context
stringlengths 545
71.9k
| questionsrc
stringlengths 16
10.2k
| question
stringlengths 11
563
|
|---|---|---|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule .
|
at what temperature can a rock be turned into gas ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens .
|
what is the 4th state of matter ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil .
|
what happens in the case of plasma ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ?
|
0 , does heat affect the energy ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q .
|
is bose-einstein condensate a part of the 5 states of matter ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma .
|
should n't you have drawn 2 circles ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens .
|
may i know the comparision between solid , liquid and gas based on the states of matter ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
and that 's right here . this is the heat of vaporization . and the same idea is happening .
|
if the molecules have a tendency to `` want '' to be around each other , would there be a horizontal line at the heat of fusion and vaporization if you went from high heat to low heat ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat .
|
why ca n't latent heat be measured in the thermometer ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy .
|
how can you tell the energy in particles of a substance ?
|
i think we 're all reasonably familiar with the three states of matter in our everyday world . at very high temperatures you get a fourth . but the three ones that we normally deal with are , things could be a solid , a liquid , or it could be a gas . and we have this general notion , and i think water is the example that always comes to at least my mind . is that solid happens when things are colder , relatively colder . and then as you warm up , you go into a liquid state . and as your warm up even more you go into a gaseous state . so you go from colder to hotter . and in the case of water , when you 're a solid , you 're ice . when you 're a liquid , some people would call ice water , but let 's call it liquid water . i think we know what that is . and then when it 's in the gas state , you 're essentially vapor or steam . so let 's think a little bit about what , at least in the case of water , and the analogy will extend to other types of molecules . but what is it about water that makes it solid , and when it 's colder , what allows it to be liquid . and i 'll be frank , liquids are kind of fascinating because you can never nail them down , i guess is the best way to view them . or a gas . so let 's just draw a water molecule . so you have oxygen there . you have some bonds to hydrogen . and then you have two extra pairs of valence electrons in the oxygen . and a couple of videos ago , we said oxygen is a lot more electronegative than the hydrogen . it likes to hog the electrons . so even though this shows that they 're sharing electrons here and here . at both sides of those lines , you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line . but we know because of the electronegativity , or the relative electronegativity of oxygen , that it 's hogging these electrons . and so the electrons spend a lot more time around the oxygen than they do around the hydrogen . and what that results is that on the oxygen side of the molecule , you end up with a partial negative charge . and we talked about that a little bit . and on the hydrogen side of the molecules , you end up with a slightly positive charge . now , if these molecules have very little kinetic energy , they 're not moving around a whole lot , then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules . let me draw some more molecules . when we talk about the whole state of the whole matter , we actually think about how the molecules are interacting with each other . not just how the atoms are interacting with each other within a molecule . i just drew one oxygen , let me copy and paste that . but i could do multiple oxygens . and let 's say that that hydrogen is going to want to be near this oxygen . because this has partial negative charge , this has a partial positive charge . and then i could do another one right there . and then maybe we 'll have , and just to make the point clear , you have two hydrogens here , maybe an oxygen wants to hang out there . so maybe you have an oxygen that wants to be here because it 's got its partial negative here . and it 's connected to two hydrogens right there that have their partial positives . but you can kind of see a lattice structure . let me draw these bonds , these polar bonds that start forming between the particles . these bonds , they 're called polar bonds because the molecules themselves are polar . and you can see it forms this lattice structure . and if each of these molecules do n't have a lot of kinetic energy . or we could say the average kinetic energy of this matter is fairly low . and what do we know is average kinetic energy ? well , that 's temperature . then this lattice structure will be solid . these molecules will not move relative to each other . i could draw a gazillion more , but i think you get the point that we 're forming this kind of fixed structure . and while we 're in the solid state , as we add kinetic energy , as we add heat , what it does to molecules is , it just makes them vibrate around a little bit . if i was a cartoonist , they way you 'd draw a vibration is to put quotation marks there . that 's not very scientific . but they would vibrate around , they would buzz around a little bit . i 'm drawing arrows to show that they are vibrating . it does n't have to be just left-right it could be up-down . but as you add more and more heat in a solid , these molecules are going to keep their structure . so they 're not going to move around relative to each other . but they will convert that heat , and heat is just a form of energy , into kinetic energy which is expressed as the vibration of these molecules . now , if you make these molecules start to vibrate enough , and if you put enough kinetic energy into these molecules , what do you think is going to happen ? well this guy is vibrating pretty hard , and he 's vibrating harder and harder as you add more and more heat . this guy is doing the same thing . at some point , these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations . and once that happens , the molecules -- let me draw a couple more . once that happens , the molecules are going to start moving past each other . so now all of a sudden , the molecule will start shifting . but they 're still attracted . maybe this side is moving here , that 's moving there . you have other molecules moving around that way . but they 're still attracted to each other . even though we 've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules . our vibration , or our kinetic energy for each molecule , still is n't strong enough to completely separate them . they 're starting to slide past each other . and this is essentially what happens when you 're in a liquid state . you have a lot of atoms that want be touching each other but they 're sliding . they have enough kinetic energy to slide past each other and break that solid lattice structure here . and then if you add even more kinetic energy , even more heat , at this point it 's a solution now . they 're not even going to be able to stay together . they 're not going to be able to stay near each other . if you add enough kinetic energy they 're going to start looking like this . they 're going to completely separate and then kind of bounce around independently . especially independently if they 're an ideal gas . but in general , in gases , they 're no longer touching each other . they might bump into each other . but they have so much kinetic energy on their own that they 're all doing their own thing and they 're not touching . i think that makes intuitive sense if you just think about what a gas is . for example , it 's hard to see a gas . why is it hard to see a gas ? because the molecules are much further apart . so they 're not acting on the light in the way that a liquid or a solid would . and if we keep making that extended further , a solid -- well , i probably should n't use the example with ice . because ice or water is one of the few situations where the solid is less dense than the liquid . that 's why ice floats . and that 's why icebergs do n't just all fall to the bottom of the ocean . and ponds do n't completely freeze solid . but you can imagine that , because a liquid is in most cases other than water , less dense . that 's another reason why you can see through it a little bit better . or it 's not diffracting -- well i wo n't go into that too much , than maybe even a solid . but the gas is the most obvious . and it is true with water . the liquid form is definitely more dense than the gas form . in the gas form , the molecules are going to jump around , not touch each other . and because of that , more light can get through the substance . now the question is , how do we measure the amount of heat that it takes to do this to water ? and to explain that , i 'll actually draw a phase change diagram . which is a fancy way of describing something fairly straightforward . let me say that this is the amount of heat i 'm adding . and this is the temperature . we 'll talk about the states of matter in a second . so heat is often denoted by q . sometimes people will talk about change in heat . they 'll use h , lowercase and uppercase h. they 'll put a delta in front of the h. delta just means change in . and sometimes you 'll hear the word enthalpy . let me write that . because i used to say what is enthalpy ? it sounds like empathy , but it 's quite a different concept . at least , as far as my neural connections could make it . but enthalpy is closely related to heat . it 's heat content . for our purposes , when you hear someone say change in enthalpy , you should really just be thinking of change in heat . i think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary . the best way to think about it is heat content . change in enthalpy is really just change in heat . and just remember , all of these things , whether we 're talking about heat , kinetic energy , potential energy , enthalpy . you 'll hear them in different contexts , and you 're like , i thought i should be using heat and they 're talking about enthalpy . these are all forms of energy . and these are all measured in joules . and they might be measured in other ways , but the traditional way is in joules . and energy is the ability to do work . and what 's the unit for work ? well , it 's joules . force times distance . but anyway , that 's a side-note . but it 's good to know this word enthalpy . especially in a chemistry context , because it 's used all the time and it can be very confusing and non-intuitive . because you 're like , i do n't know what enthalpy is in my everyday life . just think of it as heat contact , because that 's really what it is . but anyway , on this axis , i have heat . so this is when i have very little heat and i 'm increasing my heat . and this is temperature . now let 's say at low temperatures i 'm here and as i add heat my temperature will go up . temperature is average kinetic energy . let 's say i 'm in the solid state here . and i 'll do the solid state in purple . no i already was using purple . i 'll use magenta . so as i add heat , my temperature will go up . heat is a form of energy . and when i add it to these molecules , as i did in this example , what did it do ? it made them vibrate more . or it made them have higher kinetic energy , or higher average kinetic engery , and that 's what temperature is a measure of ; average kinetic energy . so as i add heat in the solid phase , my average kinetic energy will go up . and let me write this down . this is in the solid phase , or the solid state of matter . now something very interesting happens . let 's say this is water . so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line . what happens to a solid ? well , it turns into a liquid . ice melts . not all solids , we 're talking in particular about water , about h2o . so this is ice in our example . all solids are n't ice . although , you could think of a rock as solid magma . because that 's what it is . i could take that analogy a bunch of different ways . but the interesting thing that happens at zero degrees . depending on what direction you 're going , either the freezing point of water or the melting point of ice , something interesting happens . as i add more heat , the temperature does not to go up . as i add more heat , the temperature does not go up for a little period . let me draw that . for a little period , the temperature stays constant . and then while the temperature is constant , it stays a solid . we 're still a solid . and then , we finally turn into a liquid . let 's say right there . so we added a certain amount of heat and it just stayed a solid . but it got us to the point that the ice turned into a liquid . it was kind of melting the entire time . that 's the best way to think about it . and then , once we keep adding more and more heat , then the liquid warms up too . now , we get to , what temperature becomes interesting again for water ? well , obviously 100 degrees celsius or 373 degrees kelvin . i 'll do it in celsius because that 's what we 're familiar with . what happens ? that 's the temperature at which water will vaporize or which water will boil . but something happens . and they 're really getting kinetically active . but just like when you went from solid to liquid , there 's a certain amount of energy that you have to contribute to the system . and actually , it 's a good amount at this point . where the water is turning into vapor , but it 's not getting any hotter . so we have to keep adding heat , but notice that the temperature did n't go up . we 'll talk about it in a second what was happening then . and then finally , after that point , we 're completely vaporized , or we 're completely steam . then we can start getting hot , the steam can then get hotter as we add more and more heat to the system . so the interesting question , i think it 's intuitive , that as you add heat here , our temperature is going to go up . but the interesting thing is , what was going on here ? we were adding heat . so over here we were turning our heat into kinetic energy . temperature is average kinetic energy . but over here , what was our heat doing ? well , our heat was was not adding kinetic energy to the system . the temperature was not increasing . but the ice was going from ice to water . so what was happening at that state , is that the kinetic energy , the heat , was being used to essentially break these bonds . and essentially bring the molecules into a higher energy state . so you 're saying , sal , what does that mean , higher energy state ? well , if there was n't all of this heat and all this kinetic energy , these molecules want to be very close to each other . for example , i want to be close to the surface of the earth . when you put me in a plane you have put me in a higher energy state . i have a lot more potential energy . i have the potential to fall towards the earth . likewise , when you move these molecules apart , and you go from a solid to a liquid , they want to fall towards each other . but because they have so much kinetic energy , they never quite are able to do it . but their energy goes up . their potential energy is higher because they want to fall towards each other . by falling towards each other , in theory , they could do some work . so what 's happening here is , when we 're contributing heat -- and this amount of heat we 're contributing , it 's called the heat of fusion . because it 's the same amount of heat regardless how much direction we go in . when we go from solid to liquid , you view it as the heat of melting . it 's the head that you need to put in to melt the ice into liquid . when you 're going in this direction , it 's the heat you have to take out of the zero degree water to turn it into ice . so you 're taking that potential energy and you 're bringing the molecules closer and closer to each other . so the way to think about it is , right here this heat is being converted to kinetic energy . then , when we 're at this phase change from solid to liquid , that heat is being used to add potential energy into the system . to pull the molecules apart , to give them more potential energy . if you pull me apart from the earth , you 're giving me potential energy . because gravity wants to pull me back to the earth . and i could do work when i 'm falling back to the earth . a waterfall does work . it can move a turbine . you could have a bunch of falling sals move a turbine as well . and then , once you are fully a liquid , then you just become a warmer and warmer liquid . now the heat is , once again , being used for kinetic energy . you 're making the water molecules move past each other faster , and faster , and faster . to some point where they want to completely disassociate from each other . they want to not even slide past each other , just completely jump away from each other . and that 's right here . this is the heat of vaporization . and the same idea is happening . before we were sliding next to each other , now we 're pulling apart altogether . so they could definitely fall closer together . and then once we 've added this much heat , now we 're just heating up the steam . we 're just heating up the gaseous water . and it 's just getting hotter and hotter and hotter . but the interesting thing there , and i mean at least the interesting thing to me when i first learned this , whenever i think of zero degrees water i 'll say , oh it must be ice . but that 's not necessarily the case . if you start with water and you make it colder and colder and colder to zero degrees , you 're essentially taking heat out of the water . you can have zero degree water and it has n't turned into ice yet . and likewise , you could have 100 degree water that has n't turned into steam yeat . you have to add more energy . you can also have 100 degree steam . you can also have zero degree water . anyway , hopefully that gives you a little bit of intuition of what the different states of matter are . and in the next problem , we 'll talk about how much heat exactly it does take to move along this line . and maybe we can solve some problems on how much ice we might need to make our drink cool .
|
so what happens at zero degrees ? which is also 273.15 kelvin . let 's say that 's that line .
