Question

If the angle of elevation of a cloud from a point h metres above a lake has measure $\alpha$ and the angle of depression of its reflection in the lake has measure $\beta$, prove that the height of the cloud is $\frac{h\left(\tan \beta +\tan \alpha \right)}{\tan \beta -\tan \alpha }$.

Answer 1

Let height of the cloud from the point be ${}^{\prime }{H}^{\prime }$

In $△ABC,$

$\Rightarrow \tan \alpha =\frac{H}{AC}$

$\Rightarrow AC=\frac{H}{\tan \alpha }⟶\left(1\right)$

In $△AFC,$

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

$\Rightarrow AC=\frac{H+2h}{\tan \beta }⟶\left(2\right)$

From $\left(1\right)\&\left(2\right),$

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

$\Rightarrow H\tan \beta =H\tan \alpha +2h\tan \alpha$

$\Rightarrow H(\tan \beta -\tan \alpha )=2h\tan \alpha$

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

$\therefore$ Height of the cloud from lake surface $=H+h$

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

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

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

Hence, the answer is proved.