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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 $6 0^{\circ}$ . After some time, the angle of elevation reduces to $3 0^{\circ}$ (See figure). Find the distance travelled by the balloon during the interval.

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58 \sqrt{4} m

8 \sqrt{3} m

5 \sqrt{3} m

58 \sqrt{3} m

answer is D.

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Detailed Solution

Given,
Height of girl,

AG = 1.2 m

EF = 88.2 m

∠ AEB = 6 0^{\circ}

∠ DAC = 3 0^{\circ}

In the figure,

AG ∥ BF

AG = BF = 1.2 m

Distance travelled by the balloon

= BC

By the figure,

EF = BE + BF

∴ BE = EF - BF

Put the values,

BE = 88.2 - 1.2 = 87 m

BE = DC = 87 m

Again, by the figure,

∠ ABE ∠ ACD

are right angle

∴ ∠ ABE = ∠ ACD = 9 0^{\circ}

Also,

∠ EAB = 6 0^{\circ}

BE = 87 m

Thus,

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

\tan 6 0 = \frac{87}{AB}

\Rightarrow AB = \frac{87}{\tan 6 0^{\circ}}

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

△ DAC

DC = 87 m

∠ DAC = 3 0^{\circ}

Thus,

\tan 3 0 = \frac{DC}{AC}

\Rightarrow \tan 3 0^{\circ} = \frac{87}{AC}

∴ AC = \frac{87}{\tan 30}

AC = \frac{87}{\frac{1}{\sqrt{3}}} (∵ \tan 3 0 = \frac{1}{\sqrt{3}})

AC = 87 \sqrt{3} m

It can be seen in the figure that,

AC = AB + BC

∴ BC = AC - AB

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

BC = \frac{87 \times 3 - 87}{\sqrt{3}} = \frac{174}{\sqrt{3}}

BC = \frac{174 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

BC = \frac{174 \sqrt{3}}{3}

BC = 58 \sqrt{3} m

Thus, distance travelled by the balloon

= 58 \sqrt{3} m

Hence option 4 is correct.

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