Correct Option: D

AB is the surface of lake. C’ is the reflection of cloud ‘C’.
∠CPM = 30° and ∠C'PM = 60°
Let, C M = h metre
CB = (h + 200) metre
C'B = (h + 200) metre
In ∆CMP,

tan 30° =	CM
	PM

⇒	1	=	h
	√3		PM

⇒ PM = √3 h ....(i)
In ∆PMC',

tan 60° =	C'M
	PM

⇒ tan 60° =	C'B + BM
	PM

⇒ √3 =	h + 200 + 200
	PM

⇒ PM =	h + 400	.... (ii)
	√3

From equations (i) and (ii) ,

√3 h =	h + 400
	√3

⇒ 3h = h + 400
⇒ 2h = 400 ⇒ h = 200
∴ CB = h + 200 = 400 metre
Note : If the angle of elevation of a cloud from a point h metre above a lake is a and the angle of depression of its reflection in the lake is b, then

The height of the cloud =	h( tanβ + tanα )
	( tanβ - tanα )