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. The depth of the refl

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$2500 \sqrt{3}$

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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. The depth of the reflection of the cloud in the lake is also $H$ below the surface of the lake. The distance of the cloud from the observation point is $H-h$. The distance of the reflection from the observation point is $H+h$. Let $d$ be the horizontal distance from the observation point to the point directly below the cloud. In the triangle formed by the observation point,the cloud,and the point directly below it: $\cot 15^{\circ} = \frac{d}{H-h} \Rightarrow d = (H-h)(2+\sqrt{3}) \quad \dots(i)$ In the triangle formed by the observation point,the reflection,and the point directly below it: $\cot 45^{\circ} = \frac{d}{H+h}$ $\Rightarrow 1 = \frac{d}{H+h}$ $\Rightarrow d = H+h \quad \dots(ii)$ 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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