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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°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

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Detailed Solution

Given, BH = 88.2 m, $∠ A C E = 60^{0}, ∠ B C G = 30^{0} .$

In $∆ A C E,$ we have

$\tan 60^{0} = \frac{A E}{C E} . \Rightarrow \sqrt{3} = \frac{88.2 - 1.2}{C E} . \Rightarrow C E = 29 \sqrt{3} .$

In $∆ B C G,$ we have

$\tan 30^{0} = \frac{B G}{C G} . \Rightarrow \frac{1}{\sqrt{3}} = \frac{88.2 - 1.2}{C G} . \Rightarrow C G = 87 \sqrt{3} .$

Then EG = CG - CE.

$E G = 87 \sqrt{3} - 29 \sqrt{3} = 58 \sqrt{3} .$

Therefore, the distance traveled by the balloon is $58 \sqrt{3}$ m.

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