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Q.

The angle of elevation of a jet plane from a point A on the ground is $60^{0}$ . After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to $30^{0}$ . If the jet plane is flying at a constant height, then its height is

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a

$1200 \sqrt{3} m$

b

$3600 \sqrt{3} m$

c

$1800 \sqrt{3} m$

d

$2400 \sqrt{3} m$

answer is B.

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Detailed Solution

Question Image

BE is height of where the jet flying

ED = Distance travelled by Jet in 20 seconds

$\begin{matrix} = 432 \times \frac{5}{18} \times 20 \\ = 2400 meters \\ \tan 60^{\circ} = \frac{h}{A B} \\ A B = \frac{h}{\sqrt{3}} \\ \tan 30^{\circ} = \frac{h}{\frac{h}{\sqrt{3} + 2400}} \\ \frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 2400} \\ \frac{h}{\sqrt{3}} + 2400 = h \sqrt{3} \\ h \sqrt{3} - \frac{h}{\sqrt{3}} = 2400 \\ \frac{2 h}{\sqrt{3}} = 2400 \\ h = 1200 \sqrt{3} m \end{matrix}$

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