An aeroplane flying with uniform speed horizontally 1km above the ground is observed at an elevation of 60°. After 10 sec, if the elevation is observed to be 30°, then the speed of the plane (in km/hr) is

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Question

An aeroplane flying with uniform speed horizontally 1km above the ground is observed at an elevation of 60°. After 10 sec, if the elevation is observed to be 30°, then the speed of the plane (in km/hr) is

$\frac{240}{\sqrt{3}}$

$240 \sqrt{3}$

$\frac{120}{\sqrt{3}}$

$200 \sqrt{3}$

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In ΔDPA

tan 60° = $\frac{DP}{AP}$ ⇒ AP = $\frac{1}{\sqrt{3}}$

tan 30° = $\frac{EQ}{AP + PQ} = \frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}} = \frac{1}{\frac{1}{\sqrt{3}} + PQ}$

PQ = $\frac{2}{\sqrt{3}}$ km

Speed = $\frac{Distance}{time} = \frac{\frac{2}{\sqrt{3}}}{10 \sec} = \frac{2}{\sqrt{3} (\frac{10}{60 \times 60}) hrs} = 240 \sqrt{3}$ km/hr