Question

From a point $a$ metre above the lake, the angle of elevation of a cloud is $\alpha$ and the angle of depression of its reflection is $\beta$, then height of the cloud is .......

• A. $\frac{a\sin (\alpha +\beta )}{\sin (\alpha -\beta )}m$
• B. $\frac{a\sin (\alpha +\beta )}{\sin (\beta -\alpha )}m$
• C. $\frac{a\sin (\beta -\alpha )}{\sin (\alpha +\beta )}m$
• D. None of these

Answer 1

The correct option is B $\frac{a\sin (\alpha +\beta )}{\sin (\beta -\alpha )}m$

Let $OQ$ be the surface of lake ,$P$ be the point $a$ metre above the lake ,$C$ is the point where cloud is present at height $H$ and $C^{\prime }$ is reflection of cloud under the lake.

In $△PAC,\tan \alpha =\frac{H-a}{x}$
$\Rightarrow x=(H-a)\cot \alpha$ ....(i)

In $△PA{C}^{\prime },\tan \beta =\frac{H+a}{x}$
$\Rightarrow x=(H+a)\cot \beta$ ....(ii)

From equations $\left(i\right)$ and $\left(ii\right)$, we get
$(H+a)\cot \beta =(H-a)\cot \alpha$

$H\cot \beta +a\cot \beta =H\cot \alpha -a\cot \alpha$
$H\cot \alpha -H\cot \beta =a\cot \alpha +a\cot \beta$
$\Rightarrow H=\frac{a(\cot \alpha +\cot \beta )}{\cot \alpha -\cot \beta }$
$\frac{\cot \alpha +\cot \beta }{\cot \alpha -\cot \beta }=\frac{\frac{\cos \alpha }{\sin \alpha }+\frac{\cos \beta }{\sin \beta }}{\frac{\cos \alpha }{\sin \alpha }-\frac{\cos \beta }{\sin \beta }}$

$=\frac{\cos \alpha \sin \beta +\cos \beta \sin \alpha }{\cos \alpha \sin \beta -\cos \beta \sin \alpha }=\frac{\sin (\alpha +\beta )}{\sin (\beta -\alpha )}$

$\therefore H=\frac{a\sin (\alpha +\beta )}{\sin (\beta -\alpha )}$