|
how much is a kelvin ?
|
- [ lecturer ] in previous videos we talk about the emergence of the maurya empire around 322 bce , shortly after the invasion of alexander the great , as the first truly great indian empire that unifies most of the indian subcontinent . now that empire eventually falls and the next significant empire to emerge , especially if we talk about influence on india and the world , is the gupta empire , which emerges over 500 years later . now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty . and we need to be careful , do n't confuse this chandragupta i with chandragupta maurya who founded the maurya dynasty over 600 years before the time we 're talking about . chandragupta i , it 's interesting because he is really able to gain power not , initially , through conquest , but through a marriage , he has a marriage with the princess kumaradevi and as a dowry he is given control over much of this region of north east india , this region of magadha or magadha , i am always having trouble pronouncing that , so my apologies , including the famous city of pataliputra , which even in the time of the maurya empire and before the maurya empire , this was a famous seat of power . but once he 's in control of this region then he and his successors are able to have increased conquests over india , you see in this light blue color what his son , samudragupta , was able to do and then one of samudragupta 's sons , chandragupta ii is able to conquer even more . but what makes the gupta empire distinctive is n't just that they were able to unify much , or conquer , much of india again , what really makes them distinctive is because of that unification and especially the wealth that began to flow into the capital they were able to be sponsors of significant culture and science and the arts , and that 's why historians view the gupta empire as the golden age of india . and just to get an appreciation for this , the gupta empire was , during the time of kalidasa , and he is considered to be the greatest writer ever in the sanskrit language , he is like the william shakespeare of sanskrit . and just to get a feel for some of his writing , obviously this is n't sanskrit , this is english , and this is from the recognition of sakuntala , which is one of his most famous works , `` wouldst thou the young year 's blossoms `` and the fruits of its decline `` and all by which the soul is charmed , `` enraptured , feasted , fed , `` wouldst thou the earth and heaven itself `` in one sole name combine ? `` i name thee , o sakuntala ! and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function . and so he established some of the early ideas of trigonometry . he did work in summation , he had a significant astronomical work , recognizing the rotation of the earth versus the rotation of the heavens , he had an early concept of gravity . even some of these notions of the place value system and zero and decimal notation , many of our modern notions of it are traced back to aryabhata . in other videos on the islamic golden age when we talk about folks like al-khwarizmi , a lot of his work was based on what he learned from aryabhata . beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire . this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance . now , like all empires , the gupta empire does eventually fall in the mid sixth century , it 's believed , around 540 , 550 . and one of the main causes , there 's invasions from people called the hunas , who historians believe are either the huns or a group that are closely related to the huns , it 's considered to be one of the causes of the eventual decline of the gupta dynasty .
|
and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function .
|
even though aryabhatta discovered trigonometry , gravity and pi , why is the credit still given to european scientists and mathematicians ?
|
- [ lecturer ] in previous videos we talk about the emergence of the maurya empire around 322 bce , shortly after the invasion of alexander the great , as the first truly great indian empire that unifies most of the indian subcontinent . now that empire eventually falls and the next significant empire to emerge , especially if we talk about influence on india and the world , is the gupta empire , which emerges over 500 years later . now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty . and we need to be careful , do n't confuse this chandragupta i with chandragupta maurya who founded the maurya dynasty over 600 years before the time we 're talking about . chandragupta i , it 's interesting because he is really able to gain power not , initially , through conquest , but through a marriage , he has a marriage with the princess kumaradevi and as a dowry he is given control over much of this region of north east india , this region of magadha or magadha , i am always having trouble pronouncing that , so my apologies , including the famous city of pataliputra , which even in the time of the maurya empire and before the maurya empire , this was a famous seat of power . but once he 's in control of this region then he and his successors are able to have increased conquests over india , you see in this light blue color what his son , samudragupta , was able to do and then one of samudragupta 's sons , chandragupta ii is able to conquer even more . but what makes the gupta empire distinctive is n't just that they were able to unify much , or conquer , much of india again , what really makes them distinctive is because of that unification and especially the wealth that began to flow into the capital they were able to be sponsors of significant culture and science and the arts , and that 's why historians view the gupta empire as the golden age of india . and just to get an appreciation for this , the gupta empire was , during the time of kalidasa , and he is considered to be the greatest writer ever in the sanskrit language , he is like the william shakespeare of sanskrit . and just to get a feel for some of his writing , obviously this is n't sanskrit , this is english , and this is from the recognition of sakuntala , which is one of his most famous works , `` wouldst thou the young year 's blossoms `` and the fruits of its decline `` and all by which the soul is charmed , `` enraptured , feasted , fed , `` wouldst thou the earth and heaven itself `` in one sole name combine ? `` i name thee , o sakuntala ! and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function . and so he established some of the early ideas of trigonometry . he did work in summation , he had a significant astronomical work , recognizing the rotation of the earth versus the rotation of the heavens , he had an early concept of gravity . even some of these notions of the place value system and zero and decimal notation , many of our modern notions of it are traced back to aryabhata . in other videos on the islamic golden age when we talk about folks like al-khwarizmi , a lot of his work was based on what he learned from aryabhata . beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire . this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance . now , like all empires , the gupta empire does eventually fall in the mid sixth century , it 's believed , around 540 , 550 . and one of the main causes , there 's invasions from people called the hunas , who historians believe are either the huns or a group that are closely related to the huns , it 's considered to be one of the causes of the eventual decline of the gupta dynasty .
|
the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire .
|
how did the iron pillar not rust ?
|
- [ lecturer ] in previous videos we talk about the emergence of the maurya empire around 322 bce , shortly after the invasion of alexander the great , as the first truly great indian empire that unifies most of the indian subcontinent . now that empire eventually falls and the next significant empire to emerge , especially if we talk about influence on india and the world , is the gupta empire , which emerges over 500 years later . now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty . and we need to be careful , do n't confuse this chandragupta i with chandragupta maurya who founded the maurya dynasty over 600 years before the time we 're talking about . chandragupta i , it 's interesting because he is really able to gain power not , initially , through conquest , but through a marriage , he has a marriage with the princess kumaradevi and as a dowry he is given control over much of this region of north east india , this region of magadha or magadha , i am always having trouble pronouncing that , so my apologies , including the famous city of pataliputra , which even in the time of the maurya empire and before the maurya empire , this was a famous seat of power . but once he 's in control of this region then he and his successors are able to have increased conquests over india , you see in this light blue color what his son , samudragupta , was able to do and then one of samudragupta 's sons , chandragupta ii is able to conquer even more . but what makes the gupta empire distinctive is n't just that they were able to unify much , or conquer , much of india again , what really makes them distinctive is because of that unification and especially the wealth that began to flow into the capital they were able to be sponsors of significant culture and science and the arts , and that 's why historians view the gupta empire as the golden age of india . and just to get an appreciation for this , the gupta empire was , during the time of kalidasa , and he is considered to be the greatest writer ever in the sanskrit language , he is like the william shakespeare of sanskrit . and just to get a feel for some of his writing , obviously this is n't sanskrit , this is english , and this is from the recognition of sakuntala , which is one of his most famous works , `` wouldst thou the young year 's blossoms `` and the fruits of its decline `` and all by which the soul is charmed , `` enraptured , feasted , fed , `` wouldst thou the earth and heaven itself `` in one sole name combine ? `` i name thee , o sakuntala ! and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function . and so he established some of the early ideas of trigonometry . he did work in summation , he had a significant astronomical work , recognizing the rotation of the earth versus the rotation of the heavens , he had an early concept of gravity . even some of these notions of the place value system and zero and decimal notation , many of our modern notions of it are traced back to aryabhata . in other videos on the islamic golden age when we talk about folks like al-khwarizmi , a lot of his work was based on what he learned from aryabhata . beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire . this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance . now , like all empires , the gupta empire does eventually fall in the mid sixth century , it 's believed , around 540 , 550 . and one of the main causes , there 's invasions from people called the hunas , who historians believe are either the huns or a group that are closely related to the huns , it 's considered to be one of the causes of the eventual decline of the gupta dynasty .
|
this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance .
|
are there any traces of this dynasty in the current distribution of languages in india ?
|
- [ lecturer ] in previous videos we talk about the emergence of the maurya empire around 322 bce , shortly after the invasion of alexander the great , as the first truly great indian empire that unifies most of the indian subcontinent . now that empire eventually falls and the next significant empire to emerge , especially if we talk about influence on india and the world , is the gupta empire , which emerges over 500 years later . now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty . and we need to be careful , do n't confuse this chandragupta i with chandragupta maurya who founded the maurya dynasty over 600 years before the time we 're talking about . chandragupta i , it 's interesting because he is really able to gain power not , initially , through conquest , but through a marriage , he has a marriage with the princess kumaradevi and as a dowry he is given control over much of this region of north east india , this region of magadha or magadha , i am always having trouble pronouncing that , so my apologies , including the famous city of pataliputra , which even in the time of the maurya empire and before the maurya empire , this was a famous seat of power . but once he 's in control of this region then he and his successors are able to have increased conquests over india , you see in this light blue color what his son , samudragupta , was able to do and then one of samudragupta 's sons , chandragupta ii is able to conquer even more . but what makes the gupta empire distinctive is n't just that they were able to unify much , or conquer , much of india again , what really makes them distinctive is because of that unification and especially the wealth that began to flow into the capital they were able to be sponsors of significant culture and science and the arts , and that 's why historians view the gupta empire as the golden age of india . and just to get an appreciation for this , the gupta empire was , during the time of kalidasa , and he is considered to be the greatest writer ever in the sanskrit language , he is like the william shakespeare of sanskrit . and just to get a feel for some of his writing , obviously this is n't sanskrit , this is english , and this is from the recognition of sakuntala , which is one of his most famous works , `` wouldst thou the young year 's blossoms `` and the fruits of its decline `` and all by which the soul is charmed , `` enraptured , feasted , fed , `` wouldst thou the earth and heaven itself `` in one sole name combine ? `` i name thee , o sakuntala ! and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function . and so he established some of the early ideas of trigonometry . he did work in summation , he had a significant astronomical work , recognizing the rotation of the earth versus the rotation of the heavens , he had an early concept of gravity . even some of these notions of the place value system and zero and decimal notation , many of our modern notions of it are traced back to aryabhata . in other videos on the islamic golden age when we talk about folks like al-khwarizmi , a lot of his work was based on what he learned from aryabhata . beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire . this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance . now , like all empires , the gupta empire does eventually fall in the mid sixth century , it 's believed , around 540 , 550 . and one of the main causes , there 's invasions from people called the hunas , who historians believe are either the huns or a group that are closely related to the huns , it 's considered to be one of the causes of the eventual decline of the gupta dynasty .
|
beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii .
|
i thought that elephants became rooks and camels became bishops ?
|
- [ lecturer ] in previous videos we talk about the emergence of the maurya empire around 322 bce , shortly after the invasion of alexander the great , as the first truly great indian empire that unifies most of the indian subcontinent . now that empire eventually falls and the next significant empire to emerge , especially if we talk about influence on india and the world , is the gupta empire , which emerges over 500 years later . now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty . and we need to be careful , do n't confuse this chandragupta i with chandragupta maurya who founded the maurya dynasty over 600 years before the time we 're talking about . chandragupta i , it 's interesting because he is really able to gain power not , initially , through conquest , but through a marriage , he has a marriage with the princess kumaradevi and as a dowry he is given control over much of this region of north east india , this region of magadha or magadha , i am always having trouble pronouncing that , so my apologies , including the famous city of pataliputra , which even in the time of the maurya empire and before the maurya empire , this was a famous seat of power . but once he 's in control of this region then he and his successors are able to have increased conquests over india , you see in this light blue color what his son , samudragupta , was able to do and then one of samudragupta 's sons , chandragupta ii is able to conquer even more . but what makes the gupta empire distinctive is n't just that they were able to unify much , or conquer , much of india again , what really makes them distinctive is because of that unification and especially the wealth that began to flow into the capital they were able to be sponsors of significant culture and science and the arts , and that 's why historians view the gupta empire as the golden age of india . and just to get an appreciation for this , the gupta empire was , during the time of kalidasa , and he is considered to be the greatest writer ever in the sanskrit language , he is like the william shakespeare of sanskrit . and just to get a feel for some of his writing , obviously this is n't sanskrit , this is english , and this is from the recognition of sakuntala , which is one of his most famous works , `` wouldst thou the young year 's blossoms `` and the fruits of its decline `` and all by which the soul is charmed , `` enraptured , feasted , fed , `` wouldst thou the earth and heaven itself `` in one sole name combine ? `` i name thee , o sakuntala ! and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function . and so he established some of the early ideas of trigonometry . he did work in summation , he had a significant astronomical work , recognizing the rotation of the earth versus the rotation of the heavens , he had an early concept of gravity . even some of these notions of the place value system and zero and decimal notation , many of our modern notions of it are traced back to aryabhata . in other videos on the islamic golden age when we talk about folks like al-khwarizmi , a lot of his work was based on what he learned from aryabhata . beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire . this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance . now , like all empires , the gupta empire does eventually fall in the mid sixth century , it 's believed , around 540 , 550 . and one of the main causes , there 's invasions from people called the hunas , who historians believe are either the huns or a group that are closely related to the huns , it 's considered to be one of the causes of the eventual decline of the gupta dynasty .
