to Solid State Physics 8th edition, page 194–196 ^ Charles Kittel (1996). op. cit. Wiley. p. 202. ISBN 978-0-471-11181-8. ^ Jump up to: a b Green, M. A. (1990). "Intrinsic concentration, effective densities of states, and effective mass in silicon". Journal of Applied Physics. 67 (6): 2944–2954. Bibcode:1990JAP....67.2944G. doi:10.1063/1.345414. ^ "Effective mass in semiconductors". University of Colorado Electrical, Computer and Energy Engineering. Archived from the original on 2017-10-20. Retrieved 2016-07-23. ^ Callaway, Joseph (1976). Quantum Theory of the Solid State. Academic Press. ^ Grecchi, Vincenzo; Sacchetti, Andrea (2005). "Bloch Oscillators: motion of wave-packets". arXiv:quant-ph/0506057. ^ Chaitanya K. Ullal, Jian Shi, and Ravishankar Sundararamana American Journal of Physics 87, 291 (2019); ^ Jump up to: a b c Green, M. A. (1990). "Intrinsic concentration, effective densities of states, and effective mass in silicon". Journal of Applied Physics. 67 (6): 2944–2954. Bibcode:1990JAP....67.2944G. doi:10.1063/1.345414. ^ S.Z. Sze, Physics of Semiconductor Devices, ISBN 0-471-05661-8. ^ W.A. Harrison, Electronic Structure and the Properties of Solids, ISBN 0-486-66021-4. ^ This site gives the effective masses of Silicon at different temperatures. ^ M. Schubert, Infrared Ellipsometry on Semiconductor Layer Structures: Phonons, Plasmons and Polaritons, ISBN 3-540-23249-4. ^ Schubert, M.; Kuehne, P.; Darakchieva, V.; Hofmann, T. (2016). "The optical Hall effect – model description: tutorial". Journal of the Optical Society of America A. 33 (8): 1553–68. Bibcode:2016JOSAA..33.1553S. doi:10.1364/JOSAA.33.001553. PMID 27505654. ^ Xing, G. (2017). "Electronic fitness function for screening semiconductors as thermoelectric materials". Physical Review Materials. 1 (6) 065405. arXiv:1708.04499. Bibcode:2017PhRvM...1f5405X. doi:10.1103/PhysRevMaterials.1.065405. S2CID 67790664. ^ Silveirinha, M. R. G.; Engheta, N. (2012). "Transformation electronics: Tailoring the effective mass of electrons". Physical Review B. 86 (16): 161104. arXiv:1205.6325. Bibcode:2012PhRvB..86p1104S. doi:10.1103/PhysRevB.86.161104.{{cite journal}}: CS1 maint: article number as page number (link) ^ Khamehchi, K.A. (2017). "Negative-Mass Hydrodynamics in a Spin-Orbit–coupled Bose-Einstein Condensate". Physical Review Letters. 118 (15) 155301. arXiv:1612.04055. Bibcode:2017PhRvL.118o5301K. doi:10.1103/PhysRevLett.118.155301. PMID 28452531. S2CID 44198065. References [edit] Pastori Parravicini, G. (1975). Electronic States and Optical Transitions in Solids. Pergamon Press. ISBN 978-0-08-016846-3. This book contains an exhaustive but accessible discussion of the topic with extensive comparison between calculations and experiment. S. Pekar, The method of effective electron mass in crystals, Zh. Eksp. Teor. Fiz. 16, 933 (1946). External links [edit] NSM archive Archived 2015-09-24 at the Wayback Machine | hide Authority control databases Edit this at Wikidata | | --- | | International | - GND | | National | - United States | Retrieved from " Categories: Condensed matter physics Mass Hidden categories: CS1 maint: article number as page number Articles with short description Short description is different from Wikidata Articles containing Latin-language text Webarchive template wayback links This page was last edited on 4 September 2025, at 11:47 (UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. 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Press Releases Recent Articles Media Resources Magnify Magazine Employee Benefits ARUP employees can review their employee benefits, including medical coverage and open enrollment information, long-term absence arrangements, employee assistance programs, and other resources. Wellness Center Employees can access a team of dedicated wellness coaches for private or group classes, nutrition information, recovery activities, and more. Family Health Clinic Employees and their dependents can access free primary care, occupational healthcare, immunization, mental health resources, and more at the Family Health Clinic. Learning Management Systems Login Licensure and Accreditations ARUP is a College of American Pathologists (CAP)-, Clinical Laboratory Improvement Amendments (CLIA)-, and International Organization for Standardization (ISO) 15189-accredited laboratory. Additionally, ARUP meets all regulations for labeling non-FDA approved test methods offered by ARUP. Medicare Coverage Information Learn about the restrictions around Medicare billing and reimbursement before you order an ARUP test. Public Health Reporting Learn about ARUP's policy regarding public health reporting and how we stay in compliance with state regulations. Privacy and HIPAA Review our policies and practices for HIPAA and GDPR compliance. ARUP Compliance Policy Browse Tests A-Z ABCDEFGHIJKLMNOPQRSTUVWXYZAllergens Breadcrumb ARUP Home Lab Test Directory Acanthamoeba and Naegleria Culture and Stain, CSF 3000878 Acanthamoeba and Naegleria Culture and Stain , CSF ACANT CSF Example Reports Negative Present Interface Map Interface Map COMPONENT DESCRIPTION TEST TYPE INFECTIOUS UNIT OF MEASURE NUMERIC MAP LOINC 0060245 Acanthamoeba and Naegleria Culture Resultable N 9781-6 2003035 Amoeba, Giemsa Stain Resultable N 87965-0 For questions regarding the Interface Map, please contact interface.support@aruplab.com. Download to Excel Ordering Recommendation Recommendations when to order or not order the test. May include related or preferred tests. Detect Acanthamoeba spp, Naegleria spp, and other free-living amoebae. New York DOH Approval Status Indicates whether a test has been approved by the New York State Department of Health. This test is New York state approved. Specimen Required Patient PreparationInstructions patient must follow before/during specimen collection. CollectSpecimen type to collect. May include collection media, tubes, kits, etc. CSF. Specimen PreparationInstructions for specimen prep before/after collection and prior to transport. Transfer 1 mL CSF to a sterile ARUP Standard Transport Tube (ARUP supply # 43115 ) available online through eSupply using ARUP Connect™ or contact Client Services at (800) 522-2787. (Min: 0.5 mL) Storage/Transport TemperaturePreferred temperatures for storage prior to and during shipping to ARUP. See Stability for additional info. Room temperature. Unacceptable ConditionsCommon conditions under which a specimen will be rejected. Specimens in media or preservatives. RemarksAdditional specimen collection, transport, or test submission information. StabilityAcceptable times/temperatures for specimens. Times include storage and transport time to ARUP. Ambient: 72 hours; Refrigerated: Unacceptable; Frozen: Unacceptable Methodology Process(es) used to perform the test. Qualitative Culture/Microscopy/Giemsa Stain Performed Days of the week the test is performed. Sun-Sat Reported Expected turnaround time for a result, beginning when ARUP has received the specimen. 2-10 days Reference Interval Normal range/expected value(s) for a specific disease state. May also include abnormal ranges. Negative. Interpretive Data May include disease information, patient result explanation, recommendations, or details of testing. The stain will detect free-living amoeba such as Naegleria fowleri and Balamuthia mandrillaris. The culture will detect free-living amoeba such as Acanthamoeba species and Naegleria fowleri but will NOT detect Balamuthia mandrillaris. Compliance Category Standard Note Additional information related to the test. For all other specimen types refer to Acanthamoeba and Naegleria Culture (ARUP test code 0060245) and Amoeba Calcofluor Stain (ARUP test code 0060250). For Entamoeba histolytica detection refer to Entamoeba histolytica Antigen, EIA (ARUP test code 0058001). Hotline History History of test changes published on ARUP Hotlines for the last two years N/A CPT Codes The American Medical Association Current Procedural Terminology (CPT) codes published in ARUP's Laboratory Test Directory are provided for informational purposes only. The codes reflect our interpretation of CPT coding requirements based upon AMA guidelines published annually. CPT codes are provided only as guidance to assist clients with billing. ARUP strongly recommends that clients confirm CPT codes with their Medicare administrative contractor, as requirements may differ. CPT coding is the sole responsibility of the billing party. ARUP Laboratories assumes no responsibility for billing errors due to reliance on the CPT codes published. 87081; 87207 Are you an ARUP Client? Click here for your pricing. Components Components of test | Component Test Code | Component Chart Name | LOINC | --- | 0060245 | Acanthamoeba and Naegleria Culture | 9781-6 | | 2003035 | Amoeba, Giemsa Stain | 87965-0 | Component test codes cannot be used to order tests. The information provided here is not sufficient for interface builds; for a complete test mix, please click the sidebar link to access the Interface Map. Aliases Other names that describe the test. Synonyms. Culture for Acanthamoeba, Spinal Fluid Culture for Naegleria, Spinal Fluid Detection of Acanthamoeba/Naegleria Species Naegleria Culture Stain for Balamuthia mandrillaris Acanthamoeba and Naegleria Culture and Stain, CSF We value your Feedback. Feedback I want to provide feedback regarding Comment First Name Last Name Email Client ID CAPTCHA Get new captcha! What code is in the image? Enter the characters shown in the image. This question is for testing whether or not you are a human visitor and to prevent automated spam submissions. Leave this field blank Specimen Details Clinical ARUP Consult Live Chat About About Careers News Press Releases Testing Laboratory Test Directory Choose the Right Test Resources Education Center Continuing Education Expert Videos Podcast Episodes Events Contact Us 24/7 Phone Support Sales Inquiries By continuing to browse our website, you agree to the use of cookies in accordance with our Privacy Policy. Agree © 2025 ARUP Laboratories. 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https://www.teacherspayteachers.com/Browse/Search:solving%20equations%20with%20balance
Solving Equations With Balance | TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 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Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Solving Equations With Balance 1,400+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Supports Price Format All filters Filters Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education More Not grade specific Subject Art Coloring pages Graphic arts English language arts Balanced literacy Creative writing Grammar 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They are a completely digital way to engage your students in the rigor of solving one-step equations, writing equations, and using balance scales to solve one-step (and a few two-step) equations. These activities will give your students immediate feedback as they complete them, which students really enjoy. With this purchase, you will get three writing, solving and modelling equations activities that you can sh 6 th - 8 th Algebra, Math $3.00 Original Price $3.00 Rated 5 out of 5, based on 4 reviews 5.0(4) Add to cart Wish List Mobile Math Puzzles, Balancing Algebraic Equations, Printable Logic Puzzles Created by Master Xuan Mobile math puzzles help students build the logic of balancing algebraic equations and solving systems of equations. Students will figure out the weight of the shapes for the mobile to make the mobile balance. The thinking used in these puzzles is the same as the logic used behind solving simultaneous linear equations. With colorful and visual shapes, students will be engaged. This activity has been updated and is now available in 3 different versions: Boom™ Cards, Printable, and an Easel Activ 4 th - 8 th Algebra, Math CCSS 8.EE.C.8 $3.99 Original Price $3.99 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Solving Two Step Equations with a Balance Created by Try-Angle Math Have students model solving 2-step equations with a balance. Students cross off the same number of circles (loose items) from both sides. Then, they determine how many items are in 1 box. While modeling with the balance, students will also solve the equation algebraically, using inverse operations. This activity is helpful for understanding the order of steps to solve equations. Let's Stay Connected!It's my goal to create middle school math lessons that reduce math anxiety and encourage stud 6 th - 8 th Algebra, Math CCSS 7.EE.B.4 $3.00 Original Price $3.00 Add to cart Wish List Koala Craft Balancing&Solving One Step Equations With Addition And Subtraction Created by Master Xuan Are you looking for NO PREP and DIFFERENTIATED solving one step equations with addition and subtraction worksheets for your 2nd grade students? Want to make practicing adding and subtracting within 100 with regrouping more ENGAGING? Then you will love this simple Koala craft! This math craftivity is easy to use, and the ready-made template can be used if you want a cute result without spending a lot of crafting time. This Australia animal math activity can be used as a topper to those wor 2 nd Math, Numbers CCSS 2.NBT.B.5 Also included in:Cute Animal Crafts 2nd Grade Math Review Activities, Worksheets, Alphabet Bundle $3.00 Original Price $3.00 Add to cart Wish List Balance Scale Model for Solving Equations with Algebra Tiles Created by Fun with Fibonacci This is a graphic organizer for your students to use when students are Solving Equations with Algebra Tiles. They may use it as a single use consumable (where you make several copies for each student) or in a dry erase sleeve for a more sustainable option. The intention is to use the organizer with algebra tiles, with positive terms and negative terms separated, allowing students to work in the concrete stage of learning. If your students' learning progresses, there is room for them to move from 5 th - 8 th Algebra, Applied Math, Math Test Prep CCSS 5.OA.A.1 , 5.OA.A.2 , 5.OA.B.3 +15 $2.00 Original Price $2.00 Add to cart Wish List Solving Equations with the Balance Model | Digital Easel & Slides Created by JTreasures Snatch this product as it introduces how to solve equations with the balance model. Not only does it explain it, but it aims to achieve this by the Gradual Release model. With multiple guided steps, students gain the confidence to achieve this independently! This product is created on Google Slides and TPT's Easel, both formats are compatible with Google Classroom. Just post the link and it's ready to go! To use it with the Easel platform, click the "Open Digital Activity" button 4 th - 6 th Algebra, Basic Operations, Math CCSS 6.EE.A.2a Also included in:Intro to Inequalities and Equations Bundle | Google Slides & Easel $4.25 Original Price $4.25 Rated 4 out of 5, based on 2 reviews 4.0(2) Add to cart Wish List Solving Equations With Variables on Both Sides Balance Beams Activity 1 Created by Easy as Pi Learning Incorporated Shape Balance Challenge- The Power of Weighted Equations Embark on an exciting scientific journey with "Shape Balance Challenge," where students become weight detectives, analyzing tied balance beams and unraveling the mysteries of unknown weights. This unique educational tool seamlessly integrates the thrill of balancing shapes with the intellectual challenge of solving equations in a dynamic and engaging setting. Join Mackenzie in her scientific endeavor, where she uses balance beams and vario 7 th - 10 th Algebra, Math CCSS 8.EE.C.7 , 8.EE.C.7b , 8.EE.C.8c Also included in:Solving Equations With Variables on Both Sides BUNDLE $4.99 Original Price $4.99 Add to cart Wish List Solving Two Step Equations Activity with Cool Backtracking Model &Balancing Created by Mr Conceptual Understanding This digital lesson activity is on 7th Grade Math solving two-step linear equations using reverse operations by backtracking. This interactive activity on solving 2 step equations has a bit of guided notes with a video example and scaffolded examples that students can put in notes. A neat visual model to make solving 2-step equations understandable, easy and fun! Contains positive and negative rational numbers - fractions and decimals. Great for all students, and also for students with IEPs (sp 7 th - 9 th Algebra, Math CCSS 7.EE.B.3 , 7.EE.B.4 , 7.EE.B.4a +2 Also included in:Solving Two Step Equations Activity BUNDLE | Backtrack & Algorithm | 2-Step $3.00 Original Price $3.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Balancing Equations With Addition and Subtraction Problem Solving Games Created by Little Geniuses This hands-on interactive resource includes terrific game boards that allow your students hours of practice and fun. Children flip dominoes and THINK about the math sentence they are building. In the first set of game activities, players are competing to build the greatest or least accumulative total (depending on the goal of the game) and the dominoes are an integral part to making this an engaging and motivating activity. In the more advanced set of games, children are making decisions wh 1 st - 3 rd Basic Operations, Numbers CCSS K.CC.A.1 , K.CC.A.2 , K.CC.A.3 +21 $4.75 Original Price $4.75 Add to cart Wish List Balancing Equations&Solving Unknown {Sorts, Stations & Around the Room} Created by Do Good for First - Jillian Duguid Balancing equations and finding equality in number sentences is one of the most challenging concepts each year! It's a skill that needs a lot of practice to master. Make the challenge of balancing equations and solving for the unknown fun! These activities could be used in a guided math group, station or independently! Activities included:★ "Find a Balance" ~ Matching number sentences with equal sums ★ Balanced and unbalanced number sentence sort (true/false pineapples in color and b/w) ★ Inde 1 st - 2 nd Applied Math, Basic Operations CCSS 1.OA.B.4 , 1.OA.D.7 , 1.OA.D.8 Also included in:MEGA MATH (GROWING) BUNDLE! Google Slides & Seesaw • Printables • QR Code Hunts! $3.00 Original Price $3.00 Rated 4.9 out of 5, based on 92 reviews 4.9(92) Add to cart Wish List Balancing Equations: solving unknown, EQUAL, equivalence, addition & subtraction Created by Sam Nowak - Fun With Firsties My firsties were struggling with the concept of balancing equations...so I scrapped the program I was following and created my own! And, as always, sprinkled some FUN in with lots of meaningful learning ;) This jam-packed mini unit is great way for kiddos to get some extra practice with a sometimes challenging concept; Balancing equations can be quite tricky for some littles and the variety of activities in this little pack is sure to help them practice and master this skill. If your students K - 3 rd Basic Operations, Math, Numbers $4.25 Original Price $4.25 Rated 4.88 out of 5, based on 73 reviews 4.9(73) Add to cart Wish List Solving One-Step Equations with a Scale GOOGLE SLIDES DISTANCE LEARNING Created by Math in Demand Students will be solving 10 problems on one-step equations in the form of x + a = b where a and b are integers. Students will balance the equations on a scale by dragging "x", "+1", and "-1" boxes. There are 2 different ways that you can implement my activity into your classroom: (1) Google Slides and (2) PowerPoint. You can upload this resource into Google Classroom. Please watch the video! 