At a point $A$, $20 \; m$ above the level of water in a lake, the angle of elevation of a cloud is $30^\circ$. The angle of depression of the reflection of the cloud in the lake at $A$ is $60^\circ$. Find the distance of the cloud from $A$.
$OO' =$ Water surface
$A =$ Point of observation at a height of $20 \; m$ above the water surface
$C = $ Position of the cloud
$AC = $ Distance of the cloud from point $A$
$C' =$ Reflection of the cloud in the water
Then, $\;$ $O'C = O'C'$
Draw $AB \perp O'C$
Then, $O'B = OA = 20 \; m$
Let $BC = x$, $\;$ then $O'C' = 20 + x$
In $\triangle ABC'$, $\;$ $\dfrac{BC'}{AB} = \tan 60^\circ = \sqrt{3}$
i.e. $\;$ $AB = \dfrac{BC'}{\sqrt{3}} = \dfrac{BO' + O'C'}{\sqrt{3}} = \dfrac{20 + 20 + x}{\sqrt{3}}$
i.e. $\;$ $AB = \dfrac{40 + x}{\sqrt{3}}$ $\;\;\; \cdots \; (1)$
In $\triangle ABC$, $\;$ $\dfrac{BC}{AB} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
i.e. $\;$ $AB = BC \sqrt{3} = x \sqrt{3}$ $\;\;\; \cdots \; (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$,
$\dfrac{40 + x}{\sqrt{3}} = x \sqrt{3}$
i.e. $\;$ $40 + x = 3 x$ $\implies$ $x = 20$
In $\triangle ABC$, $\;$ $\dfrac{x}{AC} = \sin 30^\circ = \dfrac{1}{2}$
$\implies$ $AC = 2x = 2 \times 20 = 40$
$\therefore \;$ Distance of the cloud from point $A$ $= 40 \; m$