The lower window of a house is at a height of $ 2 \mathrm{~m} $ above the ground and its upper window is $ 4 \mathrm{~m} $ 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.

Given:

The lower window of a house is at a height of \( 2 \mathrm{~m} \) above the ground and its upper window is \( 4 \mathrm{~m} \) 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.

To do:

We have to find the height of the balloon above the ground.

Solution:

Let the height of the balloon from above the ground be $H\ m$.

From the figure,

$AB = ED =2\ m$
$BC = EF = 4\ m$

$GF = DG – (FE + ED)$
$= H – (4 + 2)$
$= H – 6\ m$

$\angle \mathrm{GCF}=30^{\circ}$

$\angle \mathrm{GBE}=60^{\circ}$

Let $AD = BE = CF = x\ m$

In $\Delta \mathrm{GBE}$,

$\tan 60^{\circ}=\frac{\mathrm{GE}}{\mathrm{BE}}$

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

$\Rightarrow x=\frac{\mathrm{H}-2}{\sqrt{3}}$.................(i)

In $\Delta \mathrm{GCF}$,

$\tan 30^{\circ}=\frac{\mathrm{GF}}{\mathrm{CF}}$

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

$\Rightarrow x=\sqrt{3}(\mathrm{H}-6)$............(ii)

From equations (i) and (ii),

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

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

$\Rightarrow 3 \mathrm{H}-18=\mathrm{H}-2$

$\Rightarrow 2 \mathrm{H}=16$

$\Rightarrow \mathrm{H}=8\ m$
Therefore, the height of the balloon above the ground is $8\ m$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

375 Views

Kickstart Your Career

Get certified by completing the course

Get Started