Question

A $1.2m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is ${60}^o$. After some time, the angle of elevation reduces to ${30}^o$.Find the distance traveled by the balloon during the interval.

Answer 1

Given:$1.2$m tall girl sees a balloon.

So,$AG=1.2$m

Also,$AG$ and $BF$ are parallel

$BF=AG=1.2$m

And the balloon is at a height of $88.2$m

So,$EF=88.2$m

Girl sees balloon first at ${60}^{\circ }$

So, $\angle EAB={60}^{\circ }$

After traveling, the angle of elevation becomes ${30}^{\circ }$

So,$\angle DAC={30}^{\circ }$

Distance travelled by the balloon$=BC$

Now, $BE=EF-BF=88.2-1.2$m

$\therefore BE=87$m

Also,$BE=DC=87$m

Here,$\angle ABE={90}^{\circ }$ and $\angle ACD={90}^{\circ }$

In right angled $△EBA$

$\tan A=\frac{BE}{AB}$

$\Rightarrow \tan {60}^{\circ }=\frac{87}{AB}$

$\Rightarrow \sqrt{3}=\frac{87}{AB}$

$\therefore AB=\frac{87}{\sqrt{3}}$m

In right angled $△DAC$

$\tan A=\frac{CD}{AC}$

$\Rightarrow \tan {30}^{\circ }=\frac{87}{AC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{87}{AC}$

$\therefore AC=87\sqrt{3}$m

Now,$AC=AB+BC$

$\Rightarrow 87\sqrt{3}=\frac{87}{\sqrt{3}}+BC$

$\Rightarrow BC=87\sqrt{3}-\frac{87}{\sqrt{3}}$

$\Rightarrow BC=87(\sqrt{3}-\frac{1}{\sqrt{3}})$

$\Rightarrow BC=87\left(\frac{3-1}{\sqrt{3}}\right)$

$\Rightarrow BC=87\times \frac{2}{\sqrt{3}}$

$\Rightarrow BC=87\times \frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow BC=87\times \frac{2\sqrt{3}}{3}$

$\Rightarrow BC=29\times 2\sqrt{3}=58\sqrt{3}$

Hence, the distance travelled by balloon$=58\sqrt{3}$m