The angle of elevation of a stationary cloud | vedclass

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(C) Let $h = 2500 \ m$ be the height of the observation point above the lake. Let $H$ be the height of the cloud above the lake surface. The distance

Vedclass · Accepted Answer

$2500 \sqrt{3}$

Vedclass · Answer

(C) Let $h = 2500 \ m$ be the height of the observation point above the lake. Let $H$ be the height of the cloud above the lake surface. The distance of the cloud from the observation point is $H-h$. The distance of the reflection of the cloud from the observation point is $H+h$.Let $x$ be the horizontal distance from the observation point to the cloud.In the triangle formed by the observation point, the cloud, and the horizontal line, we have $\cot 15^{\circ} = \frac{x}{H-h} \Rightarrow x = (H-h)(2+\sqrt{3}) \quad (i)$In the triangle formed by the observation point, the reflection of the cloud, and the horizontal line, we have $\cot 45^{\circ} = \frac{x}{H+h} \Rightarrow x = H+h \quad (ii)$Equating $(i)$ and $(ii)$:$(H-h)(2+\sqrt{3}) = H+h$$H(2+\sqrt{3}) - h(2+\sqrt{3}) = H+h$$H(2+\sqrt{3}-1) = h(2+\sqrt{3}+1)$$H(1+\sqrt{3}) = h(3+\sqrt{3})$$H = h \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = h\sqrt{3}$Given $h = 2500 \ m$, so $H = 2500\sqrt{3} \ m$.

The angle of elevation of a stationary cloud from a point $2500 \ m$ above a lake is $15^{\circ}$ and from the same point the angle of depression of its reflection in the lake is $45^{\circ}$. The height (in metres) of the cloud above the lake, given that $\cot 15^{\circ}=2+\sqrt{3}$, is

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