# Simply!!

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

Let $P$ be the position of the cloud and $B$ be its reflection in the lake.

$\text{Then PN = BN}$

$\angle POM = \alpha , \angle BOM = \beta$

$\text{Let } PM = x$

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

$\text{In}$ $\Delta ABC,$

$\tan\alpha = \dfrac{PQ}{QM} = \dfrac{x}{AN}$

$\implies AN = x.\cot\alpha...............\boxed{1}$

$\text{In}$ $\Delta OMB$

$\tan\beta = \dfrac{DM}{OM} = \dfrac{h+2x}{AN}$

$\implies AN = (x+2h)\cot\beta.............\boxed{2}$

$\text{Equating}$ $\boxed{1}$ $\text{and}$ $\boxed{2}$

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

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

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

$\implies \dfrac{x(\tan\beta - \tan\alpha)}{\tan\alpha . \tan\beta} = \dfrac{2h}{\tan\beta}$

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

$\text{Now height of the cloud is given by PN = x+h}$

$\implies \text{Height of Cloud} = \dfrac{2h\tan\alpha}{\tan\beta - \tan\alpha} + h$

$\implies \text{Height of Cloud} = \dfrac{2h\tan\alpha + h(\tan\beta - \tan\alpha)}{\tan\beta - \tan\alpha}$

$\implies\boxed{\boxed{ \text{Height of Cloud} = \dfrac{h(\tan\alpha + \tan\beta)}{\tan\beta - \tan\alpha}}}$

$\displaystyle \huge{Hence}$ $\huge{Proved}$ Note by Sai Ram
3 years, 1 month ago

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