| The angle of elevation of a stationary cloud from a point 25 m above a lake is 15

^{o}and the angle of depression of its image in the lake is 45^{o}. The height of the cloud above the lake level isA. 25 m

B.

C. 50 m

D.

### Right Answer is: B

#### SOLUTION

PQ is the lake and A is the point of observation

Given AP =25 m, angle BAD = and angle DAC=

Let BD=x

Then DQ =AP = 25 m and QC = BD+DQ (Because Image will form at distance 25 + x From the surface of lake) = X+25

Also, DC = DQ + QC = 25 +x + 25 = 50 + x

Now in triangle ADC

⇒

⇒

Also

In triangle ABD

⇒

⇒

⇒

⇒

⇒

⇒

Now BQ = BD + DQ =