ANGLES OF ELEVATION AND DEPRESSION
Angle of depression
Angle of elevation
The figure shows a student looking at national flag at Merdeka Square. The line EF is perpendicular to the vertical flagpole at F, therefore EF is the horizontal line through the point of observation E. ET is the line of sight of the object T from E. The angle between EF and ET,< FET is known as the angle of elevation.
T (object)
F
E Horizontal line
ANGLE OF ELEVATION ?
The angle of elevation is the angle between the horizontal line and the line of sight of an object when the object is above the horizontal line.
A (object)
O
Angle of elevation Horizontal line
B
F The figure shows a base ball fan watching a base ball match from a pavilion. The angle between the horizontal line EF and the line of sight EB, < FEB, is known as the angle of depression B
Horizontal line
E
ANGLE OF DEPRESSION ?
The angle of depression is the angle between the horizontal line and the line of sight of an object when the object is below the horizontal line
O
Horizontal line Angle of depression
Y
X
P
Horizontal line Angle of depression
Q
R
Angle of elevation Horizontal line
S
< PQR = < QRS (Alternate angles). Angle of elevation and the corresponding angle of depression are equal.
How to solve problems involving angle of elevation and angle of depression ?
Sketch a diagram to represent the given situation Solve the problem using trigonometric ratios and/or Pythagoras¶ Theorem
Step 1 Step 2
TRIGONOMETRIC RATIO
OPPOSITE SIDE (O)
Sin Cos Tan
ADJACENT SIDE (A)
H = A H = O A
= O
PYTHOGORAS THEOREM
a2 + b2 = c2 b a
Solving Problems
Adriel works as a cattle herder. While looking over his cattle from the top of a hill, his eye, which is 1.5 m above the ground, made an angle of depression of 540 to the nearest cattle. The distance between the bottom of a hill and the nearest cattle is 165 m. Find, in m, the height of the hill.
K 1.5 m L 540
J
540 M 165 m N
Let the cattle = N. Height of the hill =LM
K 1.5 m L 540
J
540 M 165 m N
In ¨ KMN, Tan < MNK = KM MN Tan 540 = KM 165
K 1.5 m L 540
J
540 M 165 m N KM = 165 x Tan 540 KM = 165 x 1.376 KM = 227.04 m LM = KM - KL = 227.04 ± 1.5 = 225.54 m
V tower
500 m
P
R
In the diagram, VP is a tower 500 m tall. From V, the angle of depression of point R is 650 . Calculate the distance, in m, between P and R.
V 650 500 m
650 P R
V 650 500 m
650 P In ¨ VPR, Tan < 650 = VP PR = 500 PR R
V 650 500 m
650 P PR = 500 tan 650 = 233.15 m R
In Diagram 12, J and L are two points on a horizontal plane. LK is a vertical pole.
K
8m
J
12 m DIAGRAM 12
L
The angle of depression of point J from vertex K is A B C D 26° 24' 28° 46' 32° 24' 33° 41'
K
U
8m
TanU =
8 12
U = 33041¶
U
J
12 m
L
In Diagram 13, PQ and RS are two vertical poles on a horizontal plane. The angle of elevation of P from R is 35° .
P
tm 15 m Q DIAGRAM 13
R 4m S
Calculate the value of t. A B C D 10.5 14.5 16.5 20.5