LESSON 6
Angles of Elevation and D
sion
LESSON 6
Angles of Elevation and Depression
Now we get a chance to apply all of our newly acquired skills to real-life applications, otherwise known as word problems. Let’s look at some elevation and depression problems. I first encountered these in a Boy Scout handbook many years ago.
There was a picture of a tree, a boy, and several lines.
Example 1
tree
How tall is the tree?
5'
11'
30'
Separating the picture into two triangles helps to clarify our ratios.
θ
11
5
X
θ
41
We could write this as a proportion (two ratios), 5 = x ,
11
41
and solve for X.
ANGLES OF ELEVATION AND DEPRESSION - LESSON 6
59
We can also use our trig abilities.
5 = .4545
From the “boy” triangle: tan θ =
11
θ = 24.44º
x
From the large triangle: tan 24.44º =
41
(41) (.4545) = x
18.63 = x
Solve for X.
The tree is 18.63'.
When working these problems, the value of the trig ratio may be rounded and
recorded, and further calculations made on the rounded value. You may also keep
the value of the ratio in your calculator and continue without rounding the intermediate step. This may yield slightly different final answers. These differences are
not significant for the purposes of this course.
It is pretty obvious that an angle of elevation measures up and an angle of
depression measures down. One of the keys to being a good problem solver is to
draw a picture using all the data given. It turns a one-dimensional group of words
into a two-dimensional picture.
Figure 1
depression
elevation
We assume that the line where the angle begins is perfectly flat or horizontal.
Example 2
A campsite is 9.41 miles from a point directly below the mountain
top. If the angle of elevation is 12º from the camp to the top of the
mountain, how high is the mountain?
top
mountain
campsite
60
12º
9.41 mi
LESSON 6 - ANGLES OF ELEVATION AND DEPRESSION
PRECALCULUS
You can see a right triangle with the side adjacent to the 12º angle
measuring 9.41 miles. To find the height of the mountain, or the side
opposite the 12º angle, the tangent is the best choice.
height
9.41 mi
(9.41) ( tan 12º ) = height
tan 12º =
(9.41)(.2126) = height
2 miles = height
Example 2
At a point 42.3 feet from the base of a building, the angle of elevation
of the top is 75º. How tall is the building?
height
42.3'
(42.3)(tan 75º ) = height
tan 75º =
building
(42.3) (3.7321) = height
157.87 ' = height of the building
75º
42.3'
Practice Problems 1
1. How far from the door must a ramp begin in order to rise three feet
with an 8º angle of elevation?
2. An A-frame cabin is 26.23 feet high at the center, and the angle the
roof makes with the base is 53º15'. How wide is the base?
PRECALCULUS
ANGLES OF ELEVATION AND DEPRESSION - LESSON 6
61
Solutions 1
1. 8º
X
3
2.
tan 8º = 3
x
x tan 8º = 3
3
x=
tan 8º
3
x=
.1405
x = 21.35 ft
62
26.23
LESSON 6 - ANGLES OF ELEVATION AND DEPRESSION
53º15'
X
X
53º 15" = 53.25º
tan 53.25º = 26.23
x
26.23
x=
tan 53.25º
x = 23.26
1.3392
x = 26.23
1.3392
x = 19.59
2x = 39.18 ft
PRECALCULUS
6A
Answer the questions.
1.Isaac’s camp is 5,280 feet from a point directly beneath Mt. Monadnock.
What is the hiking distance along the ridge if the angle of elevation is 25º 16'?
2. How many feet higher is the top of the mountain than his campsite?
Express as a fraction.
3. csc
q =
6. csc
a=
α
4. sec
q =
7. sec
a=
2 31
4
θ
5. cot
q =
8. cot
a=
6 3
Express as a decimal.
9. sin
q =
12. sin
a=
10. cos
q =
13. cos
a=
11. tan
q =
14. tan
a=
PRECALCULUS Lesson 6A
53
LESSON 6A
15. Use your answers from #9–11 to find the measure of q.
16. Use your answers from #12–14 to find the measure of
a.
Solve for the lengths of the sides and the measures of the angles.
17.
α
A
12
35.4º
B
18.
29º
59
D
α
C
19.
F
42.66º
100
E
α
20.
G
α
47
H
41º32'10''
54
PRECALCULUS
6B
Answer the questions.
1. The side of a lake has a uniform angle of elevation of 15º 30'. How far up the side
of the lake does the water rise if, during the flood season, the height of the lake
increases by 7.3 feet?
2. A building casts a shadow of 110 feet. If the angle of elevation from that point to
the top of the building is 29º 3', find the height of the building.
