An aeroplane is moving one kilometer high from west to east horizontally from a point on the ground due south the angle of elevation of the aeroplane is 60° and after 10 second the angle of elevation of the aeroplane is observed as 30°. Then the speed (in km/ hr) of the areoplane is

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485km/ hr

$240 \sqrt{3} km / hr$

523km/ hr

364km/ hr

answer is A.

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Detailed Solution

$\begin{array}{l} f r o m t h e d i a g r a m \\ ΔOCA \Rightarrow \cot 60^{\circ} = \frac{x}{1} \\ x = \frac{1}{\sqrt{3}} \\ ΔODB, \cot 30 = \frac{x + y}{1} \\ x + y = \sqrt{3} \end{array}$
$y = \frac{2}{\sqrt{3}}$
In 10 secdistance covered $y = \frac{2}{\sqrt{3}} km$
In 3600sec distance covered
$\frac{3600}{10} \times \frac{2}{\sqrt{3}} = 240 \sqrt{3} km / hr$
Speed of the plane = $240 \sqrt{3} km / hr$

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