The angle of elevation of a stationery cloud from a point $ 2500 \mathrm{~m} $ above a lake is $ 15^{\circ} $ and the angle of depression of its reflection in the lake is $ 45^{\circ} $. What is the height of the cloud above the lake level? $ \quad $ (Use $ \tan 15^{\circ}=0.268 $ ).

Given:

The angle of elevation of a stationery cloud from a point \( 2500 \mathrm{~m} \) above a lake is \( 15^{\circ} \) and the angle of depression of its reflection in the lake is \( 45^{\circ} \).

To do:

We have to find the height of the cloud above the lake level

Solution:

Let $A$ be the cloud, $C$ be the reflection in the lake and $B$ be the point of observation.

Let $AF=FC=h\ m$ and $BD=EF=x\ m$.

From the figure,

$\angle ABD =15^{o}, BE=DF=2500\ m$ and  $\angle DBC=45^{o}$

This implies,

$AD=h-2500\ m$

In $\vartriangle ABD$,

$tan\ 15^{o} =\frac{AD}{BD} =\frac{h-2500}{x}$

$x=\frac{h-2500}{\tan\ 15^{o}}$.........(i)

In $\vartriangle BDC$,

$tan 45^{o}=\frac{DC}{BD} =\frac{h+2500}{x}$

$1=\frac{h+2500}{x}$

$x=h+2500\ m$

Substituting $x=h+2500$ in equation (i), we get,

$h+2500=\frac{h-2500}{0.268}$

$(h+2500)0.268=h-2500\ m$

$h(1-0.268)=670+2500\ m$

$h=\frac{3170}{0.732} = 4330\ m$

Therefore, the height of the cloud above the lake level is $4300\ m$.

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Updated on: 10-Oct-2022

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