Question

The angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of its reflection in the lake is B. Prove that the height of the cloud is
$\sf \: \frac{h(tan \: \alpha + tan \: \beta )}{tan \: \alpha - tan \: \beta} \: meters$
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Answer 1

Let AB be the surface of lake and let P be a point vertically above A such that AP = h meters.

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Let C be the position of cloud and

let D be its reflection on the lake.

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Draw $\sf PQ \perp CD$

$\sf \angle \: QPC = \alpha$

$\sf \angle \: QPD = \beta$

BQ = AP = h meters

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Let CQ = x meters. Then,

BD = BC = (x + h) meters.

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From right ∆PQC, we have

$\sf \: \frac{PQ}{CQ} = cot \: \alpha$

$\sf \implies \: \frac{PQ}{x \: m} = cot \: \alpha$

$\sf \implies \: PQ = xcot \alpha \: meters$

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ-----(i)

From right ∆PQD, we have

$\sf \: \frac{PQ}{QD} = cot \: \beta$

$\sf \implies \frac{PQ}{(x + 2h) \: m} = cot \: \beta$

$\sf \implies \: PQ = (x + 2h)cot \beta \: meters$

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ-----(ii)

From (i) and (ii)

$\sf \: xcot \alpha = (x + 2h)cot \beta$

$\sf \implies \: x( \cot \alpha - cot \beta ) = 2hcot \beta$

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

$\sf \implies x( \frac{tan \: \beta - tan \: \alpha }{tan \: \beta tan \: \alpha} ) = \frac{2h}{tan \: \beta }$

$\sf \implies x = \frac{2htan \: \alpha }{(tan \: \beta - tan \alpha )}$

$\therefore$ height of cloud from the surface of lake

$\sf \: = (x + h) = \bigg\{\dfrac{2htan \: \alpha }{(tan \: \beta - tan \: \alpha )} + h \bigg\} \: m$

$\sf = \sf \bigg\{\dfrac{h(tan \: \alpha + tan \: \beta )}{(tan \: \beta - tan \: \alpha )} \bigg\} \: meters$