Question

The angle of elevation of a cloud from a point 60 m above a lake is ${30}^{\circ }$ and the angle of depression of the reflection of cloud in the lake is ${60}^{\circ }$. Find the height of the cloud.

Answer 1

Let AB be the surface of the lake and let P be a point vertically above A such that AP = 60 m



Let C be the position of the cloud and let D be its reflection in the lake.

Draw $PQ\perp CD$, Then,

$\angle QPC={30}^{\circ },\angle QPD={60}^{\circ }$,

BQ = AP = 60 m

Let $CQ=xmetres$. Then,

$BD=BC=(x+60)m$

From right $\Delta PQC$, we have

$\frac{PQ}{CQ}=\cot {30}^{\circ }=\sqrt{3}$

$\Rightarrow \frac{PQ}{x}=\sqrt{3}$

$\Rightarrow PQ=x\sqrt{3}m$ -----------(i)

From right $\Delta PQD$, we have

$\frac{PQ}{QD}=\cot {60}^{\circ }=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{PQ}{(x+60+60)m}=\frac{1}{\sqrt{3}}$

$\Rightarrow PQ=\frac{(x+120)}{\sqrt{3}}m$ -------(ii)

Equating the values of PQ from (i) and (ii), we get

$x\sqrt{3}=\frac{(x+120)}{\sqrt{3}}m$

$\Rightarrow 3x=x+120$

$\Rightarrow 2x=120$

$\Rightarrow x=60$

$\therefore$ height of the cloud from the surface of the lake $=BC$

$=(60+x)m=(60+60)m=120m$

Hence, the height of the cloud from the surface of the lake is 120 meters.

Alernative Method

In $\Delta ABE$, we have

$\tan {30}^{\circ }=\frac{h}{x}$

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

$\Rightarrow x=h\sqrt{3}\dots \dots \left(1\right)$

In $\Delta BDE$, we have

$\tan {60}^{\circ }=\frac{120+h}{x}$

$\Rightarrow \sqrt{3}=\frac{120+h}{h\sqrt{3}}$

$\Rightarrow 3h=120+h$

$\Rightarrow 2h=120$

$\Rightarrow h=60m$

Hence, height of cloud above the lake $=60+h=60+60=120m$