6 th - 7 th Algebra, Basic Operations, Math Also included in:DIGITAL MIDDLE SCHOOL MATH GOOGLE SLIDES DISTANCE LEARNING RESOURCES $2.00 Original Price $2.00 Rated 4.84 out of 5, based on 19 reviews 4.8(19) Add to cart Wish List Balanced Equations - Mixed Operations Boom Cards SOL 3.CE.2g Created by VATeacherofAllThings Practice identifying and creating equivalent equations that use multiplication, division, addition and/or subtraction using these 30 interactive, self-checking digital task cards. This deck aligns with 2023 Virginia math SOL standard 3.CE.2g. Students will:identify true equations (+, -, x and/or ÷) that use the equal or unequal signcomplete a balanced equation (+, -, x and/or ÷) by filling in the correct missing numbercomplete an equation (+, -, x and/or ÷) with an equal or unequal signsolve a 2 nd - 4 th Basic Operations, Math, Math Test Prep Also included in:Multiplication & Division Boom Cards Bundle SOL 3.CE.2 $2.75 Original Price $2.75 Rated 4.79 out of 5, based on 14 reviews 4.8(14) Add to cart Wish List Balancing Equations Addition and Subtraction Created by Spivey Sparks 1st and 2nd grades enjoy the challenge of balancing sums and differences equations! Problem solving is incorporated in these differentiated addition and subtraction task cards. Students solve for one, two, or three different addend or subtrahends. This Set is part of a Bundle:Differentiated 1st & 2nd Grades Math BundleThe 4 Levels of difficulty have been designed to meet the needs of all learners in your classroom.This pack Includes: • Teacher Directions • Game Boards (2 options- color & 1 st - 2 nd Mental Math CCSS 1.NBT.C.4 , 2.NBT.B.5 , 2.NBT.B.6 +4 Also included in:1st and 2nd Grade Math Centers $4.00 Original Price $4.00 Rated 4.87 out of 5, based on 42 reviews 4.9(42) Add to cart Wish List Math Jenga Game Cards for Balancing Equations with Addition and Subtraction Created by Life Between Summers This fun and engaging color Jenga game is perfect for spicing up math centers, review, or test prep. It contains problems for practice and review with BALANCING EQUATIONS, including understanding the meaning of the equal sign, and determining if addition and subtraction equations are true or false. These cards are to be used with this color Jenga set: Lewo Wooden Stacking Board Games Building Blocks for Kids. Game can be played in partners or a small group. Students take turns rolling the colo 1 st Basic Operations, Math CCSS 1.OA.B.3 , 1.OA.B.4 , 1.OA.D.7 +1 Also included in:Jenga 1st Grade Math Games for Centers or Review MEGA Growing Bundle + Holidays $3.00 Original Price $3.00 Rated 4.71 out of 5, based on 7 reviews 4.7(7) Add to cart Wish List Solving Add & Multiply Equations Variables on One Side | Pre-Algebra - With Ans Created by Distinguished opportunity teacher Pre Algebra Equations (One Side)- Pre-algebra - One Side Equations- Solving equations- Inverse operations- Simplifying expressions- Algebra practice- Math worksheets- Balance methodHelp students build a strong foundation in algebra with this comprehensive resource focused on Solving One-Sided Equations. Perfect for introducing PSre-algebra concepts, this set of worksheets guides learners step-by-step through the process of Solving Equations with variables on one side. With clear instructions 6 th - 10 th Algebra, Math, Other (Math) Also included in:Pre-Algebra Equations Bundle: on One Side | Add, Subtract, Multiply & Divide $4.10 Original Price $4.10 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List PBL Solving One Step Equations Learning Task using Balance Scales Created by School Sisters Math Engage your students in discovering how to solve one step equations using inverse operations! In this low floor high ceiling task, students will use a balance scale to help them make sense of how to solve an equation with inverse operations. This task will help students learn and understand concepts from this standard: ✅Solve real-world and mathematical problems by writing and solving equations of the form x + a = b and ax = b for cases in which a, b and x are all non-negative rational numbe 6 th - 8 th Math CCSS 6.EE.B.5 , 6.EE.B.7 Also included in:7th Grade Low Floor High Ceiling Math Tasks Introduce Concepts BTC Style PBL $2.25 Original Price $2.25 Rated 5 out of 5, based on 4 reviews 5.0(4) Add to cart Wish List Algebra Action Folder Game {balancing equations&solving for x} Created by forsuchtime Students learning the beginnings of hands-on algebra will love this math center game. They will be balancing problems and solving for "x" as they move from start to finish in this interactive folder game. Example Problems: (In the game, students are given a balanced equation with what number x equals. Students determine if the situation is true or false) 2x+5=3x+1 4x=2x+4 4x-x+2=x+2x-x+7 2(x+3)=10 Perfect to accompany the "Hands on Equations" curriculum! 3 rd - 5 th Algebra, Math $2.00 Original Price $2.00 Rated 4.77 out of 5, based on 58 reviews 4.8(58) Add to cart Wish List Solving Equations with Hanger Models Project-Digital and Printable Created by The Math Blender Connect the concrete to the abstract! Hanger models help students visualize balancing and solving equations. These models are built by STUDENTS! Students build the hanger models and then participate in a gallery walk to represent and solve the equation hanger models of their peers. Hands-on and engaging, this activity needs no differentiation because students can design their own model and create simple or complex models. Included in this resource: 1. Many shapes to create the hanger models ( 6 th - 8 th Algebra, Math CCSS 6.EE.B.7 , 7.EE.B.4a , 8.EE.C.7 $2.00 Original Price $2.00 Rated 4.75 out of 5, based on 4 reviews 4.8(4) Add to cart Wish List Algebraic Equations with Pan Balances Digital Boom Cards Distance Learning Created by Teacher Gems These algebraic equations with pan balances digital Boom Cards™ are a great way to practice problem-solving while learning algebra. Challenge your students to think outside the box with these algebraic equations: pan balance digital task cards! A great way to practice critical thinking skills! There are 40 task cards of varying difficulty perfect for grades 4-6. Cards start out fairly easy and get increasingly more difficult as students progress through the deck (cards). Students type their answ 4 th - 6 th Algebra $4.00 Original Price $4.00 Rated 4.8 out of 5, based on 5 reviews 4.8(5) Add to cart Wish List Writing &Solving Two Step Equations - Balancing Equations Task Cards Created by Tessa Maguire Do your students struggle with understanding two step equations? These Balance the Equation task cards help students see the mathematical operations through models. Each balance models both multiplication and addition for a two step equation. Students build their algebraic thinking with equivalence by working through balance the equation activities. By integrating algebraic concepts in elementary classrooms, students develop math reasoning and computation skills while applying number sense, logi 3 rd - 5 th Algebra, Arithmetic, Mental Math CCSS 3.OA.A.1 , 3.OA.A.4 , 3.OA.D.8 +1 Also included in:Equations, Patterns, Input Output Function Tables, & Equality with Balances $3.00 Original Price $3.00 Rated 5 out of 5, based on 3 reviews 5.0(3) Add to cart Wish List Balancing / Equivalent Equations with 2-Digit Numbers | Digital | Google Created by Miss Cabrera's Corner Looking for an activity to help students understand equivalency? This is the resource for you! This digital resource includes 6 slides to help students understand the meaning of an equal sign! Students will use their problem-solving skills to create equivalent equations involving addition and subtraction. These slides are interactive and allow students to drag and drop the number sentences to help them problem solve. Each slide has 5 equivalent equations to create. 2 nd - 5 th Basic Operations, Math $2.00 Original Price $2.00 Rated 4.6 out of 5, based on 5 reviews 4.6(5) Add to cart Wish List Solving Linear Equations Assessment Bundle | Quizzes, Test, Checkpoints, Review Created by Maths360 Need ready-to-use, editable assessments for your solving linear equations (whole number solutions) unit? This bundle has everything you need to evaluate student understanding while providing meaningful, differentiated feedback.This resource includes checkpoints, quizzes, a review assignment, and a unit test plus rubrics and answer keys. The assessments are designed to be flexible for different grading systems, including standards-based grading, and to support diverse learners with choice and tie 4 th - 12 th, Adult Education Algebra, Other (Math) Also included in:Grade 7 Math Assessment Bundle | Checkpoints, Quizzes, Reviews, Tests, Rubrics $7.49 Original Price $7.49 Add to cart Wish List Halloween Algebra Coloring Pages | Solving Equations Practice Worksheet Bundle Created by Math Rocks Eh Looking for a fun, no-prep way to keep your students engaged in math practice this October? This Halloween Algebra Coloring Bundle is perfect for reviewing and practicing solving equations while adding a festive twist to your classroom. Students solve equations, match their answers to colors, and reveal a spooky Halloween picture — all while reinforcing key Algebra 1 skills! This set includes 4 progressively more challenging activities so you can build skills step by step and keep every learn 7 th - 9 th Algebra, Coloring Pages, Math $4.50 Original Price $4.50 Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 1,400+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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https://en.wikipedia.org/wiki/Particle_Data_Group
Jump to content Particle Data Group Čeština Deutsch Español Français Italiano 日本語 Polski Português Русский Türkçe Edit links From Wikipedia, the free encyclopedia International collaboration of particle physicists The Particle Data Group (PDG) is an international collaboration of particle physicists that compiles and reanalyzes published results related to the properties of particles and fundamental interactions. It also publishes reviews of theoretical results that are phenomenologically relevant, including those in related fields such as cosmology. The PDG currently publishes the Review of Particle Physics and its pocket version, the Particle Physics Booklet, which are printed biennially as books, and updated annually via the World Wide Web. In previous years, the PDG has published the Pocket Diary for Physicists, a calendar with the dates of key international conferences and contact information of major high energy physics institutions, which is now discontinued. PDG also further maintains the standard numbering scheme for particles in event generators, in association with the event generator authors. Review of Particle Physics [edit] The Review of Particle Physics (formerly Review of Particle Properties, Data on Particles and Resonant States, and Data on Elementary Particles and Resonant States) is a voluminous, 1,200+ page reference work which summarizes particle properties and reviews the current status of elementary particle physics, general relativity and Big Bang cosmology. Usually singled out for citation analysis, it is currently the most cited article in high energy physics, being cited more than 2,000 times annually in the scientific literature (as of 2009[update]). The Review is currently divided into three sections: Particle Physics Summary Tables—Brief tables of particles: gauge and higgs bosons, leptons, quarks, mesons, baryons, constraints for the search for hypothetical particles and violation of physical laws. Reviews, Tables and Plots—Review of fundamental concepts from mathematics and statistics, table of Clebsch-Gordan coefficients, periodic table of elements, table of electronic configuration of the elements, brief table of material properties, review of current status in the fields of Standard Model, Cosmology, and experimental method of particle physics, and with tables of fundamental physical and astronomical constants (many from CODATA and the Astronomical Almanac). Particle Listings—Comprehensive version of the Particle Physics Summary Tables, with all significant measurements fully referenced. A condensed version of the Review, with the Summary Tables, a significantly shortened Reviews, Tables and Plots, and without the Particle Listings, is available as a 300-page, pocket-sized Particle Physics Booklet. The history of the Review of Particle Physics can be traced back to the 1957 article Hyperons and Heavy Mesons (Systematics and Decay) by Murray Gell-Mann and Arthur H. Rosenfeld, and the unpublished update tables for its data with the title Data for Elementary Particle Physics (University of California Radiation Laboratory Technical Report UCRL-8030) that were circulated before the actual publication of the original article. In 1963, Matts Roos independently published a compilation Data on Elementary Particles and Resonant States. On his suggestion, the two publications were merged a year later into the 1964 Data on Elementary Particles and Resonant States. The publication underwent three renamings thereafter: in 1965 to Data on Particles and Resonant States, in 1970 to Review of Particle Properties, and in 1996 to the present form Review of Particle Physics. Starting in 1972, the Review no longer appeared exclusively in Reviews of Modern Physics, but also in Physics Letters B, European Physical Journal C, Journal of Physics G, Physical Review D, and Chinese Physics C (depending on the year). Past editions of the Review of Particle Physics [edit] | Year | Title | Reference | --- | 1957–1963 | Hyperons and Heavy Mesons (Systematics and Decay) | M. Gell-Mann & A. H. Rosenfeld, Annu. Rev. Nucl. Sci. 7, 407 (1957). | | | Data for Elementary Particle Physics | University of California Radiation Laboratory Technical Report UCRL-8030 (unpublished). | | | Data on Elementary Particles and Resonant States, November 1963; Tables of Elementary Particles and Resonant States | M. Roos, Nucl. Phys. 52, 1 (1964); M. Roos, Rev. Mod. Phys. 35, 314 (1963). | | 1964 | Data on Elementary Particles and Resonant States | A. H. Rosenfeld et al., Rev. Mod. Phys. 36, 977 (1964). | | 1965 | Data on Particles and Resonant States | A. H. Rosenfeld et al., Rev. Mod. Phys. 37, 633 (1965). | | 1967 | Data on Particles and Resonant States | A. H. Rosenfeld et al., Rev. Mod. Phys. 39, 1 (1967). | | 1968 | Data on Particles and Resonant States | A. H. Rosenfeld et al., Rev. Mod. Phys. 40, 77 (1968). | | 1969 | Data on Particles and Resonant States | N. Barash-Schmidt et al., Rev. Mod. Phys. 41, 109 (1969). | | 1970 | Review of Particle Properties | A. Barbaro-Galtieri et al., Rev. Mod. Phys. 42, 87 (1970). | | 1971 | Review of Particle Properties | A. Rittenberg et al., Rev. Mod. Phys. 43, S1 (1970). | | 1972 | Review of Particle Properties | A. Barbaro-Galtieri et al., Phys. Lett. B 39, 1 (1972). | | 1973 | Review of Particle Properties | T. A. Lasinski et al., Rev. Mod. Phys. 45, S1 (1973). | | 1974 | Review of Particle Properties | A. Barbaro-Galtieri et al., Phys. Lett. B 50, 1 (1974). | | 1975 | Review of Particle Properties: Supplement to 1974 edition | A. Barbaro-Galtieri et al., Rev. Mod. Phys. 47, 535 (1975). | | 1976 | Review of Particle Properties | T. G. Trippe et al., Rev. Mod. Phys. 48, S1 (1976). | | 1977 | New Particle Searches and Discoveries: A Supplement to the 1976 Edition of "Review of Particle Properties" | C. Bricman et al. Phys. Lett. B 68, 1 (1978). | | 1978 | Review of Particle Properties | C. Bricman et al. Phys. Lett. B 75, 1 (1978). | | 1980 | Review of Particle Properties | R. L. Kelly et al., Rev. Mod. Phys. 52, S1 (1980). | | 1982 | Review of Particle Properties | M. Roos et al., Phys. Lett. B 111, 1 (1982). | | 1984 | Review of Particle Properties | C. G. Wohl et al., Rev. Mod. Phys. 56, S1 (1986). | | 1986 | Review of Particle Properties | M. Aguilar-Benítez et al., Phys. Lett. B 170, 1 (1986). | | 1988 | Review of Particle Properties | G. P. Yost et al., Phys. Lett. B 204, 1 (1988). | | 1990 | Review of Particle Properties | J. J. Hernández et al., Phys. Lett. B 239, 1 (1990). | | 1992 | Review of Particle Properties | K. Hikasa et al., Phys. Rev. D 45, S1 (1992). | | 1994 | Review of Particle Properties | L. Montanet et al., Phys. Rev. D 50, 1173 (1994). | | 1996 | Review of Particle Physics | R. M. Barnett et al., Phys. Rev. D 54, 1 (1996). | | 1998 | Review of Particle Physics | C. Caso et al., Eur. Phys. J. C 3, 1 (1998). | | 2000 | Review of Particle Physics | D. E. Groom et al., Eur. Phys. J. C 15, 1 (2000). | | 2002 | Review of Particle Physics | K. Hagiwara et al., Phys. Rev. D 66, 010001 (2002). | | 2004 | Review of Particle Physics | S. Eidelman et al., Phys. Lett. B 591, 1 (2004). | | 2006 | Review of Particle Physics | W.-M. Yao et al., J. Phys. G 33, 1 (2006). | | 2008 | Review of Particle Physics | C. Amsler et al., Phys. Lett. B 667, 1 (2008). | | 2010 | Review of Particle Physics | K. Nakamura et al. (Particle Data Group), J. Phys. G 37, 075021 (2010) | | 2012 | Review of Particle Physics | J. Beringer et al. (Particle Data Group), Phys. Rev. D 86, 010001 (2012) | | 2014 | Review of Particle Physics | K.A. Olive et al. (Particle Data Group), Chin. Phys. C 38, 090001 (2014). | | 2016 | Review of Particle Physics | C. Patrignani et al. (Particle Data Group), Chin. Phys. C 40, 100001 (2016). | | 2018 | Review of Particle Physics | M. Tanabashi et al. (Particle Data Group), Phys. Rev. D 98, 030001 (2018). | | 2020 | Review of Particle Physics | P.A. Zyla et al. (Particle Data Group), 083C01 (2020). | | 2022 | Review of Particle Physics | R.L. Workman et al. (Particle Data Group), Prog. Theor. Exp. Phys. 2022, 083C01 (2022). | | 2024 | Review of Particle Physics | S. Navas et al. (Particle Data Group), Phys. Rev. D 110, 030001 (2024). | See also [edit] CODATA References [edit] ^ "Order PDG products". Particle Data Group. Retrieved 3 September 2020. ^ C. Amsler; et al. (2008). "Review of Particle Physics" (PDF). Physics Letters B. 667 (1–5): 1–6. Bibcode:2008PhLB..667....1A. doi:10.1016/j.physletb.2008.07.018. hdl:1854/LU-685594. PMID 10020536. S2CID 227119789. Archived from the original (PDF) on 2020-09-07. Retrieved 2019-11-30. ^ T. Brooks (2008). "The 50 topcited papers from SPIRES in 2007". Symmetry Magazine. Retrieved 2009-10-08. ^ "INSPIRE HEP: Most cited papers since 2016". ^ M. Gell-Mann; A. H. Rosenfeld (1957). "Hyperons and Heavy Mesons (Systematics and Decay)" (PDF). Annual Review of Nuclear Science. 7: 407–478. Bibcode:1957ARNPS...7..407G. doi:10.1146/annurev.ns.07.120157.002203. hdl:2027/mdp.39015086417295. ^ A. H. Rosenfeld; et al. (1964). "Data on Elementary Particles and Resonant States" (PDF). Reviews of Modern Physics. 36 (4): 977–1004. Bibcode:1964RvMP...36..977R. doi:10.1103/RevModPhys.36.977. ^ A. H. Rosenfeld (1975). "The Particle Data Group: Growth and Operations-Eighteen Years of Particle Physics". Annual Review of Nuclear Science. 25: 555–598. Bibcode:1975ARNPS..25..555R. doi:10.1146/annurev.ns.25.120175.003011. ^ M. Roos (1964). "Data on elementary particles and resonant states, November 1963". Nuclear Physics. 52: 1–24. Bibcode:1964NucPh..52....1R. doi:10.1016/0029-5582(64)90671-6. ^ M. Roos (1963). "Tables of Elementary Particles and Resonant States". Reviews of Modern Physics. 35 (2): 314–323. Bibcode:1963RvMP...35..314R. doi:10.1103/RevModPhys.35.314. External links [edit] Particle Data Group official site and electronic edition of Review of Particle Physics 2018 Photo of the 2004 Review of Particle Physics First edition of the wallet card from the Particle Data Group, 1958 Particle Physics Booklet, current version Particle Physics Booklet, July 2010 Particle Physics Booklet, 2014 Particle Physics Booklet, 2018 | Authority control databases | | --- | | International | ISNI VIAF | | National | United States France BnF data | | Other | | Retrieved from " Categories: Particle physics Physical constants Hidden categories: Articles with short description Short description is different from Wikidata Articles containing potentially dated statements from 2009 All articles containing potentially dated statements
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https://www.echemi.com/community/what-is-the-role-of-sulphuric-acid-in-nitration-of-benzene_mjart2204201480_483.html
 Community Sign in | Join free Home > Community > What is the role of sulphuric acid in nitration of benzene? Downvote Chemical reactions Molecules Physical chemistry Atoms Science Acids Chemistry Organic chemistry Posted by Omar N. What is the role of sulphuric acid in nitration of benzene? The source of the nitronium ion is through the protonation of nitric acid by sulfuric acid, which causes the loss of a water molecule and formation of a nitronium ion. Chris Harrington Follow Following The source of the nitronium ion is through the protonation of nitric acid by sulfuric acid, which causes the loss of a water molecule and formation of a nitronium ion. More Upvote VOTE Downvote Nitration of Benzene is an example of ELECTROPHILIC AROMATIC SUBSTITUTION. It can take place in the presence of nitrating mixture ( HNO3 + H2SO4) . Sulphuric acid has an important role to play in this reaction ,as it protonates nitric acid which leads to loss of water molecule and results in formation of NITRONIUM ION ( the ELECTROPHILE) . This NITRONIUM ION further attacks the benzene ring and hence undergoes nitration. Chris Robinson Follow Following Nitration of Benzene is an example of ELECTROPHILIC AROMATIC SUBSTITUTION. It can take place in the presence of nitrating mixture ( HNO3 + H2SO4) . Sulphuric acid has an important role to play in this reaction ,as it protonates nitric acid which leads to loss of water molecule and results in formation of NITRONIUM ION ( the ELECTROPHILE) . This NITRONIUM ION further attacks the benzene ring and hence undergoes nitration. More Upvote VOTE Downvote Related Posts How many ml are in 1 glass of water? 26 Upvotes · 2 Comments What is the best solution for cleaning a meth pipe? 15 Upvotes · 3 Comments What is dicyanin, and what is it used for? 14 Upvotes · 3 Comments How do you prepare 0.5M of an HCl solution? 14 Upvotes · 8 Comments Is it safe to use expired salt packets from neti pots for nasal irrigation? 19 Upvotes · 2 Comments How do I prepare a 1M KOH solution? 16 Upvotes · 8 Comments How do I make a 1N H2SO4 solution? 15 Upvotes · 10 Comments Wholesale About ECHEMI Encyclopedia Contact Us Local Mall Market Price & Insight Marketing Kit Trade Data ECHEMI Group Join Us For Buyers For Suppliers Copyright@Qingdao ECHEMI Digital Technology Co., Ltd.