Express as a fraction.
3. csc
q =
6. csc
a=
4. sec
q =
7. sec
a=
5. cot
q =
8. cot
a=
α
11
θ
Express as a decimal.
10
9. sin
q =
12. sin
a=
10. cos
q =
13. cos
a=
11. tan
q =
14. tan
a=
PRECALCULUS Lesson 6B
4.6
55
LESSON 6B
15. Use your answers from #9–11 to find the measure of q.
16. Use your answers from #12–14 to find the measure of
a.
Solve for the lengths of the sides and the measures of the angles.
17.
J
α
12
18º
K
18.
M
α
L
59
29º
19.
67º
N
10.25
θ
P
20.
6
θ
2 13
Q
α
56
PRECALCULUS
6C
Answer the questions.
1. From a point 120 feet from the base of a church, the angles of elevation of the top
of the building and the top of a cross on the building are 38º and 43º respectively.
Find the height to the top of the cross. (The ground is flat.)
2. Find the height of the building as well as the height of the cross by itself.
Express as a fraction.
3. csc
q =
6. csc
a=
4. sec
q =
7. sec
a=
α
5. cot
q =
8. cot
a=
15
7.1
θ
13.2
Express as a decimal.
9. sin
q =
12. sin
a=
10. cos
q =
13. cos
a=
11. tan
q =
14. tan
a=
PRECALCULUS Lesson 6C
57
LESSON 6C
Results for #15 and 16 may vary slightly from the solutions, depending on when steps were rounded.
15. Use your answers from #9–11 to find the measure of q.
16. Use your answers from #12–14 to find the measure of
a.
Solve for the lengths of the sides and the measures of the angles.
17.
S
40º
R
25
α
18.
α
T
88
36.2º
U
19.
α
V
150
51.9º
W
20.
α
95
X
θ
7
58
PRECALCULUS
6D
Answer the questions.
1. A campsite is 12.88 miles from a point directly below Mt. Adams. If the angle
of elevation is 15.5º from the camp to the top of the mountain, how high is
the mountain?
2. At a point 60.7 feet from the base of a building, the angle of elevation from that
point to the top is 64.75º. How tall is the building?
Express as a fraction.
3. csc
q =
6. csc
a=
4. sec
q =
7. sec
a=
5. cot
q =
8. cot
a=
α
25
θ
Express as a decimal.
18.33
9. sin
q =
12. sin
a=
10. cos
q =
13. cos
a=
11. tan
q =
14. tan
a=
PRECALCULUS Lesson 6D
X
59
LESSON 6D
15. Use your answers from #9–11 to find the measure of q.
16. Use your answers from #12–14 to find the measure of
a.
Solve for the lengths of the sides and the measures of the angles.
17.
2.24
α
2
Y
θ
18.
10.5
α
Z
49.2º
A
19.
29.07º
B
56
α
C
20.
α
D
10
θ
14
60
PRECALCULUS
6H
Here are some more applications of trig functions. In some of these you may need to find a missing side,
and in others a missing angle.
Use the skills you have learned so far to answer the questions. Always begin by making a drawing and labeling
the known information.
1. A girl who is 1.6 meters tall stands on level ground. The elevation of the sun is 60°
above the horizon. What is the length of her shadow?
2.If the girl in #1 casts a shadow that is one meter long, what is the elevation
of the sun?
3. A stairway forms an angle with the floor from which it rises. This angle may be called
the angle of inclination. What is the angle of inclination of a stairway if the steps
have a tread of 20 centimeters and a rise of 16 centimeters?
Some problems will require more of your algebra skills. There are some examples of these on the next
page. The first one is done for you.
PRECALCULUS Honors 6H
61
Honors 6H
4. An observation balloon is attached to the ground at point A. On a level with A and
in the same straight line, the points B and C were chosen so that BC equals 100
meters. From the points B and C, the angle of elevation of the balloon is 40º and 30º
respectively. Find the height of the balloon.
First, make a drawing. There’s not enough information
to find x using either the angle at B or the angle at C.
However, we can make two equations using x and y.
x
40º
A
y
30º
B 100m C
x
x
Equation 2 tan 30º =
Equation 1 tan 40º =
y
y + 100
Replace tan 40º with its ratio and solve for x in Equation 1.
.8391 = x or x = .8391y
y
Replace tan 30º with its ratio in in Equation 2. .5774 =
x
y + 100
Substitute value of x from Equation 1 in Equation 2. .5774 =
.8391y
y + 100
Solve for y.
