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Applications of
Trigonometric
Functions
Solving Right Triangles
Solving a right triangle means finding the missing lengths of its
sides and the measurements of its angles. We will label right
triangles so that side a is opposite angle A, side b is opposite angle
B, and side c is the hypotenuse opposite right angle C.
B
c
a
A
b
C
Text Example
B
Solve the right triangle shown.
c
a
34.5º
A
b = 10.5
C
Solution We begin by finding the the measure of angle B. We do not need
a trigonometric function to do so. Because C = 90º and the sum of a
triangle’s angles is 180, we see that A + B = 90º. Thus,
B = 90º – A = 90º – 34.5º = 55.5º.
Now we need to find a. Because we have a known angle, and unknown
opposite side, and a known adjacent side, we use the tangent function.
tan34.5º = a/10.5
Now we solve for a.
A = 10.5tan34.5=7.22
Text Example cont.
B
Solve the right triangle shown.
c
55.5º
7.22
34.5º
A
b = 10.5
Solution
Finally, we need to find c. Because we have a known angle, a known
adjacent side, and an unknown hypotenuse, we use the cosine function.
cos34.5 = 10.5/c
c=10.5/cos34.5 = 12.74
In summary, B = 55.5º, a = 7.22, and c = 12.74.
C
Example
Example 1:
A bridge is to be constructed across a small river from post A
to post B. A surveyor walks 100 feet due south of post A. She
sights on both posts from this location and finds that the angle
between the posts is 73. Find the distance across the river from
post A to post B.
x
Post B
Post A
Use a calculator to find
tan 73o = 3.27.
100 ft.
○
x
opp
73
3.27 = tan 73=
=
adj
100
It follows that x = 327.
The distance across the river from post A to post B is 327 feet.
Text Example
Use the figure to find: a. the bearing from O to B. b. the bearing from O to A.
N
B
40º
A
20º
W
75º
O
C
Solution
E
D
25º
S
a. To find the bearing from O to B, we need the acute angle between the
ray OB and the north-south line through O. The measurement of this
angle is given to be 40º. The figure shows that the angle is measured
from the north side of the north-south line and lies west of the northsouth line. Thus, the bearing from O to B is N 40º W.
Text Example cont.
Use the figure to find: a. the bearing from O to B. b. the bearing from O to A.
N
B
40º
A
20º
W
75º
O
C
E
D
25º
Solution
S
b. To find the bearing from O to A, we need the acute angle between the
ray OA and the north-south line through O. This angle is specified by
the voice balloon in the figure. The figure shows that this angle
measures 90º – 20º, or 70º. This angle is measured from the north side
of the north-south line. This angle is also east of the north-south line.
This angle is also east of the north-south line. Thus the bearing from O
to A is N 70º E.
Example
A boat leaves the entrance of a harbor and travels 40 miles on a
bearing of S64E. How many miles south and east did the boat
travel?
S64E
40
miles
a
cos 64 
a
b
sin 64 
40
40 cos 64  a
17.5  a
b
40
40 sin 64  b
36  b
Angle of Elevation and Angle of Depression
When an observer is looking upward, the angle formed
by a horizontal line and the line of sight is called the:
angle of elevation.
line of sight
object
angle of elevation
horizontal
observer
When an observer is looking downward, the angle formed
by a horizontal line and the line of sight is called the:
horizontal
angle of depression
line of sight
object
observer
angle of depression.
Example 2:
A ship at sea is sighted by an observer at the edge of a cliff
42 m high. The angle of depression to the ship is 16. What
is the distance from the ship to the base of the cliff?
observer
cliff
42 m
horizontal
16○ angle of depression
line of sight
16○
d
42
= 146.47.
tan 16
The ship is 146 m from the base of the cliff.
d=
ship
Example 3:
A house painter plans to use a 16 foot ladder to reach a spot
14 feet up on the side of a house. A warning sticker on the
ladder says it cannot be used safely at more than a 60 angle
of inclination. Does the painter’s plan satisfy the safety
requirements for the use of the ladder?
ladder
house
14
16
sin  =
= 0.875
14
16
θ
Next use the inverse sine function to find .
 = sin1(0.875) = 61.044975
The angle formed by the ladder and the ground is about 61.
The painter’s plan is unsafe!
Simple Harmonic Motion
An object that moves on a coordinate axis is in
simple harmonic motion if its distance from the
origin, d, at time t is given by either
d = a cos  t or d = a sin  t.
The motion has amplitude |a|, the maximum
displacement of the object from its rest position.
The period of the motion is 2/ , where  > 0.
The period gives the time it takes for the motion to
go through one complete cycle.
Frequency of an Object in Simple
Harmonic Motion
• An object in simple harmonic motion given
by d=acost or d = asint has frequency f
given by f = /2, >0.
• Equivalently, f = 1/period.
Example
• A mass moves in simple harmonic motion
described by the following equation, with t
measured in seconds and d in centimeters.
Find the maximum displacement, the
frequency, and the time required for one
cycle.
d  8 cos

3
t
Example cont.
• Since a=8, the maximum displacement is 8
cm.

d  8 cos
a 8
3
t
Example cont.
• The frequency is 1/6 cm per second.
d  8 cos


3
t

3


1
3
f 


2 2 6
Example cont.
• The time required for one cycle is 6
seconds.

d  8 cos t
3


3
period 
2


2

3
6