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(C) Let the height of the aeroplane be $h = 1 \ km$. Let the positions of the aeroplane be $D$ and $E$ at two different times. Let $A$ be the point of

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$240 \sqrt{3}$

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(C) Let the height of the aeroplane be $h = 1 \ km$. Let the positions of the aeroplane be $D$ and $E$ at two different times. Let $A$ be the point of observation on the ground. Let $P$ and $Q$ be the points on the ground directly below $D$ and $E$ respectively. So,$DP = EQ = 1 \ km$.In $\Delta DAP$,$	an 60^{\circ} = \frac{DP}{AP}$ $\Rightarrow \sqrt{3} = \frac{1}{AP}$ $\Rightarrow AP = \frac{1}{\sqrt{3}} \ km$.In $\Delta EAQ$,$	an 30^{\circ} = \frac{EQ}{AQ}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{AP + PQ}$ $\Rightarrow AP + PQ = \sqrt{3}$.Substituting $AP = \frac{1}{\sqrt{3}}$,we get $\frac{1}{\sqrt{3}} + PQ = \sqrt{3}$ $\Rightarrow PQ = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$.The distance $PQ$ is covered in $10 \ s$. Therefore,speed $= \frac{	ext{Distance}}{	ext{Time}} = \frac{2/\sqrt{3} \ km}{10 \ s} = \frac{2}{10\sqrt{3}} \ km/s$.To convert into $km/h$,multiply by $3600$: $	ext{Speed} = \frac{2}{10\sqrt{3}} 	imes 3600 = \frac{720}{\sqrt{3}} = \frac{720\sqrt{3}}{3} = 240\sqrt{3} \ km/h$.

An aeroplane flying with uniform speed horizontally $1 \ km$ above the ground is observed at an elevation of $60^{\circ}$. After $10 \ s$,if the elevation is observed to be $30^{\circ}$,then the speed of the plane (in $km/h$) is

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