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Trigonometry
Heights and Distances
1. Line of sight
2. Angle of Elevation
3. Angle of depression
1. Determine the angle of elevation of the sun when the length of the shadow of
a pole is 3 times the height of the pole.
Solution: tan = = = = tan 30
The angle of elevation of the sun = 30
2. An observer 1.5 m tall is 28.5m away from a tower 30m high. Determine the angle of elevation of the top of the tower from his eye.
Solution: CE = CD – CE
= 30 – 1. = 28.
In a Rt. 1e^ CEA
tan = = 1 = tan 45
Angle of elevation from the top of the tower = 45
AB x 1 BC 3 x 3
Pole x
B C
A
3 x
E A
C
28.5m 1.5m 1.5m
28 .5m
D B
30m
m observer
3. A kite is flying at a height of 75 metres from the level ground, attached to a string inclined at 60 to the horizontal. Find the length of string.
Solution: Let x be the length of the string.
sin 60 =
x = 75 x = x
= 50 x 1. 732 = 86.6 m
4. A tree 12 m high is broken by the wind in such a way that its top touches the ground
and makes an angle radians with the ground. At what height from the bottom of
the tree is broken by the wind?
Solution: Let x be the height from the bottom of the tree
cos 45 = =
12 – x = x 2
12 = x 2 + x
12 = x ( 2 + 1)
x = x =
x = 12 (0.414)
x = 4.968 m
x
3 75 2 x
2 150 3 3 3 3
150 x 3 3
? 75m x
60
x AB 12 – x AC
1 x 2 12 – x
45 45 x 12 – x
45
A
B x C
7. From the tower 80 metres high, the angle of depression of a car on the ground is found to be 30 12 . Find the distance of the car from the foot of the tower.
Solution: tan 30 12 =
tan 30 12 0.
The distance of the car
from the foot of the tower = 137.46 m
8. The shadow of a tower, when the angle of elevation of the sun is 45 is found to be 10 metres longer than when it is 60 . Find the height of the tower.
Solution: Let x be the height of the tower
In a Rt 1e^ ACD
tan 60 =
y =
DB = DA + AB
= y + 10
= + 10
In a Rt 1e^ CDB x x x x + 10 3 3 3
1 =
30 12
Tower 80m
30 12 car
A
B x C
x = = = 137.46 m
x
C
D y A 10m B
x
60 45
tan 45 = + 10
x y
x y
x 3
x 3
x 3 x + 10 3
x + 10 3 = x 3
10 3 = x 3 – x
10 3 = x ( 3 – 1)
= x
x = x =
= 15 + 5 3 = 15 + 8.66 = 23.66 m The height of the tower = 23.66 m
9. The angle of elevation of a tower at a point is 45 . After going 20 metres towards the foot of the tower the angle of elevation of the tower becomes 60 . Calculate the height of the tower.
Solution:
In a Rt. CDB tan 60 =
y =
In a Rt. ADC tan 45 =
x + 20 3 = 3 x
20 3 = 3 x – x
x y
x y
x 3
x y + 20
x x 3 x 3
C
y B 20m
x
60 45 D A
x = 3
In 1e^ ABP, tan 60 =
y = 3 z ... (1)
In CDP, tan 30 =
3 + z = 3 y … (2)
sub (1) in (2); 3 + z = 3 ( 3 x ) 3 + z = 3 z
z = … (3)
sub (3) in (1); y = 3
11. From the top of a tree, the angle of depression of an object on the horizontal ground is found to be 60 . On descending 20 ft from the top of the tree the angle of depression of the object is found to be 30 . Find the height of the tree.
Solution: In a Rt. 1e^ BCD tan 30 =
y = 3 x
In a Rt 1e^ ABD tan 60 = = =
y z y z
y 3 + z 1 y 3 3 + z
x y
1 x 3 y
AB x + 20 20 + x BD y 3 x
3 ( 3 x ) = 20 + x 3 x = 20 + x 3 x – x = 20 2 x = 20 x = 10 The height of the tree AB = AC + BC AB = 20 + 10 AB = 30 ft
12. The angle of elevation of a cloud from a point 60 m above a lake is 30 and angle of depression of the reflection of cloud in the lake is 60 . Find the height of the cloud.
Solution: In a 1e^ FBD tan 60 =
BF = ... (1)
Above the lake
In a Rt. 1e^ ABF tan 30 =
= h – 60
BF = 3 ( h – 60 ) ... (2)
Equations (1) = (2)
20 + x 3 x A
20
x
B y D
C 30
30 60
60
F 60 60m 60
E
A
h – 60
B h
C Lake level h
D
BD
BF
60 + h BF
60 + h 3
AB
BF
1 h – 60 3 BF
BF 3
cloud
22.17 = 40 – x
x = 40 – 22.
= 17.83 m
The height of the temple is 17.83 m
14. From an aeroplane flying horizontally above a straight road, the angles of depression of two consecutive kilometer stones on the road are observed to be 45 and 30 respectively. Find the height of the aeroplane above the road when the kilometer stones are (i) on the same side of the vertical through the aeroplane
Solution: In a Rt. 1e^ tan 45 =
x = y
In a Rt. 1e^ tan 30 = = = ( x = y)
x + 1 = 3 x
1 = 3 x – x
1 = x ( 3 – 1) 1 = x (.732)
1.366 = = x
The height of the aeroplane above the road = 1.366 km
x
45 30
road (kilometer stones)
x y x y
x x y + 1 x +
1 x 3 x + 1
C
D y A 1 km B
(ii) on the opposite sides of the vertical through the aeroplane.
Solution: tan 45 =
1 =
y = x
tan 30 = =
=
1 – x = 3 x
1 = ( 3 + 1) x
x = =
= 0.366 km
15. A man standing in the field observes a flying bird in his north at an angle of elevation of 30 and after 2.5 minutes he observes the bird in his South at an angle of elevation of 60 . If the bird flies in straight line all along at a height of 60 3 metres, find its speed.
Solution: tan 30 =
60 = x
tan 60 =
180 = y
45 30
x 45 30 A y 1 – y B
x y x y
x x 1 – y 1 – x 1 x 3 1 – x
x 60 3
1 x 3 60 3
y 60 3
N
x
W 30 E 60 3 60 y
S
17. Two vertical lamp – posts of equal height stand on either side of a roadway which is 50 m wide. At a point in the roadway, between the lamp – posts, the elevations of the tops of the lamp – posts are 60 and 30 . Find the height of each lamp – post and the position of this point.
Solution: tan 30 =
y = ... (1)
tan 60 =
y = (50 – x ) 3 ... (2)
= (50 – x ) 3 (from (1))
x = (50 – x ) 3 = 150 – 3 x 4 x = 150 x = 37.
y =
= 21.65 m
A C
y y
P
B x D
y 50 – x
y x
1 y 3 x
y 50 – x
x 3
x 3
37.5 x 1. 3
30 60 road (50 – x )
50 m
18. A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30 . Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45 . Both the boys are on opposite sides of both the kite. Find the length of the string that the second boy must have so that the two kites meet.
Solution: = sin 30
= x 100 = 50 m
BF = BE – EF
= 50 – 10 = 40 m In a 1e BCF
sin 45 =
BC = 40 2
= 40 (1.414) = 56. 56 m
Length of the string that the second boy must have = 56.56 m
B
100 m F 45 C
10 10 m 30
A E D
BE
AB
BE 1
AB 2
BF
BC
2 BC
