| "text": "The 30°-60°-90° triangle. Topics in trigonometry.\nThe Topics | Home\n8\nTHE 30°-60°-90° TRIANGLE\nTHERE ARE TWO special triangles in trigonometry. One is the 30°-60°-90° triangle. The other is the isosceles right triangle. They are special because with simple geometry we can know the ratios of their sides, and therefore solve any such triangle.\nTheorem. In a 30°-60°-90° triangle the sides are in the ratio\n1 : 2 : .\nWe will prove that below.\nNote that the smallest side, 1, is opposite the smallest angle, 30°; while the largest side, 2, is opposite the largest angle, 90°. (Theorem 6). (For, 2 is larger than . Also, while 1 : : 2 correctly corresponds to the sides opposite 30°-60°-90°, many find the sequence 1 : 2 :\neasier to remember.)\nThe cited theorems\nare from the Appendix, Some theorems of plane geometry.\nHere are examples of how we\ntake advantage of knowing those ratios. First, we can evaluate the\nfunctions of 60° and 30°.\nExample 1. Evaluate cos 60°.\nAnswer. For any problem involving a 30°-60°-90° triangle, the student should not use a table. The student should sketch the triangle and place the ratio numbers.\nSince the cosine is the ratio of the adjacent side to the hypotenuse, we can see that\ncos 60° = ½.\nExample 2. Evaluate sin 30°.\nAnswer. According to the property of cofunctions, sin 30° is equal to cos 60°. sin 30° = ½.\nOn the other hand, you can see that directly in the figure above.\nProblem 1. Evaluate sin 60° and tan 60°.\nTo see the answer, pass your mouse over the colored area. To cover the answer again, click \"Refresh\" (\"Reload\").\nThe sine is the ratio of the opposite side to the hypotenuse.\nsin 60° =\n2\n= ½.\nLesson 5 of Algebra\nThe tangent is ratio of the opposite side to the adjacent.\ntan 60° =\n1\n= .\nProblem 2. Evaluate cot 30° and cos 30°.\nThe cotangent is the ratio of the adjacent side to the opposite.\nTherefore, on inspecting the figure above, cot 30° =\n1\n= .\nOr, more simply, cot 30° = tan 60°.\nProblem 1\nAs for the cosine, it is the ratio of the adjacent side to the hypotenuse. Therefore,\ncos 30° =\n2\n= ½.\nBefore we come to the next Example, here is how we relate the sides and angles of a triangle:\nIf an angle is labeled capital A, then the side opposite will be labeled small a. Similarly for angle B and side b, angle C and side c.\nExample 3. Solve the right triangle ABC if angle A is 60°, and side AB is 10 cm.\nSolution. To solve a triangle means to know all three sides and all three angles. Since this is a right triangle and angle A is 60°, then the remaining angle B is its complement, 30°.\nAgain, in every 30°-60°-90° triangle, the sides are in the ratio 1 : 2 : , as shown on the left.\nWhen we know the ratios of the sides, then to solve a triangle we do not require the trigonometric functions or the Pythagorean theorem. We can solve it by the method of similar figures.\nNow, the sides that make the equal angles are in the same ratio. Proportionally,\n2 : 1 = 10 : AC.\n2 is two times 1. Therefore 10 is two times AC. AC is 5 cm.\nThe side adjacent to 60°, we see, is always half the hypotenuse.\nAs for BC—proportionally,\n2 :\n= 10 : BC.\nTo produce 10, 2 has been multiplied by 5. Therefore,\nwill also be multiplied by 5. BC is 5\ncm.\nIn other words, since one side of the standard triangle has been multiplied by 5, then every side will be multiplied by 5.\n1 : 2 :\n= 5 : 10 : 5.\nCompare Example 11 here.\nAgain: When we know the ratio numbers, then to solve the triangle the student should use this method of similar figures,\nnot the trigonometric functions.\n(In Topic 10, we will solve right triangles whose ratios of sides we do not know.)\nProblem 3. In the right triangle DFE, angle D is 30° and side DF is 3 inches. How long are sides d and f ?\nThe student should draw a similar triangle in the same orientation. Then see that the side corresponding to\nwas multiplied by .