An aeroplane flying with uniform speed horizo | vedclass

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(C) Let the position of the observer be $A$. The aeroplane is at height $h = 1 \ km$. Let the initial position be $D$ and the final position be $E$. I

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$240 \sqrt{3}$

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(C) Let the position of the observer be $A$. The aeroplane is at height $h = 1 \ km$. Let the initial position be $D$ and the final position be $E$. 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} = \frac{1}{AP + PQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{\frac{1}{\sqrt{3}} + PQ}$. Solving for $PQ$: $\frac{1}{\sqrt{3}} + PQ = \sqrt{3}$ $\Rightarrow PQ = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$. The distance covered by the plane is $PQ = \frac{2}{\sqrt{3}} \ km$ in $10 \ s$. Speed $= \frac{	ext{Distance}}{	ext{Time}} = \frac{2/\sqrt{3} \ km}{10 \ s} = \frac{2}{\sqrt{3}} 	imes \frac{1}{10} \ km/s$. Converting to $km/h$: $	ext{Speed} = \frac{2}{10\sqrt{3}} 	imes 3600 \ km/h = \frac{7200}{10\sqrt{3}} = \frac{720}{\sqrt{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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