|
now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty .
|
what present day countries were part of the gupta dynasty ?
|
- [ lecturer ] in previous videos we talk about the emergence of the maurya empire around 322 bce , shortly after the invasion of alexander the great , as the first truly great indian empire that unifies most of the indian subcontinent . now that empire eventually falls and the next significant empire to emerge , especially if we talk about influence on india and the world , is the gupta empire , which emerges over 500 years later . now , let 's zoom in on our timeline to get a deeper appreciation of the gupta empire . it 's believed that its start was with sri gupta , he started the gupta dynasty around 240 and it 's disputed where they emerged , it mighta been in that region or in that region , there 's different accounts of where the gupta dynasty initially emerged , but even in the early 300s they really had a control of a few small kingdoms . it was n't until the reign of chandragupta i that it becomes a significant dynasty . and we need to be careful , do n't confuse this chandragupta i with chandragupta maurya who founded the maurya dynasty over 600 years before the time we 're talking about . chandragupta i , it 's interesting because he is really able to gain power not , initially , through conquest , but through a marriage , he has a marriage with the princess kumaradevi and as a dowry he is given control over much of this region of north east india , this region of magadha or magadha , i am always having trouble pronouncing that , so my apologies , including the famous city of pataliputra , which even in the time of the maurya empire and before the maurya empire , this was a famous seat of power . but once he 's in control of this region then he and his successors are able to have increased conquests over india , you see in this light blue color what his son , samudragupta , was able to do and then one of samudragupta 's sons , chandragupta ii is able to conquer even more . but what makes the gupta empire distinctive is n't just that they were able to unify much , or conquer , much of india again , what really makes them distinctive is because of that unification and especially the wealth that began to flow into the capital they were able to be sponsors of significant culture and science and the arts , and that 's why historians view the gupta empire as the golden age of india . and just to get an appreciation for this , the gupta empire was , during the time of kalidasa , and he is considered to be the greatest writer ever in the sanskrit language , he is like the william shakespeare of sanskrit . and just to get a feel for some of his writing , obviously this is n't sanskrit , this is english , and this is from the recognition of sakuntala , which is one of his most famous works , `` wouldst thou the young year 's blossoms `` and the fruits of its decline `` and all by which the soul is charmed , `` enraptured , feasted , fed , `` wouldst thou the earth and heaven itself `` in one sole name combine ? `` i name thee , o sakuntala ! and all at once is said . '' beyond literature and writing you have significant contributions to science , most notably aryabhata , he 's known for a very accurate approximation of pi , but even more important , a recognition that was an approximation and that he potentially recognized the irrationality of pi , one of the first to do so . the word sine , the trig function , is derived from aryabhata 's word for that function . and so he established some of the early ideas of trigonometry . he did work in summation , he had a significant astronomical work , recognizing the rotation of the earth versus the rotation of the heavens , he had an early concept of gravity . even some of these notions of the place value system and zero and decimal notation , many of our modern notions of it are traced back to aryabhata . in other videos on the islamic golden age when we talk about folks like al-khwarizmi , a lot of his work was based on what he learned from aryabhata . beyond the sciences , and once again , this is just a sample of all that happened during this period , you have the significant hindu epics , the mahabharata , the ramayana , the puranas , get written down and formalized , you can say they were canonized . the game of chess or the early version of the game of chess was invented , called chaturanga , and they had horsemen , which were the knights , they had infantry , which were the pawns , they had elephants , which eventually turned into bishops , but as it migrated into persia , the muslim world , and then into europe , it became our modern game of chess . famously , there is this iron pillar that is now in new delhi that is traced back to the time of the gupta empire and is believed to the reign of chandragupta ii . and what 's amazing about this , this is a pillar that 's over 20 feet high , made out of raw iron , and over 1,500 , 1,600 years it has n't corroded , and it has inscriptions on it that help historians point to the gupta empire . this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance . now , like all empires , the gupta empire does eventually fall in the mid sixth century , it 's believed , around 540 , 550 . and one of the main causes , there 's invasions from people called the hunas , who historians believe are either the huns or a group that are closely related to the huns , it 's considered to be one of the causes of the eventual decline of the gupta dynasty .
|
this was some of the coinage of the gupta empire . so the big takeaway here is , this was india 's golden age , the classical period of india , a lot of modern hinduism and indian culture can be traced back to this time period . but it is n't just its influence on india , in other videos we talk about the islamic golden age , and much of that golden age , which emerges two , three , 400 years after the time , after the gupta empire falls , much of that work is based on the discoveries and the work that is collected during the time of the guptas and then that becomes a bridge , eventually , to the european renaissance .
|
was india the only one ?
|
so let 's see if we can find the derivative with respect to x , with either x times the cosine of x . and like always , pause this video and give it a go on your own before we work through it . so when you look at this you might say , `` well , i know how to find `` the derivative with e to the x , '' that 's infact just e to the x . and let me write this down . we know a few things . we know the derivative with respect to x of e to the x. e to the x is e to the x . we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ? well as you can imagine , this might involve the product rule . and let me just write down the product rule generally first . so if we take the derivative with respect to x of the first expression in terms of x , so this is , we could call this u of x times another expression that involves x . so u times v of x . this is going to be equal to , and i 'm color-coding it so we can really keep track of things . this is going to be equal to the derivative of the first expression . so i could write that as u prime of x times just the second expression not the derivative of it , just the second expression . so times v of x and then we have plus the first expression , not its derivative , just the first expression . u of x times the derivative of the second expression . times the derivative of the second expression . so the way you remember it is , you have these two things here , you 're going to end up with two different terms . in each of them , you 're going to take the derivative of one of them , but not the other one , and then the other one you 'll take the derivative of the other one , but not the first one . so , the derivative of u times v is u prime times v , plus u times v prime . when you just look at it like that , it seems a little bit abstract and that might even be a little bit confusing , but that 's why we have a tangible example here and i color-coded intentionally . we could say that u of x is equal to e to the x . and v of x is equal to the cosine of x . so v of x is equal to cosine of x . and if u of x is equal to e to the x , we know that the derivative of that with respect to x is still e to the x . that 's one of the most magical things in mathematics . one of the things that makes e so special . so u prime of x is still equal to e to the x . and v prime of x , we know as negative sine of x . negative sine of x , and so , what 's this going to be equal to ? this is going to be equal to the derivative of the first expression . so , the derivative of e to the x which is just , e to the x , times the second expression , not taking it 's derivative , so times cosine of x . plus the first expression , not taking its derivative , so e to the x , times the derivative of the second expression . so , times the derivative of cosine of x which is negative sine . negative sine of x . and it might be a little bit confusing , because e to the x is its own derivative . this right over here , you can view this as this was the derivative as e to the x which happens to be e to the x . that 's what 's exciting about that expression , or that function . and then this is just e to the x without taking it 's derivative - they are of course , the same thing . but anyway , now we can just simplify it . this is going to be equal to ... we could write either as e to the x times cosine of x , times cosine of x minus e to the x. e to the x times sine of x . times sine of x . or , if you want , you could factor out an e to the x . this is the same thing as e to the x times cosine of x minus sine of x. cosine of x minus sine of x . so hopefully this makes the product rule a little bit more tangible . and once you have this in your tool belt , there 's a whole broader class of functions and expressions that we can start to differentiate .
|
we know a few things . we know the derivative with respect to x of e to the x. e to the x is e to the x . we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ?
|
why is the 1st derivative of ln ( x ) = 1/x ?
|
so let 's see if we can find the derivative with respect to x , with either x times the cosine of x . and like always , pause this video and give it a go on your own before we work through it . so when you look at this you might say , `` well , i know how to find `` the derivative with e to the x , '' that 's infact just e to the x . and let me write this down . we know a few things . we know the derivative with respect to x of e to the x. e to the x is e to the x . we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ? well as you can imagine , this might involve the product rule . and let me just write down the product rule generally first . so if we take the derivative with respect to x of the first expression in terms of x , so this is , we could call this u of x times another expression that involves x . so u times v of x . this is going to be equal to , and i 'm color-coding it so we can really keep track of things . this is going to be equal to the derivative of the first expression . so i could write that as u prime of x times just the second expression not the derivative of it , just the second expression . so times v of x and then we have plus the first expression , not its derivative , just the first expression . u of x times the derivative of the second expression . times the derivative of the second expression . so the way you remember it is , you have these two things here , you 're going to end up with two different terms . in each of them , you 're going to take the derivative of one of them , but not the other one , and then the other one you 'll take the derivative of the other one , but not the first one . so , the derivative of u times v is u prime times v , plus u times v prime . when you just look at it like that , it seems a little bit abstract and that might even be a little bit confusing , but that 's why we have a tangible example here and i color-coded intentionally . we could say that u of x is equal to e to the x . and v of x is equal to the cosine of x . so v of x is equal to cosine of x . and if u of x is equal to e to the x , we know that the derivative of that with respect to x is still e to the x . that 's one of the most magical things in mathematics . one of the things that makes e so special . so u prime of x is still equal to e to the x . and v prime of x , we know as negative sine of x . negative sine of x , and so , what 's this going to be equal to ? this is going to be equal to the derivative of the first expression . so , the derivative of e to the x which is just , e to the x , times the second expression , not taking it 's derivative , so times cosine of x . plus the first expression , not taking its derivative , so e to the x , times the derivative of the second expression . so , times the derivative of cosine of x which is negative sine . negative sine of x . and it might be a little bit confusing , because e to the x is its own derivative . this right over here , you can view this as this was the derivative as e to the x which happens to be e to the x . that 's what 's exciting about that expression , or that function . and then this is just e to the x without taking it 's derivative - they are of course , the same thing . but anyway , now we can just simplify it . this is going to be equal to ... we could write either as e to the x times cosine of x , times cosine of x minus e to the x. e to the x times sine of x . times sine of x . or , if you want , you could factor out an e to the x . this is the same thing as e to the x times cosine of x minus sine of x. cosine of x minus sine of x . so hopefully this makes the product rule a little bit more tangible . and once you have this in your tool belt , there 's a whole broader class of functions and expressions that we can start to differentiate .
|
we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ?
|
when using the product rule , do both functions need to be functions of x ( when i 'm taking the derivative with respect to x ) ?
|
so let 's see if we can find the derivative with respect to x , with either x times the cosine of x . and like always , pause this video and give it a go on your own before we work through it . so when you look at this you might say , `` well , i know how to find `` the derivative with e to the x , '' that 's infact just e to the x . and let me write this down . we know a few things . we know the derivative with respect to x of e to the x. e to the x is e to the x . we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ? well as you can imagine , this might involve the product rule . and let me just write down the product rule generally first . so if we take the derivative with respect to x of the first expression in terms of x , so this is , we could call this u of x times another expression that involves x . so u times v of x . this is going to be equal to , and i 'm color-coding it so we can really keep track of things . this is going to be equal to the derivative of the first expression . so i could write that as u prime of x times just the second expression not the derivative of it , just the second expression . so times v of x and then we have plus the first expression , not its derivative , just the first expression . u of x times the derivative of the second expression . times the derivative of the second expression . so the way you remember it is , you have these two things here , you 're going to end up with two different terms . in each of them , you 're going to take the derivative of one of them , but not the other one , and then the other one you 'll take the derivative of the other one , but not the first one . so , the derivative of u times v is u prime times v , plus u times v prime . when you just look at it like that , it seems a little bit abstract and that might even be a little bit confusing , but that 's why we have a tangible example here and i color-coded intentionally . we could say that u of x is equal to e to the x . and v of x is equal to the cosine of x . so v of x is equal to cosine of x . and if u of x is equal to e to the x , we know that the derivative of that with respect to x is still e to the x . that 's one of the most magical things in mathematics . one of the things that makes e so special . so u prime of x is still equal to e to the x . and v prime of x , we know as negative sine of x . negative sine of x , and so , what 's this going to be equal to ? this is going to be equal to the derivative of the first expression . so , the derivative of e to the x which is just , e to the x , times the second expression , not taking it 's derivative , so times cosine of x . plus the first expression , not taking its derivative , so e to the x , times the derivative of the second expression . so , times the derivative of cosine of x which is negative sine . negative sine of x . and it might be a little bit confusing , because e to the x is its own derivative . this right over here , you can view this as this was the derivative as e to the x which happens to be e to the x . that 's what 's exciting about that expression , or that function . and then this is just e to the x without taking it 's derivative - they are of course , the same thing . but anyway , now we can just simplify it . this is going to be equal to ... we could write either as e to the x times cosine of x , times cosine of x minus e to the x. e to the x times sine of x . times sine of x . or , if you want , you could factor out an e to the x . this is the same thing as e to the x times cosine of x minus sine of x. cosine of x minus sine of x . so hopefully this makes the product rule a little bit more tangible . and once you have this in your tool belt , there 's a whole broader class of functions and expressions that we can start to differentiate .
|
we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ?
|
what if i need to differentiate xy with respect to x ( where y is a function of x ) ?