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https://www.youtube.com/watch?v=zIcINxhxFH4
🔄 Trimerization Reactions (6 Questions in Last 5 Years) || Co-Pro series || IIT-JEE | NEET ✅ IITian explains by Unacademy 220000 subscribers 476 likes Description 8234 views Posted: 24 May 2022 Acetylene or Ethyne can be converted to benzene through the Trimerization Mechanism. When we Pass acetylene through a red hot iron tube at a temperature of 873k. The molecules of acetylene will be trimerized to give benzene. This reaction is highly exothermic and Self Sustaining. In this Video, MKA Sir will explain this and similar type of mechanism in a very detailed manner. ✅ Combat Scholarship: ✅ Brainanza Link: ✅ Envision 2.0 Course Link: h ✅ Evolve Batch Link: Use Code IITIANEXPLAINS for maximum discounts and avail of this offer. ✅ Ace Batch Link: ✅ Evolve 2.O Batch for JEE Main and Advanced 2023 : Use Code IITIANEXPLAINS for Maximum Discounts. Follow me on Unacademy for all Courses: Know about MKA Sir: IITian Explains Website: MKA Sir 161 comments Transcript: एक मैटल होस्ट नहीं स्टार्टिंग टेंप्रेचर फेज़ पे इलेक्ट्रॉनिक्स यहां यह इलेक्ट्रॉन से और मेटल से तो छोटे फिर दो इलेक्ट्रॉन वह एक यह कितना परसेंट है 40 पर सेंट है मतलब पॉइंट्स और ठीक है अब देखो यहां से चार हेलो एवरीवन यू ऑल वेरी वेलकम टू आईटी एक्सप्लेंस आईटी एक्सप्लेंस में आप सबका बहुत बहुत स्वागत है मैं आपको अपना इनके सर तो गाइस जैसे कि आप देख रहे हो हम एक बड़े कमाल का को प्रोसेस में वीडियो लेकर आएं हैं मैंने कुछ रिएक्शन आपको दिया है जो बहुत ही इंपोर्टेंट है और आज भी हम इस सीरीज को आगे बढ़ाते हुए इस टाइम राइजेशन रिएक्शंस लेकर आएं और इसको यहां तुझे रिएक्शन देख रहे हैं जहां पर तीन को मेंटेन सचिन भउजी लमका फॉर्मेशन कर रहे हैं या इससे मिलते-जुलते दो-तीन इक्वेशंस आपने रिसेंटली देखे होंगे अगर मैं आपको बहुत ज्यादा एडवांस 2021 मतलब कुछ महीने पहले अभी यहीं का पेपर है उसमें भी आपको एक फ्राइडे संरक्षण देखा था इसके अलावा 2019 के पेपर में मिला है आपको मेंस के पेपर में भी दिखाए बहुत फ्रिक्वेंटली पूछा जाता है कभी आपने सोचा इसका मैकेनिज़्म भी होता है क्या कि मतलब ऐसे ही बस एरो घुमा घुमा के दिखा दिया और यह बन गया उसको मैकेनिज़्म भी ID या इससे कैसे बहुत बढ़िया क्वेश्चंस बन सकते हैं यही और निकले क्या इसमें बहुत इंपोर्टेंट है वह कौन सी चीज है जो आपको हर हाल में देखकर जानी है आज हम इसके ऊपर बात कर रहे होंगे ठीक है तो हमारे सारे के सारे स्टूडेंट बिल्कुल रेडी रहेगा मचाएंगे आज कुछ बहुत ही शानदार टॉपिक के साथ आई तो स्वागत है उसका आईटी एक्सप्लेंस में स्टार्ट करते हैं आगे बढ़ने से पहले थोड़ा आपको अपने बारे में बता देता हूं मैं आपको अपने नेम के सभी वन 10 साल से पठारों 3 साल हो गए अभी तक पढ़ते होंगे 1000 से आपको दे दिया है इसके अलावा मैंने ऑर्गेनिक केमिस्ट्री पर एक भी लिखी जिसमें दो लाख से डाउनलोड है अगर आप चाहे तो इसको डाउनलोड कर सकते डब्लू डब्लू सेक्स डॉट कॉम पर इसके अलावा आ मुझे अलग-अलग और मैं इसे बेस्ट टीचिंग अवॉर्ड मिला है 11 बेस्ट टीचिंग अवॉर्ड मिले रिसेंटली मुझे अनअकैडमी की तरफ से भी टीचिंग अवॉर्ड मिला है तो यह आप लोगों का प्यार है कि आप इतना बढ़िया से आपने चीजों को समझा लें ताकि जब लास्ट जैसा लास्ट राउंड में हमारे 326 बच्चे आईआईटी में पढ़ रहे और 15 बच्चे आईआईटी एनआईटी में सिलेक्ट हो है तो चलिए यहां पर इस उम्मीद के साथ कि इस बार में बहुत शानदार रिजल्ट आप लोग लाकर दिखाओगे सेक्शन आपके स्टार्ट करते हैं वहीं अब हम बात कर रहे हैं टाइम रजिस्ट्रेशन रिएक्शन की पहले आपको यह समझना होगा यह है क्या मतलब बेसिकली यह है क्या जगह ट्राई का मतलब तुम समझ रहे हो ट्राय का मतलब हुआ थ्रील मर इस मार्ग का मतलब होता है पार्ट जो मर्जी हो तेरा यह पार्ट्स को बोला जाता है मतलब तीन पार्ट्स का आपस में कंबाइन हो जाना थ्री पार्ट्स का आपस में कंबाइन हो जाना अपनी सी को बोल रहे होते हैं टाइम राइजेशन मतलब अगर मैं आपको ऐसे लिखो कि यहां से थ्री लास्ट ई एक्स है और वह है अगर यह आपस में कंबाइन होकर अतिथि एक्स और अतिवाद का फॉर्मेशन कर रहा होता है तो यहीं बोला जाता है ट्राय ग्रेजुएशन और बनने वाले जो प्रोडक्ट होते इसको हम बोलते हैं ट्राई में अब देखो इसके एग्जांपल्स के थ्रू मैं आपको समझा लूंगा और साथ कैटेलिस्ट का रोल भी मैं आपको बताऊंगा यह क्या होता है यहां पर आपको एक मैटल कैटेलिस्ट की जरूरत पड़ती है या और आपको एक मैटल कैटेलिस्ट की रिक्वायरमेंट होती है मैं एक्सप्लेन का यह कुछ तो इसके साथ आपको लिए चलते जैसे एक्सांपल तो आप सब जानते हो कि जब ऐसे ट्रेन के तीन रिक्वेस्ट को ओपन मिला रहे होते हैं तो इससे बन रहा होता है सिक्स सिक्स मतलब एंजिन की बात कर रहे हैं और आप इसका जुगाड़ वाला मेथड भी जानते हो डुबो लें मेथड से क्या मतलब है मेरा मतलब आप ऐसा वाला मेथड जानते हो कि यहां पर ट्रिपल दे दिया यहां ट्रिपल दे दिया यह ट्रिपल दे दिया यह आगे यह आगे और यह आगे घुमाया इस तरीके से और बेंजीन बन गया अब देखो बेटा यह जो मैसेज देख रहे हो यह कोई मतलब यह को मेकेनिज्म नहीं है यह जस्ट लाइक ए जो है जिससे आपको समझाया जा सके रियल बिहाइंड सीन क्या है वह मैक्स प्लेयर करूंगा जस्ट अभी तो तुम यह भी क्वेश्चन देखते होंगे कई बार की 303 ले लिया सिट्रिक फल डोंट सी एच ले गया और इसके लिए जो हम यहां पर मैटल यूज करते हैं यूज करते हैं 512 डिग्री सी का टेंप्रेचर लेते हैं ऐसी लेते हैं 500 डियर टेंप्रेचर लेते यहां पर भी टाइम रजिस्ट्रेशन हो जाती है और टाइम राजस्थान के रूप में स्टाइल इन फॉर्मेशन होता है यहां पर जो चीज बन रही है यह बहन जी और यह जो चीज बन रही है यह मैसेज स्टाइलिंग ठीक है 135 कि बहन जी अच्छा इसी तरीके से मान लो अगर 323 लेता हूं 1113 लेता लेता हूं ठीक है और यह के साथ 500 में इसको करता हूं इससे क्या होगा इससे कुछ ऐसा मिथाइल बेंजीन हो ा पी लिखा गया था बचा यह वाला क्वेश्चन यह समझ वांट्स 2021 में पूछा गया सवाल है ठीक है यह वाले जो दो क्वेश्चन से यह भी प्रीवियस ईयर क्वेश्चंस के अंदर आईआईटी-जेईई में आ चुके हैं इसके अलावा इस और भी बहुत कुछ हम यूज कर सकते हैं जैसे कि मान लीजिए कि अगर मैं यहां से पीएचसी ट्रिपल डोंट सी एस टी यूज करूंगा ठीक है इसके अतिरिक्त बैलेंस तो यहां पर ज्यादा ने 51 पर सेंट के आसपास कुछ प्रोडक्ट बने का यहां से ऐसे ही 500 डिग्री सी लेते हैं और इससे अगर तुम आ रहे होंगे तो यहां से सिलाई का यहां से सिलाई का यह सिर्फ इतना ही लाएगा यहां से मिठाई खिलाएगा इस तरफ से मिठाई खिलाएगा और इस तरफ से मिठाई जाएगा इसका फॉर्मेशन और होता है अब कोई एक्शन की कुछ खास बातें मैं आपसे शेयर करना चाहूंगा जो कि बड़े इंपोर्टेंट है यह जो एक्शंस होते हैं यह हाई व्यवहारिक होते हैं इस लेसन पर एक बार आपने स्टार्ट कर दिया ना तो यह बहुत ज्यादा एग्जॉथर्मिक होते रिजन क्या है कि इन कंपाऊंडके स्टैबिलिटी की बात करते थे वह बहुत ज्यादा स्टेबल यूनिट्स नहीं है कि आपको लगता है सिर्फ तीन बहुत ज्यादा स्टेबल होते नहीं यह प्ले लिस्ट स्टेबल है लेकिन जब इसके तीन यूनिट आप उसमें कंबाइन होते तो बेंजीन बनाते हैं और चीन के बारे में आपको पता है कि यह बहुत ही ज्यादा स्टेपल होता है ड्यू टू एरोमेटिक नेचर तो एरोमेटिक होने के कारण इकाई के स्टैबिलिटी आ जाती है जिससे बहुत ज्यादा एनर्जी रिलीज होती है अगर आप देखिएगा यह 102 किलो जून तक का गिफ्ट खरीदने की रखता है इससे आपको आईडिया लग सकता है कि रिएक्शंस कितना ज्यादा ग्रोथ में और ऑफिस जाने के लिए तैयार होता है यू जस्ट नीड नॉट सूटेबल मेटल कैटेलिस्ट f5 डिग्री सी ठीक है अच्छा दूसरा मैं आपको यही बताऊंगा यह जो रस होता है वह कौन हो सकते हैं ताकि मैं यहां पर यूज किया जाता है इस जनरली आयरन होता है अब होता है आपको बताऊंगा कंफीग्रेशन फॉर स्टूडेंट्स अच्छा होता है इसके अलावा यहां पर आप यूज कर सकते हो उसमें हम वो एसबी यूज किया जा सकता है इसके अलावा यहां इरीडियम यूज कर सकते हैं या टंगस्टन या प्लैटिनम भी यूज कर सकते हैं यह सारे मेटल्स में कुछ और इंटरनेट है आजकल कुछ नए मेटल्स भी आपको दिखते हैं जहां पर आयरन ओर निकल का ऑइल यूज किया जाता है दिस इज गोल्ड पॉसिबल इस यह भी संभावित है तो मैं आपको बता दूं मैंने क्या बोला मैंने आपको यह बोला कि टाइम राइजेशन का मतलब होता है ट्राय मतलब तीन मतलब पाठ और मजहब मिल जाना 3 पार्ट्स का आपस में मिलकर एक यूनिट को प्रौम करना यह ट्रेन ग्रेजुएशन बोला जाता है टाइम रिएक्शन होता यह लोग जब हम इस प्रोसेस होता है क्योंकि इसमें जो यूनिट कंबाइन हो रहे होते वह बहुत ज्यादा स्टेबल नहीं होते और जब यह आपस में कंबाइन हो जाते तो जनरली स्ट्राइक ब्राजील के बाद है रोमांटिक जाती है और इसकी वजह से हाईली अनस्टेबल प्रोडक्ट बनता है इसकी जो गिफ्ट से फ्री एनर्जी चेंज है वह सौ बयालीस किलोजूल्स तक पौर मूल तक पहुंच सकते हैं जो कि बड़ी 19 पास है इसके अलावा यह भी है कि यहां पर सूटेबल मेटल कैटल की जरूरत होती है जो कि जनरली आयरन होता है कई बार आयरन निकल का कटलेट होता है कई बार आपको रोड़ा में रुचि नहीं होता है कई वरीय बटन जैसे मैटल हार्डनेस नहीं स्टार्टिंग टेंप्रेचर फेज़ सी और एक बार अगर टेंप्रेचर आपने अचीव कर लिया था डॉक्टर ने इसके बाद लगातार इरेक्शंस ऑटोमेटिकली चलती रहती है को बार बार एनर्जी प्रोवाइड करने की जरूरत नहीं होती मतलब यह सेल्फ सस्टेनिंग प्रोसेस के जैसा काम करता है अब यहां कुछ इंपोर्टेंट रिएक्शन है जिसे इसलिए अपने बंजर बना लिया जैसे आपने मिठाई ले सिट्रिन से मतलब प्रोफाइल से आपने मैसेज स्टाइल इन बना लिया है उसी तरीके से ब्लूटूथ साइंस अपने एग्जाम मिथाइल बेंजीन बना लिया है और भी बहुत सारे रिएक्शन से सब के सब टाइम राजस्थान है अब यह सब आपको पता होना चाहिए कैसे कैसे होते हैं रोल क्या-क्या है मेकेनिज्म क्या किया वे पहली बार आपको अपने चैनल पर आज मेकेनिज्म बताऊंगा इसका प्रपोज मेकैनिज्म क्या है ठीक है आइए शुरू करते हैं इसके मेकेनिज्म की तरफ बहुत बहुत सारी चीज़ें हो सकती है जैसे मैंने आपको बताया था यह जो आपने यहां पर यह मैसेज भेजिए पिम नहीं है अब जो मैं आपको बता रहा हूं इसका एक्चुअल मेकैनिज्म है हालांकि काफी पुराना है scientists का नाम है पर रेप केसेस में यह मेकेनिज्म ऑलरेडी प्रपोज्ड कर दिया था अभी प्रमाणिक यह मैकेनिज़्म काम करता है तो इसमें हुआ क्या था बेसिकली काम करता है तो यहां पर जो बेस्ट कैटेलिस्ट जो एनसीआरटी में देखते हो अगर मैं एनसीईआरटी के हिसाब से बात करूं तो बेस्ट कैटेलिस्ट आपको दिखता है वह आयरन दिखता है वह क्या दिखता है आयरन ऐसी दिखता है ग्रेट अब सुन मेरी बात को अब यहां पर हम क्या करने जा रहे हैं अगर आयरन के इलेक्ट्रॉनिक कंफीग्रेशन की बात करें तो यह रिड्यूस होता है फॉरेस्ट होता है अब जरा सा इसके डू पर विचार करो ध्यान से देखना इसके डिश को इसके में 123 456 है यह साथ रेडी है और साइकिल एडिसन के लिए टाइम राजस्थान के लिए इसका जो 3D है यही रिस्पॉन्सिबल होता है वह कैसे रिस्पांसिबल होगा समझना बहुत ही इंट्रेस्टिंग बात है देखो सबसे पहले क्या होता है कि यह माल के चलते हैं कुछ इस तरीके से आपको 111 दिख रहा है यहां पर कि यहां पर आ रहा है और यहां पर मालूम 111 है यहां आ रहा है यहां आ रहा है और यहां पर आपका मैटल है मैंने यहां पर आयरन ले लिया है मैं इसको किसी और मेडल से किसी से विजय प्रदान करता हूं तो मैं जैंटल आइज नहीं करता हूं मैं एक दृढ़ सेंस में मैटल हम लेकर चलता हूं तो यह क्या करते हैं इस सरफेस के पास आता है ना तो मैटल जब इसके पास पहुंचता है तो आपको पता होगा कि जेंटली जिसका प्राइवेटाइजेशन करते हैं वह एक लाइन होता है चाहे वह इसलिए होगा यह प्रोफाइल होगा यह विटामिन होगा या इमीडीएटली इन होगा एक कोई ना कोई आपका क्या होगा एक इसकी हाइब्रिड साइज वाला कार्टून को रखने वाला पॉइंट जरूर होगा तो यह क्या करेगा यहां से इस तरीके से इस तरीके से कम होता है और यह एक इलेक्ट्रॉन जो हैं यह इसके वेकेंट डी और विटल को अप्रोच करते हैं इस वाले वेस्ट इंडीज और बटन को तो यह एक इलेक्ट्रॉन और एक इलेक्ट्रॉन इस मेटल के डी में जाकर फिट होना स्टार्ट होगा ऐसे और ऐसे ही ना एक आयरन मैटल जो होगा चार्ट तरह का इस तरीके से कोंबिनेशन डे को यूज कर आ सकता है अवेलेबल करा सकता है ठीक है तो पेयरिंग वहां से है अब आप टर्ड देखो यह कुछ इस तरीके का बनता हुआ नजर आया देखो यहां से आ रहा है यहां से आ रहे यात्रियों को 112 है यहां से 112 है यहां से आ रहे हैं यहां से आर है यहां से मैटर है और इस तरीके से यहां से लिंकेज आपको नजर आ रहा है इस मेटल के साथ ठीक है इसको बोला जाता है मैटल और बेल का फल कि साइकिल रोक पेंटा दाए इन कि इंटरमीडिएट यह जो है एक इंटरमीडिएट बनता है जो कि मेटल ऑयल या प्रिंटर डायन इंटरमीडिएट है यह रिपीट 1948 में बोला था अब होता क्या है कि यह वाले जो मेटल्स होते हैं ना यह पीछे की तरफ चले जाते हैं जैसे कि मानो इस तरीके से देखिए इस तरीके से चले जाते हैं यह कुछ आपको ऐसा नजर आता है ध्यान से देखना मैं थोड़ा सा एक्सप्लेन कर रहा हूं समझ जाओ कुछ इस तरीके से यह वाले जो है ना कुछ इस तरीके से पीछे की तरफ यह मैटल है आप इसको सामने ना समझे यह पीछे की तरफ है यहां से लगा हुआ आर यहां से और यहां से और यहां से आ रहे हैं और सामने की ओर रहता है अब रोल में देखो यह तीसरा वाला यूनिट है यहां से आर लगा हुआ है यहां से और लगाओ अब क्या करता है देखिए यहां पर दो अभी भी डे जो है वह खाली हैं 2 डी में जो यह जो आप देख रहे हो यह जो आपने और कलर ब्लू कलर वाले वह उसके वाले रेस्ट्रॉन्ट्स गए थे अब देखो बच्चा क्या होगा यह अपना इलेक्ट्रोंस को फिर से इस डीके अप्रोच में लाता है अब देखो जब डीके फेज में के फिर से कंबाइन होता है ना तो कुछ इंट्रस्टिंग फैक्ट यहां पर होगा अब देखो जब दो इलेक्ट्रॉन इनकमिंग होंगे तो जो चार्ट्स डेंसिटी यहां बढ़ रही थी वह चार्ट दिन सिटी-2 इलेक्ट्रॉनिक को एक सेलेक्ट कर देगी बाहर निकाल देगी तो जब यह तो अप्रोच करेंगे ना तो वह तो इलेक्ट्रॉन फिर से बाहर की तरफ निकल जाएंगे आर्य व समय यह नोट इन आउट मेटल के साथ चलता रहेगा और फाइनली यह दो इलेक्ट्रॉन इन दो इलेक्ट्रॉन के साथ जो इसके साथ यह इलेक्ट्रॉनिक्स के साथ यह स्टोंस के साथ आएगा और मेटल एडिट इस डिश कंडीशन में आउट हो जाएगा फिर से बोल है क्या बोला देखो मैं कल का काम है कि है प्रोवाइड करना इनकमिंग इलेक्ट्रोंस को जो कि मेडिकल में आया था शुरुआत में दो इलेक्ट्रॉन मेटल में गए अभी जैसे पहुंचेगा अपने दो इलेक्ट्रोंस को देगा वह दो इलेक्ट्रॉनिक्स फूटेगा तो अब जो होगा यह दो उस मेटल के साथ चलेंगे इलेक्ट्रॉनिक्स अब फिर से वह इसको कट करेंगे आप एक बार यह जाएगा उसके विकेट बटन के पास या Enfield इलेक्ट्रोंस के पास और एक बार यह जाएगा एक बार यह एक बार वह एक बार यह एक बार वह यह कॉन्टिन्यूस्ली चलता रहेगा और ऐसा तो है नहीं कि कौन इसे चलाते रहेंगे तो प्रोडक्ट बन जाएगा नहीं एक ऐसा टाइम का यह जो दो इलेक्ट्रॉन होंगे अब मेटल उसके पास ना जाकर यह इस तरीके से टूटे यह देखो कैसे टूटेंगे देखो यहां से आर है यहां से आ रहे हैं यहां से देखो डबल रोल है यहां डबल रोल है यहां आ रहा है यहां आ रहे हैं यहां आ रहा है अब है क्या होगा कि यह इलेक्ट्रोंस यहां यह इलेक्ट्रॉन से और मेटल से तो छोटे थे तो इलेक्ट्रॉन वह एक यहां और एक यहां आखिर कंबाइन हो जाएंगे और ऐसा क्या हो रहा होगा यहां से अब यह ऐसा कुछ बन रहा होगा देखो झाला गया यहां डबल था यहां डबल भाई यहां से और यहां से और यहां से और यहां से और यहां से आर हो गया और नेट अलग हो गया अब यहां पर 1000 गांव में दिया जाता है कि मैटल अलग कैसे हो जाता है को एक सिंगल एक ही के साथ तो यह होने का यह कैटलिस्ट के जैसा काम करता ना कि यह पॉइंट के जैसा तो इसका काम यही है कि उसको कट करके इलेक्ट्रॉनिक उधर देना मतलब पकड़ना और छोड़ना पकड़ना और छोड़ना तो मटर बार-बार पकड़ता है और एक साइड लपेटा इंटरमीडिएट बनाता है दूसरा यूनिट आया उसके साथ टाइम रजिस्ट्रेशन करा दिया और निकल लिया है फिर यह दूसरे के पास गया फिर से सेंटर लाइन बना या फिर उसके साथ-साथ एडिसन किया और निकाल लिया है मैटल का काम ही यही है यह मेटल ऑयल साइकिल प्रिंटर डायन इंटरमीडिएट बनाना प्रोडक्शन को आगे मूव करना हम टिकट होता है मैंने सिंपलीफाइड लैंग्वेज में समाया मैं को अनम्यूट कर दो मैंने क्या बोला था तो मैंने यह बोला कि जो सेंडल मेंटल होते हैं वह जिंदगी वह चूस किए जाते हैं जिसके पास दी में पूरी तरीके से भरा हुआ इलेक्ट्रॉन ना हो मतलब बेसिक होंगे 207 हो गई डेथ हो गई इस तरह के आते डिटेल वाले नहीं चलेंगे ठीक है हां बाकि को सोच समझ कर लेते बॉलिंग और फील्डिंग और जैसे बहुत सारे फिजिकल खाकर में रिस्पांसिबल होते हैं अब हम क्या करते हैं मेटल सबसे पहले आता है और यह क्या करता है मैटल पर्स और हल्का टाइल्स था कि वो प्रिंटर टाइम का फॉर्मेशन करता है यह जो आप यहां पर देख रहे हैं अब यह दूसरे फिर से यूनिट को अपने पास आता है और जो इसने इलेक्ट्रॉन पहले से ही कट किए थे उसको छोड़ता है और फिर वह और विटल्स दूसरे के साथ कनेक्ट हो तो यह फ्रीक्वेंसी लगातार चलती रहती है किसके किसके बीच में यह जो महल का जो यह वाला आया था आपके पास साइकिल एप एंड शाइन यूनिट के बीच में और एक एक्स्ट्रा बचा हुआ यूनिट के बीच में और फाइनली यह साइज हो जाता है ट्रेन ग्रेजुएशन हो जाता है वह जिनका मैसेज प्लेन का जो भी फॉर्मेशन करते हैं यह था इसका एक्चुअल में कैंसर अब आप लोग यह को जानने की जरूरत है क्या अगर दाउद 7 है बच्चों का यह होता कैसे तो इज्जत उसको मैंने एक्सप्लेन कर दिया अब मैं आपको बताऊंगा इसके ऊपर बेस्ड क्वेश्चंस और आप लोग बहुत ध्यान से देखना बड़ा ही इंट्रेस्टिंग है और इसको इधर बातें हैं अ बहुत ध्यान से देखो पहला ही क्वेश्चंस एडवांस 2021 का है पहला सवाल और रीसेंट कि अभी आया कुछ महीने पुराना क्वेश्चन है इससे ज्यादा नेट क्वेश्चन अभी अवेलेबल नहीं है आपके एग्जाम के बाद यह बिल्कुल होगा तो बोला गया कि मैग्नीशियम कार्रवाई है उस पर वाटर डाला गया है तो पी बनता है फॉर ग्राम इन दिनों पर बहुत सारे रिएक्शन सोएं तो मैं आपको बता दूं सबसे पहले कि मैग्नीशियम कार्रवाई पर जब बर्ड लोगे ना तो यह प्रोफाइल का फॉर्मेशन करता हालात एलियन भी बनता है तो हम प्रोवाइड नहीं लेते क्योंकि आगे जाकर को साइड रिएक्शन वगैरह कराया गया ठीक है तो यहां से जो चीज बन रही होगी वह यह बन रही होगी ch3 लूसी 111 टीच ठीक है यह बात हवा लोग यह कैसे बन गया सर यह बता तो यह कुछ ऐसा है सी 111 सीएसई इसके ऊपर - है और इसके ऊपर 3 - 4 तो बैलेंस करने के लिए 2mg टू प्लस यहां पर लगा हुआ है ठीक है इस बात को दिमाग मे रख अब तो यहां से 4 ग्राम बना आपके पास छोड़ ग्राम यह आया 100 ग्राम यह है ठीक है अब आ गया अब आगे परसेंटेज बड़ा इंपोर्टेंट रोल प्ले करेगा अब देखिए यहां से nh2 - यह nh2 - क्या करेगा हाइड्रोजन को खिंचेगा और यह - क्या करेगा Mi के ऊपर एस्से टू कराकर इसको हटा देगा तो यहां से जो आपका संयोग बन रहा है क्यों बन रहा है ch3cooh डोंट सी और यहां से यार सीरियसली ठीक है यह हमारे पास हो गया ब्लूटूथ ऑन अब इसके बाद रेड हॉट आयरन ट्यूब है ऐसे ज़मीन है मैं भी परसेंटेज ने लखनऊ में परसेंटेज मिलूंगा अब जब इसका टाइम रजिस्ट्रेशन करेंगे तो मैंने जस्ट आपको थोड़ी देर पहले बोला है इसके तीन यूनिट आपस में मिलेंगे कितने फुट में लगे तीन मिनट और इस से क्या बन रहा होगा एग्जाम मिथाइल बेंजीन का फॉर्मेशन होगा इस तरीके से यहां से MB मीडिया पहले इसको फोल्ड कर लेता हूं सवाल में खोला गया पहले एक वैल्यू बताइए फिर बाइक है ना मैं पहले एक निकालता हूं फिर मैं आऊंगा अब देखो इसको ऐसे सॉल्व किया जब मैं इसको अपने तरीके से सॉल्व करने की कोशिश करता हूं आप एक काम पर मार्क्स लेकर मत चलो मांस लेकर चलोगे कोंप्लिकेटेड हो जाएगा मूड लेकर चलो ठीक है तो सबसे पहले बताओ 4 ग्राम यह है और यह जो हमारे पास आया है वह यह आया तो एक दो तीन तो भारत इन 3640 इसका जो मांस हो गया पीड़िता मॉलिक्यूलर मास व है 40 ग्राम और गिवन मांस कितने 4 ग्राम तो नंबर ऑफ मौलिकता हो गए 440 मतलब एक वायरस बाद अब जीरो प्वाइंट अव्वल मगर यह जो हमारे पास है यह है पॉइंट वन मोंठ ओल्ड ठीक है मूल अव्वल अपन चलेंगे पॉइंट्स अनमोल है अब गया हो अब देखो स्पीड से यह डैंड्रफ को यहां से यहां एक मॉल से एक मॉल का फॉर्मेशन हो रहा है पर जो यहां पर कनवर्ज़न रेस्क्यू है वह सिर्फ 75परसेंट है तो देखो पॉइंट वन से पॉइंट वन तो बनेगा नहीं क्योंकि 75परसेंट है कन्वर्ट हो रहा है 75परसेंट इंजेक्शन फॉर पर जा रही है तो हम यह बोल सकते हैं कि इसका जो फॉर्मेशन होगा वह 0.1 प्वाइंट्स 153 फोर्थ इतने मोल यह फॉर्म हो रहे होंगे ठीक आगे अब रेड हॉट लुक में यह 40 पर सेंट एफिशिएंसी के साथ आगे बढ़ रहा है तो मतलब आप जो बना हमारा एक्स ग्राम आया भी एक्स कैसे फ़ाइल करें सोचो जरा देखो अगर 3 से यह तो एक बनेगा और एक लेंगे तो एक बटा तीन बनेगा तो इतना लेंगे तो इतना वाइट विनेगर समझ में यह बात नहीं समझ में बात क्यों का अगर मैं बोलूं कि क्यों का एक ही मूल था मतलब हमारे पास बिट्टू आई मगर एक मॉल था यह कितना बड़ा होगा बनवाई फ्री बनाएगा तो कि तृणमूल से ट्राय ग्रेजुएशन होता है तो तो यहां पर इतना बोल दिया है तो कितना आ रहा होगा 123 आ रहा होगा तो अगर मैं इसको फॉर्म करने की कोशिश और कितना बना होगा कितना अनमोल बना होगा तो अगर हम इसका मोल निकाल लें तो हमारे पास क्या हो जाएगा 0.1 इंडोर 384 इतना तो था ही और हमारे पास one-third इसका आ रहा है क्योंकि आपको पता है कि तीन लगेगा और यह एक है तो बनवाई थी ने दोनों हाथ कर देंगे अ कितना परसेंट है 40 पर कि है मतलब पॉइंट्स और ठीक है अब देखो यहां से फोर फोर यहां से पॉइंट वन कैंसिल हो गया यहां से 343 कैंसिल हो गया मतलब हमारे पास आ गया जीरो प्वाइंट कितना गया जीरो वन मॉल बन गया हमारे पास ठीक है 0.01 मोड अगर तो इसका मोलर मास निकाल कर देखोगे तो इसका मोलर मास जो है वह 1662 आएगा चेक करके देख लेना तो हम से एक पूछा गया कितना ग्राम