Hypotenuse opposite Adjacent - Radford University

Section 6.4 Trigonometry
Right Triangles

Hypotenuse opposite


Adjacent
Trigonometric Ratios

sin  opp . csc  hyp.
hyp. opp.

cos  adj. sec  hyp.
hyp. adj.

tan  opp. cot  adj.
adj. opp.

Example 1

Given the following right triangles find these trigonometric ratios sin A, cos A, and tan A

B

5
3

A 4C

sin a  opp .  3
hyp. 5

cos a  adj.  4
hyp. 5

tan A  opp.  3
adj. 4

Example 2
Use the following figure to answer questions #1-3

B
c

a

AbC

1) Find sin B , if a  5 and b  12

B

c
12

A 5C

c2  a2  b2
c 2  52  122
c 2  169
c  169  c  13

B

c = 13

12

A 5C

sin B  12
13

2) Find b , if cos A  .7 and c  12
B

12
b

A aC c2  a2  b2
122  8.4  b2
cos A  a 144  70.56  b2
12 b2  73.44

0.7  a b  73.44  8.6
12

a  .712  8.4

3) Find sin B , if b  a 3
B

2a 30
b =a 3

60 C
Aa

sin B  sin 30  a  1
2a 2

Example 4
Solve the following right triangle

B

10

34 C
A

Find the value of the missing angle first.

B  90  34  56

B

56

10
a

34 C
Ab

sin 34  a cos 34  b
10 10

.559  a .829  b
10 10

a  10.559 b  10.829

a  5.59 b  8.29

Example 5

Solve the following right triangle

B

11 7

AC

Find angle A
sin A  7

11
sin A  .636

A  sin 1.636  39.5

Find B
b  90  39.5  50.5

Find b

112  7 2  b2
121  49  b2
121  49  b2
b2  72
b  72  8.5

Example 6

A construction worker is standing 50 feet from the base of the tower with an angle of elevation
of 27 degrees. Find the height of the tower?

?

27
50 ft

cos27  h

50
.891  h

50

h  .89150

h  44.6 ft

Example 7
Find the area of the given triangle

14
32
18

Solution:

14
h

32

18

sin 32  h
14

h  14sin 32
h  7.42 units

A  1 bh  1 7.4218  66.8 square units

22

Example 8

A surveyor at point P wished to measure the distance PQ in the drawing that follows. The
surveyor sights point Q, makes a right angle at P and steps off 100 feet to R, and again sights Q.
Find PQ.

R

64

100 ft

P Q

tan 64  PQ
100

2.050  PQ
100

PQ  2.050 100
PQ  205