7.7– Solve Right Triangles

Inverse Trig Ratios: To find the measure of an angle, take the inverse of the trig function. A 7 . 10 sin A° = C B A° = sin-1 (7/10) BC AB = mA° sin-1 A° = 44.4° AC AB = mA° cos-1 BC AC = mA° tan-1

Steps to solving with inverse trig: 1. 2. Label all sides with reference angle What two sides do you know?

30 tan A° = 20 A H A° = tan-1 (30/20) A° = 56.3° O SOH – CAH – TOA 1. Use a calculator to approximate the measure of A to the nearest tenth of a degree. 30 20 tan A° = A H A° = tan-1 (30/20) A° = 56.3° O SOH – CAH – TOA

O 14 sin A° = 26 A A° = sin-1 (14/26) H A° = 32.6° SOH – CAH – TOA 1. Use a calculator to approximate the measure of A to the nearest tenth of a degree. O 14 26 sin A° = A A° = sin-1 (14/26) H A° = 32.6° SOH – CAH – TOA

O 10 cos A° = 16 A A° = cos-1 (10/16) H A° = 51.3° SOH – CAH – TOA 1. Use a calculator to approximate the measure of A to the nearest tenth of a degree. O 10 16 cos A° = A A° = cos-1 (10/16) H A° = 51.3° SOH – CAH – TOA

H O SOH – CAH – TOA A P PQ QR x . 22 y . 22 sin 37° = cos 37° = 2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places. H O SOH – CAH – TOA A P PQ QR x . 22 180 – 90 – 37 y . 22 sin 37° = cos 37° = mP = 53° 22  sin 37° = x 22  cos 37° = y 17.57 = y 13.24 = x

O SOH – CAH – TOA A H ST U UT x . 15 y . 15 cos 70° = sin 70° = 2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places. O SOH – CAH – TOA A H ST U UT x . 15 180 – 90 – 70 y . 15 cos 70° = sin 70° = mU = 20° 15  cos 70° = x 15  sin 70° = y 5.13 = x 14.10 = y

O A SOH – CAH – TOA H P N PN 17 12 c2 = a2 + b2 tan P° = 2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places. O A SOH – CAH – TOA H P N PN 17 12 c2 = a2 + b2 tan P° = 180 – 90 – 54.8 c2 = 122 + 172 c2 = 144+ 289 P° = tan-1(17/12) mN = 35.2° c2 = 433 P° = 54.8°

H A SOH – CAH – TOA O U E UM c2 = a2 + b2 15 18 sin U° = 2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places. H A SOH – CAH – TOA O U E UM c2 = a2 + b2 15 18 sin U° = 180 – 90 – 56.4 182 = a2 + 152 U° = sin-1(15/18) 324 = a2 + 225 mE = 33.6° 99 = a2 U° = 56.4°

O SOH – CAH – TOA H A S U UM c2 = a2 + b2 7.5 31.3 cos S° = 2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places. O SOH – CAH – TOA H A S U UM c2 = a2 + b2 7.5 31.3 cos S° = 180 – 90 – 76.1 31.32 = a2 + 7.52 979.69= a2 + 56.25 S° = cos-1(7.5/31.3) mU = 13.9° 923.44 = a2 S° = 76.1°

HW Problem #14 O SOH – CAH – TOA A H D F EF c2 = a2 + b2 3 9 7.7 485-487 5-7, 11-14, 21-25 #14 O SOH – CAH – TOA A H D F EF c2 = a2 + b2 3 9 180 – 90 – 70.5 cos D° = 92 = a2 + 32 mF = 19.5° 81= a2 + 9 D° = cos-1(3/9) 72 = a2 D° = 70.5°