बनेगा तो 0.01 108 मतलब 1.62 ग्राम एक्स बनेगा है नमक हम लास्ट में फिर से रिपीट कर दूंगा रोकना है ना थोड़ा इंतजार करना अब आ जाओ वाले पार्ट की तरफ उपन्यास करते हैं वहीं देखो क्या होता है यह बहुत इंटरेस्टिंग समझना जरा सा को यहां पर 4 ग्राम से स्टार्ट किया गया है और यहां से 1 इंच प्लस और यह मतलब आप समझ रहे हो यह कुछ अगर हम इसके साथ-साथ जो पॉइंट है इसके साथ होंगे तो यह बना रहा यह देखिए यहां से ch3 है सीरियल अब जरा सोचो क्या सोचने आपको यह जो हमारे पास है कि तथा पॉइंट अव्वल यह पॉइंट और लिखा हुआ 100 पर्सेंट है इधर आ जाए तो मतलब पॉइंट से यह पॉइंट अव्वल है अब इसको थोड़ा हमारे पास ठीक है इसके साथ ही आपको पता होगा डॉल डॉल डॉल कि इसके अतिरिक्त यहां से आपके पास यहां से यहां से आ रहा हूं ठीक है यह तो यह पॉइंट और इसमें और ही चाहिए एक तो बना होगा चुके एक से सुना ध्यान से दो से एक बनता है तो एक से आधा बनेगा तो पॉइंट अव्वल टू यहां से हो गया चुके केवल आठ परसेंट है तो मेन 2018 ले रहे होंगे तो यहां पर हमारे पास आया पॉइंट 1222 आयाम और लेफ्ट के वजह से और आप पॉइंट यहां पर एटी परसेंट की बने अब इन्हें मुश्किल है यह कौन सा करेक्शन है फ्रॉम यह हमारे पास क्लोरोफॉर्म रिएक्शन है अब हम के साथ सोचो क्या होगा दिमाग पर जोर डालो यह वाला पार्ट हट जाएगा हटे गा समझ रहें और किस में चला जाएगा यह चला जा रहा होगा क्लोरोफॉर्म में और यह जो ड्यू बनाया यह डिवाइस करेगा वेयरिंग ब्रा इस की बात की जा रही है तो ch3cooh 30 इंच यहां से सीओएच ले सकते हैं ना मैं सीओएआई ने लेकर भी किया था दो आंसर माने गए थे तो मैं मोची लेकर चलना अब यह कितना बन रहा होगा वह देखता सोचो एक से एक आ रहा है और इसका कितना हमारे पास है एटी परसेंट है तो अब हम को लेकर चलते हैं तो यह हो जाएगा 0.1 व्हाय दोएस इन 2018 इतना था अब एक बन रहा है हमारे पास और कितना बैठ करके 120 हमारे पास मतलब 2108 इतना पता है तो आप इसका तो इसका कर देंगे ठीक है यहां से कट कर देंगे यह पॉइंट और इसको सॉल्व करके आप के आस पास आना चाहिए 3.2 ग्राम के आसपास यह आना चाहिए तो बेसिकली बन रहा है 6 ग्राम बन रहा है ग्राम बन रहा है मैं आपको सेंड करता हूं मैं यहां पर क्या दिया गया है ना आपको इसका मूल गिवर है पहले तो आप सारे के सारे क्वेश्चंस रॉक लीजिए इसके बाद आप सोचिए कि किससे क्या आ रहा होगा और इसको यहां से अगर आप देखोगे तो एक से एक गया है तो यहां पर हम कुछ भी 12131 वाइफ को डीलिंग सीधा 121 है जितना मॉल था उतना मिल जाएगा पर 75परसेंट है तो पॉइंट से 1534 कर देंगे अब तीन के साइकिल हैं जिससे एक बनता है तो एक के सेक्शन से बनवाई थी बनेगा इसीलिए हमने क्या-क्या आगे जाकर 123 कर दिया है अब हम कैलकुलेट कर दिया मास्क सवेर मोड निकला मोड़ को कंवर्ट कर दिया मांस निकला यहां पर हम क्या कि यह कि देखो यह 4 ग्राम यह पॉइंट वन मॉल कुछ रोक कर आए तो एक से एक बन रहा है तो हम कुछ नहीं किया सौ परसेंट है तो सीधा वैसे का वैसे नीचे है पॉइंट 1.1 जब हम ब्राइटर में लेंगे एंड आल मैं तो यह हमारे पास बना अभी अस्सी परसेंट के आसपास पूरा नहीं है तो इसको 2 से ध्यान से सुनना दो के मिलने से एक बन रहा था तो एक के मिलने से एक बटा दो देगा तो पॉइंट बन के मिले से पॉइंट 152 बनेगा तो पॉइंट वन बाय हो चुके मात्रा 80 पर सेंट इन वर्जन है तो 108 ले लिया गया आइए फिर से एनओसी में 80परसेंट जा रहा है तो एक से एक जा रहा है तो हम मास्क तो कुछ करेंगे नहीं चुके पॉइंट पस्त प्रेजेंट फिर से तो पॉइंट से अगेन इनटू हो गया अब ऐसा करने से हम सॉल्व करके देख लिए मांस हमारा कितना आ गया और जो कंपाउंड का मॉडल मार्क था उस फॉर थे एनुअल डे 3.2 ग्राम आ गया हालांकि 3.2 से 3.9 तक आंसर लिया गया था कई लोगों ने लगाकर सॉल्व कर रहे थे दोनों बयान से पलट माने गए थे किलर हुआ आपका ट्राय ग्रेजुएशन का बड़ा ही इंपोर्टेंट रोल है नेक्स्ट मूव करते हुए एडिट क्वेश्चन कर पाते ID फाइनल प्रोडक्ट की बड़े कमाल का क्वेश्चन एकदम फर्स्ट क्वेश्चन है और सॉल्व करते हैं सबसे पहले कैल्शियम कार्रवाई करे अब आप जानते हैं कैल्शियम कार्बाइड इसको बोला जाता है हम इसके ऊपर क्या कर रहे हो वाटर डाल लें पानी डाल रहे हैं यह पानी डाल दे यहां से कौन सी गैस निकली ऐसे सीन गैस निकली बोली तो 30 इसको गैस देखिए अब सिचुएटेड टू आयरन के प्रिंट्स में हिट करेंगे ट्राय रिएक्शन होगा और टाइम है जिससे क्या बनेगा सीख एक बस अब बेंजीन बनेगा तो जो हमारे पास आ गया वहीं आ गया अब नाइट्रिक एसिड से आ रहा होगा इस ट्रॉफी के साथ मिलने से नाइट्रोम यहां से इलेक्ट्रोफिलिक रोमांटिक सब्सीट्यूशन रिएक्शन हुई और यहां से क्या बन गया नाइट्रोबेंजीन बन गया ऐसी अश्लील के साथ लव इस करेक्शन हो गया और रिडक्शन होने के बाद क्या बन गया एनी लिया गया इन इंग्लिश को ब्रोमीन वाटर के साथ लाओ तो यहां पर आपको पता है इन योर पर्सनल ग्रूमिंग वाटर में जाने के बाद ट्राई प्रोमो डेरिवेटिव्स देते हैं बिहार ब्यास ब्यास यह बन गया अच्छा इंट्रस्टिंग स्टेप है अतिशय दी अब सीरियल ये रिश्ता क्या है यह बहुत ही स्ट्रांग बेचती - को जनरेट करें यह जो है बहुत ही स्ट्रांग बेस है वह इससे क्या हो रहा है तो इससे क्या होगा कुछ ऐसा कि यहां से आपके पास nh2 है ठीक है अच्छा यहां से लगा हुआ यहां से और यहां से आप बहुत स्ट्रोंग हो पता है कि ब्लास्ट हो गया ब्यूटाइल लिखते हो गया मिठाई लीजिए होंगे यह सब क्या करते हैं अमीन के हाइड्रोजन को निकाल सकते हैं अब देखो यह वाला हाइड्रोजन जो है इसको यह खींच लेगा कौन सी स्त्री - स्त्री - इसको खींच लेगा तो अब देखो यह कुछ ऐसा हो जाएगा अब लुट यहां से आ जाएगा एन एच - यहां से बीच यहां से बीच यहां से भी अब आपको पता है क्या पता है कि बहुत जबरदस्त प्लसएम् लगाता है यह अंदर की तरफ आएगा यह आगे जाएगा यह आ गया आएगा अब देखिए यहां पर डाइजोन हम प्रोवाइड लिखा हुआ है अब हम क्या कर रहे होंगे पी एच ए एन 111 टेंथ प्लस यहां से हमला करेगा यह आगे की तरफ मुड़ जाएगा अब देखो कुछ ऐसा होगा कैसा हो रहा होगा देखो यहां से जरा ध्यान से देखना बहुत ध्यान से देख ना पड़े इंट्रस्टिंग स्टेप है इस तरह के क्वेश्चंस बहुत कम आपको देखने को मिलते हैं जनरली अच्छी-अच्छी टेस्ट सीरीज में दिखेंगे अब यहां से हमारे पास हो गया है डबल रोल है ने इस अब देखो यह अंदर आ गया यहां से लाकर लगा दिया यहां से आग लगा दी आर यहां से लाकर लगा बिहार और यहां से यहां से डबल 110 और यहां पर यह जो अब जरनल क्या होता था यहां पर हाइड्रोजन हुआ करता था और वह Bigg Boss अंदर आकर रोमांटिक को फिर से ला देता था पर अभी यह ना बहुत बुरी तरीके से फंस गया है क्या से क्योंकि अभी यहां पर मिठाई जैसे नहीं आई क्वेश्चन है ही नहीं तो यह प्योर डबल रोल है और ये ग्रुप है बता आंसर आगे आपके पास एनएच बिहार बिहार बिहार ग्रुप लगा हुआ सौदा करेक्ट आंसर आईएस वास्तव में सबसे पहले सफल बनाएं तमाम रजिस्टर करके बेंजिन था कि नाइट्रोबेंजीन है डिडक्शन कराकर इन बनाएं दुर्गेश वगैरह बनाएं सी इलायची हाइड्रोजन निकाले अंदर की तरफ पैरों से हमला कर दिया और सुबह ट्रैफिक बहुत इसलिए करें पैरों सेटिंग क्लोज बाकी ऑप्शन दिखेगा यहां पर भी अ गायब कर दिया तो ऐसे यह कम से कम मैटर सेट कर दिया तो इससे क्या काम से और यहां यहां पर क्या कराया गया है यहां पर क्या कराया गया है डीआर बीआर बीआर कि दोनों ऑप्शन सैंपल है वह तो यहां पर चीफ लगा दिया गया लगना चाहिए यहां पर और यहां पर बिहार बहुत ही गलत है तो TV गलत है चीज थे करेक्ट आंसर आगे बात समझ में आ थे किलर हो रहा है बहुत शानदार शानदार क्वेश्चन चलिए देखते हैं ओके अब नेक्स्ट क्वेश्चन की तरफ आते हैं देखो जाएगी एडवांस में बैक टू बैक क्वेश्चन पूछे गए मैंने पहला क्वेश्चन तुमको 2101 का कराया था और अभी वाला क्वेश्चन हम आपको चाहिए एडवांस 2018 11 हैं इस चीज आप समझ रहे हो कि यह क्वेश्चंस कितने फ्रिक्वेंटली एस्क्ड क्वेश्चंस हैं तो कांग्रेस 204 है और एसीटोन है अब आपको मैं हल्की हमारा आज का टॉपिक नहीं है फिर भी हम आपको समझा देते हैं यह स्टोंस के साथ क्या होता है इसके 351 आपस में मिलते हैं और कांग्रेस टो सपोर्ट हाइड्रेट कर देता है ठीक है अब इसको इसका टेस्ट भी है मैं आपको शॉट भी दिखा देता हूं आप लोग देखो यहां से सीईओ यहां से ch3 है ठीक है यहां से सील तो यहां से ch3 है ठीक है यहां से सीईओ और यहां से ch3 अब आप एक काम करो क्या करो यहां से हाइड्रोजन लोग यहां से हाइड्रोजन दो स्टोर निकालो यहां से हाइड्रोजन तो स्टोर निकालो यहां से एडिसन को स्टोर निकालो अजय को एक 1144 तीन टेबल बोर्ड 10th बोर्ड को आपने मिला दिया उसमें यहां से यह मिठाई यहां से मिठाई यहां से मिसाइल और यह क्या बन गया मिसाइल का फॉर नेशन कैन सी आईटी में भी आपको दिख जाएगा ठीक है 135 प्राइस आफ यह तो यह अब यहां अभी जस्ट मैंने थोड़ी देर पहले अगर आप देखोगे को मैंने समझाया था तो यहां से आप ले लो कि यहां से ट्रिपल है यहां से ट्रिपल है यहां से ट्रिपल है ठीक है यहां से मिठाई लोग गया यहां से मिठाई लोग यहां से मिठाई लोग यह इधर आएगा यह इधर आएगा इधर आएगा तीन लोग आपस में कंबाइन कर दो शॉर्टकट हालांकि मैंने लगा दिया है ठीक है तो यहां से भी मैसेज चाहिए निबंध है जी हां यहां पर भी क्या हो रहा है रजिस्ट्रेशन हो रहा है कि यहां से टाइम रजिस्ट्रेशन और है तो MB टाइम राजस्थान अभिनव राजस्थान आ गया अब यहां से B.Ed 2nd ओवर 60 से और थोड़ा लाइन है तो B.Ed 2nd चक्र होगा प्रोमो फॉर कलेक्शन का यह वाला पॉइंट प्रोमो फॉर प्रोमो फॉर ब्रदर फॉर और गया अब देखो ऐसा करने से क्या बना यहां से आ गया सीओएआई ने यहां से भी आ गया सीओएआई ने यहां से भी आ गया सीओ एनए 120 था वह गायब हो गया ठीक है समझ में आ रही बाद अच्छा बोला हाइड्रोलाइसिस करा दी जिसका तो फिर से एसिड बन जाएंगे हालांकि को इस नहीं बनता है सीधा ही लेना चाहिए ना तो बाद में जाकर फिर से सॉफ्ट बनेगा इसके बाद थोड़ा लाइन के साथ इसको आठ कीजिए थोड़ा लाइन बोला कि यह ओवैसी वो ही आप में से कई बच्चे बोलते हैं यह सारा थोड़ा लेट क्यों लेते हुए मैं वह इस YouTube लिखे को जरूर और इसे थोड़ा टाइम लिखा है तो एग्जाम में सोचना पड़ेगा आपको यह सब तो यह हो गया यह हो गया यह और बच गया क्या जी सिर्फ और सिर्फ बच गया बेंजीन सो ऐसी वाले केस में ओनली आपको जो नजर आया है वह नजर आया है कि बेनजीर तो यहां पर गया काम से यहां को मैसेज डिलीट पूछा था यह ख़त्म है आवाज जैन कमल प्रमेय क्या कहलाता है आपने एनसीआरटी में भी देखा होगा कि लेमन संरक्षण कराता है अब जो यह कैलीबरेशन रिडक्शन है यह सीएमएचओ को सीएचसी में सीएओ सीएओ फ्री में सीओ श्रेणी में कंवर्ट करेगा बेसिकली यह कारगिल रोसिए Store में कनवर्ट करता तो क्या बन गया यहां से भी 135 पोजिशंस पर यहां से क्या लग गया है मिठाई लग गया तो डीज़ल टो करेक्ट करेक्ट आंसर क्या हो गया 1 बीएमडी तैयार हुआ सबका अरे भाई क्या हुआ सबका देख लो देख लो कितना खूबसूरत सवाल थाना थाना खूबसूरत वाला आगे बढ़ने से पहले जबरदस्त आपको बात बताना चाहेंगे जैसा कि आपको पता है हमारे गुड शानदार ऑफर आया हुआ है लास्ट चांस से आपके पास 24 बंद काटी ही प्लस का साइक्लोनिक सर्कुलेशन लेने के लिए पुराने प्राइस पर यहां पर आप 14% तक का डिस्काउंट ले सकते हो ठीक है ना जिया अब देखो पहले जो आपको प्लस और सीबीएसई का मिलाकर इतना पढ़ता था वह 9755 में मिल रहा है और जो आपको यह 1.4 मिलता वो भी जाकर बहुत घटकर यह इतना पहुंच गया है दोनों मिला कि आपको 1.38 मिलता था विन्यास मिल रहा था बहुत कम है तो 24 महीने का सब्सक्रिप्शन ले सकते हो और जब भी तुम हमारे किसी भी अ यही के बीच में और कोई भी मैच में ऐसा तो है नहीं कि एक सब्सक्रिप्शन लेकर मेरा एक क्लास देख सकते हो सारे टीचर के पास हम देख सकते हो हंड्रेड आफ टीचर से क्लास हंड्रेड आफ कोर्स डू ट्रेडिंग सेशंस आफ थे डाउट फीचर बहुत कुछ रैंक प्रिडिक्टर एवरीथिंग यहां पर कोड यूज करना है आईटी एक्सप्लेंस व्हाय चैनल आप इसको यूज करके तुरंत डिस्काउंट आपको मिल जाएगा मैक्सिमम पॉसिबल डिस्काउंट आपको यहां से मिल सकता है ठीक है अच्छा कुछ हमारे बैठ चल रही वॉल्व बीच अमेरिकी आज भी क्लास है आज मैं पता नहीं आप इस वीडियो को देख रहे हैं पर आज मेरी एक क्लास 07:15 पीएम में एक क्लास है बहुत जबरदस्त गिरावट अभी चल रहा है लूजिंग चलता है बहुत शानदार चल रहा है तो आप लोग अगर आना चाहे तो बिल्कुल आप सकते हैं मतलब मेरी क्लास तो दिल्ली ही दो से तीन क्लासेस मिनिमम होती है मिनिमम कि मैं बात करो और बाकी सब मिलाकर तो 17 18 घंटे कुछ न कुछ चलता ही रहता है तो बैठे हैं हंसकर बैटरी प्रभाव और यह इस तरीके से चलता है कि महत्व है फिजिक्स है फिजिकल है इनऑर्गेनिक है और दैनिक है हर एक सब्जेक्ट एक रहस्य में चलता है टेस्ट हो गए हर एक चीज रिकॉग्नाइज थे स्ट्रक्चर धोखे से ज्यादा मजा आ जाता है तो कोर्सेज में वह आपको यहां मिलता है इसके अलावा इस बार हाई यही का जो मेघा फ्रॉम बैठे उन्नत इस तारीख को यह मैंने बनाया है तो आप लोग जरूर फॉर्म बट को लीजिएगा यह बहुत जबरदस्त प्यूरीफाइड लाइव स्कॉलरशिप टेस्ट है अब जो बच्चे इकोनॉमिकली उस लेवल पर नहीं है कि सब्सक्रिप्शन को बाइक कर सके लेकिन ऊपर वाले का दिया हुआ दिमाग है टैलेंट है अगर नॉलेज है आपके पास ज्ञान है बैलेंस है तो आपका रिकॉर्डिंग स्वागत करता हूं आप लोग जरूर आइए को डाइड इन स्पेन से उसके उन्नत तारीख को नई आईडी बनाइए और नई आईडी के थ्रू आप लोग मिक्स छोड़ आए और मचाएं ठीक है इसमें अगर कम बेड में अच्छा इसके अलावा एक इंपोर्टेंट बात और है कि सर प्रीवियस ईयर क्वेश्चंस हमारे कहां से कर सकते तो पी वाई क्योंकि भी चीज चल रही है वह भी आप लोग जरूर चेक आउट कीजिए को डिटेल में एक्सप्लेन से यह एवरीडे 8:00 चलती है 8pm सैटरडे आप जब भी देख रहे होंगे चल रहा होगा इसके अलावा त्यौहार का रैंक प्रिडिक्टर इस पर आप अपने इसको के हिसाब से रैंक को फिट कर सकते यह सब फ्री के फीचर से इसमें को पैसे नहीं लगते ठीक है ना तो और इसके अलावा हमारा डांस और लक्ष जो है न एकेडमिक पर चल रहा है यह मंडे में फ्राइडे को एवरीडे 7:00 चलता है तो आप लोग चेक कीजिए इसको एक बार ठीक है ना हम्म बस यही है आज के लिए बस इतना ही अपना ख्याल रखिए और बहुत सारी चीजें आपके पास है और आप लोग कमेंट करके बताइएगा फेवरेट क्वेश्चन सांसद वैसे मेरा फेवरेट क्वेश्चन आज का वनडे था और यह वाला मैकेनिज़्म जो था यह मेरा शानदार आज मुझे मजा आया इसमें और जब मैं ठीक है थैंक यू वेरी मच मुलाकात आपसे बहुत जल्दी होती तब तक अपना ख्याल रखिए बाय टेक केयर
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Lecture 4.2. Jordan form March 31, 2020 Today we finally address deficient matrices, that is those which are non-diagonalizable, which do not have enough eigenvectors to span the space. A typical example is the following m × m matrix Jm(λ1) =           λ1 1 0 . . . 0 0 0 λ1 1 . . . 0 0 0 0 λ1 . . . 0 0 . . . . . . . . . . . . . . . . . . 0 0 0 . . . λ1 1 0 0 0 . . . 0 λ1           . We have λ1 on the main diagonal, 1 on the diagonal above the main, all other entries are 0. Such a matrix is called a Jordan block of size m with eigenvalue λ1. Its characteristic polynomial is (λ1 −λ)m, so the only eigenvalue is λ1, and the eigenspace corresponding to this eigenvalue is 1-dimensional. It is spanned by an eigenvector (1, 0, . . . , 0)T. Check all this! Look how this matrix maps the vectors of the standard basis: Ae1 = λ1e1, Ae2 = λe2 + e1, Ae3 = λ1e3 + e2, and so on. So e1 is an eigenvector, and the rest of ej are called generalized eigenvectors. Definition. Let A be a square matrix, and λ be an eigenvalue and v ̸= 0 an eigenvector, so that Av = λv. Then the first generalized eigenvector attached to v is a solution v1 of the equation Av1 = λv1 + v. 1 The second generalized eigenvector v2 attached to v is a solution of Av2 = λv2 + v1. and so on. Generalized eigenvectors form a chain v1, v2, v2, . . . , vk such that Avj+1 = λvj+1 + vj. Jordan’s Theorem. For every linear operator L in a (complex) finite-dimensional space there is a basis consisting of eigenvectors and generalized eigenvectors. It is called the Jordan basis for L. I recall that two matrices A and B are called similar if there is a non-singular matrix C such that A = CBC−1. Geometrically, this means they they represent the same linear operator, and encode it with respect to two different bases. The matrix of the operator with respect to its Jordan basis has a Jordan form which consists of diagonal blocks, each block is a Jordan block. Corollary. Every square matrix is similar to its Jordan form. Two matrices are similar if and only if they have the same Jordan form (up to permutation of Jordan blocks). If A is a matrix, and J is its Jordan form, then A = BJB−1, where B is the matrix whose columns are eigenvectors and generalized eigen-vectors. Examples. Here are all possible Jordan forms for n = 2: λ1 0 0 λ2 ! , λ1 1 0 λ1 ! . The first of these is diagonalizable, it has two Jordan blocks of size 1. The second has one Jordan block of of size 2. 2 Next we list all possible Jordan forms for n = 3:    λ1 0 0 0 λ2 0 0 0 λ3   ,    λ1 1 0 0 λ1 0 0 0 λ2   ,    λ1 1 0 0 λ1 1 0 0 λ1   . Jordan forms obtained by permutation of the blocks are considered the same. First matrix has 3 blocks of size 1, second has 2 blocks of sizes 2 and 1, and the third matrix has one block of size 3. First matrix is diagonalizable (λ1, λ2, λ3 are not necessarily distinct!): it has three linearly independent eigenvectors. The second has two linearly independent eigenvectors, one with eigen-value λ1 and one with eigenvalue λ2 (again λ1 and λ2) are not necessarily distinct!). It also has one generalized eigenvector attached to the true eigen-vector corresponding to λ1. The third matrix has only one eigenvector (up to proportionality) and to it two generalized eigenvectors are attached. In general, suppose we have some eigenvalue λ. The dimension of the eigenspace corresponding to λ is the number of Jordan blocks with this λ on the main diagonal. To each of these eigenvector, some number of generalized eigenvectors can be attached. The sum of sizes of all Jordan blocks with eigenvalue λ is the multiplicity of λ as a root of the characteristic equation. So, for example, this matrix         2 1 0 0 0 0 2 1 0 0 0 0 2 0 0 0 0 0 2 1 0 0 0 0 2         has characteristic equation λ5 −32 = 0. The only eigenvalue is λ = 2. The eigenspace corresponding to this eigenvalue has dimension 2. So we have two linearly independent eigenvectors, they are in fact e1 and e4. In addition we have generalized eigenvectors: to e1 correspond two of them: first e2 and second e3. To the eigenvector e4 corresponds a generalized eigenvector e5. To find the Jordan form and the Jordan basis for some matrix, you do the following: a) find eigenvalues. 3 b) for each eigenvalue, find a basis of the eigenspace. If the sum of the dimensions of eigenspaces is n, the matrix is diagonalizable, and your eigenvectors make a basis of the whole space. c) if not, try to find generalized eigenvectors v1, v2, . . . by solving (A − λI)v1 = v, for an eigenvector v, then, if not enough, (A −λI)v2 = v1, and so on. You need only one solution of each of these equations. These generalized eigenvectors will be attached to those λ for which the number of true linearly independent eigenvectors is less than the multiplicity of λ as the root of the characteristic equation. d) Jordan’s Theorem guarantees you that eventually you will find suf-ficient number of generalized eigenvectors, so that they, together with true eigenvectors make a basis of the whole space. This is the Jordan basis. Example 1. (Probl. 40, p. 305) Which pairs of these matrices are always similar (for all a, b, c, d) and which are not (for some a, b, c, d)? A = a b c d ! , B = b a d c ! , C = c d a b ! , D = d c b a ! . Answer: A and D are similar. B and C are similar. All other pairs may not be similar for some a, b, c, d. Explanation: A and D have the same characteristic polynomial. Also B and C have the same characteristic polynomial. But these two polynomials are different, and it is easy to find a, b, c, d for which they are numerically different. Now A is similar to D because a b c d ! = 0 1 1 0 ! d c b a ! 0 1 1 0 ! . Multiplication on 0 1 1 0 ! , which is self-inverse, from the left interchanges rows and multiplication on the same from the right interchanges columns. Similarly B is similar to C. Example 2. Find the Jordan form and the Jordan basis for the following matrix    0 1 0 −4 4 0 −2 1 2   . 4 Solution. The characteristic polynomial is (2−λ)3, so the only eigenvalue is 2. By the standard procedure we find two linearly independent eigenvec-tors: (0, 0, 1)T and (1, 2, 0)T. None of them has a generalized eigenvector, so one has to try some linear combination of them, for example (1, 2, 1)T. To this eigenvector, there is a generalized eigenvector (0, 1, 0)T. So the Jordan form and a Jordan basis can be taken as J =    2 0 0 0 2 1 0 0 2   , and B =    0 1 0 0 2 1 1 1 0   . The main use of the Jordan form is for solving differential and difference equations with deficient matrices. One can compute the powers and expo-nential of the Jordan block (see the book, p. 300-301). The k-th power is upper triangular and has the following form: on the main diagonal stand λk. on the diagonal above the main stand kλk−1, on the next diagonal stand k(k−1) 2 λk−2, and so on. On the diagonal number m ≤k above the main diagonal stand k! m!