The angle of elevation of a cloud from a point h metre above a lake is θ.

Solution:

Let  be the surface of the lake and  be the point of observation. So  .

The given situation can be represented as,

Here, is the position of the cloud and  is the reflection in the lake. Then .

Let  be the perpendicular from  on . Then  and .

Let , , then  and 

Here, we have to find the height of cloud.

So we use trigonometric ratios.

In ,

$\Rightarrow \tan \theta=\frac{C M}{P M}$

$\Rightarrow \tan \theta=\frac{x}{y}$

$\Rightarrow y=\frac{x}{\tan \theta}$   ...........(1)

Again in $\triangle P M C^{\prime}$,

$\Rightarrow \tan 45^{\circ}=\frac{C^{\prime} M}{P M}$

$\Rightarrow 1=\frac{C^{\prime} B+B M}{P M}$

$\Rightarrow \mathrm{I}=\frac{x+h+h}{y}$

$\Rightarrow y=x+2 h$

$\Rightarrow \frac{x}{\tan \theta}=x+2 h$ [Using (1)]

$\Rightarrow \frac{x-x \tan \theta}{\tan \theta}=2 h$

$\Rightarrow x=\frac{2 h \tan \theta}{(1-\tan \theta)}$

Now,

$\Rightarrow C B=h+x$

$\Rightarrow C B=h+\frac{2 h \tan \theta}{(1-\tan \theta)}$

$\Rightarrow C B=\frac{h(1-\tan \theta)+2 h \tan \theta}{1-\tan \theta}$

$\Rightarrow C B=\frac{h(1+\tan \theta)}{1-\tan \theta}=h \tan \left(45^{\circ}+\theta\right) \quad\left[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\right]$

Hence the correct answer is $a$.