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Trigonometry Sec. MathHands.com Márquez Solving Right Triangles Solving Right Triangles This is an important milestone in our trigonometric journey. Solving triangles is one if not the central question in trigonometry. We have already had some success at this using, pythagoras, the 30-60 Theorem, 45-45 Theorems, the almost-triangles, and the 23-67 Theorem. We are now ready to fulfill one of the great trigonometric promises, to measure buildings, trees, mountains, starts, planets, anything by just looking at it... and measuring an angle or twoa . This can be done in many ways. We will illustrate two, one with just common sense and another solving equations. The general idea is to compare a known side with one that is unknown. The unknown side can be determined so long as the ration is known. For example, recall that on a 30-60 Triangle, the small to hypothenuse ratio is 50%, and suppose the hypothenuse is known to be 14 ft, then since the ratio is 50% the small one must be 7 feet. We are now ready to repeat this sort of reasoning, not just for 30-60 or 45-45 but for all right triangles. Example 1 Example 2 The known side is the hypothenuse, 14ft., the unknow side is the opposite of 30◦ , s. The function that describes the opposite to hypothnuse ratio is the sine function. It says that the sin 30◦ = .5, i.e. the opposite to hyp ratio is 50%, thus the opposite is 50% of 14 ft., which is 7ft. Using the equations the solution would look something like this: sin 30◦ 14ft 15 ◦ s .5 = = .5(14 f t) = 7 ft = s 14 f t s 14 f t s The known side is the side adjacent to the angle 15◦ , 8 ft., the unknow side is the opposite side, s. The function that describes the opposite to adjacent ratio is the tangent function. The tangent function says that the opposite to adjacent ration for such a triangle is tan 15◦ ≈ .278 = 27.8%, thus the opposite is about 27.8% of 8 ft., which is approximately 2.2 ft. Using the equations the solution would look something like this: tan 15◦ = s 15 ◦ .278 ≈ 8 ft .278(8 f t) ≈ s 2.2 f t ≈ Example 3 15 ◦ cos 15◦ 8 s Example 4 The known side is adjacent to the angle 15◦ , 8 ft., the unknow side is the hypothenuse, r. The function that describes the ratio between adjacent and hypothenuse is the cosine function. It says that the cos 15◦ ≈ .966, i.e. the adjacent side is approximately 96.6% of the hypothenuse. Thus the hypothenuse is 8ft divided by 96.6% which is approximately 8.3 ft. Using the equations the solution would look something like this: r s 8 ft s 8 ft s = .966 ≈ (.966)(s) ≈ s ≈ 8 s 8 s 8 8(.966) = 8.3 This time we look at the same triangle as example 3, but from an diferent point of view. This time we look at it from the other corner, the 75◦ -corner. From here, the known side is the side opposite, while the unknown is still the hypothenuse. The sine function was born to describe such a ratio, opp/hyp. Note we should expect the same answer since we calculated it in example 3. r 8 ft 75 ◦ sin 75◦ r sin 75◦ 8 r = 8 = r = r ≈ 8 sin 75◦ 8 = 8.3 .966 a some artistic liberty was taken in making this statement c 2007 MathHands.com c 2007 MathHands.com Trigonometry Sec. MathHands.com Márquez 1. Solve ALL sides of the right triangle. y r y r 4.5 39◦ 5 25◦ 2. Solve ALL sides of the right triangle. r y r 7. 5 x 10◦ 7 55◦ 3. Solve ALL sides of the right triangle. r 8. 5 x 20◦ y r 8 15◦ 4. Solve ALL sides of the right triangle. r 9. 2 x 70◦ y r 15 17◦ r 12 5. Solve ALL sides of the right triangle. 10. y r 80◦ x 37◦ 10 6. Solve ALL sides of the right triangle. r 11. 50◦ 15 x c 2007 MathHands.com Trigonometry Sec. MathHands.com Márquez r 12. 17 x 23◦ river 17◦ 25◦ 10 yd. r 13. 49◦ 8 16. (a) Determine the distance from the tree trunk to the opposite side of the river as shown. x 14. Just by looking at him, can you figure out how big Diego’s head is? See picture for angles & measurements. river 20◦ 23◦ 5 yd. 40◦ 49◦ (b) (con’t from previous exercise) How far away is the tree from the river? Use the diagram below. 2 in. d river 18 ◦ 15. Determine the distance from the tree trunk to the opposite side of the river as shown. c 2007 MathHands.com