An aeroplane flying at a height of 3000 m
above the ground passes vertically above another plane at an
instant when the angles of elevation of the two planes from
the same point on the ground are $60 °$ and $45 °$ respectively.
The height of the lower plane from the ground is

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$500 m$

$1000 \sqrt{3} m$

$1000 / \sqrt{3} m$

$1500 (\sqrt{3} + 1) m$

answer is A.

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Detailed Solution

Let $Q$ and $P$ be the position of the two planes
and $A$ be the point on the ground such that $∠ Q A O = 60 °,$
$∠ P A O = 45 °$ (Fig. 27.17) and $O P,$ the height of lower plane
be $h$

Also $O Q = 3000 m$

Then $O A = O P = h$

and $O Q = h \tan 60^{\circ} = 3000$

$\Rightarrow h = 3000 \cot 60^{\circ} = \frac{3000}{\sqrt{3}} = 1000 \sqrt{3} m$ .

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