1. 7.7– Solve Right Triangles

2. 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

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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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 =
c2 =
P° = tan-1(17/12)
mN = 35.2°
c2 = 433
P° = 54.8°

10. 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 = a
U° = sin-1(15/18)
324 = a
mE = 33.6°
99 = a2
U° = 56.4°

11. 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 = a
979.69= a
S° = cos-1(7.5/31.3)
mU = 13.9°
= a2
S° = 76.1°

12. HW Problem #14 O SOH – CAH – TOA A H D F EF c2 = a2 + b2 3 9
7.7
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°