If the angle of elevation of …

A is the cloud.
D is a point h metres above the lake {tex}\angle{/tex}ADB = {tex}\alpha {/tex}
A' is the reflection of A in the lake  {tex}\angle{/tex}ADB = {tex}\beta {/tex}
EC is the water level
In right triangle ABD,
tan {tex}\alpha {/tex} = {tex}\frac{{AB}}{{DB}}{/tex} ..........(1)
In right triangle ABD, tan {tex}\alpha {/tex} = {tex}\frac{{AB}}{{DB}}{/tex}.......(1)
tan {tex}\beta {/tex} = {tex}\frac{{A'B}}{{DB}} = \frac{{A'C + CB}}{{DB}}{/tex}

= {tex}\frac{{AC + CB}}{{DB}}{/tex}
{tex}\because {/tex} AC = A'C (by the law of reflection)
= {tex}\frac{{AB + BC + CB}}{{DB}} = \frac{{AB + h + h}}{{DB}} = \frac{{AB + 2h}}{{DB}}{/tex}  (2)
Dividing (1) by (2), we get {tex}\frac{{\tan \alpha }}{{\tan \beta }} = \frac{{AB}}{{AB + 2h}}{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{{\tan \alpha }}{{\tan \beta }} = \frac{{AB + 2h}}{{AB}}{/tex} ......By inverted
{tex}\Rightarrow {/tex} {tex}\frac{{\tan \beta }}{{\tan \alpha }} = 1 + \frac{{2h}}{{AB}}{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{{2h}}{{AB}} = \frac{{2h\tan \alpha }}{{\tan \beta - \tan \alpha }}{/tex} ........(3)
In right triangle ABD,cosec {tex}\alpha {/tex} = {tex}\frac{{AD}}{{AB}}{/tex}
{tex}\Rightarrow {/tex} AD = AB cosec {tex}\alpha {/tex}
{tex}\Rightarrow {/tex} AD = {tex}\frac{{2h\tan \alpha \cos ec\alpha }}{{\tan \beta - \tan \alpha }}{/tex}.........From (2)
{tex}\Rightarrow {/tex} AD = {tex}\frac{{2h\frac{{\sin \alpha }}{{\cos \alpha }}.\frac{1}{{\sin \alpha }}}}{{\tan \beta - \tan \alpha }}{/tex}
{tex}\Rightarrow {/tex} AD = {tex}\frac{{2h\sec \alpha }}{{\tan \beta - \tan \alpha }}{/tex}
Hence, the distance of the cloud from the point of observation is {tex}\frac{{2h\sec \alpha }}{{\tan \beta - \tan \alpha }}{/tex}