\nLesson 26 of Algebra\nTherefore, each side will be multiplied by . Side d\nwill be 1 = . Side f will be 2.\nProblem 4. In the right triangle PQR, angle P is 30°, and side r is 1 cm. How long are sides p and q ?\nThe side corresponding to 2 has been divided by 2. Therefore, each side must be divided by 2. Side p will be ½, and side q will be ½.\nProblem 5. Solve the right triangle ABC if angle A is 60°, and the hypotenuse is 18.6 cm.\nThe side adjacent to 60° is always half of the hypotenuse -- therefore, side b is 9.3 cm.\nBut this is the side that corresponds to 1. And it has been multiplied by 9.3. Therefore, side a will be multiplied by 9.3. It will be 9.3 cm.\nProblem 6. \tProve: The area A of an equilateral triangle whose side is s, is\nA = ¼s2.\nThe area A of any triangle is equal to one-half the sine of any angle times the product of the two sides that make the angle. (Topic 2, Problem 6.)\nIn an equilateral triangle each side is s , and each angle is\n60°. Therefore,\nA = ½ sin 60°s2.\nSince sin 60° = ½,\nProblem 1\nA = ½· ½ s2 = ¼s2.\nProblem 7. \tProve: The area A of an equilateral triangle inscribed in a circle of radius r, is\nA =\n34\nr2.\nThe three radii divide the triangle into three congruent triangles.\nSide-side-side\nHence each radius bisects each vertex into two 30° angles. If we extend the radius AO, then AD is the perpendicular bisector of the side CB.\nTheorem 2\nTriangle OBD is therefore a 30-60-90 triangle. If we call each side of the equilateral triangle s, then in the right triangle OBD,\n½s r\n=\ncos 30° = ½.\nTherefore,\ns = r\nso that\ns2 = 3r2.\nNow,\nthe area\nA of an equilateral triangle is\nA = ¼s2.\nProblem 6\nTherefore,\nA = ¼s2 =\n¼· 3r2\n=\n34\nr2.\nThat is what we wanted to prove.\nProblem 8. Prove: The angle bisectors of an equilateral triangle meet at a point that is two thirds of the distance from the vertex of the triangle to the base.\nLet ABC be an equilateral triangle, let AD, BF, CE be the angle bisectors of angles A, B, C respectively; then those angle bisectors meet at the point P such that AP is two thirds of AD.\nFirst, triangles BPD, APE are congruent.\nFor, since the triangle is equilateral and BF, AD are the angle bisectors, then angles PBD, PAE are equal and each\n30°;and the side BD is equal to the side AE, because in an equilateral triangle the angle bisector is the perpendicular bisector of the base.\nTheorem 2\nAngles PDB, AEP then are right angles and equal.\nTherefore,\nAngle-side-angle\ntriangles BPD, APE are congruent.\nNow,\nBPPD\n= csc 30° =\n2.\nProblem 2\nTherefore, BP = 2PD.\nBut AP = BP, because triangles APE, BPD are conguent, and those are the sides opposite the equal angles.\nTherefore, AP = 2PD.\nTherefore AP is two thirds of the whole AD.\nWhich is what we wanted to prove.\nThe proof\nHere is the proof that in a 30°-60°-90° triangle\nthe sides are in the ratio 1 : 2 : . It is based on the fact that a 30°-60°-90° triangle is half of an equilateral triangle.\nDraw the equilateral triangle ABC. Then each of its equal angles is 60°. (Theorems 3 and 9)\nDraw the straight line AD bisecting the angle at A into two 30° angles. Then AD is the perpendicular bisector of BC (Theorem 2). Triangle ABD\ntherefore is a 30°-60°-90° triangle.\nNow, since BD is equal to DC, then BD is half of BC.\nThis implies that BD is also half of AB, because AB is equal to BC. That is,\nBD : AB = 1 : 2\nFrom the Pythagorean theorem, we can find the third side AD:\nAD2 + 12\n=\n22\nAD2\n=\n4 − 1 = 3\nAD\n=\n.\nTherefore in a 30°-60°-90° triangle the sides are in the ratio 1 : 2 : ; which is what we set out to prove.\nCorollary.\nThe square drawn on the height of an equalateral triangle is three fourths of the square drawn on the side.\nNext Topic: The Isosceles Right Triangle\nThe Topics | Home\nPlease make a donation to keep TheMathPage online.Even $1 will help.\nCopyright © 2022\nLawrence Spector\nQuestions or comments?\nE-mail: teacher@themathpage.com",
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