|
so let 's see if we can find the derivative with respect to x , with either x times the cosine of x . and like always , pause this video and give it a go on your own before we work through it . so when you look at this you might say , `` well , i know how to find `` the derivative with e to the x , '' that 's infact just e to the x . and let me write this down . we know a few things . we know the derivative with respect to x of e to the x. e to the x is e to the x . we know how to find the derivative cosine of x . the derivative with respect to x of cosine of x is equal to negative sine of x . but , how do we find the derivative of their product ? well as you can imagine , this might involve the product rule . and let me just write down the product rule generally first . so if we take the derivative with respect to x of the first expression in terms of x , so this is , we could call this u of x times another expression that involves x . so u times v of x . this is going to be equal to , and i 'm color-coding it so we can really keep track of things . this is going to be equal to the derivative of the first expression . so i could write that as u prime of x times just the second expression not the derivative of it , just the second expression . so times v of x and then we have plus the first expression , not its derivative , just the first expression . u of x times the derivative of the second expression . times the derivative of the second expression . so the way you remember it is , you have these two things here , you 're going to end up with two different terms . in each of them , you 're going to take the derivative of one of them , but not the other one , and then the other one you 'll take the derivative of the other one , but not the first one . so , the derivative of u times v is u prime times v , plus u times v prime . when you just look at it like that , it seems a little bit abstract and that might even be a little bit confusing , but that 's why we have a tangible example here and i color-coded intentionally . we could say that u of x is equal to e to the x . and v of x is equal to the cosine of x . so v of x is equal to cosine of x . and if u of x is equal to e to the x , we know that the derivative of that with respect to x is still e to the x . that 's one of the most magical things in mathematics . one of the things that makes e so special . so u prime of x is still equal to e to the x . and v prime of x , we know as negative sine of x . negative sine of x , and so , what 's this going to be equal to ? this is going to be equal to the derivative of the first expression . so , the derivative of e to the x which is just , e to the x , times the second expression , not taking it 's derivative , so times cosine of x . plus the first expression , not taking its derivative , so e to the x , times the derivative of the second expression . so , times the derivative of cosine of x which is negative sine . negative sine of x . and it might be a little bit confusing , because e to the x is its own derivative . this right over here , you can view this as this was the derivative as e to the x which happens to be e to the x . that 's what 's exciting about that expression , or that function . and then this is just e to the x without taking it 's derivative - they are of course , the same thing . but anyway , now we can just simplify it . this is going to be equal to ... we could write either as e to the x times cosine of x , times cosine of x minus e to the x. e to the x times sine of x . times sine of x . or , if you want , you could factor out an e to the x . this is the same thing as e to the x times cosine of x minus sine of x. cosine of x minus sine of x . so hopefully this makes the product rule a little bit more tangible . and once you have this in your tool belt , there 's a whole broader class of functions and expressions that we can start to differentiate .
|
or , if you want , you could factor out an e to the x . this is the same thing as e to the x times cosine of x minus sine of x. cosine of x minus sine of x . so hopefully this makes the product rule a little bit more tangible .
|
i get how y is a function , but how is x a function ?
|
( piano music ) steven : this is steven zucker . juliana : and juliana kreinik , and we are discussing lyonel feininger 's cathedral of the future from 1919 . steven : this was one of the primary images of the bauhaus right ? juliana : it was the starting image . steven : was its intent actually marketing ? it was a brochure . juliana : this was a graphic work that was intended to be in poster format as a kind of announcement of the opening of the school of this new bauhaus . steven : it 's fabulous . okay , so now first it 's a woodcut and it 's rendered in a way that makes it feel very rough , not polished . juliana : i always think of it as a very stark , graphic work in its most raw sense because of all the lines and it 's black and white and there 's a bit of texture that you can almost see the impression from the wood , but then you also see all these angular sharp lines that come out . you see something emerging from within that . this is the cathedral of the future . steven : but it 's so not what we think of when we think of the bauhaus . when i think of the bauhaus , i think of it as something that 's completely concerned with the technology of the present . right ? juliana : yes . steven : and concerned with , to some extent , craft . not concerned with religious , but in fact early on the bauhaus was actually concerned , maybe not with religious but with spiritual . juliana : definitely the spiritual . this definitely has a sense that you 're looking at something other worldly . but do n't forget of course this is a german school . so the sense of the spiritual and the cathedral they 're all kind of connected , and art making is part of that . steven : so , this is 1919 . germany is going through tremendous transition at this moment , right ? juliana : yes . steven : is this just at the moment of the revolution in germany ? what 's happening ? juliana : everything is happening in terms of post war chaos and reformation , rebuilding culture , rebuilding buildings , rebuilding a sort of sense of purpose . steven : it 's interesting how the bauhaus then becomes this fulcrum for the way in which the spiritual transforms itself into the technological or into the industrial . juliana : yes . technology and spirituality and redemption all of those sort of themes come into play after world war i . there 's such a sense of decimation and hopelessness , but here we have this image where there 's sort of a sense of optimism , something that your sort of looking up at and looking forward to . we could really talk about what the cathedral represents to the bauhaus and really what is it about building ? what does the word bauhaus even mean ? steven : well the bauhaus refers to the small building or workshop that would be just outside of a large medieval building campaign perhaps for a cathedral in fact . so , you have actually the shop where the masons are doing their work outside and it is a place , i think , as it was originally understood by gropius , the artists would come together where there would be a unity of arts so that all of the crafts were in a sense working in unison to the betterment of that culture or the betterment of society . juliana : you know , we should talk about who gropius was . steven : okay . so , gropius was an architect . juliana : right . steven : and he was the first director , really the founder of the bauhaus . juliana : and so that 's why we see that architecture is featured so prominently . steven : if you look at this diagram of the curriculum of the bauhaus architecture is in the center and it is really positioned as the summation of all of the arts . juliana : because it 's perceived in having elements of all of art making . you have design . you have craftsmanship , color . you have form . you have space , all these different dimensionalities . everything that 's featured within architecture . steven : so , this may actually answer a question for me because one of the things that i was thinking about when i was looking at this print is why would feininger choose a woodcut if it seems so historical . really what we 're talking about is a kind of return to a unity of the arts . so , the bauhaus is looking towards a kind of promise and looking forward , it is also very much trying to retrieve a kind of lost ideal . juliana : definitely . the graphic piece , it 's a woodcut . so you 're working with your hands to carve out the design , and then it 's a cathedral coming into view . steven : and there 's a directness because of the quality of the woodcut that is very immediate and we can really feel , in a sense , the tactile nature of the block from which it was carved . juliana : but you know what 's also so smart if you think about technology , is what reproduces well ? a woodcut . it 's meant for reproduction . so , this is sort of looking forward to technological reproduction where they 're looking to advertise . it 's a marketing device . what 's going to work best in order to show everybody what their new school is about and what kind of ideology they 're trying to promote ? steven : it 's interesting , and that notion of the technological promise and the sense of exploration is also , i think , referred to in a crystalline quality . juliana : yes . steven : if you look at kind of the expressionist artchitecture that 's being produced at this time in germany when actually there was an economic crisis and very little money , much of it was paper architecture . that is to say really fanciful designs but very often looking to this notion of the crystal light , the future as expressed through a series of sort of the drama and play of contrasting light and shadow . and i think we see that clearly rendered here . juliana : these themes resonate throughout the time period of the bauhaus , just 1919 to 1933 . so , at this point gropius starts this school . steven : and then it 's sort of remade of course and gropius will truly remake it when he gets to dessau . juliana : yes . steven : ... and is able to design the buildings in which it 's housed . juliana : and then the coursework will grow and expand and will have all kinds of changes , but this really is the primary or original piece . steven : yeah . in a sense it 's the kind of founding document . juliana : founding document . that 's the word . foundational document even . steven : yeah . juliana : ... is what this is probably best considered as . steven : by the time we get to the end of the bauhaus 's run in the early 1930s the spiritual is , to some extent , excised . juliana : well yeah , the spiritual transitions into something else , into notions of spiritual technology and its connections . steve : oh interesting . interesting . juliana : so , yes . i think the idea of what one is connected with spiritually changes . steven : terrific . thanks . juliana : thank you .
|
what does the word bauhaus even mean ? steven : well the bauhaus refers to the small building or workshop that would be just outside of a large medieval building campaign perhaps for a cathedral in fact . so , you have actually the shop where the masons are doing their work outside and it is a place , i think , as it was originally understood by gropius , the artists would come together where there would be a unity of arts so that all of the crafts were in a sense working in unison to the betterment of that culture or the betterment of society .
|
is there any meaning to the cube shape on the pediment in the bottom-center of the building ?
|
( piano music ) steven : this is steven zucker . juliana : and juliana kreinik , and we are discussing lyonel feininger 's cathedral of the future from 1919 . steven : this was one of the primary images of the bauhaus right ? juliana : it was the starting image . steven : was its intent actually marketing ? it was a brochure . juliana : this was a graphic work that was intended to be in poster format as a kind of announcement of the opening of the school of this new bauhaus . steven : it 's fabulous . okay , so now first it 's a woodcut and it 's rendered in a way that makes it feel very rough , not polished . juliana : i always think of it as a very stark , graphic work in its most raw sense because of all the lines and it 's black and white and there 's a bit of texture that you can almost see the impression from the wood , but then you also see all these angular sharp lines that come out . you see something emerging from within that . this is the cathedral of the future . steven : but it 's so not what we think of when we think of the bauhaus . when i think of the bauhaus , i think of it as something that 's completely concerned with the technology of the present . right ? juliana : yes . steven : and concerned with , to some extent , craft . not concerned with religious , but in fact early on the bauhaus was actually concerned , maybe not with religious but with spiritual . juliana : definitely the spiritual . this definitely has a sense that you 're looking at something other worldly . but do n't forget of course this is a german school . so the sense of the spiritual and the cathedral they 're all kind of connected , and art making is part of that . steven : so , this is 1919 . germany is going through tremendous transition at this moment , right ? juliana : yes . steven : is this just at the moment of the revolution in germany ? what 's happening ? juliana : everything is happening in terms of post war chaos and reformation , rebuilding culture , rebuilding buildings , rebuilding a sort of sense of purpose . steven : it 's interesting how the bauhaus then becomes this fulcrum for the way in which the spiritual transforms itself into the technological or into the industrial . juliana : yes . technology and spirituality and redemption all of those sort of themes come into play after world war i . there 's such a sense of decimation and hopelessness , but here we have this image where there 's sort of a sense of optimism , something that your sort of looking up at and looking forward to . we could really talk about what the cathedral represents to the bauhaus and really what is it about building ? what does the word bauhaus even mean ? steven : well the bauhaus refers to the small building or workshop that would be just outside of a large medieval building campaign perhaps for a cathedral in fact . so , you have actually the shop where the masons are doing their work outside and it is a place , i think , as it was originally understood by gropius , the artists would come together where there would be a unity of arts so that all of the crafts were in a sense working in unison to the betterment of that culture or the betterment of society . juliana : you know , we should talk about who gropius was . steven : okay . so , gropius was an architect . juliana : right . steven : and he was the first director , really the founder of the bauhaus . juliana : and so that 's why we see that architecture is featured so prominently . steven : if you look at this diagram of the curriculum of the bauhaus architecture is in the center and it is really positioned as the summation of all of the arts . juliana : because it 's perceived in having elements of all of art making . you have design . you have craftsmanship , color . you have form . you have space , all these different dimensionalities . everything that 's featured within architecture . steven : so , this may actually answer a question for me because one of the things that i was thinking about when i was looking at this print is why would feininger choose a woodcut if it seems so historical . really what we 're talking about is a kind of return to a unity of the arts . so , the bauhaus is looking towards a kind of promise and looking forward , it is also very much trying to retrieve a kind of lost ideal . juliana : definitely . the graphic piece , it 's a woodcut . so you 're working with your hands to carve out the design , and then it 's a cathedral coming into view . steven : and there 's a directness because of the quality of the woodcut that is very immediate and we can really feel , in a sense , the tactile nature of the block from which it was carved . juliana : but you know what 's also so smart if you think about technology , is what reproduces well ? a woodcut . it 's meant for reproduction . so , this is sort of looking forward to technological reproduction where they 're looking to advertise . it 's a marketing device . what 's going to work best in order to show everybody what their new school is about and what kind of ideology they 're trying to promote ? steven : it 's interesting , and that notion of the technological promise and the sense of exploration is also , i think , referred to in a crystalline quality . juliana : yes . steven : if you look at kind of the expressionist artchitecture that 's being produced at this time in germany when actually there was an economic crisis and very little money , much of it was paper architecture . that is to say really fanciful designs but very often looking to this notion of the crystal light , the future as expressed through a series of sort of the drama and play of contrasting light and shadow . and i think we see that clearly rendered here . juliana : these themes resonate throughout the time period of the bauhaus , just 1919 to 1933 . so , at this point gropius starts this school . steven : and then it 's sort of remade of course and gropius will truly remake it when he gets to dessau . juliana : yes . steven : ... and is able to design the buildings in which it 's housed . juliana : and then the coursework will grow and expand and will have all kinds of changes , but this really is the primary or original piece . steven : yeah . in a sense it 's the kind of founding document . juliana : founding document . that 's the word . foundational document even . steven : yeah . juliana : ... is what this is probably best considered as . steven : by the time we get to the end of the bauhaus 's run in the early 1930s the spiritual is , to some extent , excised . juliana : well yeah , the spiritual transitions into something else , into notions of spiritual technology and its connections . steve : oh interesting . interesting . juliana : so , yes . i think the idea of what one is connected with spiritually changes . steven : terrific . thanks . juliana : thank you .