(k −m)!λk−m, the terms of the binomial formula. The exponential eJt of the Jordan block is also upper triangular: on the main diagonal: eλt, on the next above: teλt, then on the next: (t2/2)eλt, and so on. on the m-th diagonal above the main: (tm/m!)eλt. Exercise. Prove these facts. In practice, one does the following. Consider a differential equation x′ = Ax. Let λ be an eigenvalue of A and v an eigenvector. To it corresponds a solution x0(t) = eλtv. 5 let v1 be the first generalized eigenvector. To it corresponds the solution x1(t) = eλt(tv + v1). Indeed x′ 1 = λteλtv + eλt(v + λv1). Plugging this to the equation and dividing by eλt we obtain λtv + (v + λv1) = tAv + Av1, and this is true since Av = λv and Av1 = λv1 + v. Similarly, if there is a second generalized eigenvector, the form of the third solution is x2(t) = (t2/2)eλtv + teλtv1 + eλtv2. And in general, for the (m + 1)st generalized eigenvector, xm(t) = (tm/m!)eλtv + (tm−1/(m −1)!)eλtv1 + . . . + vm. Exercise. Verify this. Effect of the Jordan form on stability. When Re λ < 0, eλt →0 as t →∞. This does not change if we multiply the exponent on any power of t. Therefore when all eigenvalues have negative real part, the system x′ = Ax has stable equilibrium at 0, no matter whether A is diagonalizable or not. However, when Re λ = 0, then eλt is bounded, but it oscillates. Multipli-cation on t gives an unbounded function. So when all eigenvalues have non-positive real part, this is not sufficient for stability. If those eigenvalues with Re λ = 0 are not deficient (have as many linearly independent eigenvectors as their multiplicities as roots of the characteristic equation) then the system is stable. However if there is at least one eigenvalue with Re λ = 0 which is deficient (has fewer eigenvectors than its multiplicity) then the system is not stable: each generalized eigenvector will give an unbounded solution. Remark. If a matrix is known only approximately, for example, if its entries are results of some measurements, and computation indicates that the characteristic equation has a multiple root, one cannot conclude whether this root is indeed multiple, and if it is, whether the root is deficient or not (whether there are as many eigenvectors as its multiplicity, or less). Many computer systems have a command called Jordan, which is supposed to find the Jordan form, but one has to be very careful with them: it is easy to deceive these programs, even with 2 × 2 matrices! 6
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2.1.3: More about Constant of Proportionality - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 2.1: Representing Proportional Relationships with Tables 2: Introducing Proportional Relationships { } { "2.1.1:_One_of_These_Things_is_Not_Like_the_Others" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.1.2:_Introducing_Proportional_Relationships_with_Tables" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.1.3:_More_about_Constant_of_Proportionality" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "2.01:_New_Page" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.02:_New_Page" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.03:_New_Page" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.04:_New_Page" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2.5:_Let\'s_Put_it_to_Work" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sat, 15 Aug 2020 03:29:46 GMT 2.1.3: More about Constant of Proportionality 38103 38103 admin { } Anonymous Anonymous 2 false false [ "article:topic", "license:ccby", "authorname:im" ] [ "article:topic", "license:ccby", "authorname:im" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Pre-Algebra 4. Pre-Algebra I (Illustrative Mathematics - Grade 7) 5. 2: Introducing Proportional Relationships 6. 2.1: Representing Proportional Relationships with Tables 7. 2.1.3: More about Constant of Proportionality Expand/collapse global location Pre-Algebra I (Illustrative Mathematics - Grade 7) Front Matter 1: Scale Drawings 2: Introducing Proportional Relationships 3: Measuring Circles 4: Proportional Relationships and Percentages 5: Rational Number Arithmetic 6: Expressions, Equations, and Inequalities 7: Angles, Triangles, and Prisms 8: Probability and Sampling 9: Putting it All Together Back Matter 2.1.3: More about Constant of Proportionality Last updated Aug 15, 2020 Save as PDF 2.1.2: Introducing Proportional Relationships with Tables 2.2: Representing Proportional Relationships with Equations Page ID 38103 Illustrative Mathematics OpenUp Resources ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Lesson 1. Summary 2. Glossary Entries Practice Lesson Let's solve more problems involving proportional relationships using tables. Exercise 2.1.3.1: Equal Measures Use the numbers and units from the list to find as many equivalent measurements as you can. For example, you might write “30 minutes is 1 2 hour.” You can use the numbers and units more than once. 1 1 2 0.3 centimeter 12 40 24 meter 0.4 0.01 1 10 hour 60 3 1 3 6 feet 50 30 2 minute 2 5 inch Exercise 2.1.3.2: Centimeters and Millimeters There is a proportional relationship between any length measured in centimeters and the same length measured in millimeters. Figure 2.1.3.1 There are two ways of thinking about this proportional relationship. If you know the length of something in centimeters, you can calculate its length in millimeters. Complete the table. What is the constant of proportionality? | length (cm) | length (mm) | --- | | 9 | | | 12.5 | | | 50 | | | 88.49 | | Table 2.1.3.1 If you know the length of something in millimeters, you can calculate its length in centimeters. Complete the table. What is the constant of proportionality? | length (mm) | length (cm) | --- | | 70 | | | 245 | | | 4 | | | 699.1 | | Table 2.1.3.2 How are these two constants of proportionality related to each other? Complete each sentence: To convert from centimeters to millimeters, you can multiply by __. To convert from millimeters to centimeters, you can divide by _ _or multiply by __. Are you ready for more? How many square millimeters are there in a square centimeter? How do you convert square centimeters to square millimeters? How do you convert the other way? Exercise 2.1.3.3: Pittsburgh to Phoenix On its way from New York to San Diego, a plane flew over Pittsburgh, Saint Louis, Albuquerque, and Phoenix traveling at a constant speed. Complete the table as you answer the questions. Be prepared to explain your reasoning. Figure 2.1.3.2: A United States map with 5 segments representing the distance a plane flew. The segments are New York to Pittsburgh, Pittsburgh to Saint Louis, Saint Louis to Albuquerque, Albuquerque to Phoenix and Phoenix to San Diego. By United States Census Bureau. Public Domain. American Fact Finder. Source. | segment | time | distance | speed | --- --- | | Pittsburgh to Saint Louis | 1 hour | 550 miles | | | Saint Louis to Albuquerque | 1 hour 42 minutes | | | | Albuquerque to Phoenix | | 330 miles | | Table 2.1.3.3 What is the distance between Saint Louis and Albuquerque? How many minutes did it take to fly between Albuquerque and Phoenix? What is the proportional relationship represented by this table? Diego says the constant of proportionality is 550. Andre says the constant of proportionality is 9 1 6. Do you agree with either of them? Explain your reasoning. Summary When something is traveling at a constant speed, there is a proportional relationship between the time it takes and the distance traveled. The table shows the distance traveled and elapsed time for a bug crawling on a sidewalk. Figure 2.1.3.3: Table with 2 columns and 4 rows of data. The columns are: distance traveled (cm) and elapsed time (sec). The table has the ordered pairs (the fraction 3 over 2 comma 1), (1 comma the fraction 2 over 3), (3 comma 2) and (10 comma the fraction 20 over 3). Each pair has an arrow pointing from the value in the first column to the value in the second column. Times the fraction 2 over 3 is below the table. We can multiply any number in the first column by 2 3 to get the corresponding number in the second column. We can say that the elapsed time is proportional to the distance traveled, and the constant of proportionality is 2 3. This means that the bug’s pace is 2 3 seconds per centimeter. This table represents the same situation, except the columns are switched. Figure 2.1.3.4: Table with 2 columns and 4 rows of data. The columns are: elapsed time (sec) and distance traveled (cm). The table has the ordered pairs (1 comma the fraction 3 over 2), (the fraction 2 over 3 comma 1), (2 comma 3) and (the fraction 20 over 3 comma 10). Each pair has an arrow pointing from the value in the first column to the value in the second column. Times the fraction three over 2 is below the table. We can multiply any number in the first column by 3 2 to get the corresponding number in the second column. We can say that the distance traveled is proportional to the elapsed time, and the constant of proportionality is 3 2. This means that the bug’s speed is 3 2 centimeters per second. Notice that 3 2 is the reciprocal of 2 3. When two quantities are in a proportional relationship, there are two constants of proportionality, and they are always reciprocals of each other. When we represent a proportional relationship with a table, we say the quantity in the second column is proportional to the quantity in the first column, and the corresponding constant of proportionality is the number we multiply values in the first column to get the values in the second. Glossary Entries Definition: Constant of Proportionality In a proportional relationship, the values for one quantity are each multiplied by the same number to get the values for the other quantity. This number is called the constant of proportionality. In this example, the constant of proportionality is 3, because 2⋅3=6, 3⋅3=9, and 5⋅3=15. This means that there are 3 apples for every 1 orange in the fruit salad. | number of oranges | number of apples | --- | | 2 | 6 | | 3 | 9 | | 5 | 15 | Table 2.1.3.4 Definition: Equivalent Ratios Two ratios are equivalent if you can multiply each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, 8:6 is equivalent to 4:3, because 8⋅1 2=4 and 6⋅1 2=3. A recipe for lemonade says to use 8 cups of water and 6 lemons. If we use 4 cups of water and 3 lemons, it will make half as much lemonade. Both recipes taste the same, because and are equivalent ratios. | cups of water | number of lemons | --- | | 8 | 6 | | 4 | 3 | Table 2.1.3.5 Definition: Proportional Relationship In a proportional relationship, the values for one quantity are each multiplied by the same number to get the values for the other quantity. For example, in this table every value of p is equal to 4 times the value of s on the same row. We can write this relationship as p=4⁢s. This equation shows that s is proportional to p. | s | p | --- | | 2 | 8 | | 3 | 12 | | 5 | 20 | | 10 | 40 | Table 2.1.3.6 Practice Exercise 2.1.3.4 Noah is running a portion of a marathon at a constant speed of 6 miles per hour. Complete the table to predict how long it would take him to run different distances at that speed, and how far he would run in different time intervals. | time in hours | miles traveled at 6 miles per hour | --- | | 1 | | | 1 2 | | | 1 1 3 | | | | 1 1 2 | | | 9 | | | 4 1 2 | Table 2.1.3.7 Exercise 2.1.3.5 One kilometer is 1000 meters. Complete the tables. What is the interpretation of the constant of proportionality in each case? | meters | kilometers | --- | | 1,000 | 1 | | 250 | | | 12 | | | 1 | | Table 2.1.3.8 The constant of proportionality tells us that: | kilometers | meters | --- | | 1 | 1,000 | | 5 | | | 20 | | | 0.3 | | Table 2.1.3.9 The constant of proportionality tells us that: What is the relationship between two constants of proportionality? Exercise 2.1.3.6 Jada and Lin are comparing inches and feet. Jada says that the constant of proportionality is 12. Lin says it is 1 12. Do you agree with either of them? Explain your reasoning. Exercise 2.1.3.7 The area of the Mojave desert is 25,000 square miles. A scale drawing of the Mojave desert has an area of 10 square inches. What is the scale of the map? (From Unit 1.2.6) Exercise 2.1.3.8 Which of these scales is equivalent to the scale 1 cm to 5 km? Select all that apply. 3 cm to 15 km 1 mm to 150 km 5 cm to 1 km 5 mm to 2.5 km 1 mm to 500 m (From Unit 1.2.5) Exercise 2.1.3.9 Which one of these pictures is not like the others? Explain what makes it different using ratios. Figure 2.1.3.5 (From Unit 2.1.1) This page titled 2.1.3: More about Constant of Proportionality is shared under a CC BY license and was authored, remixed, and/or curated by Illustrative Mathematics. Back to top 2.1.2: Introducing Proportional Relationships with Tables 2.2: Representing Proportional Relationships with Equations Was this article helpful? Yes No Recommended articles 2.1.2: Introducing Proportional Relationships with Tables 2.1.1: One of These Things is Not Like the Others 2.2: Representing Proportional Relationships with Equations 2.3: Comparing Proportional and Nonproportional Relationships 2.4: Representing Proportional Relationships with Graphs Article typeSection or PageAuthorIllustrative MathematicsLicenseCC BY Tags This page has no tags. © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. 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https://www.r-bloggers.com/2021/10/pearson-correlation-in-r/
Pearson correlation in R [This article was first published on R tutorials – Statistical Aid: A School of Statistics, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here) Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. The Pearson correlation coefficient, sometimes known as Pearson’s r, is a statistic that determines how closely two variables are related. Its value ranges from -1 to +1, with 0 denoting no linear correlation, -1 denoting a perfect negative linear correlation, and +1 denoting a perfect positive linear correlation. A correlation between variables means that as one variable’s value changes, the other tends to change in the same way. Creating or Importing data into R Let’s import data into R or create some example data as follows: set.seed(150) data <- data.frame(x = rnorm(50, mean = 50, sd = 10), random = sample(c(-10:10), 50, replace = TRUE)) data$y <- data$x + data$random set.seed(150) data <- data.frame(x = rnorm(50, mean = 50, sd = 10), random = sample(c(-10:10), 50, replace = TRUE)) data$y <- data$x + data$random set.seed(150) data <- data.frame(x = rnorm(50, mean = 50, sd = 10), random = sample(c(-10:10), 50, replace = TRUE)) data$y <- data$x + data$random If we want to calculate the Pearson’s correlation of x and y in data, we can use the following code: correlation <- cor(data$x, data$y, method = 'pearson') Checking the results: > correlation 0.9025428 correlation <- cor(data$x, data$y, method = 'pearson') Checking the results: > correlation 0.9025428 correlation <- cor(data$x, data$y, method = 'pearson') Checking the results: > correlation 0.9025428 From the above result, we get that Pearson’s correlation coefficient is 0.90, which indicates a strong correlation between x and y. Interpretation of Pearson Correlation Coefficient The value of the correlation coefficient (r) lies between -1 to +1. When the value of – r=0; there is no relation between the variable. r=+1; perfectly positively correlated. r=-1; perfectly negatively correlated. r= 0 to 0.30; negligible correlation. r=0.30 to 0.50; moderate correlation. r=0.50 to 1 highly correlated. A common misconception about the Pearson correlation is that it provides information on the slope of the relationship between the two variables being tested. This is incorrect, the Pearson correlation only measures the strength of the relationship between the two variables. To illustrate this, consider the following example: set.seed(150) xvalues <- rnorm(50, mean = 50, sd = 10) random <- sample(c(10:30), 50, replace = TRUE) data <- data.frame(x = rep(xvalues, 2), random = rep(random, 2), category = rep(c("One","Two"), each = 50)) data$y[data$category=="One"] <- 20 + data$x[data$category=="One"]/data$random[data$category=="One"] data$y[data$category=="Two"] <- 20 + data$x[data$category=="Two"]/(5data$random[data$category=="Two"]) correlation.one <- cor(data$x[data$category=="One"], data$y[data$category=="One"], method = 'pearson') correlation.two <- cor(data$x[data$category=="Two"], data$y[data$category=="Two"], method = 'pearson') set.seed(150) xvalues <- rnorm(50, mean = 50, sd = 10) random <- sample(c(10:30), 50, replace = TRUE) data <- data.frame(x = rep(xvalues, 2), random = rep(random, 2), category = rep(c("One","Two"), each = 50)) data$y[data$category=="One"] <- 20 + data$x[data$category=="One"]/data$random[data$category=="One"] data$y[data$category=="Two"] <- 20 + data$x[data$category=="Two"]/(5data$random[data$category=="Two"]) correlation.one <- cor(data$x[data$category=="One"], data$y[data$category=="One"], method = 'pearson') correlation.two <- cor(data$x[data$category=="Two"], data$y[data$category=="Two"], method = 'pearson') set.seed(150) xvalues <- rnorm(50, mean = 50, sd = 10) random <- sample(c(10:30), 50, replace = TRUE) data <- data.frame(x = rep(xvalues, 2), random = rep(random, 2), category = rep(c("One","Two"), each = 50)) data$y[data$category=="One"] <- 20 + data$x[data$category=="One"]/data$random[data$category=="One"] data$y[data$category=="Two"] <- 20 + data$x[data$category=="Two"]/(5data$random[data$category=="Two"]) correlation.one <- cor(data$x[data$category=="One"], data$y[data$category=="One"], method = 'pearson') correlation.two <- cor(data$x[data$category=="Two"], data$y[data$category=="Two"], method = 'pearson') The Pearson correlation coefficient of these two sets of x and y values is exactly the same: correlation.one 0.462251 correlation.two 0.462251 correlation.one 0.462251 correlation.two 0.462251 ``` correlation.one 0.462251 correlation.two 0.462251 ``` However, when we plot these x and y values on a chart, the relationship looks very different: library(ggplot2) gg <- ggplot(data, aes(x, y, colour = category)) gg <- gg + geom_point() gg <- gg + geom_smooth(alpha=0.3, method="lm") print(gg) library(ggplot2) gg <- ggplot(data, aes(x, y, colour = category)) gg <- gg + geom_point() gg <- gg + geom_smooth(alpha=0.3, method="lm") print(gg) library(ggplot2) gg <- ggplot(data, aes(x, y, colour = category)) gg <- gg + geom_point() gg <- gg + geom_smooth(alpha=0.3, method="lm") print(gg) Learn Data Science and Machine Learning Data Analysis Using R/R Studio Import data into R Principal component analysis (PCA) code Canonical correlation analysis (CCA) code Independent component analysis (ICA) code Cluster Analysis using R One-way ANOVA using R Two-way ANOVA using R Paired sample t-test using R Random Forest in R Chi-square test using R The post Pearson correlation in R appeared first on Statistical Aid: A School of Statistics. 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https://scholarworks.uni.edu/cgi/viewcontent.cgi?article=1907&context=hpt
