An aeroplane is moving one kilometer high from west to east horizontally. From a point on the ground the angle of elevation of the aeroplane is 60⁰, and after 10 second, the angle of elevation of the aeroplane is observed as 30⁰. Then the speed of the aeroplane is

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$240 \sqrt{3} km / hr$

364 km/hr

485 km/hr

523 km/hr

answer is A.

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Detailed Solution

$T a n 60^{0} = \frac{1}{x}, T a n 30^{0} = \frac{1}{x + y} \Rightarrow y = \frac{2}{\sqrt{3}}$
$S p e e d = \frac{distance}{t i m e} = \frac{2}{\sqrt{3}} \frac{60.60}{10} = 240 \sqrt{3}$
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