Question

If the angle of elevation of a cloud from a point H metres above a lake is alpha and angle of depression of its reflection in the lake is beta , prove that height of the cloud is H(tan $\beta$ +tan $\alpha$ ) / tan$\beta$ -tan  $\alpha$. THIS QUESTION IS REALLY VERY IMPORTANT FROM EXAM'S POINT OF VIEW.[HIGHER ORDER THINKING SKILLS QUESTION]

Answer 1

Here it is solved more clearly and in detail ...




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Answer 2

Proof:

Note: Refer the image attached for pictorial reference.

As per the let us assume that P is the point which is at H meters distance from the lake,

The cloud point is C and the reflection point of the cloud is C`

Let us also assume that the distance of the cloud from the lake is X meters,

So we can say $B C=B C^{\prime}=X$

As $B M=A P=h$,

Hence $C M=X-h and M C^{*}=X+h$

So in$\Delta C P M, \frac{C M}{P M}=\tan \alpha$

$P M=\frac{C M}{\tan \alpha}=P M=\frac{X-h}{\tan \alpha}$……… (i)

The same way for$\Delta C^{*} P M, \frac{C^{\prime} M}{P M}=\tan \beta$

$P M=\frac{C^{\prime} M}{\tan \beta}=P M=\frac{X+h}{\tan \beta}$……………. (ii)

So from (i) and (ii) we can say,

$\frac{H(\tan \beta+\tan \alpha)}{\tan \beta-\tan \alpha}$

$(X-h) \tan \beta=(X+h) \tan \alpha$

$X \tan \beta-h \tan \beta=X \tan \alpha+h \tan \alpha$

$X(\tan \beta-\tan \alpha)=h(\tan \beta+\tan \alpha)$

$X=\frac{h(\tan \beta+\tan \alpha)}{(\tan \beta-\tan \alpha)}$

Hence proved that the height of the cloud is $\frac{H(\tan \beta+\tan \alpha)}{\tan \beta-\tan \alpha}$