Problems involving Angle of Depression
In this section, we try to solve problems when Angles of
depression are given.
Angle of Depression and Angle of Elevation are equal become they
are alternative angles.
A player sitting on the top of a tower of height 20 m observes the
angle of epression of a ball lying on the ground as 60° . Find the distance between the foot of the
tower and the ball. (√3 = 1.732)
Let BC be the height of the tower and A be the position of the ball lying on the ground. Then,
BC = 20 m and ∠XCA = 60° = ∠CAB
Let AB = x metres.
In right triangle ABC,
tan 60° = BC/AB
Hence, the distance between the foot of the tower and the ball is
11.54 m.
Example 6.27
The horizontal distance between two buildings is 140 m. The angle of depression of the top of
the first building when seen from the top of the second building is 30° . If the height of the first building is 60 m,
find the height of the second building. (√3 = 1.732)
Solution
The height of the first building AB = 60 m.
Now, AB = MD = 60 m Let the height of the second building
CD = h. Distance BD = 140 m
Now, AM = BD = 140 m
From the diagram,
∠XCA = 30° = ∠CAM
CM = 80.78
Now, h = CD = CM +MD = 80.78+60 =
140.78
Therefore the height of the second building is 140.78 m
Example 6.28
From the top of a tower 50 m high, the angles of depression of the top and bottom of a tree are
observed to be 30° and 45° respectively. Find the height of the tree. (√3
= 1.732)
Solution
The height of the tower AB = 50 m
Let the height of the tree CD = y and BD = x
From the diagram, ∠XAC = 30° = ∠ACM and ∠XAD = 45° = ∠ADB
Therefore, height of the tree = CD = MB = AB
−AM = 50 −28. 85 = 21. 15 m
Example 6.29
As observed from the top of a 60 m high light house from the sea level, the angles of depression
of two ships are 28° and 45° . If one ship is exactly behind the other on
the same side of the lighthouse, find the distance between the two ships. (tan 28° =0.5317)
Solution
Let the observer on the lighthouse CD be at D.
Height of the lighthouse CD = 60 m
From the diagram,
∠XDA = 28° = ∠DAC and
∠XDB = 45° = ∠DBC
Distance between the two ships AB = AC −BC = 52.85 m
Example 6.30
A man is watching a boat speeding away from the top of a tower.
The boat
makes an angle of
depression of 60° with the man’s eye when at a distance of 200 m from the
tower. After 10 seconds, the angle of depression becomes 45°. What is the
approximate speed of the boat (in km / hr), assuming that it is sailing in
still water? (√3 = 1.732)
Solution
Let AB be the tower.
Let C and D be the positions of the boat.
From the diagram,
∠XAC = 60° = ∠ACB and
∠XAD = 45° = ∠ADB , BC = 200 m
In right triangle ABC, tan 60° = AB/BC
gives √3 = AB /200
we get AB = 200√3 ...(1)
In right triangle ABD, tan 45° = AB/BD
Gives 1 = 200√3 / BD [by (1) ]
we get, BD = 200√3
Now, CD = BD − BC
CD = 200√3 − 200 = − 200(√3
– 1) = 146 4.
It is given that the distance CD is covered in 10 seconds.
That is, the distance of 146.4 m is covered in 10 seconds.
Therefore, speed of the boat = distance / time
= 146.4 / 10 = 14. 64 m/s
gives 14. 64 × (3600/1000) km/hr = 52. 704 km/hr
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