Simply!!

Let ANAN be the surface of the lake and OO be the point of observation such that OA=hOA=h meters.

Let PP be the position of the cloud and BB be its reflection in the lake.

Then PN = BN\text{Then PN = BN}

POM=α,BOM=β\angle POM = \alpha , \angle BOM = \beta

Let PM=x\text{Let } PM = x

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

In\text{In} ΔABC,\Delta ABC,

tanα=PQQM=xAN\tan\alpha = \dfrac{PQ}{QM} = \dfrac{x}{AN}

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

In\text{In} ΔOMB\Delta OMB

tanβ=DMOM=h+2xAN\tan\beta = \dfrac{DM}{OM} = \dfrac{h+2x}{AN}

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

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

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

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

    x(1tanα1tanβ)=2htanβ\implies x(\dfrac{1}{\tan\alpha} - \dfrac{1}{\tan\beta}) = \dfrac{2h}{\tan\beta}

    x(tanβtanα)tanα.tanβ=2htanβ\implies \dfrac{x(\tan\beta - \tan\alpha)}{\tan\alpha . \tan\beta} = \dfrac{2h}{\tan\beta}

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

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

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

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

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

Hence\displaystyle \huge{Hence} Proved\huge{Proved}

Note by Sai Ram
2 years, 8 months ago

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What is the question that you are answering?

If it's a solution to an existing problem, please post it directly.

Calvin Lin Staff - 2 years, 8 months ago

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