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An aeroplane flying with uniform speed horizontally one $\mathrm{km}$ above the ground is observed at an elevation of $60^{\circ}$. After $10 \mathrm{~s}$ if the elevation is observed to be $30^{\circ}$, then the speed of the plane (in $\mathrm{km} / \mathrm{h}$ ) is
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Verified Answer
The correct answer is:
$240 \sqrt{3}$
$\begin{array}{llll}
\tan 60^{\circ} & =\frac{D P}{A P} \\
\Rightarrow \sqrt{3} & =\frac{1}{A P} \\
\Rightarrow A P & =\frac{1}{\sqrt{3}} & {[\because E Q=D P=1]}
\end{array}$
In $\Delta E A Q$,
$\begin{aligned}
\tan 30^{\circ} & =\frac{E Q}{A P+P Q} \\
\Rightarrow \frac{1}{\sqrt{3}} & =\frac{1}{\frac{1}{\sqrt{3}}+P Q} \Rightarrow \frac{1}{\sqrt{3}}+P Q=\sqrt{3} \\
\Rightarrow P Q & =\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}} \mathrm{~km} \\
\therefore \text { Speed of plane } & =\frac{\text { Distance }}{\text { Time }}=\frac{2 / \sqrt{3}}{\frac{10}{60 \times 60}}=\frac{2}{\sqrt{3}} \times \frac{3600}{10} \\
& =240 \sqrt{3} \mathrm{~km} / \mathrm{h}
\end{aligned}$
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