Published Time: Wed, 31 Jan 2024 14:30:29 GMT University of Northern Iowa University of Northern Iowa UNI ScholarWorks UNI ScholarWorks Honors Program Theses Student Work 2024 Proportional Reasoning: Understanding Student Thinking with Proportional Reasoning: Understanding Student Thinking with Direct and Inverse Proportions Direct and Inverse Proportions Emma Marshall University of Northern Iowa Let us know how access to this document benefits you Copyright ©2024 Emma Marshall Follow this and additional works at: Recommended Citation Recommended Citation Marshall, Emma, "Proportional Reasoning: Understanding Student Thinking with Direct and Inverse Proportions" (2024). Honors Program Theses . 908. This Open Access Honors Program Thesis is brought to you for free and open access by the Student Work at UNI ScholarWorks. It has been accepted for inclusion in Honors Program Theses by an authorized administrator of UNI ScholarWorks. For more information, please contact scholarworks@uni.edu . Offensive Materials Statement: Materials located in UNI ScholarWorks come from a broad range of sources and time periods. Some of these materials may contain offensive stereotypes, ideas, visuals, or language. PROPORTIONAL REASONING: UNDERSTANDING STUDENT THINKING WITH DIRECT AND INVERSE PROPORTIONS A Thesis Submitted in Partial Fulfillment of the Requirements for the Designation University Honors Emma Marshall University of Northern Iowa May 2024 This Study by: Emma Marshall Entitled: Proportional Reasoning: Understanding Student Thinking with Direct and Inverse Proportions has been approved as meeting the thesis requirement for University Honors Approved by: Olof Steinthorsdottir, Honors Thesis Advisor Dr. Jessica Moon Asa, Director, University Honors Program 1 I. Title Proportional Reasoning: Understanding Student Thinking with Direct and Inverse Proportions II. Purpose The topic of proportions in mathematics is often misunderstood by students, yet proportional reasoning is the foundation for higher-level math. The purpose of this study is to understand how students are currently solving direct and inverse proportion problems and how the degree of difficulty increases when solving inverse proportion problems. There is little research done about student thinking with inverse proportions. Primary data has been collected from students and will be used to analyze students’ thinking about inverse proportion as well as a comparative look between direct and inverse proportional reasoning. By understanding students' thinking about proportions, educators can better prepare lessons that will strengthen the students’ conceptual understanding rather than just procedural knowledge of direct and inverse proportions. III. Literature Review Proportional reasoning is a critical foundation of mathematics for students as they progress through the school grade levels and into the real world. A proportion is a relationship between two ratios that can be set equivalent to each other (Christou & Philippou, 2002). There are two main types of proportional relationships, direct and inverse. A direct proportion is defined as when one quantity increases, the other quantity increases by a steady rate of change. Lobato and Ellis defined direct proportion saying, “When two quantities are related proportionally, the ratio of one quantity to the other is invariant as the numerical values of both 2 quantities change by the same factor” (2010, p. 11). Inverse proportions define the opposite. As one quantity increases, the other decreases at a steady rate of change. Another way to define inverse relationships is “changing one quantity by a factor of k means that the other quantity needs to be changed by a factor of 1/k,” (Markworth, 2012, p. 541). Direct and inverse proportions are types of proportional relationships. Student Mathematical Development Students begin learning about proportionality in kindergarten by looking at the relationship between numbers and number pairs. As they progress, the depth to which students use qualitative relationships increases. Third through fifth graders begin learning multiplicative reasoning and patterns. The concept and teaching of proportional reasoning is beneficial because students are learning about other mathematical ideas like multiplication, constant rate of change, linear functioning, and more. According to the Iowa Department of Education, the actual term “proportions” makes a greater appearance between sixth through eighth grade (Pforts, 2010). Students start with a basic understanding of the qualitative relationship of direct and inverse proportions. Then, they begin to add numbers into those qualitative relationships, turning them into quantitative problems. Most students begin their proportion journey by trying repeated addition strategies and slowly start to make multiplicative connections with math problems that are doubling or halving the quantities given. After that, students are expected to construct a quantitative understanding of the two qualitative relationships. They must see the rate of change between or within-measure spaces and be able to apply it to the missing quantity (Ferrandez-Reinisch, 1985). Connecting mathematical ideas across grade levels is critical. Proportional reasoning allows students to construct high-level explanations of similar traits among ideas to increasingly sophisticated problems. The ability to recognize and deeply understand proportional 3 relationships is needed for future success in algebra and beyond (Seeley & Schielack, 2007). Students’ conceptual understanding of proportional relationships will be a guide as they continue to build upon this mathematical knowledge throughout schooling and potentially in their future careers. Elements that Influence Student Thinking One of the elements that influence students is the number relationships in a proportional situation. Fernández and his team found that when the ratios have a multiplicative relationship that is a non-integer, they were less successful. The data they collected had 46.4% of students correctly answer problems with integer relationships and only 34.3% of students answered correctly with non-integer problems (Fernández, Ceneida, et al., 2010). With Dr. Riehl and Dr. Steinthorsdottir's research (2019), the same goes for “enlarging” over “shrinking” problems. When the ratio can be scaled down, that is classified as a shrink problem, and when the target is larger than the given situation or “scaled up”, that is classified as an enlarge problem. Valid Direct Proportion Strategies When looking at proportional reasoning strategies, research has grouped them into four distinct, correct categories. To explain the strategies, I will use the following missing value proportion problem: Kelly is making cranberry juice for a party. Her original recipe is 3 cups of cranberry juice for every 2 cups of water. If Kelly has enough supplies for 15 cups of cranberry juice, how many cups of water should she add if she wants to keep the same taste? The first strategy is the “build-up” method. Students take the given quantities and develop a soft relationship between them. For example, as cranberry juice increases by 3 cups, the water will increase by 2 cups. They see this relationship but use repeated addition measures to get to the missing value. This could be represented in a picture, written numbers, or a ratio 4 table. Figure 1 provides an example of how students could set up the problem to find how much water is needed for 15 cups of cranberry juice. Figure 1 Ratio Table Example Cranberry (cups) 3 6 9 12 15 Water (cups) 2 4 6 8 10 In the Figure above, the student found the solution by adding another round of cranberry juice. Three cups plus another 3 cups of cranberry juice get them to 6 cups of cranberry juice. Then, they must do the same for water. Two cups of water added gets them a total of 4 cups of water. This process is repeated until the student stops at 15 cups of cranberry juice and reveals the answer of 10 cups of water as the missing value. There are many ways to display the “build-up” strategy from repeated addition to multiplicative building. Students are building their understanding using a rudimentary method. Students do not fully understand the relationship within (scalar method) or between (unit rate) measure spaces but the build-up strategy can become a transition step into more sophisticated forms of reasoning (Ercole, Frantz, & Ashline, 2011). Students used different strategies as they progressed through different grades. Student thinking tends to get more robust, losing the initial build-up strategy, their strategies become more sophisticated. Repeated addition strategies shift to proportional strategies as students transition from primary to secondary school (Fernández, Llinares, et al., 2010). In some cases, the build-up method decreases as students encounter more systematic teaching. In this instance, students’ problem-solving skills using the “build-up strategy” tend to be rewired to follow a given set of instructions. For inverse proportions, students must recognize that the relationship 5 between quantities is inverse so that as they build “down” on one of the quantities, the other is building “up.” As one quantity increases, the other shrinks. The second strategy is finding the scale factor. Students who use this method find the multiplicative relationship within two quantities of the same variable. This is often called the “within-measure space multiplicative relationship” or scalar method. With integer problems, students tended to use the within-measure space multiplicative relationship strategy to solve more often (Riehl & Steinthorsdottir, 2019). In the example, students will recognize that the two quantities covary. The students recognize that 5 groups of 3 cups of cranberry juice are equivalent to fifteen cups of cranberry juice, therefore, they get a scale factor of 5. Students would then multiply the 2 cups of water with a scale factor of 5 to get the answer of 10 cups of water. They are multiplying between the same variable which is the key to a within-measure space relationship. The third strategy involves finding the unit rate. Students who use this method look at the relationship between two quantities of different variables. This multiplicative relationship is often called “between-measure space” or the constant of proportionality. The quantities that make up a ratio covary, but the relationship between them remains invariant (Silvestre & Ponte, 2011; Langrall & Swafford, 2000). It breaks the problem down into finding how much of a single quantity would affect the second quantity. For example, students could take the 3 cups of cranberry juice and divide it by the 2 cups of water to get a unit rate of 1.5. For every 1 cup of water, there are 1.5 cups of cranberry juice. Then, they would divide the 15 cups of cranberry juice by 1.5 to get 10 cups of water. The important part of the “between-measure space” relationship is that students are solving for the unit rate between the two different variables. 6 The fourth strategy is almost entirely procedural and often fails to identify the proportional relationship between the quantities. This is the cross-multiplication method. This method is not recommended, but if taught, should be introduced once students have a solid conceptual understanding of proportions (Chapin & Anderson, 2003). Figure 2 provides an example of how students could set up the problem using a cross-multiplication method. Figure 2 Cross-Multiplication Example Then, they would multiply across the quantities to get 3x=30 and solve algebraically to get x=10 cups of water. It is an efficient algorithm, especially when dealing with noninteger numbers, but can loop students into the procedure of using the numbers without understanding what they mean (Ercole, Frantz, & Ashline, 2011). Sense-making is important before moving on to procedures like this when needed in proportion problems. Cross-multiplying is not good for students’ sense-making (Lobato, et al., 2010). Most importantly, cross multiplication cannot be applied to inverse proportions which furthers the divide in students' conceptual understanding of proportional relationships. Invalid Direct Proportion Strategies Students can correctly solve proportional reasoning problems after lots of direct instruction, but they have a much stronger ability to problem solve and use proportional reasoning skills when discovering concepts through problem-based teaching. Exploration and investigation lead to correct reasoning and justification. It takes educators carefully planning a 3 2 15 x7 sequence of problems for students to explore that will give students the best learning experience. However, no matter the preparation and teaching provided, many students still face difficulties, resulting in erroneous methods and misconceptions (Miller & Fey, 2000). It is important for educators to be aware of these common errors so that instruction can be geared toward combating them. With direct proportions, the common errors include non-proportional reasoning, miscalculations, or additive errors. (Langrall & Swafford, 2000). When students use additive comparison, they are finding the difference between the two quantities in the given ratio rather than comparing them multiplicatively. Additive reasoning is in direct contrast to proportional reasoning yet it is the most frequent error. This type of reasoning is very common among students and often begins their rudimentary problem-solving of proportions (Bright, Joyner, & Wallis, 2003). With inverse proportions, similar issues can occur, but it is also common to misinterpret the problem by solving it proportionally. In the research done by De Bock et al. (2015), when errors occurred with inverse proportion problems, it was most often due to proportional errors. Some errors cannot be seen by looking at a single problem. There can be multiple problems of varying difficulty even though the broad category of a “direct” or “inverse” problem exists for all of them. To see the true thinking of the student, teachers need to allow students to apply their knowledge in different situations, with different degrees of difficulty (Bright, Joyner, & Wallis, 2003). This will reveal commonalities of errors that educators can focus on to help the entire class succeed with direct and inverse proportions. IV. Research Question to Be Answered 8 The goal of this research is to better understand students’ proportional reasoning skills and strategies for inverse proportions and how they compare to students’ reasoning for direct proportions. How are students reasoning about inverse proportion? What common errors are occurring in inverse proportion problems? Does student thinking differ when dealing with direct versus inverse proportions? V. Methodology This study analyzed data about proportional reasoning collected by Dr. Steinthorsdottir and Dr. Riehl in their study "Route to Reason" (Riehl & Steinthorsdottir, 2014; 2019). They gave direct and inverse proportion problems to 409 5th to 8th graders from a Midwest school district in 2017. For this study, I analyzed the written work of 243 students in 7th and 8th grade on problems addressing inverse proportion. At this stage of schooling, students have worked with proportional reasoning for some time now, more so in 8th grade than in 7th grade. These two grades were selected because they have had more time to develop proportional reasoning skills (over 5th and 6th graders) which will make observation and analysis of their strategies more effective. The research done in the areas of non-integer vs integer relationships in proportions informed my decision on students’ questions to analyze. According to the research, enlarging proportion problems with an integer relationship tend to have the most correct answers from students. With this information, I chose an enlarging, non-integer inverse proportion problem. This would give me the clearest view of a strategy. Sometimes with enlarging, integer relationships, students can find a way to reach the correct answer without understanding the 9 strategy behind their work. If a student understands the multiplicative nature of proportion and can apply a multiplicative strategy, it should not matter if it is an integer or non-integer number in relation to the problem. In addition, by using an “enlarging” problem, fewer students will face confusion after multiple experiences with solving direct proportion problems about enlarging. Shrinking problems are a different category of work that I will not be looking into at this time. I chose to select a problem that was enlarging and a noninteger relationship. I wanted a problem that was more familiar to students while still being challenging. The inverse proportion problem I chose to investigate was question #8 on the instrument collected by Dr. Steinthorsdottir and Dr. Riehl. It reads, “The Green Soles Nature Club is going to clean up the stream bank. They figure it will take 12 members all working together 6 hours. They decided to recruit more helpers. If 18 members work together, how long will it take? Explain your thinking.” It is a more familiar enlarging problem with 12 members increasing to 18 members, while the noninteger, within-measure space relationships of 1.5 adds a challenge. The literature review explained multiple strategies students use when solving proportion problems. To best analyze how students are reasoning about inverse proportions, the data from the inverse problem above was separated into different categories based on the strategy used by the student. First, students’ thinking was separated into correct and erroneous answers. The strategies that resulted in a correct answer were then analyzed, using the framework of strategies research has