Question

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

Answer 1

Let $A$ be a point $hm$ above the lake $AF$ and $B$ be the position of the cloud.

Draw a line parallel to $EF$ from $A$ on $BD$ at $C$.

But, $BF=DF$

Let, $BC=m$

so, $BF=(m+h)$

$\Rightarrow$ $BF=DF=(m+h)m$

Consider $△BAC$, $\sin \alpha =\frac{BC}{AB}$

$\Rightarrow$ $\sin \alpha =\frac{m}{AB}$

$\therefore$ $AB=mcosec\alpha$ ---------- (1)

and, $AC=m\cot \alpha$

Consider $△ACD,$ $\tan \beta =\frac{CD}{AC}$

$\Rightarrow$ $\tan \beta =\frac{2h+m}{AC}$

$AC=(2h+m)\cot \beta$

Therefore, $m\cot \alpha =(2h+m)\cot \beta$

Substituting the value of m in (1) we get,

$\Rightarrow$ $AB=cosec\alpha \left[2h\cot \beta /\right(\cot \alpha -\cot \beta \left)\right]$

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