.5774(y + 100) = .8391y
.5774 y + 57.74 = .8391y
57.74 = .2617y
y = 220.6 (rounded)
Solve for x, which is the height of the balloon. x = .8391y
x = .8391 (220.6) = 185.1 m
5. Tom wished to find the width of a river. He observed a tree directly across the river
on the opposite bank. The angle of elevation to the top of the tree was 32º. Then
Tom moved directly back from the bank 50 meters and found that the angle of
elevation to the top of the tree was 21º. What is the width of the river?
6.In the side of a hill that slopes upward at an angle of 32º, a tunnel is bored sloping
downward at an angle of 12º15' from the horizontal. How far below the surface of
the hill is a point 38 meters down the tunnel?
62
PRECALCULUS
6
test
Use for #1–4: Devan stands 926 meters from
a point directly below the peak of a mountain.
The angle of elevation between Devan and the
top of the mountain is 42°.
Use for #5–8: From a point 80 meters from the
base of a building to the top of the building, the
angle of elevation is 51°. From the same point to
the top of a flag staff on the building, the angle
of elevation is 54°.
1. Which equation can be used to find
the height of the mountain (x)?
A.
B.
C.
D.
sin 42º = x/926
tan 42º = 926/x
cos 48º = 926/x
tan 42º = x/926
2. What is the height of the mountain?
A.
B.
C.
D.
833.8 m
1,028.4 m
619.6 m
1,383.9 m
3. A tower 50 meters high is built on
top of the mountain. What is the
angle of elevation from Devan’s
position to the top of the tower?
(Round decimal degrees to tenths.)
A.
B.
C.
D.
40º
43º
57º
46º
14' 44''
42'
15'
20' 08''
4. If a bird flew from Devan’s position
to the top of the mountain, how
many meters would it travel?
A.
B.
C.
D.
408.4 m
1,246.1 m
1,383.9 m
1,280 m
PRECALCULUS Test 6
5. What equation can be used to find
the combined height (y) of building
and flagpole?
A.
B.
C.
D.
y
y
y
y
=
=
=
=
80 tan 51º
80 sin 54º
80 tan 54º
tan 51º
80
6. What is the height of the building
alone?
A.
B.
C.
D.
98.8 m
110.1 m
64.8 m
58.1 m
7. What is the height of the flagpole
alone?
A.
B.
C.
D.
15.1 m
45.3 m
4.2 m
11.3 m
8. How long must a cable be in order
to stretch from the observation
point to the top of the building?
A.
B.
C.
D.
102.9 m
127.1 m
136.1 m
50.3 m
15
test 6
Use for #9–10: A car traveled a distance of 100
feet up a ramp to a bridge. The angle of elevation
of the ramp was 10°.
A. 46.21º
B. 46.12º
C. 46.35º
D. 46.4º
9. How high was the bridge above
road level?
A.
B.
C.
D.
17.4 ft
98.5 ft
10 ft
100 ft
10. What is the actual distance from
the beginning of the ramp to the
base of the bridge?
11.
A.
B.
C.
D.
575 ft
98.5 ft
89.4 ft
17.4 ft
3 is the ratio for:
3
A.
B.
C.
D.
cos 45º
cos 30º
tan 60º
tan 30º
13. 46º 21' 02'' =
14.
sin α is equal to:
cos α
A. tan α
B. cot α
C. sec α
D. csc α
15.
1 is equal to:
cos α
A.
B.
C.
D.
csc α
sec α
sin α
cos α
12. Arcsin .8192 =
16
A.
B.
C.
D.
1.22
35º
55º
.9999
PRECALCULUS
12.
12.
13.
13.
Lesson
6A
Lesson 6A
1.
5,280
D
D cos 25º 16 ' = 5,280
cos 25º 16 ' =
14.
14.
15.
15.
16.
16.
17.
17.
5,280
cos 25º 16 '
D ≈ 5,838.77 ft
D=
2.
tan 25º 16 ' =
Lesson 6A
1.
3.
4.
5.
2.
M
5,280
M = (D
5,280) ( tan 25º 16 ')
5,280
cos 25ºM
16≈' 2
=, 492.09 ft
M
D
D cos 25º 16 ' = 5,280
csc θ = 2 31 = 31
4
2
25º16' D = 5,280
cos
16 '3
31 =25º31
= 93
sec θ = 2 31 = 5,280
ft
D
≈
5
838
77
,
.