identified for direct proportion problems. The strategies are (1) Build-Up strategies, ranging from rudimentary unit by unit “build up” to more complex multiplicative build up, (2) Within Measure Space Comparison strategies, (3) Between Measure Space Comparison strategies, (4) Cross-Multiplication strategies, and (5) strategy unclear. This helped me to see what ways students were reasoning about inverse proportions. The erroneous category was 10 broken down into different answers with an invalid reasoning behind each one. I grouped the data based on an answer. I did expect errors based on additive thinking and errors where students interpret the problem as direct proportion. The category for random answers was based on whether there was no explanation given for an invalid answer or if there was only one answer that was not one of the most common errors. I further analyzed the mathematical reasons for the errors as I let it inform my results on what common errors there are in inverse proportion problems. The categories revealed where student misunderstandings are most likely taking place. Finally, research already completed by Dr. Steinthorsdottir and Dr. Riehl in their study "Route to Reason" (Riehl & Steinthorsdottir, 2014; 2019) was used in comparison with the results on student understanding of inverse proportion problems. They looked at the strategies used to solve direct proportion problems. I compared an enlarging, noninteger direct proportion problem to the enlarging, noninteger inverse proportion problem. The data for the direct proportion problem had been analyzed according to strategies. I used the analysis conducted by Dr. Steinthorsdottir and Dr. Riehl. Afterward, I synthesized my results to answer my research questions and shared what students were currently thinking about proportions. My analysis and categorization revealed the most common thinking patterns and the most common errors. I took this information to make generalizations of students' conceptual understanding of inverse proportions. VI. Results To begin, the percentage of correct answers was very low for both 7th and 8th grade data. There were 92 total 7th graders and 16% of the 7th graders answered this inverse proportion problem correctly. Similarly, there were 151 total 8th graders and 23% of the 8th graders 11 answered this inverse proportion problem correctly. There was an increase in correct answers from 7th to 8th grade, which can partially be explained by the additional year of mathematics learning for 8th graders. The data revealed a range of strategies used by the students to solve the inverse proportion problem. As explained previously, there was certain student thinking that made clear categories of strategies with direct proportions. In inverse proportional reasoning, however, not all of these same categories could be described in the same way. The analysis of strategies revealed how students were reasoning about inverse proportions. The strategies that emerged from the analysis were (1) Build-Up method, (2) Within Measure Space Relationship method, (3) Cross Multiplication Hybrid Relationship method, (3) Rate per One method, and (4) No clear explanation (see Table 1). The following sections will explain the different strategies used by students. Tab le 1 Valid Strategies for Problem 8 Build-Up Within- Measure Space Cross Multiplication Hybrid Rate per One Unclear Total Number of 7th Graders 4 (26.7%) 4 (26.7%) 3 (20%) 0 (0%) 4 (26.7%) 15 (100%) Number of 8th Graders 8 (23.5%) 6 (17.6%) 4 (11.8%) 3 (8.8%) 13 (38.2%) 34 (100%) Valid Inverse Proportion Strategies Build-Up Strategies Taking a look at the strategies used, one can see a similar percentage of 7th and 8th grade students using the “build-up” strategy to solve the problem. For 7th graders, 26.7% of the 12 students used this strategy while 23.5% of 8th graders did (see Table 1). Figure 3 is an example of a “build-up” strategy used by an 8th grader. In the strategy, the student sets up the information given to them in the problem as a proportion: for 12 members working together, it takes 6 hours. They divide the members by two and multiply the hours by two to get 6 members working for 12 hours. The student understands that this problem is an inverse proportion because they write, “If you take away members it will take longer so dividing the members in half will double the work time”. The student understands that as one quantity increases, the other quantity decreases at a constant rate of change. Finally, the student multiplies 6 members by 3 to reach 18 members, and by understanding the inverse relationship the 12 hours needs to be divided by 3 to find the answer. This is a “build-up” strategy because the student works with the “within-measure space” relationship but by breaking it down into smaller steps, “builds down” first and then “builds up”. They built the 3/2 relationship to reach their desired answer of 18 members. This, then, reveals the answer of 4 hours. Figure 3 8th Grader’s “Build-Up” Strategy Within-Measure Space Strategies 13 The strategy of the within-measure space relationship is seen when students find a scale factor of 1.5 or 3/2. One efficient way to demonstrate this is in Figure 4. The student sets up a within-measure space relationship between the 12 and 18 members and finds that 18 is 3/2 of 12. Using this scale factor inversely, the student sees that 4 is ⅔ of 6. The student represents this, however, by multiplying 12 by 1.5 and dividing the 6 hours by 1.5 to get the total of 4 hours. With the data broken into strategies, the use of within-measure space in 7th grade is higher at 26.7% while the 8th graders use it 17.6% of the time (see Table 1). One factor that might be affecting the lower use of build-up strategies is the high percentage of unclear explanations in the 8th grade data. More students might have been using this strategy mentally without writing down their thinking. These strategies exhibit students’ understanding of within-measure space relationships. Their use of the within-measure space relationship demonstrates that they are understanding the properties of an inverse relationship. Figure 4 8th Grader’s “Within-Measure Space” Strategy Between Measure Space - Unit Per One Strategies 14 A new strategy that I did not expect was found in three 8th graders, with only 8.8% of correct answers (see Table 1). They found the hourly rate per member. The student in Figure 5 knew that 12 members times 6 hours gets 72 total hours. To understand what that means, the student understands that each of the 12 members would work 6 hours which equals a total of 72 hours needed to complete the work. This gives us the rate per person. Therefore, when there are 18 members, the 72 hours of total work is divided among them to get 4 hours of work per member. This new strategy was thought-provoking as I analyzed it. Three students solved the problem using this strategy, showing that there is student thinking happening. The strategy works well for inverse proportion problems because it can be used for any problem with two quantities. For these students, it shows a deep understanding of the story problem and the mathematical principles behind the numbers. Figure 5: 8th Grader’s “Rate per Person” Strategy Between Measure Space - Cross Multiplication Hybrid Strategies As I was analyzing the between-measure space relationship, I noticed that it would not work the same way as a direct proportion would. In a direct proportion, the constant of 15 proportionality or “k” is 𝑎𝑎 𝑏𝑏 , however, an inverse proportion has the “k” as 𝑎𝑎 × 𝑏𝑏 . Finding the relationship ⅓ between the given 6 hours and the target of 18 members, which are two opposite quantities, does not have a unit that makes sense within the context of the problem. Mathematically it can be explained with what I have labeled as a hybrid cross multiplication and between-measure space strategy (see Figure 6). Figure 6 Explanation of the Mathematics Behind the Hybrid Cross Multiplication Strategy Looking between Figure 6 and the student’s work in Figure 7, the student finds ⅓ of 6 hours which is the answer of 2. Next, they subtract the 2 hours from 6 to get an answer of 4 hours. This aligns with the mathematical explanation in Figure 6 because 12 members times ⅓ gets the answer of 4 hours. 2 times 6 times ⅓ is the same equation. The only difference with Figure 7 is that the student subtracts their answer of 2 by 6 to get the answer of 4 hours while the mathematical equation multiplies that 2 by 2 hours. With this new hybrid strategy, one can see a large number of students using it; 20% of 7th graders and 11.8% of 8th graders used this strategy (see Table 1). Figure 7 8th Graders Cross Multiplication and Between-Measure Space Hybrid 16 Invalid Inverse Strategies While there was analysis of the correct answers, there is also value in analyzing the strategies of incorrect answers. After separating the answers into their categories, I followed students’ thinking through the work they wrote down with their answers. There were 92 total 7th graders and 77 invalid answers given. 83.7% of sampled 7th graders answered this inverse proportion problem using an invalid strategy (see Table 2). There were 151 total 8th graders and 117 invalid answers given. 77.5% of sampled 8th graders answered this inverse proportion problem using an invalid strategy (see Table 2). Of the common answers, I found a pattern of students still using direct proportional reasoning with the inverse proportion problem. There were also additive reasoning errors and invalid build-up strategies. Table 2 Invalid Answers for Problem 8 2 hours 3 hours 4.5 hours 9 hours 12 hours Random Total Number of 7th graders 5(6.5%) 35 (45.5%) 5 (6.5%) 14 (18.2%) 5 (6.5%) 13 (16.9%) 77 (100%) Number of 8th graders 6(5.1%) 44 (37.6%) 4 (3.4%) 39 (33.3%) 8 (6.8%) 16 (13.7%) 117 (100%) 17 The first category I looked at was the answer of 9 hours. 18.2 % of 7th graders and 33.3% of 8th graders answered with “9 hours” which is a significant amount (see Table 2). Many students found the scale factor of 3/2 or built up to 9 hours using the thinking of direct proportionality. A third of 8th graders were solving this problem as a direct proportion which is a concern. Figure 8 reveals one student’s thought process. They found the between-measure space relationship with a unit rate of 2. They divided 18 members by 2 to get their answer of “9 hours.” This reveals that the student is thinking about a direct relationship, however, if you think about the problem logically, you can see that more members would mean that the job could be done in less time. Part of the problem is that students are missing the important part of thinking about the math behind the given problem. I believe that if the problem was given with no numbers first, and they were asked to think about how more members would impact the number of hours, many students would conclude that the answer would have to be less than the current number of hours. They would be thinking about the mathematical principles behind the story problem, rather than relying on the pattern of problems that they had solved prior. Then, when students are handed the original problem, the answer of 9 would cause students to pause and reevaluate their thought process. Students finding an answer of 9 hours did not connect that this problem was inverse proportions and not a direct proportion. Figure 8 18 8th Grader’s “9 hour” Strategy The next category of answers was 3 hours. Many 7th and 8th grade students also fell into additive reasoning errors. 45.5% of 7th graders and 37.6% of 8th graders had an answer of 3 hours (see Table 2). This was the largest category of error in both grade levels. As I analyzed this common error, I saw a lot of different strategies being used, most of them were additive. It was interesting that there were so many different strategies yet they all got 3 hours. One strategy was a mix of a build-up and multiplicative strategy. The student in Figure 9 found the unit rate of 2 by dividing 12 by 6. Then the student divided 12 by 2 and 6 by 2 to get the answers of 6 and 3. They switched over to the build-up strategy and added 6 to 12 members to get 18 members. Next, they subtracted 3 hours to 6 hours to get their final answer of 3 hours. The student could see the inverse relationship a little because they understood that as one quantity increases, the other decreases which is why they added 6 to the top quantity but subtracted 3 from the bottom quantity. Another student reasoned in their response like this, “If you add half as much people to the first amount, it will take half of the time.” They are explaining that if you add 6 more members (which is half of 12) to 12 members to get 18 members, you need to divide the hours 19 by half, which is 3 hours. A commonality among students that I observed was that they were looking at the 6 extra members that were added from 12 to get 18 members. This fueled a lot of the different strategies that I saw. Figure 9 8th Grader’s “3 hour” Strategy There were a few answers that were not common, being under 7% of students but had repetition for a few students (see Table 2). The “12 hours” answer tended to be just an additive and multiplicative error as well. Many students saw an increase of 6 members between 12 and 18 members and added 6 to the number of hours as well. The answers of “2 hours” and “4.5 hours” had no consistent strategy but some were solved with an additive error as well. One interesting strategy of a student was that they built up to 24 members by doubling the 12 members. 18 is the median between 12 and 24 members. Then, they halved the 6 hours to 3 hours and found the middle of 6 and 3. This got them the answer of 4.5 hours. This student understood the concept of the problem being an inverse but did not show an understanding of the multiplicative relationship involved in the proportion. These 3 answers were found to be repeated by a few students but there was no consistency in strategies. 20 Direct vs. Inverse Comparison Using analysis already completed by Dr. Steinthorsdottir and Dr. Riehl in their study "Route to Reason," I compared the breakdown of strategies used with an enlarging, noninteger, direct proportion problem. Question #3 says “Grandma has some special plant food that comes in powder form. She uses 3 scoops to feed 9 plants. There are 11 scoops of plant food left. Using the same strength, how many plants can she feed? Explain your thinking.” 50% of 7th and 8th grade students got this direct proportion problem correct while about 20.2% of these same students got the inverse proportion problem correct. Looking at the strategies displayed in Table 3, 70.8% of students used the build-up method with the direct proportion problem. There is a dramatic decrease in the use of the build-up method with the inverse problem of 25.4% in Table 4. This reveals that the build-up method is more difficult for students to use effectively in inverse proportion problems because it requires the student to build up one quantity and decrease the other. Using additive building will not result in the correct answer, but it has to be a multiplicative building. This seems to have been more difficult for students and it goes along with Dr. Steinthorsdottir and Dr. Riehl’s research. They found that more students used additive build-up strategies (31%) than multiplicative build-up (19%). It makes sense that fewer students used the multiplicative build-up strategy with the inverse proportion problem because the additive one gave the wrong answer. Table 3 Valid Direct Proportion Strategies for Problem 3 for 7th and 8th Graders Build -Up Method Within-Measure Space Relationship Between-Measure Space Relationship Cross- Multiplication Random but correct Total Number of correct 92 8 27 2 1 130 21 answers Percentage of correct strategies 70.8% 6.2% 20.8% 1.5% 0.8% 100% Table 4 Valid Inverse Proportion Strategies in Problem 8 for 7th and 8th Graders Build-Up Method Within-Measure Space Relationship Cross Multiplication Hybrid Relationship Rate per one relationship Random but correct Total Number of correct answers 12 10 7 3 17 49 Percentage of correct strategies 24.5% 20.4% 14.3% 6.1% 34.7% 100% The direct proportion problem had 6.2% of students solve using the within-measure space relationship and 20.8% of students solve using the between-measure space relationship (see Table 3). Students chose to find the whole number unit rate of 3 by dividing 9 plants by 3 scoops of feed. The integer was more obvious than finding the within-measure space relationship of 11/3. However, the opposite was true for the inverse proportion problem. 20.4% of students used the within-measure space strategy and 14.3% used the between-measure space strategy (see Table 4). As was found earlier, although it was more obvious to see the 12 members/6 hours relationship, the inverse proportion doesn’t work directly between the quantities. Instead, it is more of a cross-multiplication relationship. Therefore, more students chose to look at the within-measure space relationship even though it was not an integer scale factor. 