See the sketches after the solutions.
1) A person looking at the top of a building sees an angle of elevation of 45 degrees, then
moves away 25 more feet and then sees an angle of elevation of 30 degrees to the top of the
building. How tall is the building?
Answer: Call x the distance from the person to the building when the person sees an angle of elevation
of 45 degrees. Call h the height of the building. Because the angle is 45 degrees, we know that h = x.
Now when the person moves back 25 more feet, the distance to the building is x + 25 = h + 25 and the
angle of elevation is 30 degrees. We have
1
h
= tan(30o ) = √
h + 25
3
So
√
3h = h + 25 and solving for h,
h= √
25
≈ 34.15
3−1
2) A drainage pipe falls 8 inches for 100 feet. What is the angle of depression?
Answer:
Convert all measurements to feet, and note that
tan(
8
12
100
) = tan(
What we want is the angle θ whose tangent is
2
3
100
) = tan(
2
)
300
2
300
So make sure your calculator is in the degree mode, not radian mode, and on a TI-83 you would enter
tan−1 (2/300) and get .38 approximately.
3) Karen is in a hot air baloon that is tethered at a height of 100 feet. How far away is her
car if the angle of depression from the balloon is 72.3 degrees?
Answer:
Call h the distance. Then h is the hypoteneuse of a right triangle, and
100
= cos(17.7o )
h
So
h=
100
≈ 104.97
cos(17.7)
4) A ramp is 12 feet long and 4 feet high. What is the angle of elevation?
Answer:
Call the angle θ. Then sin(θ) =
4
12
=
1
3
So
1
θ = sin−1 ( ) ≈ 19.47o
3
5) Mary is exactly 6 feet tall and she is standing 14 feet away from a light post. The tip of
her shadow make a 60o angle of elevation with the top of the post. How long is her shadow?
Answer: Call x the length of her shadow. Then
6
= tan(60o )
x
so
x=
6
6
= √ ≈ 3.46
tan(60o )
3
Notice that you did not need the information that she is standing 14 feet away from the light post.