If the angle of elevation of a cloud from a point $ h $ metres above a lake is $ \alpha $ and the angle of depression of its reflection in the lake be $ \beta $, prove that the distance of the cloud from the point of observation is $ \frac{2 h \sec \alpha}{\tan \beta-\tan \alpha} $.

Given:

The angle of elevation of a cloud from a point \( h \) metres above a lake is \( \alpha \) and the angle of depression of its reflection in the lake be \( \beta \).

To do:

We have to prove that the distance of the cloud from the point of observation is \( \frac{2 h \tan \alpha}{\tan \beta-\tan \alpha} \).

Solution:

Let $A$ be the cloud, $C$ be the reflection in the lake and $B$ be the point of observation.

Let $AD=y\ m$, $FC=h+y\ m$ and $BD=EF=x\ m$.

From the figure,

$\angle ABD =\alpha, BE=DF=h\ m$ and  $\angle DBC=\beta$

In $\vartriangle ABD$,

$tan\ \alpha =\frac{AD}{BD} =\frac{y}{x}$

$x=\frac{y}{\tan\ \alpha}$.........(i)

In $\vartriangle BDC$,

$tan\ \beta=\frac{DC}{BD} =\frac{y+h+h}{x}$

$x=\frac{2h+y}{\tan \beta}$.........(ii)

From (i) and (ii), we get,

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

$y\tan \beta=(y+2h)\tan \alpha$

$y(\tan \beta-\tan \alpha)=2h\tan \alpha$

$y=\frac{2h\tan \alpha}{\tan \beta-\tan \alpha}$..............(iii)

In $\vartriangle ABD$,

$sin\ \alpha =\frac{AD}{AB} =\frac{y}{AB}$

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

$=\frac{\frac{2h(\frac{\sin \alpha}{\cos \alpha})}{\tan \beta-\tan \alpha}}{\sin \alpha}$

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

Hence proved.

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Updated on: 10-Oct-2022

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