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. [4 MARKS]

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

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

$\Rightarrow \frac{PQ}{xm}=\sqrt{3}\Rightarrow PQ=x\sqrt{3}m$ -----------(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 metres.

Alernative Method,

[$\frac{1}{2}$ MARK]

In $\Delta ABE$

$tan30=\frac{h}{x}$

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

In $\Delta BDE$

$tan60=\frac{120+h}{x}$

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

$3h=120+h$

$2h=120$

$h=60m$ [1 MARK]

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

= 60 + 60

= 120 m [$\frac{1}{2}$ MARK]