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Section 6.2
Applications of Right
Triangles
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.
Objectives


Solve right triangles.
Solve applied problems involving right triangles and
trigonometric functions.
Solve a Right Triangle
To solve a right triangle means to find the lengths of all
sides and the measures of all angles.
Example
In triangle ABC, find a, b, and B, where
a and b represent the lengths of sides
and B represents the measure of angle
B. Here we use standard lettering for
naming the sides and angles of a right
triangle: Side a is opposite angle A, side
b is opposite angle B, where a and b are
the legs, and side c, the hypotenuse, is
opposite angle C, the right angle.
B
106.2
a
61.7º
A
b
C
Example (cont)
B
B  90º A  90º 61.7º  28.3º
hyp
a
sin 61.7º 

opp 106.2
a  106.2sin 61.7º
a  93.5
adj
b
cos 61.7º 

hyp 106.2
106.2
a
61.7º
A
b
C
A  61.7º
a  93.5
b  106.2 cos 61.7º
B  28.3º
b  50.3
b  50.3
C  90º
c  106.2
Example
House framers can use trigonometric functions to
determine the lengths of rafters for a house. They first
choose the pitch of the roof, or the ratio of the rise over
the run. Then using a triangle with that ratio, they
calculate the length of the rafter needed for the house.
Jose is constructing rafters for a roof with a 10/12 pitch
on a house that is 42 ft wide. Find the length x of the
rafter of the house to the nearest tenth of a foot.

Run: 12
Rise: 10
Pitch: 10/12
Example (cont)
Solution:
First find the angle 
that the rafter makes
with the side wall.
10
tan  
 0.8333
12
 ≈ 39.8º
Use the cosine function to determine the length x of
the rafter.
Example (cont)
Solution continued:
x
21 ft
cos 39.8º 
x
39.8º
21 ft
x cos 39.8º  21 ft
21 ft
x
cos 39.8º
x  27.3 ft
The length of the rafter for this house is approximately
27.3 ft.
Angle of Elevation
The angle between the horizontal and a line of sight
above the horizontal is called an angle of elevation.
Angle of Depression
The angle between the horizontal and a line of sight
below the horizontal is called an angle of depression.
Example
In Telluride, CO, there is a free gondola ride that provides
a spectacular view of the town and the surrounding
mountains. The gondolas that begin in the town at an
elevation of 8725 ft travel 5750 ft to Station St. Sophia,
whose altitude is 10,550 ft. They then continue 3913 ft to
Mountain Village, whose elevation is 9500 ft.
a) What is the angle of elevation from the town to Station
St. Sophia?
b) What is the angle of depression from Station St.
Sophia to Mountain Village?
Example (cont)
Label a drawing with the given information.
Example (cont)
a) Difference in elevation of St. Sophia to town is 10,550
ft – 8725 ft or 1825 ft. This is the side opposite the
angle of elevation .
Station St. Sophia
5750 ft
Town

1825 ft
Angle of elevation
1825 ft
sin  
 0.3174
5750 ft
Example (cont)
Using a calculator, we find that
≈ 18.5º
The angle of elevation from town to Station St. Sophia
is approximately 18.5º.
b) When parallel lines are cut by a transversal,
alternate interior angles are equal. Thus the angle
of depression, , from Station St. Sophia to
Mountain Village is equal to the angle of elevation
from Mountain Village to Station St. Sophia.
Example (cont)
Difference in elevation of Station St. Sophia and the
elevation of Mountain Village is 10,550 ft – 9500 ft, or
1050 ft.
Angle of depression
Station St. Sophia

1050 ft
3913 ft
Mountain Village
Angle of elevation
1050 ft
  15.6º
sin  
 0.2683
3913 ft
The angle of depression from Station St. Sophia to
Mountain Village is approximately 15.6º.
Bearing: First-Type
One method of giving direction, or bearing, involves
reference to a north-south line using an acute angle. For
example, N55ºW means 55º west of north and S67ºE
means 67º east of south.
Example
A forest ranger at point
A sights a fire directly
south. A second ranger
at point B, 7.5 mi east,
sights the same fire at a
bearing of S27º23´W.
How far from A is the
fire?
Example (cont)
Find the complement of 27º23´.
B  90º 27º 23
B  62º 37
B  62.62º
Since d is the side opposite 62.62º, use the tangent
function ratio to find d.
d
 tan 62.62º
7.5 mi
d  7.5 mi  tan 62.62º
d  14.5 mi
The forest ranger at point A is about 14.5 mi from the fire.
Example
In U.S. Cellular Field, the home of the Chicago White Sox
baseball team, the first row of seats in the upper deck is
farther away from home plate than the last row of seats in
the original Comiskey Park. Although there is no
obstructed view in U.S. Cellular Field, some of the fans
still complain about the present distance from home plate
to the upper deck of seats. From a seat in the last row of
the upper deck directly behind the batter, the angle of
depression to home plate is 29.9º, and the angle of
depression to the pitcher’s mound is 24.2º. Find (a) the
viewing distance to home plate and (b) the viewing
distance to the pitcher’s mound.
Example (cont)
Example
We know that 1 = 29.9º and 2 = 24.2º. The distance
form home plate to the pitcher’s mound is 60.5 ft. In the
drawing, we d1 be the viewing distance to home plate, d2
the viewing distance to the pitcher’s mound, h the
elevation of the last row, and x the horizontal distance
form the batter to a point directly below the seat in the
last row of the upper deck.
Begin by finding x.
Example (cont)
Use the tangent function with 1 = 29.9º and 2 = 24.2º:
h
h
tan 29.9º 
and tan 24.2º 
x
x  60.5
h  x tan 29.9º and h  x  60.5 tan 24.2º
x tan 29.9º  x  60.5 tan 24.2º
x tan 29.9º  x tan 24.2º  60.5 tan 24.2º
x tan 29.9º x tan 24.2º  x tan 24.2º x tan 24.2º  60.5 tan 24.2º
x tan 29.9º  tan 24.2º   60.5 tan 24.2º
60.5 tan 24.2º
 216.5
x
tan 29.9º  tan 24.2º
Example (cont)
Then find d1 and d2 using the cosine function:
216.5
cos 29.9º 
d1
and
216.5  60.5
cos 24.2º 
d2
216.5
d1 
cos 29.9º
and
277
d2 
cos 24.2º
d1  249.7
and
d2  303.7
The distance to home plate is about 250 ft, and the
distance to the pitcher’s mound is about 304 ft.