Question

Question 18
The lower window of a house is at a height of 2m above the ground and its upper window is 4m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be ${60}^{\circ }$ and ${30}^{\circ }$, respectively. Find the height of the balloon above the ground.

Answer 1

Let the height of the balloon from above the ground is H.

A and $OP=w_{2}R=w_{1}Q=x$

Given that, height of lower window from above the ground $=w_{2}P=2m=OR$

Height of upper window from above the lower window $=w_1{w}_2=4m=QR$

$\therefore BQ=OB-(QR+RO)$

$=H-(4+2)$

$=H-6$

And, $\angle B{w}_{1}Q={30}^{\circ }$

$\Rightarrow \angle B{w}_{2}R={60}^{\circ }$

Now, in $\Delta B{w}_{2}R$,

$tan{60}^{\circ }=\frac{BR}{w_{2}R}=\frac{BQ+QR}{x}$

$\Rightarrow \sqrt{3}=\frac{(H-6)+4}{x}$

$\Rightarrow x=\frac{H-2}{\sqrt{3}}\dots \dots \left(i\right)$

And in $\Delta B{w}_{1}Q,$

$tan{30}^{\circ }=\frac{BQ}{w_{1}Q}$

$tan{30}^{\circ }=\frac{H-6}{x}=\frac{1}{\sqrt{3}}$

$\Rightarrow x=\sqrt{3}(H-6)\dots \dots \left(ii\right)$

From Eq. (i) and (ii),

$\sqrt{3}(H-6)=\frac{(H-2)}{\sqrt{3}}$

$\Rightarrow 3(H-6)=H-2$
$\Rightarrow 3H-18=H-2$

$\Rightarrow 2H=16\Rightarrow H=8$

So, the height of the balloon above the ground is 8 m.