|
we could really talk about what the cathedral represents to the bauhaus and really what is it about building ? what does the word bauhaus even mean ? steven : well the bauhaus refers to the small building or workshop that would be just outside of a large medieval building campaign perhaps for a cathedral in fact . so , you have actually the shop where the masons are doing their work outside and it is a place , i think , as it was originally understood by gropius , the artists would come together where there would be a unity of arts so that all of the crafts were in a sense working in unison to the betterment of that culture or the betterment of society .
|
it almost looks like the cube has replaced a cross , perhaps suggesting that the bauhaus is creating a church of creation rather than a church of devotion ?
|
( piano music ) man : we 're in the museum of the duomo in florence and we 're looking at a donatello . it 's not marble . it 's not bronze . it 's wood . it looks so frail . it 's a sculpture of mary magdalene . woman : it 's a very difficult sculpture to look at because it 's ugly . mary magdalene is shown as a hermit with her hands about to be clasped in prayer and she 's old and wrinkled . her body is exposed to us . she 's got these muscular arms . man : but thin also . woman : the skin on her chest and neck , and her face look like the skin of an old woman . it 's difficult to look at . man : i think it 's difficult because there is a whole series of contrast that we 're not used to being shown in sculpture . you have a body that clearly was once very beautiful . she 's got high cheek bones . she 's tall and graceful . but you 're right . the body has weathered . this life has taken its toll . and it 's almost as if she 's wasted away . all that 's left is a pure spirituality . the kind of pure faith . she 's clothed only in her very long hair , which is one of her attributes . when we see a woman by , for instance , christ 's feet with long red hair we know that 's mary magdalene . here that hair has grown out . in fact , even her belt is actually her hair wrapped around her . i think it 's so interesting the choice of materials . i started out by saying this is wood . there 's something about the organic quality of wood and its frailty , it 's ability to soften , to rot , that seems somehow appropriate here . woman : actually i 'm amazed that it 's in as good condition as it is given the fact that it 's wood . man : now if you look you can see that it 's been painted . it 's been guilded . you can see that long hair had been painted red and gold and there are traces of that , that are still left . there 's a look on her face , which is so intense and yet at the same time almost as if she has left this earth already . woman : i see this very much as part of donatello 's interest in the specifics of an individual , just like we saw with habakkuk . this really intense specificity that 's so different than the gothic and the medieval and deep sympathy for humanity that comes out of humanism of the early renaissance . man : his sympathy is infectious . there 's however kind of power here as well . her hands are so long and so elegant , they almost create a kind of cathedral as she brings them together . she is the church itself in some ways . she 's an enormously important figure . she sees christ first when he is resurrected . she is the figure that sees him crucified , that does n't run away as so many of his followers did . woman : and she 's someone who makes a very direct choice to leave a worldly life , to leave a life of the sensuality of the world for the spiritual . man : think about how important that is as a message in florence in the fifteenth century when you have a culture that has put enormous emphasis on material luxury , and here is somebody that functions as a conscience to the city . you know , there 's something else that 's interesting though . this is a late sculpture by donatello . he 's left behind the proportional accuracy of say saint mark . this is a figure that is almost gothic again in the length of her body . there is a willingness here to put front and center this spirituality and the symbolism of the figure as opposed to emphasizing the anatomical accuracy . woman : although she still stands in contropposto still there is that attention to the body that can only come from the early renaissance . man : it is donatello in a sense playing loose with his own rules . ( piano music )
|
in fact , even her belt is actually her hair wrapped around her . i think it 's so interesting the choice of materials . i started out by saying this is wood .
|
do you think her worried looking face along with her hopeful posture was being sculpted because it was during the moment jesus was being crucified ?
|
( piano music ) man : we 're in the museum of the duomo in florence and we 're looking at a donatello . it 's not marble . it 's not bronze . it 's wood . it looks so frail . it 's a sculpture of mary magdalene . woman : it 's a very difficult sculpture to look at because it 's ugly . mary magdalene is shown as a hermit with her hands about to be clasped in prayer and she 's old and wrinkled . her body is exposed to us . she 's got these muscular arms . man : but thin also . woman : the skin on her chest and neck , and her face look like the skin of an old woman . it 's difficult to look at . man : i think it 's difficult because there is a whole series of contrast that we 're not used to being shown in sculpture . you have a body that clearly was once very beautiful . she 's got high cheek bones . she 's tall and graceful . but you 're right . the body has weathered . this life has taken its toll . and it 's almost as if she 's wasted away . all that 's left is a pure spirituality . the kind of pure faith . she 's clothed only in her very long hair , which is one of her attributes . when we see a woman by , for instance , christ 's feet with long red hair we know that 's mary magdalene . here that hair has grown out . in fact , even her belt is actually her hair wrapped around her . i think it 's so interesting the choice of materials . i started out by saying this is wood . there 's something about the organic quality of wood and its frailty , it 's ability to soften , to rot , that seems somehow appropriate here . woman : actually i 'm amazed that it 's in as good condition as it is given the fact that it 's wood . man : now if you look you can see that it 's been painted . it 's been guilded . you can see that long hair had been painted red and gold and there are traces of that , that are still left . there 's a look on her face , which is so intense and yet at the same time almost as if she has left this earth already . woman : i see this very much as part of donatello 's interest in the specifics of an individual , just like we saw with habakkuk . this really intense specificity that 's so different than the gothic and the medieval and deep sympathy for humanity that comes out of humanism of the early renaissance . man : his sympathy is infectious . there 's however kind of power here as well . her hands are so long and so elegant , they almost create a kind of cathedral as she brings them together . she is the church itself in some ways . she 's an enormously important figure . she sees christ first when he is resurrected . she is the figure that sees him crucified , that does n't run away as so many of his followers did . woman : and she 's someone who makes a very direct choice to leave a worldly life , to leave a life of the sensuality of the world for the spiritual . man : think about how important that is as a message in florence in the fifteenth century when you have a culture that has put enormous emphasis on material luxury , and here is somebody that functions as a conscience to the city . you know , there 's something else that 's interesting though . this is a late sculpture by donatello . he 's left behind the proportional accuracy of say saint mark . this is a figure that is almost gothic again in the length of her body . there is a willingness here to put front and center this spirituality and the symbolism of the figure as opposed to emphasizing the anatomical accuracy . woman : although she still stands in contropposto still there is that attention to the body that can only come from the early renaissance . man : it is donatello in a sense playing loose with his own rules . ( piano music )
|
it 's not bronze . it 's wood . it looks so frail .
|
so how is the wood still there ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 .
|
would that be considered an `` improper '' fraction because there are square roots in the denominators ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right .
|
why is it important to write the names of similar triangles in a certain order ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
is n't triangle abc the same as triangle cab ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
do the letters matter in any triangle or can you put any letter on a triangle ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides .
|
6 is n't 27 the longest side of the triangle ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides .
|
what would that squiggly line be called ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
if it is proved by sss similarity postulate that triangle abc is similar to triangle edf then does it mean that angle a is equal to angle e ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
if it is proved by sss similarity postulate that triangle abc is similar to triangle edf then does it mean that angle a is equal to angle e ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides .
|
aka if another triangle is double of another triangle does that mean their similar 100 % ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles .
|
that means that the triangles must be congruent , so if two triangles are congruent are n't they also similar ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 .
|
( besides identifying the labeled point/line segments ) , i mean is the hypotenuse the only side with an appelation ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles .
|
on the third problem , why cant you use the two small triangles and say segment bd is the same because of they are on the same triangle ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same .
|
how is 3 root 3 ( fg ) considered as the longest side ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same .
|
how did 3 root 3 became bigger than 6 ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same .
|
and how did 18 root 3 became bigger than 27 ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same .
|
sal multiplies the 1/3root3 by root3/root3 , how is he able to do that ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy .
|
what is a good way to find the corresponding sides of one triangle to another ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here .
|
why is 27 < 18 square roots of 3 ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc .
|
how did sal come to that conclusion ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel .
|
9 , how would knowing , that an angle in the smaller triangle is a right angle , help us make statements about similarity ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
why do you have to make sure that triangle ace ~ triangle bcd are triangle ace ~ triangle bcd and not triangle cea ~ triangle dbc ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right .
|
how are the triangles similar ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides .
|
why is side xt taken in the triangle xst in the above video ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 .
|
why dont we take the side xz ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side .
|
how do you figure out if the ratios of corresponding sides are constant ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same .
|
what does `` sas '' mean ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor .
|
what is congruent and ratio ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
how would we prove that triangle xyz ~ triangle xts in a two-column statement/reason proof ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right .
|
how to determine similar triangles by their angles ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right .
|
i mean in the case where the triangles are turned or flipped ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle .
|
is triangle abc~def is same as acb~dfe ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 .
|
4 , would n't it be possible to use pythagorean theorem to determine the other side to check for similarity ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here .
|
so if the triangles have the same scaling factor does that mean they 're automatically sss ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well .
|
why is 18root3 longer than 27 ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc .
|
what does this symbol mean -- -- - > ~ ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same .
|
how did 3 x sqrt3 x sqrt3 become a 3 ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 .
|
can someone explain how the three fractions are equal ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc .
|
how do you solve the proportion involving similar figures using variables ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides .
|
how do we write that 2 lines are proportional in a proof ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order .
|
do n't shapes have to have to have the same angles to be similar ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 .
|
9 how is xy/xt=x3/x5 i kind of did n't know what happen ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides .
|
how is efg similiar to hig ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right .
|
do these similarity postulates only apply to triangles or to other shapes as well ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc .
|
what is the basic proportionality theorem ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel .
|
9 , how would knowing , that an angle in the smaller triangle is a right angle , help us make statements about similarity ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right .
|
with the second problem at around are n't the different sides of the two triangles going up by different factors , meaning that they 're not actually similar ... ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same .
|
how is 18 root 3 larger than 27 ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles .
|
what is the difference between similarity and congruence ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides .
|
can you explain the side-splitter and triangle-angle-bisector theorems ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 .
|
would that be considered an `` improper '' fraction because there are square roots in the denominators ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 .
|
so would ace ~ dcb be incorrect ?
|
what i want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar , using some of the postulates that we 've set up . so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel . we know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent . so that angle is going to correspond to that angle right over there . and we 're done . we have one angle in triangle aec that is congruent to another angle in bdc , and then we have this angle that 's obviously congruent to itself that 's in both triangles . so both triangles have a pair of corresponding angles that are congruent , so they must be similar . so we can write , triangle ace is going to be similar to triangle -- and we want to get the letters in the right order . so where the blue angle is here , the blue angle there is vertex b . then we go to the wide angle , c , and then we go to the unlabeled angle right over there , bcd . so we did that first one . now let 's do this one right over here . this is kind of similar , but it looks , just superficially looking at it , that yz is definitely not parallel to st . so we wo n't be able to do this corresponding angle argument , especially because they did n't even label it as parallel . and so you do n't want to look at things just by the way they look . you definitely want to say , what am i given , and what am i not given ? if these were n't labeled parallel , we would n't be able to make the statement , even if they looked parallel . one thing we do have is that we have this angle right here that 's common to the inner triangle and to the outer triangle , and they 've given us a bunch of sides . so maybe we can use sas for similarity , meaning if we can show the ratio of the sides on either side of this angle , if they have the same ratio from the smaller triangle to the larger triangle , then we can show similarity . so let 's go , and we have to go on either side of this angle right over here . let 's look at the shorter side on either side of this angle . so the shorter side is two , and let 's look at the shorter side on either side of the angle for the larger triangle . well , then the shorter side is on the right-hand side , and that 's going to be xt . so what we want to compare is the ratio between -- let me write it this way . we want to see , is xy over xt equal to the ratio of the longer side ? or if we 're looking relative to this angle , the longer of the two , not necessarily the longest of the triangle , although it looks like that as well . is that equal to the ratio of xz over the longer of the two sides -- when you 're looking at this angle right here , on either side of that angle , for the larger triangle -- over xs ? and it 's a little confusing , because we 've kind of flipped which side , but i 'm just thinking about the shorter side on either side of this angle in between , and then the longer side on either side of this angle . so these are the shorter sides for the smaller triangle and the larger triangle . these are the longer sides for the smaller triangle and the larger triangle . and we see xy . this is two . xt is 3 plus 1 is 4 . xz is 3 , and xs is 6 . so you have 2 over 4 , which is 1/2 , which is the same thing as 3/6 . so the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle , for both triangles , the ratio is the same . so by sas we know that the two triangles are congruent . but we have to be careful on how we state the triangles . we want to make sure we get the corresponding sides . and i 'm running out of space here . let me write it right above here . we can write that triangle xyz is similar to triangle -- so we started up at x , which is the vertex at the angle , and we went to the shorter side first . so now we want to start at x and go to the shorter side on the large triangle . so you go to xts . xyz is similar to xts . now , let 's look at this right over here . so in our larger triangle , we have a right angle here , but we really know nothing about what 's going on with any of these smaller triangles in terms of their actual angles . even though this looks like a right angle , we can not assume it . and if we look at this smaller triangle right over here , it shares one side with the larger triangle , but that 's not enough to do anything . and then this triangle over here also shares another side , but that also does n't do anything . so we really ca n't make any statement here about any kind of similarity . so there 's no similarity going on here . there are some shared angles . this guy -- they both share that angle , the larger triangle and the smaller triangle . so there could be a statement of similarity we could make if we knew that this definitely was a right angle . then we could make some interesting statements about similarity , but right now , we ca n't really do anything as is . let 's try this one out , this pair right over here . so these are the first ones that we have actually separated out the triangles . so they 've given us the three sides of both triangles . so let 's just figure out if the ratios between corresponding sides are a constant . so let 's start with the short side . so the short side here is 3 . the shortest side here is 9 square roots of 3 . so we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here , is 3 square roots of 3 over the next longest side over here , which is 27 . and then see if that 's going to be equal to the ratio of the longest side . so the longest side here is 6 , and then the longest side over here is 18 square roots of 3 . so this is going to give us -- let 's see , this is 3 . let me do this in a neutral color . so this becomes 1 over 3 square roots of 3 . this becomes 1 over root 3 over 9 , which seems like a different number , but we want to be careful here . and then this right over here -- if you divide the numerator and denominator by 6 , this becomes a 1 and this becomes 3 square roots of 3 . so 1 over 3 root 3 needs to be equal to square root of 3 over 9 , which needs to be equal to 1 over 3 square roots of 3 . at first they do n't look equal , but we can actually rationalize this denominator right over here . we can show that 1 over 3 square roots of 3 , if you multiply it by square root of 3 over square root of 3 , this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 , times 3 is 9 . so these actually are all the same . this is actually saying , this is 1 over 3 root 3 , which is the same thing as square root of 3 over 9 , which is this right over here , which is the same thing as 1 over 3 root 3 . so actually , these are similar triangles . so we can actually say it , and i 'll make sure i get the order right . so let 's start with e , which is between the blue and the magenta side . so that 's between the blue and the magenta side . that is h , right over here . i 'll do it like this . triangle e , and then i 'll go along the blue side , f. actually , let me just write it this way . triangle efg , we know is similar to triangle -- so e is between the blue and the magenta side . blue and magenta side -- that is h. and then we go along the blue side to f , go along the blue side to i , and then you went along the orange side to g , and then you go along the orange side to j . so triangle efj -- efg is similar to triangle hij by side-side-side similarity . they 're not congruent sides . they all have just the same ratio or the same scaling factor . now let 's do this last one , right over here . let 's see . we have an angle that 's congruent to another angle right over there , and we have two sides . and so it might be tempting to use side-angle-side , because we have side-angle-side here . and even the ratios look kind of tempting , because 4 times 2 is 8 . 5 times 2 is 10 . but it 's tricky here , because they are n't the same corresponding sides . in order to use side-angle-side , the two sides that have the same corresponding ratios , they have to be on either side of the angle . so in this case , they are on either side of the angle . in this case , the 4 is on one side of the angle , but the 5 is not . so because if this 5 was over here , then we could make an argument for similarity , but with this 5 not being on the other side of the angle -- it 's not sandwiching the angle with the 4 -- we ca n't use side-angle-side . and frankly , there 's nothing that we can do over here . so we ca n't make some strong statement about similarity for this last one .
