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An aeroplane flying at a height of $300 \mathrm{~m}$ above the ground passes vertically above another plane at an instant when the angles of elevation of two planes from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. What is the height of the lower plane from the ground? $\quad$
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The correct answer is:
$100 \sqrt{3} \mathrm{~m}$
Let $\mathrm{P}$ and $\mathrm{Q}$ be the positions of two aeroplanes when $\mathrm{Q}$ is vertically below $\mathrm{P}$ and $\mathrm{OP}=300 \mathrm{~m}$ Let the angles of elevation of $\mathrm{P}$ and $\mathrm{Q}$ at a point $\mathrm{A}$ on the ground be $60^{\circ}$ and $45^{\circ}$ respectively. - In $\Delta \mathrm{AOQ}$
$\tan 45^{\circ}=\frac{\mathrm{OQ}}{\mathrm{OA}} \Rightarrow \mathrm{OA}=\mathrm{OQ}$
In $\Delta \mathrm{AOP}$
$\tan 60^{\circ}=\frac{\mathrm{OP}}{\mathrm{OA}}=\frac{300}{\mathrm{OA}}=\sqrt{3}$
$\Rightarrow \mathrm{OA}=\frac{300}{\sqrt{3}}=100 \sqrt{3}$
Hence, $\mathrm{OQ}=100 \sqrt{3} \mathrm{~m}$
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