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(B) Let the height of the aeroplane be $h = 4500 	ext{ m}$.Let the initial position be $A$ and the final position be $B$.At point $A$, the angle of e

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(B) Let the height of the aeroplane be $h = 4500 	ext{ m}$.Let the initial position be $A$ and the final position be $B$.At point $A$, the angle of elevation is $60^{\circ}$. Let the distance from the point on the ground to the point directly below $A$ be $x$.$	an(60^{\circ}) = \frac{4500}{x} \implies \sqrt{3} = \frac{4500}{x} \implies x = \frac{4500}{\sqrt{3}} = 1500\sqrt{3} 	ext{ m}$.At point $B$, the angle of elevation is $30^{\circ}$. Let the total distance from the point on the ground to the point directly below $B$ be $y$.$	an(30^{\circ}) = \frac{4500}{y} \implies \frac{1}{\sqrt{3}} = \frac{4500}{y} \implies y = 4500\sqrt{3} 	ext{ m}$.The distance covered by the aeroplane is $d = y - x = 4500\sqrt{3} - 1500\sqrt{3} = 3000\sqrt{3} 	ext{ m}$.The time taken is $t = 30 	ext{ s}$.Speed $= \frac{	ext{Distance}}{	ext{Time}} = \frac{3000\sqrt{3}}{30} = 100\sqrt{3} 	ext{ m/s}$.

The angle of elevation of an aeroplane from a point on the ground is $60^{\circ}$. After flying for $30$ seconds, the angle of elevation changes to $30^{\circ}$. If the aeroplane is flying at a constant height of $4500 \text{ m}$, what is the speed (in $\text{m/s}$) of the aeroplane (in $\sqrt{3}$)?

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