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# A $1.2\ m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2\ m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^o$. After sometime, the angle of elevation reduces to $30^o$ (see figure). Find the distance travelled by the balloon during the interval.

"

Given:

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of \( 88.2 \mathrm{~m} \) from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is \( 60^{\circ} \). After some time, the angle of elevation reduces to \( 30^{\circ} \).

To do:

We have to find the distance travelled by the balloon during the interval.

Solution:

Let $x$ be the horizontal distance between the girl and the balloon initially and $y$ be the horizontal distance between the girl and the balloon finally.

Therefore,

Initially

$tan\ 60^o=\frac{88.2-1.2}{x}$

$\sqrt3=\frac{87}{x}$

$x=\frac{87}{\sqrt3}$

$x=\frac{87\sqrt3}{\sqrt3\times\sqrt3}$

$x=\frac{87\sqrt3}{3}$

$x=29\sqrt3$

Finally,

$tan\ 30^o=\frac{88.2-1.2}{y}$

$\frac{1}{\sqrt3}=\frac{87}{y}$

$y=87\sqrt3$

Therefore,

The distance travelled by the balloon during the interval$=y-x$

$=87\sqrt3-29\sqrt3$

$=58\sqrt3$

The distance travelled by the balloon during the interval is $58\sqrt3\ m$.

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