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AnGLeS oF eLeVATion AnD DepreSSion - LeSSon 659
Now we get a chance to apply all of our newly acquired skills to real-life applica- tions, otherwise known as word problems. Let’s look at some elevation and depres- sion 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?
Separating the picture into two triangles helps to clarify our ratios.
11
θ^5
X
We could write this as a proportion (two ratios), 115 = 41 x^ ,
and solve for x.
LeSSon 6
Angles of Elevation and Depression
LeSSon 6
AnGLeS oF eLeVATion AnD D Sion
60 LeSSon^ 6 - AnGLeS^ oF^ eLeVATion^ AnD DepreSSion^ precALcULUS
We can also use our trig abilities.
From the “boy” triangle: tan^ θ =^115 =.^4545 θ = 24.44º
From the large triangle: tan 24 44. º = 41 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 inter- mediate 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
elevation
depression
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?
campsite
top
mountain
12º 9.41 mi
( )( ) =
x
x
62 LeSSon^ 6 - AnGLeS^ oF^ eLeVATion^ AnD DepreSSion^ precALcULUS
Solutions 1
tan. º.
tan. º
x
x
xx
x
x
x
. ft
X
8º^3
X
X
tan º
tan º
tan º
. ft
x
x
x
x
x
PRECALCULUS LESSon 6A 53
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.
- csc q = 6. csc a =
- sec q = 7. sec a =
- cot q = 8. cot a =
Express as a decimal.
- sin q = 12. sin a =
- cos q = 13. cos a =
- tan q = 14. tan a =
PRECALCULUS LESSon 6B 55
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.
- csc q = 6. csc a =
- sec q = 7. sec a =
- cot q = 8. cot a =
Express as a decimal.
- sin q = 12. sin a =
- cos q = 13. cos a =
- tan q = 14. tan a =
LESSon 6B
56 PRECALCULUS
- Use your answers from #9–11 to find the measure of q.
- 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.
J α
K
L
M
P
67º N
Q
LESSon 6C
58 PRECALCULUS
Results for #15 and 16 may vary slightly from the solutions, depending on when steps were rounded.
- Use your answers from #9–11 to find the measure of q.
- 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.
R
S
U
T
W
150 V
X
PRECALCULUS LESSon 6D 59
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.
- csc q = 6. csc a =
- sec q = 7. sec a =
- cot q = 8. cot a =
Express as a decimal.
- sin q = 12. sin a =
- cos q = 13. cos a =
- tan q = 14. tan a =
X
PRECALCULUS HonoRS 6H 61
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.
6H
HonoRS 6H
62 PRECALCULUS
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. Equation 1 tan 40º = xy Equation 2 tan 30º = (^) y +x 100 Replace tan 40º with its ratio and solve for x in Equation 1. .8391 = xy or x = .8391y Replace tan 30º with its ratio in in Equation 2. .5774 = (^) y + x 100 Substitute value of x from Equation 1 in Equation 2. .5774 = .8391y + 100 y 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?
x
A y B 100mC
TEST 6
16 PRECALCULUS
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°.
- How high was the bridge above road level?
A. 17.4 ft B. 98.5 ft C. 10 ft D. 100 ft
- What is the actual distance from the beginning of the ramp to the base of the bridge?
A. 575 ft B. 98.5 ft C. 89.4 ft D. 17.4 ft
- 33 is the ratio for:
A. cos 45º B. cos 30º C. tan 60º D. tan 30º
- Arcsin .8192 =
A. 1. B. 35º C. 55º D..
- 46º 21' 02'' =
A. 46.21º
B. 46.12º
C. 46.35º
D. 46.4º
- (^) cossin^ αα is equal to:
A. tan α
B. cot α
C. sec α
D. csc α
- 1
cos α
is equal to:
A. csc α B. sec α C. sin α D. cos α
precALcULUS
LeSSon 6A - LeSSon 6A
SoLUTionS 281
Lesson 6ALesson 6A
- cos º ' , cos º ' , ,
D
D
D 25 16^2280
cos º ' ,. tan º ' (^) ,
D ft M M
= ,, tan º ' ,. csc
( )(^ )
= =
M ≈ ft
4
θ ..
