Question

The angle of elevation of a cloud from a point $x$ $m$ above a lake is $\theta$ and the angle of depression of its reflection in the lake is ${45}^o$. The height of the cloud is

• A. $x\tan ({45}^o-\theta )$
• B. $x\tan ({45}^o+\theta )$
• C. $\frac{1}{x}\cot ({45}^o-\theta )$
• D. $\frac{1}{x}\cot ({45}^o+\theta )$

Answer 1

The correct option is B $x\tan ({45}^o+\theta )$

Let $A$ be the position of the cloud and $B$ be the reflection of the cloud in lake.

Let $E$ be the points $xmetres$ above the lake making angle of angle of elevation $\theta$ with $A$ and angle of depression ${45}^{\circ }$ with $B$

Let $h$ be the height of the cloud

Now, from the figure

$EF=DC=x$

$AC=BC=h$

$AD=h-x$

$BD=h+x$

In $\Delta AED$

$\tan \theta =\frac{AD}{ED}=\frac{h-x}{ED}$ ------- $\left(1\right)$

In $\Delta EDB$

$\tan {45}^{\circ }=\frac{BD}{ED}$

$ED=h+x$ ----- $\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\tan \theta =\frac{h-x}{h+x}$

$\begin{array}{c}h\left(\tan \theta -1\right)=-x\left(\tan \theta +1\right)\\ \\ h=\frac{x\left(1+\tan \theta \right)}{1-\tan \theta }\\ \\ h=\frac{x\left(\tan {45}^{\circ }+\tan \theta \right)}{1-\tan {45}^{\circ }\tan \theta }\left(1=\tan {45}^{\circ }\right)\\ \\ =x\tan \left({45}^{\circ }+\theta \right)\end{array}$

Thus the height of the cloud is $x\tan \left({45}^{\circ }+\theta \right)$

Hence,

Option $B$ is the correct answer.