|
so over here , i have triangle bdc . it 's inside of triangle aec . they both share this angle right over there , so that gives us one angle . we need two to get to angle-angle , which gives us similarity . and we know that these two lines are parallel .
|
doesnt the square for the 3rd triangle tell you its a right angle ?
|
: we have 4 graphs here and then 4 function definitions . what i want you to do is pause this video and think about which of these graphs map up to which of these function definitions . i 'm assuming you 've given a go at it . let 's go through each of these and think about what their graphs would look like . one thing that i like to do , because it 's just a simple thing to do , is just think about what happens when x is equal to 0 . especially when x is an exponent like this , you would have , let me just write this down , y of 0 is going to be equal to 2 minus 1/3 to the 0 power . that 's equal to 2 minus 1 . 1/3 to the 0 is 1 , which is equal to 1 . in which of these graphs , when x is equal to 0 , do we have y equaling 1 ? here , x equals 0 , y equals negative 1 . here , x equals 0 , y equals negative 1 , or it looks like negative 1 . here , x equals 0 , y looks like 1 , so this is a candidate . here , x is equal to 0 , y also looks like 1 , so these last two seem like a candidate for this function definition right over here . now let 's think about the behavior of this function . let 's think about what happens when x approaches a very , very large number . when x approaches a very large number , let 's just imagine y of 1,000 , and 1,000 really is n't that large of a number , but let 's just ... that 's going to be 2 minus 1/3 to the 1,000th power . well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 . let me write this down . this part right over here is going to be close to 0 , close to 0 . another way to think about it is as x increases , this part approaches 0 . this is very close to 0 . this thing , y of 1,000 , y of 1,000 will be close to 2 . or another way of thinking about it , as x gets larger and larger and larger , this part is going to get closer and closer and closer to 0 , so you 're going to have 2 minus something that 's closer and closer and closer to 0 , so as x gets larger , y is going to approach 2 . which of these two have that behavior ? it 's clearly this one on the right . as x gets larger and larger and larger , we see that y is getting closer and closer and closer and closer to 2 . this one right over here , we could say , is y is equal to 2 minus 1/3 to the x power . we could also think about its behavior as x gets smaller , as x gets more and more and more negative . so 1/3 to a very negative number right over here , that 's the same thing as 3 to a very positive value . as x gets more and more and more negative , this will be essentially 3 to a more and more positive value and you 're subtracting it from 2 , so y is going to become more and more and more negative . we see that as x becomes more and more negative , y becomes more and more negative . this , once again , is consistent . now let 's think about this function right over here . here , we see y is equal to 1/2 to the x minus 2 . we could , first of all , just think about what y of 0 is . y of 0 is going to be 1/2 to the 0 power minus 2 , which is equal to 1 minus 2 , which is equal to negative 1 . both of these would be candidates for this function right over here . when x is equal to 0 , y is negative 1 . when x is equal to 0 , y is negative 1 . but now let 's think about the behavior of this function . as x becomes larger and larger and larger values , what is y going to approach ? just like we saw over here , you have a fraction . you have 1/2 being raised to larger and larger and larger values . let 's think about this . as this gets raised to larger and larger values , this part is going to approach 0 . 1/2 times 1/2 times 1/2 times 1/2 , that 's going to approach 0 fairly quickly . as this approaches 0 , y is going to approach negative 2 . as x gets larger and larger and larger , 1/2 to the x approaches 0 , and so y is going to approach negative 2 from above . let 's see where we see that . that looks like this one right over here . once again , we said these are our two candidates . when x is 0 , y is negative 1 . here , we see as x gets larger and larger and larger , y is approaching negative 2 because this part is becoming smaller and smaller and smaller values . this one is that one right over there . you could also think about its behavior as x becomes more and more negative . as x becomes more and more negative , that 's like raising 2 to a positive value , so 2 to a positive value , you see that as x becomes more negative , y becomes larger and larger and larger . all right , we got two left . y equals 2 to the x . this might be the simplest of all , y equaling 2 to the x power . when x is equal to 0 , y should be equal to 1 , and we see that 's this graph right over here , and this is your most basic type of exponential function . as x increases , y increases . this is your classic exponential graph shape . as x approaches more and more negative values , as x approaches more and more negative values , raising 2 to a very large negative value , so imagine when x is ... imagine y of negative 10 . this is n't even that negative . that 's going to be 2 to the negative 10 power , which is the same thing as 1/2 to the 10th power . as x becomes more and more and more negative , this expression is going to get closer and closer and closer to 0 . this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit . this is where order of operations really matter . when you see negative 3 to the x , it might be a little confusing . you 're , like , `` is it negative , `` is it the whole negative 3 to the x power , `` or is it negative 3 to the x ? '' here , we just have to remind ourselves order of operations , exponentials are the top priority right after a parentheses . you would actually do your exponential first . you take 3 to the x , and it 's going to be the negative of that . it 's essentially going to be , it 's going to be your classic exponential function , but because of this negative , you 're going to flip it over the x-axis , and that 's what this has right over here . as x gets larger and larger and larger , 3 to the x is going to become a much larger and larger value but then we 're taking the negative of it , so y is going to become smaller and smaller and smaller . likewise , as x is more and more and more negative , 3 to the x is going to approach 0 . when x is equal to 0 , 3 to the 0 is 1 , but you have the negative out front , we see y is negative 1 . this is y is equal to negative 3 to the x power .
|
this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit . this is where order of operations really matter .
|
what is the deductive reasoning in 4th graph ?
|
: we have 4 graphs here and then 4 function definitions . what i want you to do is pause this video and think about which of these graphs map up to which of these function definitions . i 'm assuming you 've given a go at it . let 's go through each of these and think about what their graphs would look like . one thing that i like to do , because it 's just a simple thing to do , is just think about what happens when x is equal to 0 . especially when x is an exponent like this , you would have , let me just write this down , y of 0 is going to be equal to 2 minus 1/3 to the 0 power . that 's equal to 2 minus 1 . 1/3 to the 0 is 1 , which is equal to 1 . in which of these graphs , when x is equal to 0 , do we have y equaling 1 ? here , x equals 0 , y equals negative 1 . here , x equals 0 , y equals negative 1 , or it looks like negative 1 . here , x equals 0 , y looks like 1 , so this is a candidate . here , x is equal to 0 , y also looks like 1 , so these last two seem like a candidate for this function definition right over here . now let 's think about the behavior of this function . let 's think about what happens when x approaches a very , very large number . when x approaches a very large number , let 's just imagine y of 1,000 , and 1,000 really is n't that large of a number , but let 's just ... that 's going to be 2 minus 1/3 to the 1,000th power . well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 . let me write this down . this part right over here is going to be close to 0 , close to 0 . another way to think about it is as x increases , this part approaches 0 . this is very close to 0 . this thing , y of 1,000 , y of 1,000 will be close to 2 . or another way of thinking about it , as x gets larger and larger and larger , this part is going to get closer and closer and closer to 0 , so you 're going to have 2 minus something that 's closer and closer and closer to 0 , so as x gets larger , y is going to approach 2 . which of these two have that behavior ? it 's clearly this one on the right . as x gets larger and larger and larger , we see that y is getting closer and closer and closer and closer to 2 . this one right over here , we could say , is y is equal to 2 minus 1/3 to the x power . we could also think about its behavior as x gets smaller , as x gets more and more and more negative . so 1/3 to a very negative number right over here , that 's the same thing as 3 to a very positive value . as x gets more and more and more negative , this will be essentially 3 to a more and more positive value and you 're subtracting it from 2 , so y is going to become more and more and more negative . we see that as x becomes more and more negative , y becomes more and more negative . this , once again , is consistent . now let 's think about this function right over here . here , we see y is equal to 1/2 to the x minus 2 . we could , first of all , just think about what y of 0 is . y of 0 is going to be 1/2 to the 0 power minus 2 , which is equal to 1 minus 2 , which is equal to negative 1 . both of these would be candidates for this function right over here . when x is equal to 0 , y is negative 1 . when x is equal to 0 , y is negative 1 . but now let 's think about the behavior of this function . as x becomes larger and larger and larger values , what is y going to approach ? just like we saw over here , you have a fraction . you have 1/2 being raised to larger and larger and larger values . let 's think about this . as this gets raised to larger and larger values , this part is going to approach 0 . 1/2 times 1/2 times 1/2 times 1/2 , that 's going to approach 0 fairly quickly . as this approaches 0 , y is going to approach negative 2 . as x gets larger and larger and larger , 1/2 to the x approaches 0 , and so y is going to approach negative 2 from above . let 's see where we see that . that looks like this one right over here . once again , we said these are our two candidates . when x is 0 , y is negative 1 . here , we see as x gets larger and larger and larger , y is approaching negative 2 because this part is becoming smaller and smaller and smaller values . this one is that one right over there . you could also think about its behavior as x becomes more and more negative . as x becomes more and more negative , that 's like raising 2 to a positive value , so 2 to a positive value , you see that as x becomes more negative , y becomes larger and larger and larger . all right , we got two left . y equals 2 to the x . this might be the simplest of all , y equaling 2 to the x power . when x is equal to 0 , y should be equal to 1 , and we see that 's this graph right over here , and this is your most basic type of exponential function . as x increases , y increases . this is your classic exponential graph shape . as x approaches more and more negative values , as x approaches more and more negative values , raising 2 to a very large negative value , so imagine when x is ... imagine y of negative 10 . this is n't even that negative . that 's going to be 2 to the negative 10 power , which is the same thing as 1/2 to the 10th power . as x becomes more and more and more negative , this expression is going to get closer and closer and closer to 0 . this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit . this is where order of operations really matter . when you see negative 3 to the x , it might be a little confusing . you 're , like , `` is it negative , `` is it the whole negative 3 to the x power , `` or is it negative 3 to the x ? '' here , we just have to remind ourselves order of operations , exponentials are the top priority right after a parentheses . you would actually do your exponential first . you take 3 to the x , and it 's going to be the negative of that . it 's essentially going to be , it 's going to be your classic exponential function , but because of this negative , you 're going to flip it over the x-axis , and that 's what this has right over here . as x gets larger and larger and larger , 3 to the x is going to become a much larger and larger value but then we 're taking the negative of it , so y is going to become smaller and smaller and smaller . likewise , as x is more and more and more negative , 3 to the x is going to approach 0 . when x is equal to 0 , 3 to the 0 is 1 , but you have the negative out front , we see y is negative 1 . this is y is equal to negative 3 to the x power .