9
6 3
3 3 3 3 ft
3
M
= º6163' == 3 3
cot
tanθ25
4
52
,280
93
M = (5,280) ( tan 25º 16 ')
9
M ≈ 2, 492.09 ft
7. sec α = 31
3. csc θ = 2 231 = 31
2
2 43
8. cot α = 2
31
4. sec θ = 9 = 31 = 31 3 = 93
9
6 3
3 3 3 3 3
9. sin θ = 4 ≈ .3592
3=3 3
5. cot θ = 26 31
4
2
10. cos θ = 6 3 ≈ .9333
93
6. csc α = 2 31
9
11. tan θ = 4 ≈ .3849
3
7. sec α = 6 31
2
6
3 ≈ .9333
12. sin α = 2 3
8. cot α = 2 31
9
13. cos α = 44 ≈ .3592
9. sin θ = 2 31≈ .3592
2 31
14. tan α = 6 3 ≈ 2.5981
10. cos θ = 64 3 ≈ .9333
2 31
15. arcsin .3592 ≈ 21.05º
4 ≈ .3849
tan θ = .9333
11.
16. arcsin
≈ 68.96º
6 3
17.
tan54. º = B
12. sin α = 6 3 12
≈ .9333
2B31
= (12) ( tan54.6º ) ≈ 16.89
13. cos α = 4 ≈12
.3592
sin352.431
º=
A
6.43º =≈ 12
A sin
14. tan
α =35
2.5981
4
A =≈ 2112
15. arcsin .3592
.05º ≈ 20.72
sin35.4º
16. arcsin .9333 ≈ 68.96º
α = 90 − 35.4º = 54.6º
17.
tan
54.6º = B
PR
eCal cs
ulus
18.
in 61º = D12
59
B = (12) ( tan54.6º ) ≈ 16.89
D = (59)(sin 61º ) ≈ 51.6
sin35.4º = 12
cos 61º = c A
59
A sin35.4º = 12
c = (59) (cos 61º ) ≈ 28.6
A º== 6112
º .4º ≈ 20.72
α = 90º −29
sin35
6.
18.
18.
csc α =
19.
19.
20.
20.
6 3
6 3 .9333
sinα
=
sinα = 6 3 ≈
≈ .9333
2
2 31
31
4
cos
cos α
3592
α=
= 4 ≈
≈ ..3592
2
2 31
31
L e ss o n 6 A - LESSON 6A
6 3 2.5981
tan
tan α
α=
= 643 ≈
≈ 2.5981
4
ar
arc
csin
sin ..3592
3592 ≈
≈ 21
21..05
05ºº
arcsin
arcsin ..9333
9333 ≈
≈ 68
68..96
96ºº
B
B
tan
tan5
54
4..6
6ºº =
= 12
12
( )) ((tan54.6º )) ≈ 16.89
B
= (12
12 tan54.6º ≈ 16.89
B=
12
sin
sin35
35..4
4ºº =
= 12
A
A
3
5
.
4
º
=
12
sin
A
A sin35.4º = 12
12
A
≈
20..72
72
A=
= sin12
≈ 20
sin35
35..4
4ºº
α = 90 − 35.4º = 54.6º
α = 90 − 35.4º = 54.6º
D
s
D
siin
n 61
61ºº =
= 59
59
(59)(
) ≈ 51
D
)(sin
D=
= (59
sin 61
61ºº ) ≈
51..6
6
c
c
cos
= 59
cos 61
61ºº =
59
(59)) ((co
c
s 61
61ºº )) ≈
28..6
6
= (59
c=
≈ 28
cos
90ºº −
29ºº =
61ºº
α
α=
= 90
−29
= 61
F
F
tan 47
47..34
34ºº =
= 100
tan
100
(100)(
)
F
)(tan
F=
= (100
tan 47
47..34
34ºº )
F
F≈
≈ 108
108..52
52
100
sin
sin 42
42..66
66ºº =
= 100
e
e
e
sin
42
.
66
º
1
00
=
e sin 42.66º = 100
100
100
e=
=
≈ 147.57
e
sin 42.66º ≈ 147.57
sin 42.66º
α=
= 90
90ºº −
−42
42..66
66ºº =
= 47
47..34
34ºº
α
G
an 41
41..54
54ºº =
= G
ttan
47
47
tan 41
G
= (( 47
47)) (( tan
41..54
54ºº )) ≈
41..64
64
G=
≈ 41
47
cos 41
41..54
54ºº =
= 4H7
cos
H
H cos
cos 41
= 47
41..54
54ºº =
47
H
47
47
H=
=
≈ 62.79
H
cos
41
54ºº ≈ 62.79
cos 41..54
α=
= 90
90ºº −
−41
41ºº 32
32 ''10
10 "" =
= 48
48ºº 27
27 ''50
50 ""
α
θ
≈
41
.
54
º
θ ≈ 41.54º
s o lut i o n s
281