Question

The angle of elevation of a stationary cloud from a point $2500$ m above a lake is ${15}^0$ and from the same point the angle of depression of its reflection in the lake is ${45}^o$. The height (in meters) of the cloud above the lake, given that $\cot {15}^o=2+\sqrt{3}$, is

• A. $2500$
• B. $2500\sqrt{2}$
• C. $2500\sqrt{3}$
• D. $5000$

Answer 1

The correct option is C $2500\sqrt{3}$

Let $EB=x$m, then

$\tan {15}^0=\frac{h}{x}$

$\Rightarrow x=h\cot {15}^0$ ...(1)

and from triangle EBD, we can get $\tan {45}^o=\frac{h+5000}{x}$

$\to x\cdot 1=h+5000$ ...(2)

From (1) and (2)

$\Rightarrow h\cot {15}^0=h+5000$
$\Rightarrow h(2+\sqrt{3})=h+5000$

$\Rightarrow h=\frac{5000}{\sqrt{3}+1}=\frac{5000(\sqrt{3}-1)}{3-1}=2500(\sqrt{3}-1)m$

$\therefore$ height of cloud is $2500+2500(\sqrt{3}-1)$ $=2500\sqrt{3}$m