D.N.A.
1. Find the sin B, cos B,
and tan B.
11
C
5
6
A
2. Find the value of x.
B
14
12 
A
x
C
B
Find x. Round to the nearest tenth.
1)
B
45
A
x
41 
x
15 
C
A
15
2)
C
B
A
3)
17
C
x
34 
B
8 – 5: Angles of Elevation and
Depression
pp. 464 - 470
• Solve problems involving angles of elevation.
• Solve problems involving angles of depression.
• angle of elevation
• angle of depression
Standard 19.0 Students use trigonometric functions
to solve for an unknown length of a side of a right
triangle, given an angle and a length of a side. (Key)
Angle of Elevation
CIRCUS ACTS At the circus, a person
in the audience at ground level watches
the high-wire routine. A 5-foot-6-inch tall
acrobat is standing on a platform that is
25 feet off the ground. How far is the
audience member from the base of the
platform, if the angle of elevation from
the audience member’s line of sight to
the top of the acrobat is 27°?
Make a drawing
Angle of Elevation
Make a drawing
S o:h C a:h T o:a
= Opposite
Adjacent =
opp. 30.5
Tan 27 

adj.
x
.5095 30.5

1
x
.5095 x  30.5
30.5
x
 59.9
.5095
DIVING At a diving competition, a 6-foot-tall diver
stands atop the 32-foot platform. The front edge of
the platform projects 5 feet beyond the ends of the
pool. The pool itself is 50 feet in length. A camera is
set up at the opposite end of the pool even with the
pool’s edge. If the camera is angled so that its line of
sight extends to the top of the diver’s head, what is
the camera’s angle of elevation to the nearest
degree?
A.
B.
C.
D.
A
B
C
D
Make a drawing
Diver = 6’
S o:h C a:h T o:a
opp. 38
Tan x 

adj. 45
Tan x  0.8444
Platform
height =
32’
Opp = 38’
 0.8391
mx  40
Adj = 45’
Pool width = 50’
DIVING At a diving competition, a 6-foot-tall diver
stands atop the 32-foot platform. The front edge of
the platform projects 5 feet beyond the ends of the
pool. The pool itself is 50 feet in length. A camera is
set up at the opposite end of the pool even with the
pool’s edge. If the camera is angled so that its line of
sight extends to the top of the diver’s head, what is
the camera’s angle of elevation to the nearest
degree?
A. 37°
B. 35°
0%
D
0%
C
0%
B
D. 50°
0%
A
C. 40°
A.
B.
C.
D.
A
B
C
D
Angle of Depression
From her treehouse, Joan can look directly into her
bedroom window. The angle of depression from the
treehouse to the base of her house is
. 42The
 tree
is 10 meters from the base of the house. How far is
the treehouse from the base of the house?
adj.
Cos 42 
hyp
10
Cos 42 
x
10
10
x

 13.5m
cos 42 .7431
adj.
hyp.
Angle of Depression
A wheelchair ramp is 3 meters long and inclines at 6°.
Find the height of the ramp to the nearest tenth of a
centimeter.
A 0.3 cm
opp
Sin 6 
hyp
B 31.4 cm
C 31.5 cm
x
0.1045 
3
D 298.4 cm
 x
3(.1045)   3
 3
Opposite x
Multiply each side by 3.
Simplify.
Answer: The height of the ramp is about 0.314 meters,
or 0.314(100) = 31.4 centimeters. The answer
is B.
A roller coaster car is at one of its highest points. It
drops at a 63° angle of depression for 320 feet. How
long of a vertical distance was the drop?
A. 145 ft
0%
B. 628 ft
C. 359 ft
D. 285 ft
1.
2.
3.
4.
A
B
C
D
A
B
C
D
• Homework:
• pp 466 – 470, problems 1 – 10, 27, and
29 – 35
Indirect Measurement
Vernon is on the top deck of a cruise ship and
observes two dolphins following each other directly
away from the ship in a straight line. Vernon’s
position is 154 meters above sea level, and the
angles of depression to the two dolphins are 35° and
36°. Find the distance between the two dolphins to
the nearest meter.
x
35
154
Indirect Measurement
ΔMLK and ΔMLJ are right triangles. The distance
between the dolphins is JK or JL – KL. Use the right
triangles to find these two lengths.
Because
are horizontal lines, they are parallel.
Thus,
and
because they
are alternate interior angles. This means that mMJL  36
and mMKL  35.
Indirect Measurement
Vernon is on the top deck of a cruise ship and
observes two dolphins following each other directly
away from the ship in a straight line. Vernon’s
position is 154 meters above sea level, and the
angles of depression to the two dolphins are 35° and
36°. Find the distance between the two dolphins to
the nearest meter.
x
35
154
x
35
154
154
Tan 35 
x
154
.7002 
x
154
x
 219.94 m
.7002
xm
219.94m
y
36
154
y
36
154
154
Tan 36 
y
154
.7265 
y
154
y
 211.98 m
.7265
219.93 m
211.98 m
Answer: The distance between the dolphins is JK – KL.
JL – KL ≈ 219.93 – 211.96, or about 8 meters.
Madison looks out her second-floor window, which is
15 feet above the ground. She observes two parked
cars. One car is parked along the curb directly in
front of her window, and the other car is parked
directly across the street from the first car. The
angles of depression of Madison’s line of sight to the
cars are 17° and 31°. Find the distance between the
1.
A
two cars to the nearest foot.
A. 14 ft
2.
3.
4.
0%
B. 24 ft
C. 37 ft
D. 49 ft
A
B
C
D
B
C
D