The angle of elevation of a …

Let us suppose that A be the point of observation and let B be the position of the cloud .

Suppose  PQ be the surface of lake.
Then C is the reflection of cloud and AP {tex}= 2500\ meter{/tex}
{tex}\angle BAD = 15^\circ {/tex} and {tex}\angle DAC = 45^\circ {/tex}
Let us suppose that BD = x
Then {tex}DQ = AP = 2500{/tex} meter and {tex}QC = BQ = BD + DQ = 2500 + x{/tex}
Also {tex}DC = DQ + QC = 2500 + 2500 + x = 5000 + x{/tex}
Now, In {tex}\Delta ADC{/tex},
{tex}\tan 45^\circ = \frac{{DC}}{{AD}}{/tex}
{tex} \Rightarrow 1 = \frac{{5000 + x}}{{AD}}{/tex}
{tex} \Rightarrow AD = 5000 + x{/tex} ................. (i)
{tex}\tan 15^\circ = \tan \left( {45^\circ - 30^\circ } \right) = \frac{{\tan 45^\circ - \tan 30^\circ }}{{1 + \tan 45^\circ\tan 30^\circ }} = \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{{\sqrt 3 }}}} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}{/tex}
Again, In {tex}\Delta ABD{/tex},
{tex}\tan 15^\circ = \frac{{BD}}{{AD}}{/tex}
{tex} \Rightarrow \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{x}{{5000 + x}}{/tex}
{tex} \Rightarrow \left( {5000 + x} \right)\left( {\sqrt 3 - 1} \right) = x\left( {\sqrt 3 + 1} \right){/tex}
{tex} \Rightarrow 5000\sqrt 3 - 5000 + \sqrt 3 x - x = \sqrt 3 x + x{/tex}
{tex} \Rightarrow 5000\left( {\sqrt 3 - 1} \right) = 2x{/tex}
{tex} \Rightarrow x = \frac{{5000}}{2}\left( {\sqrt 3 - 1} \right){/tex}
{tex} \Rightarrow x = 2500\left( {\sqrt 3 - 1} \right){/tex}
Now,
{tex}BQ = BD + DQ{/tex}
= x + 2500
{tex} = 2500\left( {\sqrt 3 - 1} \right) + 2500{/tex}
{tex} = 2500\sqrt 3 - 2500 + 2500{/tex}
{tex} = 2500\sqrt 3 {/tex}
Thus,  the height of the cloud above the lake level is {tex}2500\sqrt 3 m{/tex}.