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An aero plane flying horizontally $1 k m$ above the ground is observed at an elevation of $60 °$ . If after $10$ seconds, the elevation is observed to be $30 °$ , then the uniform speed of the aero plane per hour is

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240 \sqrt{3} k m / h r

120 \sqrt{3} k m / h r

80 \sqrt{3} k m / h r

160 \sqrt{3} k m / h r

answer is A.

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Detailed Solution

In $Δ O A B, \tan 60 ° = \frac{A B}{O A}$

$\sqrt{3} = \frac{1}{O A}$

$\Rightarrow O A = \frac{1}{\sqrt{3}}$

In $Δ O C D, \tan 30 ° = \frac{C D}{O C}$

$\frac{1}{\sqrt{3}} = \frac{1}{O A + A C}$

$O A + A C = \sqrt{3}$

$\frac{1}{\sqrt{3}} + A C = \sqrt{3}$

$A C = \frac{2}{\sqrt{3}} k m$

Speed of Aero plane $= \frac{A C}{10 \sec} = \frac{\frac{2}{\sqrt{3}} \times 3600}{10} k m / h$

$= \frac{720}{\sqrt{3}} = 240 \sqrt{3} km/hr$

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