22 Different strategies were more common with direct vs inverse proportion problems which was an interesting finding. The build-up strategy was the most common for the direct problem by a lot while the build-up strategy and within-measure space were used in similar frequency with inverse problems. There was also 2.5 times the amount of correct answers for the direct proportion problem than the inverse proportion problem. There is less of a conceptual understanding of inverse proportion relationships than direct proportion relationships. VII. Application After analyzing the data on inverse proportion relationships and comparing it to the direct proportion problems, it became evident that there is less conceptual understanding of inverse relationships than direct proportion relationships. The implication of these findings is a need for teachers to include more inverse proportion problems in their teaching. I believe that if it was introduced more to students with an intentionality of teaching the difference between the two, students would more quickly recognize the differences when they encounter different types of proportion problems. One idea I can envision is to have students work with story problems that do not yet include numbers. For example, I would give students the same problem as problem 8 but take out the 18 members and instead describe that more workers were coming in. Teachers could ask students questions that would cause them to think deeply about the mathematics going on before giving them a chance to solve. Such questions could include, “What is going on in this story problem?” “What do we know?” “What do we want to know?” 23 “How might more members change the amount of hours needed to complete the work? Why?” “If we were told new members came in and it took 10 hours to finish the work, do you think that there were more members or fewer members? Why?” Allowing time for students to think critically about the problem before solving it would help them to recognize that it is an inverse relationship, causing them to pause on the potential desire to jump into solving every problem like a direct proportion. Another idea for the betterment of both direct and inverse problem-solving is working with multiplicative relationships. Helping students to see the reasoning behind scale factor and unit rate will bring a concrete understanding of both proportional relationships. Asking students questions like, “What would it look like to have just 1 of a quantity? How would that affect the other quantity?” This would guide students into multiplicative thinking. For the teachers of the 409 midwestern students that were tested, they can use that data to shape their future instruction. Finding where students currently are with proportional reasoning can give an educator the information needed to push students to the next level of mathematical reasoning. I believe that with the information gathered from this data, many teachers can broaden their teaching to include inverse proportional reasoning using effective proportion-solving strategies. VIII. Conclusion After all of the research and analysis of student thinking with direct and inverse proportions, I have found the most common strategies used to solve inverse proportion problems. The “build-up” method and “within-measure space” relationship paralleled the direct proportion strategies that were previously found in other research. The “rate per one” method and the “cross 24 multiplication and between-measure space hybrid” were unique in the data collected from an inverse proportion problem. The common errors found in inverse proportion problems were additive reasoning errors and solving the problem using a direct proportional relationship. Finally, I compared my inverse proportion strategies with a direct proportion problem to see the similarities and differences between the two. The “build-up” method was the most commonly used among the students for both problems. In addition, inverse proportions are more difficult for students than direct proportions. With about a 30% difference in correct answers between the direct proportion problem and the inverse proportion problem, the data proves that inverse proportion problems tend to be more difficult for students to solve. These results show the importance for educators to be including more inverse proportion problems in their teaching so that students can become more familiar with them. Teachers should also provide experiences with inverse proportion problems that develop their mathematical thinking, focusing on the conceptual understanding of proportions. This will help students in the future as they progress in their educational goals and careers. Deeply understanding mathematical concepts over procedural work is what creates bright minds and effective problem-solvers in the future. Throughout my research, I was better able to understand students’ proportional reasoning strategies for inverse and direct proportions. 25 VII. References Bright, G. W., Joyner, J. M., & Wallis, C. (2003). Assessing Proportional Thinking, Mathematics Teaching in the Middle School , 9(3), 166-172. Chapin, S. H., & Anderson, N. C. (2003). Crossing the Bridge to Formal Proportional Reasoning, Mathematics Teaching in the Middle School , 8(8), 420-425. Retrieved Mar 15, 2023, from Christou, C., & Philippou, G. (2002). Mapping and development of intuitive proportional thinking. Journal of Mathematical Behavior , 20 , 321-336. De Bock, D., Van Dooren, W. & Verschaffel, L. (2015). Students’ Understanding of Proportional, Inverse Proportional, and Affine Functions: Two Studies of the Role of External Representations International Journal of Science and Mathematics Education 13 (Suppl 1), 47–69 Ercole, L. K., Frantz, M., & Ashline, G. (2011). Multiple Ways to Solve Proportions, Mathematics Teaching in the Middle School , 16 (8), 482-490. Fernández, Ceneida & Ciscar, Salvador & Dooren, Wim & Bock, Dirk & Verschaffel, Lieven. (2010). Effect of number structure and nature of quantities on secondary school students proportional reasoning. Studia Psychologica. 53. Fernandez, C., Llinares, S., Modestou, M., Gagatsis, A. (2010). Proportional Reasoning: How Task Variables Influence the Development of Students’ Strategies from Primary to Secondary School. Acta Didactica Universitatis Comenianae Mathematics , 10 , 1-18. 26 Ferrandez-Reinisch, A.-M. (1985). The Acquisition of Inverse Proportionality: A Training Experiment. Journal for Research in Mathematics Education , 16 (2), 132–140. Langrall, C.W., & Swafford, J. (2000). Three Balloons for Two Dollars: Developing Proportional Reasoning. Mathematics Teaching in the Middle School, 6 . Lobato J. Ellis A. B. Charles R. I. & Zbiek R. M. (2010). Developing Essential Understanding of Ratios Proportions and Proportional Reasoning for Teaching Mathematics in Grades 6- National Council of Teachers of Mathematics. Markworth, K. A. (2012). Proportioning Cats and Rats, Mathematics Teaching in the Middle School , 17 (9), 538-543. Miller, J. L., & Fey, J. T. (2000). Take Time for Action: Proportional Reasoning, Mathematics Teaching in the Middle School , 5(5), 310-313. Pforts, A. (2010, November). Grade 6: Mathematics . Grade 6 | Mathematics | Iowa Department of Education. Retrieved March 15, 2023, from Pforts, A. (2010, November). Grade 7: Mathematics . Grade 7 | Mathematics | Iowa Department of Education. Pforts, A. (2010, November). Grade 8: Mathematics . Grade 8 | Mathematics | Iowa Department of Education. 27 Riehl, S. M. & Steinthorsdottir, O. B. (2019). Missing-value proportion problems: The effects of number structure characteristics. Investigations in Mathematics Learning, 11(1). 56-68. DOI: 10.1080/19477503.2017.1375361 Riehl, S. M., & Steinthorsdottir, O. B. (2014). Revisiting Mr. Tall and Mr. Short, Mathematics Teaching in the Middle School , 20 (4), 220-228. Seeley, C., & Schielack, J. F. (2007). A Look at the Development of Ratios, Rates, and Proportionality, Mathematics Teaching in the Middle School , 13 (3), 140-142. Silvestre, Ana & Ponte, João. (2011). Missing value and comparison problems: What pupils know before the teaching of proportion.
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https://www.ncbi.nlm.nih.gov/books/NBK52770/
Part V: Phototransduction and Photoreceptor Synaptic Pathways - Webvision - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. Kolb H, Fernandez E, Jones B, et al., editors. Webvision: The Organization of the Retina and Visual System [Internet]. Salt Lake City (UT): University of Utah Health Sciences Center; 1995-. Webvision: The Organization of the Retina and Visual System [Internet]. Show details Kolb H, Fernandez E, Jones B, et al., editors. Salt Lake City (UT): University of Utah Health Sciences Center; 1995-. Contents Search term < PrevNext > Part V: Phototransduction and Photoreceptor Synaptic Pathways Phototransduction in Rods and Cones Glutamate and Glutamate Receptors in the Vertebrate Retina Bipolar Cell Pathways in the Vertebrate Retina GABA C Receptors in the Vertebrate Retina S-Potentials and Horizontal Cells Copyright: © 2025 Webvision . All copyright for chapters belongs to the individual authors who created them. However, for non-commercial, academic purposes, images and content from the chapters portion of Webvision may be used with a non-exclusive rights under a Attribution, Noncommercial 4.0 International (CC BY-NC) Creative Commons license. Cite Webvision, as the source. Commercial applications need to obtain license permission from the administrator of Webvision and are generally declined unless the copyright owner can/wants to donate or license material. Use online should be accompanied by a link back to the original source of the material. All imagery or content associated with blog posts belong to the authors of said posts, except where otherwise noted. Bookshelf ID: NBK52770 Contents < PrevNext > Share on Facebook Share on Twitter Views PubReader Print View Cite this Page PDF version of this title (236M) Related Items in Bookshelf All Reference Works All Textbooks Recent Activity Clear)Turn Off)Turn On) Part V: Phototransduction and Photoreceptor Synaptic Pathways - WebvisionPart V: Phototransduction and Photoreceptor Synaptic Pathways - Webvision Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close Kolb H, Fernandez E, Jones B, et al., editors. Webvision: The Organization of the Retina and Visual System [Internet]. Salt Lake City (UT): University of Utah Health Sciences Center; 1995-. Part V: Phototransduction and Photoreceptor Synaptic Pathways. Available from: Making content easier to read in Bookshelf Close We are experimenting with display styles that make it easier to read books and documents in Bookshelf. Our first effort uses ebook readers, which have several "ease of reading" features already built in. The content is best viewed in the iBooks reader. You may notice problems with the display of some features of books or documents in other eReaders. Cancel Download Share Share on Facebook Share on Twitter URL
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https://dsp-book.narod.ru/DEM1/toc.htm
Table of Contents Electronic and Electrical Engineering Detection, Estimation, and Modulation Theory, Part I Author(s):Harry L. Van Trees ISBN:0471221082(Electronic) 0471095176(Print) Copyright © 2001 John Wiley & Sons, Inc. i-xviFrontmatter and Index Harry L. Van Trees PDF Full Text (Size: 2242K) Published Online:22 Jan 2002 Chapter 1: 1-18Introduction Harry L. Van Trees PDF Full Text (Size: 1732K) Published Online:22 Jan 2002 Chapter 2: 19-165Classical Detection and Estimation Theory(1) Harry L. Van Trees PDF Full Text (Size: 11745K) Published Online:22 Jan 2002 Chapter 2: 19-165Classical Detection and Estimation Theory(2) Harry L. Van Trees PDF Full Text (Size: 11745K) Published Online:22 Jan 2002 Chapter 2: 19-165Classical Detection and Estimation Theory(3) Harry L. Van Trees PDF Full Text (Size: 11745K) Published Online:22 Jan 2002 Chapter 3: 166-238Representations of Random Processes(1) Harry L. Van Trees PDF Full Text (Size: 5908K) Published Online:22 Jan 2002 Chapter 3: 166-238Representations of Random Processes(2) Harry L. Van Trees PDF Full Text (Size: 5908K) Published Online:22 Jan 2002 Chapter 4: 239-422Detection of Signals - Estimation of Signal Parameters(1) Harry L. Van Trees PDF Full Text (Size: 15517K) Published Online:22 Jan 2002 Chapter 4: 239-422Detection of Signals - Estimation of Signal Parameters(2) Harry L. Van Trees PDF Full Text (Size: 15517K) Published Online:22 Jan 2002 Chapter 4: 239-422Detection of Signals - Estimation of Signal Parameters(3) Harry L. Van Trees PDF Full Text (Size: 15517K) Published Online:22 Jan 2002 Chapter 4: 239-422Detection of Signals - Estimation of Signal Parameters(4) Harry L. Van Trees PDF Full Text (Size: 15517K) Published Online:22 Jan 2002 Chapter 4: 239-422Detection of Signals - Estimation of Signal Parameters(5) Harry L. Van Trees PDF Full Text (Size: 15517K) Published Online:22 Jan 2002 Chapter 4: 239-422Detection of Signals - Estimation of Signal Parameters(6) Harry L. Van Trees PDF Full Text (Size: 15517K) Published Online:22 Jan 2002 Chapter 5: 423-466Estimation of Continuous Waveforms Harry L. Van Trees PDF Full Text (Size: 3711K) Published Online:22 Jan 2002 Chapter 6: 467-622Linear Estimation(1) Harry L. Van Trees PDF Full Text (Size: 12802K) Published Online:22 Jan 2002 Chapter 6: 467-622Linear Estimation(2) Harry L. Van Trees PDF Full Text (Size: 12802K) Published Online:22 Jan 2002 Chapter 6: 467-622Linear Estimation(3) Harry L. Van Trees PDF Full Text (Size: 12802K) Published Online:22 Jan 2002 Chapter 6: 467-622Linear Estimation(4) Harry L. Van Trees PDF Full Text (Size: 12802K) Published Online:22 Jan 2002 Chapter 7: 623-633Discussion Harry L. Van Trees PDF Full Text (Size: 1423K) Published Online:22 Jan 2002 635-670Appendix: A Typical Course Outline Harry L. Van Trees PDF Full Text (Size: 2808K) Published Online:22 Jan 2002 671-682Glossary Harry L. Van Trees PDF Full Text (Size: 913K) Published Online:22 Jan 2002 683-686Author Index Harry L. Van Trees PDF Full Text (Size: 400K) Published Online:22 Jan 2002
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https://www.mathspanda.com/ASFM/Lessons/Invariant_points_and_invariant_lines_LESSON.pdf
www.mathspanda.com Invariant points and invariant lines Starter 1. A square undergoes a shear, axis invariant, mapping . The point is a vertex of the square before the shear. Find the new coordinates of the vertex. 2. Find the matrix which transforms and . 3. Find the values of and such that: (a) (b) Notes Invariant points An invariant point is a point that is unaffected by a transformation. Therefore, for matrix transformations this means: . Is it possible for a matrix transformation to have no invariant points? No, the origin is always an invariant point as E.g. 1 By letting , form and solve the simultaneous equations of to show that is the equation of invariant points. E.g. 2 Using the equations formed from , state what happens when: (a) (b) Working: (a) When : Substituting in : So the origin is the unique invariant point. When , , and , we have a line of invariant points passing through the origin. Either we have a line of invariant points passing through the origin or the origin is the unique invariant point. When solving equations, either: Simultaneous equations are identical invariant points lie on line passing through origin …or… Simultaneous equations are different the origin is the unique invariant point x− (0, 1) →(−4, 1) (6, 2) ( 4 3) →( 9 10) ( 2 1) →( 5 6) x y ( 2 3 4 −1) ( x y) = ( x y) ( −1 2 1 0) ( x y) = ( x y) M ( x y) = ( x y) M ( 0 0) = ( 0 0) M = ( a b c d) M ( x y) = ( x y) y = ( a −c −1 d −b −1)x ( a b c d) ( x y) = ( x y) a = 1 d = 1 ax + by = x a = 1 x + by = x ⇒ y = 0 y = 0 cx + dy = y x = 0 a ≠1 d ≠1 a −c ≠1 d −b ≠1 ⇒ ⇒ Page of 1 4 www.mathspanda.com E.g. 3 Find the invariant points under the transformation . Working: The equations are equal so the invariant points under the transformation are all the points lying on the line i.e. all points of the form . E.g. 4 Find the invariant points under the transformation . Invariant lines A line is an invariant line under a transformation if the image of a point on the line is also on the line. It does not mean that every point on the line must map onto itself. N.B. Any line of invariant points is also an invariant line. There are two options to consider: invariant lines through the origin and invariant lines not through the origin. Invariant lines through the origin An invariant line through the origin has the form . Any point on this line is of the form So if is an invariant line: By expanding the matrix, equations involving and can be formed. By rearranging, can be eliminated to form an equation where . Since and must be the same line we can solve the equation to find the values of that gives invariant points. E.g. 5 Find the equation of any invariant lines through the origin of the transformation whose matrix is . Working: Any point on the invariant line has coordinates of the form Substituting : Since the gradient must be the same as : & So and are the invariant lines passing through the origin. ( 4 1 6 3) ( 4 1 6 3) ( x y) = ( x y) 4x + y = x ⇒ y = −3x 6x + 3y = y ⇒ y = −3x ( 4 1 6 3) y = −3x (k, −3k) ( 3 4 1 2) y = m x (k, mk) y = m x M ( k mk) = ( x y) x y k y = f (m)x y = m x y = f (m)x m = f (m) m ( 2 3 0 −1) (k, mk) ( 2 3 0 −1) ( k mk) = ( x y) x = 2k + 3mk ⇒ x = k(2 + 3m) ⇒ k = x 2 + 3m y = −mk k = x 2 + 3m y = ( −m 2 + 3m)x y = m x m = −m 2 + 3m 3m2 + 3m = 0 ⇒ m(m + 1) = 0 ⇒ m = 0 m = −1 y = 0 y = −x Page of 2 4 www.mathspanda.com E.g. 6 Find the invariant lines of the matrix which pass through the origin. Invariant lines not passing through the origin An invariant line not through the origin has the form , where . Any point on this line is of the form So if is an invariant line: Again expand the matrix and form equations involving and . By rearranging, can be eliminated to form an equation where . The lines and must be the same. By equating coefficients of and we get: and Solve the equation to find values of . Check whether these values satisfy : If they do, is an invariant line If not, is only satisfied when so is an invariant line. E.g. 7 7 Find the invariant lines of the matrix . Working: Any point on the invariant line is of the form . Substituting : This image point must lie on the line , so equating coefficients of and : : or Now substitute into the equation formed by equating coefficients of to see if it satisfies the equation. : : So is an invariant line. : is only true when . So is the other invariant line. ( 3 1 2 4) y = m x + c c ≠0 (k, mk + c) y = m x + c M ( k mk + c) = ( x y) x y k y = f (m)x + g(m)c y = m x + c y = f (m)x + g(m)c x y m = f (m) 1 = g(m) m = f (m) m m− 1 = g(m) y = m x + c 1 = g(m) c = 0 y = m x ( 2 1 2 3) (k, mk + c) ( 2 1 2 3) ( k mk + c) = ( x y) x = 2k + mk + c ⇒ x = k(2 + m) + c ⇒ k = 1 2 + m x − 1 2 + m c y = 2k + 3mk + 3c ⇒ y = k(2 + 3m) + 3c k = 1 2 + m x − 1 2 + m c y = ( 1 2 + m x − 1 2 + m c)(2 + 3m) + 3c y = ( 2 + 3m 2 + m )x + (3 −2 + 3m 2 + m )c y = m x + c x c x m = 2 + 3m 2 + m ⇒ m2 −m −2 = 0 ⇒ (m −2)(m + 1) = 0 ⇒ m = 2 m = −1 c c 3 −2 + 3m 2 + m = 1 m = 2 3 −2 + 3 × 2 2 + 2 = 3 −8 4 = 1 ✓ y = 2x + c m = −1 3 −2 + 3 × (−1) 2 + (−1) = 3 −−1 1 ≠1 ∴(3 −2 + 3m 2 + m )c = c c = 0 y = −x Page of 3 4 www.mathspanda.com E.g. 8 Find any invariant lines of the matrix Types of transformations and invariant lines Reflections Rotations Enlargements Any line that is perpendicular Only rotations of about Any line passing through to the mirror line will be an the origin have an invariant the origin is an invariant invariant line. line. In such cases line i.e. is invariant Video A: Invariant points and lines Video B: Invariant points and lines Solutions to Starter and E.g.s Exercise p87 3D Qu 1ac, 2ace, 3-5 Summary An invariant point is a point that is unaffected by a transformation i.e. Invariant lines through the origin: Invariant lines not passing through the origin: ( 4 3 −3 −2) 180∘ y = k x y = k x M ( x y) = ( x y) M ( k mk) = ( x y) M ( k mk + c) = ( x y) Page of 4 4