7.7– solve right triangles
DESCRIPTION
7.7– Solve Right Triangles. 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. - PowerPoint PPT PresentationTRANSCRIPT
7.7– Solve Right Triangles
Inverse Trig Ratios:
To find the measure of an angle, take the inverse of the trig function.
sin A° = 7 .
10
A° = sin-1 (7/10)
A° = 44.4°
A
BC
sin-1 BCAB
= mA°
cos-1 ACAB
= mA°
tan-1 BCAC
= mA°
Steps to solving with inverse trig:
1.
2.
Label all sides with reference angle
What two sides do you know?
1. Use a calculator to approximate the measure of A to the nearest tenth of a degree.
SOH – CAH – TOA
tan A° = 3020
A° = tan-1 (30/20)
A° = 56.3°
A H
O
1. Use a calculator to approximate the measure of A to the nearest tenth of a degree.
SOH – CAH – TOA
sin A° = 1426
A° = sin-1 (14/26)
A° = 32.6°
A
H
O
1. Use a calculator to approximate the measure of A to the nearest tenth of a degree.
SOH – CAH – TOA
cos A° = 1016
A° = cos-1 (10/16)
A° = 51.3°
A
H
O
PQ QRP
2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places.
SOH – CAH – TOA
sin 37° = x .
22
22 sin 37° = x
13.24 = x
180 – 90 – 37
mP = 53°
cos 37° = y .
22
22 cos 37° = y
17.57 = y
A
HO
ST UTU
SOH – CAH – TOA
cos 70° = x .
15
15 cos 70° = x
5.13 = x
180 – 90 – 70
mU = 20°
sin 70° = y .
15
15 sin 70° = y
14.10 = y
AH
O
2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places.
PNP
SOH – CAH – TOA
c2 = a2 + b2
c2 = 122 + 172
c2 = 144+ 289c2 = 433
20.81c
180 – 90 – 54.8
mN = 35.2°
N
tan P° = 1712
P° = tan-1(17/12)
P° = 54.8°
AH
O
2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places.
UMU
SOH – CAH – TOA
180 – 90 – 56.4
mE = 33.6°
E
sin U° = 1518
U° = sin-1(15/18)
U° = 56.4°
c2 = a2 + b2
182 = a2 + 152
324 = a2 + 22599 = a2
3 11 a
A
H
O
2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places.
UMS
SOH – CAH – TOA
180 – 90 – 76.1
mU = 13.9°
U
cos S° = 7.531.3
S° = cos-1(7.5/31.3)
S° = 76.1°
c2 = a2 + b2
31.32 = a2 + 7.52
979.69= a2 + 56.25923.44 = a2
30.39 a
AH
O
2. Solve the right triangle. Approximate angles to the nearest tenth of a degree and sides to two decimal places.
7.7 485-487 5-7, 11-14, 21-25
HW Problem
#14
EFD
SOH – CAH – TOA
180 – 90 – 70.5
mF = 19.5°
F
cos D° = 39
D° = cos-1(3/9)
D° = 70.5°
c2 = a2 + b2
92 = a2 + 32
81= a2 + 9 72 = a2
6 2 a
AH
O