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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
4 days, 5 hours 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 · 3 days, 2 hours ago

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