At a point A, 20 meter above the level of water in a lake, the angle of elevation of a cloud is $30^{o}$. The angle of depression of the reflection of the cloud in the lake, at a A is $60^{o}$. Find the distance of the cloud from A.

Given: At a point A, 20 meter above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at a A is 60°. 

To do: To Find the distance of the cloud from A.

Solution:

Let AB be the surface of the lake and P be the point of observation such that

$AP = 20 meter$. Let C be the position of the cloud and C’ be its reflection in the lake.

Then $CB = C'B$. Let we assume PM be perpendicular from P on CB.

$\angle CPM=30^{o}$

And $\angle C'PM = 60^{o} $

Let $CM = h$. Then $CB = h + 20$ and $C’B = h + 20$.

In $\vartriangle CMP$ we have,

$tan30^{o}=\frac{CM}{PM}$

$\Rightarrow \frac{1}{\sqrt{3}} =\frac{h}{PM}$

$\Rightarrow PM=h\sqrt{3} \ \ ..........( 1)$

In $\vartriangle PMC'$ :

$tan60^{o}=\frac{C'M}{PM}$

$\Rightarrow \sqrt{3} =\frac{C'B+BM}{PM}$

$\Rightarrow \sqrt{3} =\frac{h+20+20}{PM}$

$\Rightarrow PM=\frac{h+40}{\sqrt{3}} ...........( 2)$

On comparing$( 1)$ and $( 2)$ , we get

$h\sqrt{3} =\frac{h+40}{\sqrt{3}}$

$\Rightarrow h\sqrt{3} \times \sqrt{3} =h+40$

$\Rightarrow 3h=h+40$

$\Rightarrow 2h=40$

$\Rightarrow h=\frac{40}{2}$

$\Rightarrow h=20\ m$

Now, $CB =CM+MB=h+20+20+20=40$.

Hence, The height of the cloud from the surface of the lake is 40 meter.
 

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Updated on: 10-Oct-2022

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