sec cot csc
θ θ
αα α α θ
sec cot sin. c
oos. tan. sin
θ θ α
cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
. º. º s
A ≈
α
- iin º sin º. cos º co
= (^ )(^ )
= (^ )
D
D
c c
ss º. º º º
( ) = − =
α
D
5,280 ft
M
Lesson 6A
- cos º ' , cos º ' , ,
D
D
D 25 16^2280
cos º ' ,. tan º ' (^) ,
D ft M M
= ,, tan º ' ,. csc
( )(^ )
= =
M ≈ ft
4
θ ..
sec cot csc
θ θ
αα α α θ
sec cot sin. c
oos. tan. sin
θ θ α
cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
A = ≈
Lesson 6A
- cos º ' , cos º ' , ,
D
D
D 25 16^2280
cos º ' ,. tan º ' (^) ,
D ft M M
= ,, tan º ' ,. csc
( )(^ )
= =
M ≈ ft
4
θ ..
sec cot csc
θ θ
αα α α θ
sec cot sin. c
oos. tan. sin
θ θ α
cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
A = ≈
- sinα = (^) 2 316 3 ≈.. cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
. º. º s
A ≈
α
- iin º sin º. cos º co
= (^ )(^ )
= (^ )
D
D
c c
ss º. º º º tan. º
( ) = − = = =
α
- F F ((^ )(^ )
=
tan. º . sin. º sin. º
F
e e
e = = − =
sin. º. º. º. º
α
- ttan. º tan. º. cos. º
= (^ )(^ )
G
G ≈
H
H
H
cos. º cos. º. º º
α '' " º ' "
. º
θ ≈
- sinα = (^) 2 316 3 ≈.. cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
. º. º s
A ≈
α
- iin º sin º. cos º co
= (^ )(^ )
= (^ )
D
D
c c
ss º. º º º tan. º
( ) = − = = =
α
- F F ((^ )(^ )
=
tan. º . sin. º sin. º
F
e e
e = = − =
sin. º. º. º. º
α
- ttan. º tan. º. cos. º
= (^ )(^ )
G
G ≈
H
H
H
cos. º cos. º. º º
α '' " º ' "
. º
θ ≈
sinα =
cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
. º. º s
A ≈
α
- iin º sin º. cos º co
= (^ )(^ )
= (^ )
D
D
c c
ss º. º º º tan. º
( ) = − = = =
α
- F F ((^ )(^ )
=
tan. º . sin. º sin. º
F
e e
e = = − =
sin. º. º. º. º
α
- ttan. º tan. º. cos. º
= (^ )(^ )
G
G ≈
H
H
H
cos. º cos. º. º º
α '' " º ' "
. º
θ ≈
sinα =
cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
. º. º s
A ≈
α
- iin º sin º. cos º co
= (^ )(^ )
= (^ )
D
D
c c
ss º. º º º tan. º
( ) = − = = =
α
- F F ((^ )(^ )
=
tan. º . sin. º sin. º
F
e e
e = = − =
sin. º. º. º. º
α
- ttan. º tan. º. cos. º
= (^ )(^ )
G
G ≈
H
H
H
cos. º cos. º. º º
α '' " º ' "
. º
θ ≈
sinα =
cos. tan. ar
α α
ccsin.. º arcsin.. º tan
tan. º. sin. º sin
= (^ )(^ )
B
B
A
A
sin. º.
. º. º s
A ≈
α
- iin º sin º. cos º co
= (^ )(^ )
= (^ )
D
D
c c
ss º. º º º tan. º
( ) = − = = =
α
- F F ((^ )(^ )
=
tan. º . sin. º sin. º
F
e e
e = = − =
sin. º. º. º. º
α
- ttan. º tan. º. cos. º
= (^ )(^ )
G
G ≈
H
H
H
cos. º cos. º. º º
α '' " º ' "
. º
θ ≈
