7.7– Solve Right Triangles

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Presentation transcript:

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 sin A° = 0.7 BC AB = mA° sin sin sin-1 AC AB = mA° cos-1 A° = sin-1 (0.7) A° = 44.4° BC AC = mA° tan-1

30 tan A° = 20 tan A° = 1.5 A° = tan-1 (1.5) A° = 56.3° O A 1. Use a calculator to approximate the measure of A to the nearest tenth of a degree. 30 20 tan A° = tan A° = 1.5 A° = tan-1 (1.5) A° = 56.3° O A SOH – CAH – TOA

14 sin A° = 26 sin A° = 0.5385 A° = sin-1 (0.53) A° = 32.6° O H 1. Use a calculator to approximate the measure of A to the nearest tenth of a degree. 14 26 sin A° = sin A° = 0.5385 A° = sin-1 (0.53) A° = 32.6° O H SOH – CAH – TOA

10 cos A° = 16 cos A° = 0.625 A° = cos-1 (0.625) A° = 51.3° A H 1. Use a calculator to approximate the measure of A to the nearest tenth of a degree. 10 16 cos A° = cos A° = 0.625 A° = cos-1 (0.625) A° = 51.3° A H SOH – CAH – TOA

O H SOH – CAH – TOA P PQ QR x . 22 c2 = a2 + b2 sin 37° = 2. Solve the right triangle. O H SOH – CAH – TOA P PQ QR x . 22 180 – 90 – 37 c2 = a2 + b2 sin 37° = 222 = a2 + 13.242 mP = 53° 22  sin 37° = x 484 = a2 + 175.296 13.24 = x 308.7 = a2

O A SOH – CAH – TOA P N PN 17 12 c2 = a2 + b2 tan P° = 2. Solve the right triangle. O A SOH – CAH – TOA P N PN 17 12 c2 = a2 + b2 tan P° = 180 – 90 – 54.8 c2 = 122 + 172 tan P° = 1.42 mN = 35.2° c2 = 144+ 289 P° = tan-1 (1.42) c2 = 433 P° = 54.8°

A H SOH – CAH – TOA ST U UT x . 15 c2 = a2 + b2 cos 70° = 2. Solve the right triangle. A H SOH – CAH – TOA ST U UT x . 15 180 – 90 – 70 c2 = a2 + b2 cos 70° = 152 = a2 + 5.132 mU = 20° 15  cos 70° = x 225 = a2 + 26.32 5.13 = x 198.68 = a2

O H SOH – CAH – TOA U E UM c2 = a2 + b2 15 18 sin U° = 2. Solve the right triangle. O H SOH – CAH – TOA U E UM c2 = a2 + b2 15 18 sin U° = 180 – 90 – 56.4 182 = a2 + 152 sin U° = 0.833 mE = 33.6° 324 = a2 + 225 U° = sin-1 (0.83) 99 = a2 U° = 56.4°

A H SOH – CAH – TOA S U UT c2 = a2 + b2 7.5 31.3 cos S° = 2. Solve the right triangle. A H SOH – CAH – TOA S U UT c2 = a2 + b2 7.5 31.3 cos S° = 180 – 90 – 76.1 31.32 = a2 + 7.52 cos S° = 0.24 mU = 13.9° 979.69= a2 + 56.25 S° = cos-1 (0.24) 923.44 = a2 S° = 76.1°