Question

The angle of elevation of a stationary cloud from a point $2500$m above a lake is ${15}^{th}$ and the angle of depression of its reflection in the lake is ${45}^o$. The height of the cloud above the lake level is?

• A. $2500\sqrt{3}$m
• B. $2500$m
• C. $500\sqrt{3}$m
• D. $250\sqrt{3}$m

Answer 1

The correct option is A $2500\sqrt{3}$m

REF.image

In $△AOD,tan15=\frac{AD}{OD}\Rightarrow AD=OD.tan15$

In $△ODC,tan45=\frac{DC}{OD}\Rightarrow DC=OD$

$\Rightarrow AD=DCtan15$

$\Rightarrow (h-2500)=(h+2500)\left(tan\right(45-30\left)\right)$

$\Rightarrow (h-2500)=(h+2500)\left(\frac{1{-}^1/\sqrt{3}}{1{+}^1/\sqrt{3}}\right)$

$\Rightarrow h(1-\frac{\sqrt{3}-1}{\sqrt{3}+1})=2500(1+\frac{\sqrt{3}-1}{\sqrt{3}+1})\Rightarrow h=2500\left(\frac{2\sqrt{3}}{\sqrt{3}+1}\right)\left(\frac{\sqrt{3}+1}{2}\right)$

$\Rightarrow h=2500\sqrt{3}$