The angle of elevation of a stationary cloud | vedclass

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(A) Let $H$ be the height of the cloud above the lake level and $h = 2500 \, m$ be the height of the observation point above the lake.Let $x$ be the h

Vedclass · Accepted Answer

$2500 \sqrt{3} \, m$

Vedclass · Answer

(A) Let $H$ be the height of the cloud above the lake level and $h = 2500 \, m$ be the height of the observation point above the lake.Let $x$ be the horizontal distance from the observation point to the cloud.From the angle of elevation,we have $	an(15^\circ) = \frac{H - h}{x}$,so $x = \frac{H - h}{	an(15^\circ)} = (H - h) \cot(15^\circ)$.From the angle of depression of the reflection,the depth of the reflection below the lake is $H$. The total vertical distance from the observation point to the reflection is $H + h$.Thus,$	an(45^\circ) = \frac{H + h}{x}$,so $x = \frac{H + h}{	an(45^\circ)} = (H + h) \cot(45^\circ)$.Equating the two expressions for $x$: $(H - h) \cot(15^\circ) = (H + h) \cot(45^\circ)$.Since $\cot(45^\circ) = 1$ and $\cot(15^\circ) = 2 + \sqrt{3}$,we have $(H - 2500)(2 + \sqrt{3}) = H + 2500$.$H(2 + \sqrt{3}) - 2500(2 + \sqrt{3}) = H + 2500$.$H(1 + \sqrt{3}) = 2500(3 + \sqrt{3}) = 2500 \sqrt{3}(\sqrt{3} + 1)$.$H = 2500 \sqrt{3} \, m$.

The angle of elevation of a stationary cloud from a point $2500 \, m$ above a lake is $15^\circ$ and the angle of depression of its reflection in the lake is $45^\circ$. The height of the cloud above the lake level is

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