2.6 Applying the Trigonometric Ratios

Transcription

1 2.6 Applying the Trigonometric atios FOCUS Use trigonometric ratios to solve a right triangle. When we solve a triangle, we find the measures of all the angles and the lengths of all the sides. To do this we use any of the sine, cosine, and tangent ratios. A side to A C B side opposite A opposite sin A cos A tan A opposite We can use the acronym SOH-CAH-TOA to help us remember these ratios. Example 1 Finding the Measures of All Angles Find all unknown angle measures to the nearest degree. E 5 cm 9 cm D F Solution Find the measure of D. EF is the side opposite D. DE is the side to D. So, use the tangent ratio. tan D EF tan D Substitute: EF 9 and DE 5 DE 9 tan D 5 D 61 The acute angles in a right triangle have a sum of 90. So, F 90 D F F 29 opposite We could have used the tangent ratio to find F Copyright 2011 Pearson Canada Inc.

2 Check 1. Find all unknown angle measures to the nearest degree. a) 10.5 cm H 6.4 cm G G J G Find the measure of G. HJ is the side. GH is the side. So, use the ratio. We could use the tangent ratio to check the measure of J. G G The acute angles have a sum of 90. So, J 90 J 90 J b) 7.2 cm N K M 7.8 cm K K Find the measure of K. MN is the side K. NK is the. So, use the ratio. K K The acute angles have a sum of 90. So, N 90 N 90 N We could use the cosine ratio to check the measure of N Copyright 2011 Pearson Canada Inc. 109

3 Example 2 Finding the Lengths of All Sides Find all unknown side lengths to the nearest tenth of a metre. P m Solution Find the length of P. P is the side to. is the. So, use the cosine ratio. cos P cos Substitute: 76 and 24.3 P cos 76 Multiply both sides by cos 76 P P P is about 5.9 m long. P side to 76 side opposite 24.3 m Find the length of P. P is the side opposite. is the. So, use the sine ratio. opposite sin P sin Substitute: 76 and 24.3 P sin 76 Multiply both sides by sin 76 P We could use the P Pythagorean Theorem to P is about 23.6 m long. check the side lengths Copyright 2011 Pearson Canada Inc.

4 Check 1. Find all unknown side lengths to the nearest tenth of a metre. T 18.4 m S 43 U Find the length of ST. TU is the side S. ST is the. So, use the ratio. Find the length of SU. TU is the side S. SU is the side S. S S S S ST ST is about long. SU SU is about long Copyright 2011 Pearson Canada Inc. 111

5 Example 3 Solving a Triangle Solve this triangle. Give angle measures to the nearest degree. Give side lengths to the nearest tenth of a centimetre cm B 19.2 cm A Solution C Find the measure of B. AB is the side to B. BC is the. So, use the cosine ratio. cos B AB cos B Substitute: AB 12.2 and BC 19.2 BC 12.2 cos B 19.2 B 51 The acute angles in a right triangle have a sum of 90. So, C 90 B C C 39 We know the lengths of two sides. Find the length of AC. Use the Pythagorean Theorem to find AC. AC 2 BC 2 AB 2 AC AC AC AC AC is about 14.8 cm long. We could have used the sine ratio and the exact measure of B to find AC Copyright 2011 Pearson Canada Inc.

6 Check 1. Solve this triangle. Give side lengths to the nearest tenth of a centimetre. D F cm E We know the length of one side and the measure of one acute angle. The acute angles have a sum of. So, E E E Find the length of DF. DF is the side F. EF is the. So, use the ratio. Find the length of DE. DE is the side F. EF is the. So, use the ratio. F F F F DF is about long. DE is about long. Practice 1. Which ratio would you use to find the measure of each angle? a) P b) E 18.6 cm C D 4.8 cm E 7.3 cm emember the acronym SOH CAH TOA. P 13.4 cm is the side. P is the side. DE is the. CE is the Copyright 2011 Pearson Canada Inc. 113

7 2. Which ratio would you use to find the length of each indicated side? a) GH b) MN H 2.5 cm F N 6.3 cm 51 G M 21 P HF is the. GH is the. MN is the. NP is the. 3. Find all unknown angle measures to the nearest degree. a) V b) 4.2 cm U 17.6 cm 13.1 cm 5.5 cm W S Find the measure of U. Find the measure of. U U U U The acute angles have a sum of 90. The acute angles have a sum of 90. So, W 90 W 90 W So, Copyright 2011 Pearson Canada Inc.

8 4. Find all unknown side lengths to the nearest tenth of a centimetre. 8.7 cm 48 S Find the length of S. Find the length of S. S S S S S is about long. S is about long. 5. Solve this triangle. Give angle measures to the nearest degree. Give side lengths to the nearest tenth of a centimetre. D 13.8 cm 21.5 cm E C Find the measure of E. Find the length of CD. Use the Pythagorean Theorem. E E E E CD is about long. The acute angles have a sum of 90. So, C 90 C 90 C Copyright 2011 Pearson Canada Inc. 115

9 6. The base of a ladder is on level ground 1.9 m from a wall. The ladder leans against the wall. The angle between the ladder and the ground is 65. a) How far up the wall does the ladder reach? b) How long is the ladder? Give your answers to the nearest tenth of a metre. F H m G a) The ladder reaches the wall at a height of about. b) The ladder is about long Copyright 2011 Pearson Canada Inc.