|
well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 .
|
what would the graph be if the equation were the other way : y = ( -3 ) ^x ?
|
: we have 4 graphs here and then 4 function definitions . what i want you to do is pause this video and think about which of these graphs map up to which of these function definitions . i 'm assuming you 've given a go at it . let 's go through each of these and think about what their graphs would look like . one thing that i like to do , because it 's just a simple thing to do , is just think about what happens when x is equal to 0 . especially when x is an exponent like this , you would have , let me just write this down , y of 0 is going to be equal to 2 minus 1/3 to the 0 power . that 's equal to 2 minus 1 . 1/3 to the 0 is 1 , which is equal to 1 . in which of these graphs , when x is equal to 0 , do we have y equaling 1 ? here , x equals 0 , y equals negative 1 . here , x equals 0 , y equals negative 1 , or it looks like negative 1 . here , x equals 0 , y looks like 1 , so this is a candidate . here , x is equal to 0 , y also looks like 1 , so these last two seem like a candidate for this function definition right over here . now let 's think about the behavior of this function . let 's think about what happens when x approaches a very , very large number . when x approaches a very large number , let 's just imagine y of 1,000 , and 1,000 really is n't that large of a number , but let 's just ... that 's going to be 2 minus 1/3 to the 1,000th power . well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 . let me write this down . this part right over here is going to be close to 0 , close to 0 . another way to think about it is as x increases , this part approaches 0 . this is very close to 0 . this thing , y of 1,000 , y of 1,000 will be close to 2 . or another way of thinking about it , as x gets larger and larger and larger , this part is going to get closer and closer and closer to 0 , so you 're going to have 2 minus something that 's closer and closer and closer to 0 , so as x gets larger , y is going to approach 2 . which of these two have that behavior ? it 's clearly this one on the right . as x gets larger and larger and larger , we see that y is getting closer and closer and closer and closer to 2 . this one right over here , we could say , is y is equal to 2 minus 1/3 to the x power . we could also think about its behavior as x gets smaller , as x gets more and more and more negative . so 1/3 to a very negative number right over here , that 's the same thing as 3 to a very positive value . as x gets more and more and more negative , this will be essentially 3 to a more and more positive value and you 're subtracting it from 2 , so y is going to become more and more and more negative . we see that as x becomes more and more negative , y becomes more and more negative . this , once again , is consistent . now let 's think about this function right over here . here , we see y is equal to 1/2 to the x minus 2 . we could , first of all , just think about what y of 0 is . y of 0 is going to be 1/2 to the 0 power minus 2 , which is equal to 1 minus 2 , which is equal to negative 1 . both of these would be candidates for this function right over here . when x is equal to 0 , y is negative 1 . when x is equal to 0 , y is negative 1 . but now let 's think about the behavior of this function . as x becomes larger and larger and larger values , what is y going to approach ? just like we saw over here , you have a fraction . you have 1/2 being raised to larger and larger and larger values . let 's think about this . as this gets raised to larger and larger values , this part is going to approach 0 . 1/2 times 1/2 times 1/2 times 1/2 , that 's going to approach 0 fairly quickly . as this approaches 0 , y is going to approach negative 2 . as x gets larger and larger and larger , 1/2 to the x approaches 0 , and so y is going to approach negative 2 from above . let 's see where we see that . that looks like this one right over here . once again , we said these are our two candidates . when x is 0 , y is negative 1 . here , we see as x gets larger and larger and larger , y is approaching negative 2 because this part is becoming smaller and smaller and smaller values . this one is that one right over there . you could also think about its behavior as x becomes more and more negative . as x becomes more and more negative , that 's like raising 2 to a positive value , so 2 to a positive value , you see that as x becomes more negative , y becomes larger and larger and larger . all right , we got two left . y equals 2 to the x . this might be the simplest of all , y equaling 2 to the x power . when x is equal to 0 , y should be equal to 1 , and we see that 's this graph right over here , and this is your most basic type of exponential function . as x increases , y increases . this is your classic exponential graph shape . as x approaches more and more negative values , as x approaches more and more negative values , raising 2 to a very large negative value , so imagine when x is ... imagine y of negative 10 . this is n't even that negative . that 's going to be 2 to the negative 10 power , which is the same thing as 1/2 to the 10th power . as x becomes more and more and more negative , this expression is going to get closer and closer and closer to 0 . this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit . this is where order of operations really matter . when you see negative 3 to the x , it might be a little confusing . you 're , like , `` is it negative , `` is it the whole negative 3 to the x power , `` or is it negative 3 to the x ? '' here , we just have to remind ourselves order of operations , exponentials are the top priority right after a parentheses . you would actually do your exponential first . you take 3 to the x , and it 's going to be the negative of that . it 's essentially going to be , it 's going to be your classic exponential function , but because of this negative , you 're going to flip it over the x-axis , and that 's what this has right over here . as x gets larger and larger and larger , 3 to the x is going to become a much larger and larger value but then we 're taking the negative of it , so y is going to become smaller and smaller and smaller . likewise , as x is more and more and more negative , 3 to the x is going to approach 0 . when x is equal to 0 , 3 to the 0 is 1 , but you have the negative out front , we see y is negative 1 . this is y is equal to negative 3 to the x power .
|
as x becomes more and more and more negative , this expression is going to get closer and closer and closer to 0 . this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit .
|
3 why does y= ( -3 ) ^x graph go through the y axais at one ?
|
: we have 4 graphs here and then 4 function definitions . what i want you to do is pause this video and think about which of these graphs map up to which of these function definitions . i 'm assuming you 've given a go at it . let 's go through each of these and think about what their graphs would look like . one thing that i like to do , because it 's just a simple thing to do , is just think about what happens when x is equal to 0 . especially when x is an exponent like this , you would have , let me just write this down , y of 0 is going to be equal to 2 minus 1/3 to the 0 power . that 's equal to 2 minus 1 . 1/3 to the 0 is 1 , which is equal to 1 . in which of these graphs , when x is equal to 0 , do we have y equaling 1 ? here , x equals 0 , y equals negative 1 . here , x equals 0 , y equals negative 1 , or it looks like negative 1 . here , x equals 0 , y looks like 1 , so this is a candidate . here , x is equal to 0 , y also looks like 1 , so these last two seem like a candidate for this function definition right over here . now let 's think about the behavior of this function . let 's think about what happens when x approaches a very , very large number . when x approaches a very large number , let 's just imagine y of 1,000 , and 1,000 really is n't that large of a number , but let 's just ... that 's going to be 2 minus 1/3 to the 1,000th power . well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 . let me write this down . this part right over here is going to be close to 0 , close to 0 . another way to think about it is as x increases , this part approaches 0 . this is very close to 0 . this thing , y of 1,000 , y of 1,000 will be close to 2 . or another way of thinking about it , as x gets larger and larger and larger , this part is going to get closer and closer and closer to 0 , so you 're going to have 2 minus something that 's closer and closer and closer to 0 , so as x gets larger , y is going to approach 2 . which of these two have that behavior ? it 's clearly this one on the right . as x gets larger and larger and larger , we see that y is getting closer and closer and closer and closer to 2 . this one right over here , we could say , is y is equal to 2 minus 1/3 to the x power . we could also think about its behavior as x gets smaller , as x gets more and more and more negative . so 1/3 to a very negative number right over here , that 's the same thing as 3 to a very positive value . as x gets more and more and more negative , this will be essentially 3 to a more and more positive value and you 're subtracting it from 2 , so y is going to become more and more and more negative . we see that as x becomes more and more negative , y becomes more and more negative . this , once again , is consistent . now let 's think about this function right over here . here , we see y is equal to 1/2 to the x minus 2 . we could , first of all , just think about what y of 0 is . y of 0 is going to be 1/2 to the 0 power minus 2 , which is equal to 1 minus 2 , which is equal to negative 1 . both of these would be candidates for this function right over here . when x is equal to 0 , y is negative 1 . when x is equal to 0 , y is negative 1 . but now let 's think about the behavior of this function . as x becomes larger and larger and larger values , what is y going to approach ? just like we saw over here , you have a fraction . you have 1/2 being raised to larger and larger and larger values . let 's think about this . as this gets raised to larger and larger values , this part is going to approach 0 . 1/2 times 1/2 times 1/2 times 1/2 , that 's going to approach 0 fairly quickly . as this approaches 0 , y is going to approach negative 2 . as x gets larger and larger and larger , 1/2 to the x approaches 0 , and so y is going to approach negative 2 from above . let 's see where we see that . that looks like this one right over here . once again , we said these are our two candidates . when x is 0 , y is negative 1 . here , we see as x gets larger and larger and larger , y is approaching negative 2 because this part is becoming smaller and smaller and smaller values . this one is that one right over there . you could also think about its behavior as x becomes more and more negative . as x becomes more and more negative , that 's like raising 2 to a positive value , so 2 to a positive value , you see that as x becomes more negative , y becomes larger and larger and larger . all right , we got two left . y equals 2 to the x . this might be the simplest of all , y equaling 2 to the x power . when x is equal to 0 , y should be equal to 1 , and we see that 's this graph right over here , and this is your most basic type of exponential function . as x increases , y increases . this is your classic exponential graph shape . as x approaches more and more negative values , as x approaches more and more negative values , raising 2 to a very large negative value , so imagine when x is ... imagine y of negative 10 . this is n't even that negative . that 's going to be 2 to the negative 10 power , which is the same thing as 1/2 to the 10th power . as x becomes more and more and more negative , this expression is going to get closer and closer and closer to 0 . this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit . this is where order of operations really matter . when you see negative 3 to the x , it might be a little confusing . you 're , like , `` is it negative , `` is it the whole negative 3 to the x power , `` or is it negative 3 to the x ? '' here , we just have to remind ourselves order of operations , exponentials are the top priority right after a parentheses . you would actually do your exponential first . you take 3 to the x , and it 's going to be the negative of that . it 's essentially going to be , it 's going to be your classic exponential function , but because of this negative , you 're going to flip it over the x-axis , and that 's what this has right over here . as x gets larger and larger and larger , 3 to the x is going to become a much larger and larger value but then we 're taking the negative of it , so y is going to become smaller and smaller and smaller . likewise , as x is more and more and more negative , 3 to the x is going to approach 0 . when x is equal to 0 , 3 to the 0 is 1 , but you have the negative out front , we see y is negative 1 . this is y is equal to negative 3 to the x power .
|
well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 .
|
what if like the last equation `` y=-3^x '' is then `` y=-3^-x '' ?
|
: we have 4 graphs here and then 4 function definitions . what i want you to do is pause this video and think about which of these graphs map up to which of these function definitions . i 'm assuming you 've given a go at it . let 's go through each of these and think about what their graphs would look like . one thing that i like to do , because it 's just a simple thing to do , is just think about what happens when x is equal to 0 . especially when x is an exponent like this , you would have , let me just write this down , y of 0 is going to be equal to 2 minus 1/3 to the 0 power . that 's equal to 2 minus 1 . 1/3 to the 0 is 1 , which is equal to 1 . in which of these graphs , when x is equal to 0 , do we have y equaling 1 ? here , x equals 0 , y equals negative 1 . here , x equals 0 , y equals negative 1 , or it looks like negative 1 . here , x equals 0 , y looks like 1 , so this is a candidate . here , x is equal to 0 , y also looks like 1 , so these last two seem like a candidate for this function definition right over here . now let 's think about the behavior of this function . let 's think about what happens when x approaches a very , very large number . when x approaches a very large number , let 's just imagine y of 1,000 , and 1,000 really is n't that large of a number , but let 's just ... that 's going to be 2 minus 1/3 to the 1,000th power . well , 1/3 to the 1,000 power , that 's going to be a very , very , very small number . we 're multiplying 1/3 times 1/3 , 1/3 . you think of 1,000 1/3s and multiplying them together , i 'm sure you 're going to get a number very , very close to 0 . let me write this down . this part right over here is going to be close to 0 , close to 0 . another way to think about it is as x increases , this part approaches 0 . this is very close to 0 . this thing , y of 1,000 , y of 1,000 will be close to 2 . or another way of thinking about it , as x gets larger and larger and larger , this part is going to get closer and closer and closer to 0 , so you 're going to have 2 minus something that 's closer and closer and closer to 0 , so as x gets larger , y is going to approach 2 . which of these two have that behavior ? it 's clearly this one on the right . as x gets larger and larger and larger , we see that y is getting closer and closer and closer and closer to 2 . this one right over here , we could say , is y is equal to 2 minus 1/3 to the x power . we could also think about its behavior as x gets smaller , as x gets more and more and more negative . so 1/3 to a very negative number right over here , that 's the same thing as 3 to a very positive value . as x gets more and more and more negative , this will be essentially 3 to a more and more positive value and you 're subtracting it from 2 , so y is going to become more and more and more negative . we see that as x becomes more and more negative , y becomes more and more negative . this , once again , is consistent . now let 's think about this function right over here . here , we see y is equal to 1/2 to the x minus 2 . we could , first of all , just think about what y of 0 is . y of 0 is going to be 1/2 to the 0 power minus 2 , which is equal to 1 minus 2 , which is equal to negative 1 . both of these would be candidates for this function right over here . when x is equal to 0 , y is negative 1 . when x is equal to 0 , y is negative 1 . but now let 's think about the behavior of this function . as x becomes larger and larger and larger values , what is y going to approach ? just like we saw over here , you have a fraction . you have 1/2 being raised to larger and larger and larger values . let 's think about this . as this gets raised to larger and larger values , this part is going to approach 0 . 1/2 times 1/2 times 1/2 times 1/2 , that 's going to approach 0 fairly quickly . as this approaches 0 , y is going to approach negative 2 . as x gets larger and larger and larger , 1/2 to the x approaches 0 , and so y is going to approach negative 2 from above . let 's see where we see that . that looks like this one right over here . once again , we said these are our two candidates . when x is 0 , y is negative 1 . here , we see as x gets larger and larger and larger , y is approaching negative 2 because this part is becoming smaller and smaller and smaller values . this one is that one right over there . you could also think about its behavior as x becomes more and more negative . as x becomes more and more negative , that 's like raising 2 to a positive value , so 2 to a positive value , you see that as x becomes more negative , y becomes larger and larger and larger . all right , we got two left . y equals 2 to the x . this might be the simplest of all , y equaling 2 to the x power . when x is equal to 0 , y should be equal to 1 , and we see that 's this graph right over here , and this is your most basic type of exponential function . as x increases , y increases . this is your classic exponential graph shape . as x approaches more and more negative values , as x approaches more and more negative values , raising 2 to a very large negative value , so imagine when x is ... imagine y of negative 10 . this is n't even that negative . that 's going to be 2 to the negative 10 power , which is the same thing as 1/2 to the 10th power . as x becomes more and more and more negative , this expression is going to get closer and closer and closer to 0 . this one is clearly that one . then finally , just deductive reasoning , you could say that this function is represented by that graph , but let 's reason through a little bit . this is where order of operations really matter . when you see negative 3 to the x , it might be a little confusing . you 're , like , `` is it negative , `` is it the whole negative 3 to the x power , `` or is it negative 3 to the x ? '' here , we just have to remind ourselves order of operations , exponentials are the top priority right after a parentheses . you would actually do your exponential first . you take 3 to the x , and it 's going to be the negative of that . it 's essentially going to be , it 's going to be your classic exponential function , but because of this negative , you 're going to flip it over the x-axis , and that 's what this has right over here . as x gets larger and larger and larger , 3 to the x is going to become a much larger and larger value but then we 're taking the negative of it , so y is going to become smaller and smaller and smaller . likewise , as x is more and more and more negative , 3 to the x is going to approach 0 . when x is equal to 0 , 3 to the 0 is 1 , but you have the negative out front , we see y is negative 1 . this is y is equal to negative 3 to the x power .
