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Q.

The angle of elevation of a cloud from a point 100 meter above the surface of a lake is 30° and the angle of depression of its image in the lake is 60° then height of the cloud above the lake is


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a

100 m

b

50 m

c

200 m

d

150 m 

answer is C.

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Detailed Solution

It is given that the angle of elevation of a cloud from a point 30° and is 100 meter above the surface of a lake. The angle of depression of its image in the lake is 60°.
According to the given data the figure has been drawn below.
Question ImageIn ΔABC,   tanθ= Perpendicular base  
tan 30 ° = AC BC    tan30°=h-100d
              h-100d=13          tan30°=13    .....(1)
In ΔBCD,  
tanθ= Perpendicular base  
tan 60 ° = CD BC  
 tan 60°=h+100d
               h+100d=3       .tan60°=3   ...........(2)
Dividing equation (1) by (2) , we get
 h-100h+100=13  3h-300=h+100 3h-h=300+100
 2h=400  h=200 m
Height above the lake is 200 m.
So the correct option is 3.
 
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