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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
10 months, 3 weeks ago

## Comments

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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.

Staff - 10 months, 3 weeks ago

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