Download angle of elevation

CHAPTER 6:
The Trigonometric Functions
6.1
6.2
6.3
6.4
6.5
6.6
The Trigonometric Functions of Acute Angles
Applications of Right Triangles
Trigonometric Functions of Any Angle
Radians, Arc Length, and Angular Speed
Circular functions: Graphs and Properties
Graphs of Transformed Sine and Cosine
Functions
Copyright © 2009 Pearson Education, Inc.
6.2
Applications of Right Triangles


Solve right triangles.
Solve applied problems involving right triangles and
trigonometric functions.
Copyright © 2009 Pearson Education, Inc.
Solve a Right Triangle
To solve a right triangle means to find the lengths of
all sides and the measures of all angles.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 4
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.
Copyright © 2009 Pearson Education, Inc.
B
106.2
a
61.7º
A
b
C
Slide 6.2 - 5
B
Example
Solution:
B  90º A  90º 61.7º  28.3º
hyp
a
sin 61.7º 

opp 106.2
a  106.2sin 61.7º
106.2
a
61.7º
A
b
C
a  93.5
adj
b
cos61.7º 

hyp 106.2
b  106.2 cos61.7º
b  50.3
Copyright © 2009 Pearson Education, Inc.
A  61.7º
a  93.5
B  28.3º
b  50.3
C  90º
c  106.2
Slide 6.2 - 6
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.

Rise: 10
Pitch: 10/12
Run: 12
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 7
Example
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 8
Example
Solution continued:
x
39.8º
21 ft
21 ft
cos 39.8º 
x
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 9
Angle of Elevation
The angle between the horizontal and a line of sight
above the horizontal is called an angle of elevation.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 10
Angle of Depression
The angle between the horizontal and a line of sight
below the horizontal is called an angle of depression.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 11
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?
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 12
Example
Solution
Label a drawing with the given information.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 13
Example
Solution
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
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 14
Example
Solution continued
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 15
Example
Solution continued
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
Angle of elevation
1050 ft
sin  
 0.2683
3913 ft
Mountain Village
  15.6º
The angle of depression from Station St. Sophia to
Mountain Village is approximately 15.6º.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 16
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 17
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?
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 18
Example
Solution:
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 19
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 20
Example
Solution:
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 21
Example
Solution:
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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 22
Example
Solution continued:
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
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 23
Example
Solution continued:
Then find d1 and d2 using the cosine function:
216.5
cos 29.9º 
d1
216.5  60.5
and cos 24.2º 
d2
216.5
d1 
cos 29.9º
277
and 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.
Copyright © 2009 Pearson Education, Inc.
Slide 6.2 - 24