Question : 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 metres, then what is the speed (in m/s) of an aeroplane?
Option 1: $50\sqrt3$
Option 2: $100\sqrt3$
Option 3: $200\sqrt3$
Option 4: $300\sqrt3$
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Correct Answer: $100\sqrt3$
Solution : Given: The angle of elevation of an aeroplane from a point on the ground is 60° and the height is 4500 metres. Let the base length be $B_{1}$ when the elevation is 60° and the Base length is $B_{2}$ when the elevation is 30°. $\tan \ 60° = \frac{4500}{B_{1}}$ $⇒B_{1} = \frac{4500}{\sqrt{3}}$ Now, the angle of the elevation of the aeroplane is changed to 30°. $\tan \ 30° = \frac{4500}{B_{2}}$ $⇒B_{2} = 4500\sqrt{3}$ For the speed of the aeroplane, $\text{Speed} = \frac{\text{Distance}}{\text{Time}}= \frac{B_2 \:–\: B_{1}}{30} = \frac{4500\sqrt{3} \:–\: (\frac{4500\sqrt{3}}{3})}{30} = 100\sqrt{3}$ m/s Hence, the correct answer is $100\sqrt{3}$.
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