|
as x increases , y increases . this is your classic exponential graph shape . as x approaches more and more negative values , as x approaches more and more negative values , raising 2 to a very large negative value , so imagine when x is ... imagine y of negative 10 .
|
how would the graph change ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi .
|
could somebody walk me through a detailed explanation of this problem ; what is the principal value of sin^-1 ( -1/2 ) ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants .
|
what is the difference between range and domain ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 .
|
how does sal know the triangle is a 30-60-90 ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process .
|
is the relationship between arccos and cos the same as the relationship between logarithms and exponents ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example .
|
how did sal know what are the domain and range of arccos x ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ?
|
why does sal take theta as an angle outside the triangle and not as 60 itself ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi .
|
if the range of arccos is restricted between 0 and pi , how can we get the arccos of 3pi ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top .
|
why is the cosine restricted to the upper quadrants and the sine restricted to th right quadrants and tangent as well ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process .
|
how does sam decide the restrictions ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 .
|
how is the x-value of minus 1/2 determined when sal makes a point on that specific part of the circle ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 .
|
how does he know that the triangle in `` '' is a 30 60 90 triangle ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range .
|
is the arccos function the same as secant function ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down .
|
why is the angle supplementary to 180 and not 360 ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about .
|
how does sal know that theta is the larger angle on the right and not the smaller angle on the left ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 .
|
how does sal know it is a 30 , 60 , 90 triangle ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
and that 's my axes . what 's 3 pi ? 2 pi is if i go around once .
|
5 , what if there is bigger angle like 100 pi instead of 3pi ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
so algebraically speaking , sin ( arcsin x ) = sin ( 1/sin x ) and the sins ' cancel out leaving only x ?
|
i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so i 've already made videos on the arcsine and the arctangent , so to kind of complete the trifecta , i might as well make a video on the arccosine . and just like the other inverse trigonometric functions , the arccosine is kind of the same thought process . if i were to tell you the arc , no , i 'm doing cosine , if our tell you that arccosine of x is equal to theta . this is an equivalent statement to saying that the inverse cosine of x is equal to theta . these are just two different ways of writing the exact same thing . and as soon as i see either an arc- anything , or an inverse trig function in general , my brain immediately rearranges this . my brain immediately says , this is saying that if i take the cosine of some angle theta , that i 'm going to get x . or that same statement up here . either of these should boil down to this . if i say , you know , what is the inverse cosine of x , my brain says , what angle can i take the cosine of to get x ? so with that said , let 's try it out on an example . let 's say that i have the arc , i 'm told , no , two c 's there , i 'm told to evaluate the arccosine of minus 1/2 . my brain , you know , let 's say that this is going to be equal to , it 's going to be equal to some angle . and this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2 . and as soon as you put it in this way , at least for my brain , it becomes a lot easier to process . so let 's draw our unit circle and see if we can make some headway here . so that 's my , let me see if i can draw a little straighter . maybe i could actually draw , put rulers here , and if i put a ruler here , maybe i can draw a straight line . let me see . no , that 's too hard . ok , so that is my y-axis , that is my x-axis . not the most neatly drawn axes ever , but it 'll do . let me draw my unit circle . looks more like a unit ellipse , but you get the idea . and the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle . so if we have some angle , the x-value is going to be equal a minus 1/2 . so we got a minus 1/2 right here . and so the angle that we have to solve for , our theta , is the angle that when we intersect the unit circle , the x-value is minus 1/2 . so let me see , this is the angle that we 're trying to figure out . this is theta that we need to determine . so how can we do that ? so this is minus 1/2 right here . let 's figure out these different angles . and the way i like to think about it is , i like to figure out this angle right here . and if i know that angle , i can just subtract that from 180 degrees to get this light blue angle that 's kind of the solution to our problem . so let me make this triangle a little bit bigger . so that triangle , let me do it like this . that triangle looks something like this . where this distance right here is 1/2 . that distance right there is 1/2 . this distance right here is 1 . hopefully you recognize that this is going to be a 30 , 60 , 90 triangle . you could actually solve for this other side . you 'll get the square root of 3 over 2 . and to solve for that other side you just need to do the pythagorean theorem . actually , let me just do that . let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 . so you immediately know this is a 30 , 60 , 90 triangle . and you know that because the sides of a 30 , 60 , 90 triangle , if the hypotenuse is 1 , are 1/2 and square root of 3 over 2 . and you also know that the side opposite the square root of 3 over 2 side is 60 degrees . that 's 60 , this is 90 . this is the right angle , and this is 30 right up there . but this is the one we care about . this angle right here we just figured out is 60 degrees . so what 's this ? what 's the bigger angle that we care about ? what is 60 degrees supplementary to ? it 's supplementary to 180 degrees . so the arccosine , or the inverse cosine , let me write that down . the arccosine of minus 1/2 is equal to 120 degrees . did i write 180 there ? no , it 's 180 minus the 60 , this whole thing is 180 , so this is , right here is , 120 degrees , right ? 120 plus 60 is 180 . or , if we wanted to write that in radians , you just right 120 degrees times pi radian per 180 degrees , degrees cancel out . 12 over 18 is 2/3 , so it equals 2 pi over 3 radians . so this right here is equal to 2 pi over 3 radians . now , just like we saw in the arcsine and the arctangent videos , you probably say , hey , ok , if i have 2 pi over 3 radians , that gives me a cosine of minus 1/2 . and i can write that . cosine of 2 pi over 3 is equal to minus 1/2 . this gives you the same information as this statement up here . but i can just keep going around the unit circle . for example , i could , how about this point over here ? cosine of this angle , if i were to add , if i were to go this far , would also be minus 1/2 . and then i could go 2 pi around and get back here . so there 's a lot of values that if i take the cosine of those angles , i 'll get this minus 1/2 . so we have to restrict ourselves . we have to restrict the values that the arccosine function can take on . so we 're essentially restricting it 's range . we 're restricting it 's range . what we do is we restrict it 's range to this upper hemisphere , the first and second quadrants . so if we say , if we make the statement that the arccosine of x is equal to theta , we 're going to restrict our range , theta , to that top . so theta is going to be greater than or equal to 0 and less than or equal to 2 pi . less , oh sorry , not 2 pi . less than or equal to pi , right ? where this is also 0 degrees , or 180 degrees . we 're restricting ourselves to this part of the hemisphere right there . and so you ca n't do this , this is the only point where the cosine of the angle is equal minus 1/2 . we ca n't take this angle because it 's outside of our range . and what are the valid values for x ? well any angle , if i take the cosine of it , it can be between minus 1 and plus 1 . so x , the domain for the arccosine function , is going to be x has to be less than or equal to 1 and greater than or equal to minus 1 . and once again , let 's just go check our work . let 's see if the value i got here , that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the ti-85 . we turn it on . so i need to figure out the inverse cosine , which is the same thing as the arccosine of minus 1/2 , of minus 0.5 . it gives me that decimal , that strange number . let 's see if that 's the same thing as 2 pi over 3 . 2 times pi divided by 3 is equal to , that exact same number . so the calculator gave me the same value i got . but this is kind of a useless , well , it 's not a useless number . it 's a valid , that is the answer . but it does n't , it 's not a nice clean answer . i did n't know that this is 2 pi over 3 radians . and so when we did it using the unit circle , we were able to get that answer . so hopefully , actually let me ask you , let me just finish this up with an interesting question . and this applies to all of them . if i were to ask you , you know , say i were to take the arccosine of x , and then i were to take the cosine of that , what is this going to be equal to ? well , this statement right here can be said , well , let 's say that the arccosine of x is equal to theta , that means that the cosine of theta is equal to x , right ? so if the arccosine of x is equal to theta , we can replace this with theta . and then the cosine of theta , well the cosine of theta is x . so this whole thing is going to be x. hopefully i did n't get confuse you there , right ? i 'm saying look , arccosine of x , just call that theta . now , by definition , this means that the cosine of theta is equal to x . these are equivalent statements . these are completely equivalent statements right here . so if we put a theta right there , we take the cosine of theta , it has to be equal to x . now let me ask you a bonus , slightly trickier question . what if i were to ask you , and this is true for any x that you put in here . this is true for any x , any value between negative 1 and 1 including those two endpoints , this is going to be true . now what if i were ask you what the arccosine of the cosine of theta is ? what is this going to be equal to ? my answer is , it depends on the theta . so , if theta is in the , if theta is in the range , if theta is between , if theta is between 0 and pi , so it 's in our valid a range for , kind of , our range for the product of the arccosine , then this will be equal to theta . if this is true for theta . but what if we take some theta out of that range ? let 's try it out . let 's take , so let me do one with theta in that range . let 's take the arccosine of the cosine of , let 's just do one of them that we know . let 's take the cosine of , let 's stick with cosine of 2 pi over 3 . cosine of 2 pi over 3 radians , that 's the same thing as the arccosine of minus 1/2 . cosine of 2 pi over 3 is minus 1/2 . we just saw that in the earlier part of this video . and then we solved this . we said , oh , this is equal to 1 pi over 3 . so for in the range of thetas between 0 and pi it worked . and that 's because the arccosine function can only produce values between 0 and pi . but what if i were to ask you , what is the arccosine of the cosine of , i do n't know , of 3 pi . so if i were to draw the unit circle here , let me draw the unit circle , a real quick one . and that 's my axes . what 's 3 pi ? 2 pi is if i go around once . and then i go around another pi , so i end up right here . so i 've gone around 1 1/2 times the unit circle . so this is 3 pi . what 's the x-coordinate here ? it 's minus 1 . so cosine of 3 pi is minus 1 , right ? so what 's arccosine of minus 1 ? arccosine of minus 1 . well remember , the range , or the set of values , that arccosine can evaluate to is in this upper hemisphere . it 's between , this can only be between pi and 0 . so arccosine of negative 1 is just going to be pi . so this is going to be pi . arccosine of negative , this is negative 1 , arccosine of negative 1 is pi . and that 's a reasonable statement , because the difference between 3 pi and pi is just going around the unit circle a couple of times . and so you get an equivalent , it 's kind of , you 're at the equivalent point on the unit circle . so i just thought i would throw those two at you . this one , i mean this is a useful one . well , actually , let me write it up here . this one is a useful one . the cosine of the arccosine of x is always going to be x. i could also do that with sine . the sine of the arcsine of x is also going to be x . and these are just useful things to , you should n't just memorize them , because obviously you might memorize it the wrong way , but you should just think a little bit about it , and you 'll never forget it .
|
let me just call this , i do n't know , just call this a . so you 'd get a squared , plus 1/2 squared , which is 1/4 , which is equal to 1 squared , which is 1 . you get a squared is equal to 3/4 , or a is equal to the square root of 3 over 2 .
|
is the value of sin and cos is always between -1 and 1 because the ratio of opp/hyp or adj/hyp is always equal or less than 1 ?
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.