Question

The angle of elevation of a cloud from a point ${}^{\prime }{h}^{\prime }m$ above the lake is $\alpha$ and angle of depression of reflection in the lake is $\beta$ . Prove that the height of the cloud from surface of water is $\frac{h(\tan \beta +\tan \alpha )}{\tan \beta -\tan \alpha }$.

Answer 1

Let $AN$ be the surface of the lake and $O$ be the point of observation such that $OA=hm$.

Let $P$ be the position of the cloud and $P^{\prime }$ be its reflection in the lake.

Then,

$PN=P^{\prime }N$

Let,

$OM\perp PN$

Also,

$\angle POM=\alpha$ and $\angle P^{\prime }OM=\beta$

Let,

$PM=x$

Then,

$PN=PM+MN=PM+OA=x+h$

In $\Delta POM$, we have

$\tan \alpha =\frac{PM}{OM}=\frac{x}{AN}$

$\Rightarrow AN=x\cot \alpha$ …… (1)

In $\Delta OM{P}^{\prime }$, we have

$\tan \beta =\frac{P^{\prime }M}{OM}=\frac{x+2h}{AN}$

$\Rightarrow AN=(x+2h)\cot \beta$ …… (2)

From equations (1) and (2), we have

$x\cot \alpha =(x+2h)\cot \beta$

$x(\cot \alpha -\cot \beta )=2h\cot \beta$

$x(\frac{1}{\tan \alpha }-\frac{1}{\tan \beta })=\frac{2h}{\tan \beta }$

$x\left(\frac{\tan \beta -\tan \alpha }{\tan \alpha \tan \beta }\right)=\frac{2h}{\tan \beta }$

$x=\frac{2h}{\tan \beta -\tan \alpha }$

Therefore, height of the cloud is,

$PN=x+h$

$PN=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }+h$

$PN=\frac{2h\tan \alpha +h(\tan \beta -\tan \alpha )}{\tan \beta -\tan \alpha }$

$PN=\frac{h(\tan \alpha +\tan \beta )}{\tan \beta -\tan \alpha }$

Hence, proved.