Question 55

The angle of elevation of an aeroplane from a point on the ground is 60°. After flying for 30 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a height of 4500 m, then what is the speed (in m/s) of aeroplane?

Solution

So, $$\tan60^{\circ\ }=\frac{BD}{PD}=\frac{4500}{PD}\ .$$

or, $$PD=\frac{4500}{\sqrt{3}}=1500\sqrt{3}\ m.$$

And, $$\tan30^{\circ\ }=\frac{AC}{PC}=\frac{4500}{PC}\ .$$

or, $$PC=4500\sqrt{3}\ m\ .$$

So, $$AB=CD=\left(PC-PD\right)=4500\sqrt{3}-1500\sqrt{3}=3000\sqrt{3}\ m.$$

So, Speed of plane= $$\frac{3000\sqrt{3}}{30}\ \frac{m}{\sec}=100\sqrt{3}\ \frac{m}{\sec}\ .$$

B is correct choice.

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