Question

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

Answer 1

Let PQ be the surface of the lake. A is the point vertically above P such that AP=20 m.

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

Let BC=H meters

CQ= H + 20 and BD = H + 40

Now, In $△ABD$

$tan{60}^{\circ }=\frac{BD}{AB}$

$\Rightarrow \sqrt{3}=\frac{H+20+20}{AB}$

$\Rightarrow \sqrt{3}.AB=H+40$

$\Rightarrow AB=\frac{H+40}{\sqrt{3}}...\left(i\right)$

And, In $△ABC$

$tan{30}^{\circ }=\frac{BC}{AB}$

$\frac{1}{\sqrt{3}}=\frac{BC}{AB}$

$AB=\sqrt{3}H$....(ii)

From eq.(i) and (ii)

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

$\Rightarrow 3H=H+40$

$\Rightarrow 2H=40\Rightarrow H=20$

Putting the value of H in eq.(ii) ,we get

AB=$20\sqrt{3}$

Again,in $△ABC$

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

=$(20\sqrt{3}{)}^2+(20{)}^2$

=1200+400

=1600

AC=$\sqrt{1600}$ = 40 m

Hence,the distance of cloud from A is 40 m.