If the angle of elevation of …

Let AB be the surface of the lake and P be the point of observation such that AP = 60 metres. Let C be the position of the cloud and C be its reflection in the lake. Then, CB = C'B. Let PM be perpendicular from P on CB. Then,{tex}\angle \mathrm { CPM } = 30 ^ { \circ }{/tex} and {tex}\angle C 'P M = 60 ^ { \circ }{/tex} Let CM = h. Then, CB = h + 60. Consequently, C'B = h + 60.
In {tex}\triangle \mathrm { CMP },{/tex} we have

{tex}\tan 30 ^ { \circ } = \frac { C M } { P M }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { h } { P M }{/tex} 
or, PM = √3 h........(I)
In {tex}\triangle P M C, ^ { \prime }{/tex} we have
{tex}\tan 60 ^ { \circ } = \frac { C ^ { \prime } M } { P M }{/tex}
{tex}\Rightarrow \quad \tan 60 ^ { \circ } = \frac { C B + B M } { P M }{/tex}
{tex}\Rightarrow \quad \sqrt { 3 } = \frac { h + 60 + 60 } { P M }{/tex}
{tex}\Rightarrow \quad P M = \frac { h + 120 } { \sqrt { 3 } }{/tex}.......(ii)
from equations (i) and (ii), we get
{tex}\sqrt { 3 } h = \frac { h + 120 } { \sqrt { 3 } } \Rightarrow 3 h = h + 120 \Rightarrow 2 h = 120 \Rightarrow h = 60{/tex}
Now, CB = CM + MB = h+ 60 = 60 + 60 = 120.
Hence, the height of the cloud from the surface of the lake is 120 metres.