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 \( 30^{\circ} \). Find the distance travelled by the balloon during the interval.
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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 $58\sqrt3$.
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