Download Section 5.5

5.5 – Applications of Trigonometric Ratios
Curriculum Outcomes:
D6
solve problems involving measurement using bearings and vectors
D7
determine the accuracy and precision of a measurement (optional)
D8
solve problems involving similar triangles and right triangles
D12 solve problems using the trigonometric ratios
Angle of Elevation
The angle of elevation is an angle formed between your horizontal line of sight and the line of
sight when you look upward (inclination).
eye
angle of elevation
Horizontal line of sight
Example 1: A tree casts a shadow 9 m long when the angle of elevation of the sun is 62º. How
tall is the tree?
A
The angle of elevation is
inside the triangle across
from the right angle.
x
62°
C
9m
Solution 1:
tan B 
opp
adj
tan 62  
x
9
B
x
9
1.8807 
1.8807  (9)  x
x  16.9 m
Therefore the tree is 16.9 m tall.
Angle of Depression
The angle of depression is an angle formed between your horizontal line of sight and the line of
sight when you look downward.
eye
Horizontal line of sight
angle of depression
Example 2: While standing on a cliff near Peggy’s Cove, you see a yacht at an angle of
depression of 16º. If the cliff is 24.7 m. high, how far is the yacht from the shore directly below
you?
A
16°
74°
The angle of depression
hyp
24.7 m
is at the top and outside
of the triangle
opp
and outside the
16°
B
adj
C
triangle.
Solution 2:
tan C 
tan 16  
0.2867 
0.2867  ( x) 
opp
adj
24.7
x
24.7
x
24.7
 ( x)
x
0.2867 x
24.7

0.2867
0.2867
tan A 
OR
opp
adj
tan 74  
x
24.7
3.4874 
x
24.7
3.4874  (24.7) 
x
 (24.7)
24.7
x  86.1 m
x  86.2 m
Therefore the yacht is 86.2 m from the shore.
Exercises:
1. Two surveyors are at sea level on opposite sides of Kelly’s mountain that is 780 m. high.
A marker is placed at the top such that both surveyors can see it. Each finds their angle
of elevation to the marker. One angle is 48.3º and the other is 32.5º. What would be the
length of a tunnel between the positions of the surveyors?
2. A forest ranger in a tower 128 m high, spots two fires in the same line. One fire is at an
angle of depression of 42.7º and the other at 61.9º. How far apart are the two fires?
Answers:
A
1.
let BC = x
780 m
B
let BD = y
48.3°
32.5°
x
C
In Δ ABC
tan B 
opp
adj
tan 32.5  
0.6371 
0.6371  ( x) 
780
x
780
x
780
 ( x)
x
0.6371 x
780

0.6371
0.6371
x  1224.3 m
In Δ ACD
tan D 
opp
adj
tan 48.3 
1.1223 
1.1223  ( y ) 
780
y
780
y
780
 ( y)
y
1.1223 y
780

1.1223
1.1223
x  695 m
y
D
The tunnel distance, BD , is BD = BC + CD or x + y. Therefore the tunnel distance is
(1224.3 m + 695 m) or 1.92 km.
Distance BD = 1224.3 m + 695 m
Distance BD = 1919.3 m.
Distance BD = 1.92 km.
A
2.
42.7°
61.9°
128 m
61.9°
B
C
y
x
tan C 
opp
adj
tan 42.7  
0.9228 
D
let BD = x
let BC = y
In Δ ABD
0.9228  ( x) 
42.7°
128
x
128
x
128
 ( x)
x
0.9228 x
128

0.9228
0.9228
x  138.7 m
In Δ ABC
tan D 
opp
adj
tan 61.9  
1.8728 
1.8728  ( y ) 
128
y
128
y
128
 ( y)
y
1.8728 y
128

1.8728
1.8728
x  68.3 m
The distance between the two fires, CD , is CD = BD - BC or x – y. Therefore the distance
between the two fires is (138.7 m – 68.3 m) or 70.4 m.
Distance CD = 138.7 m – 68.3 m
Distance CD = 70.4 m.