An airplane when flying at a height of $3125 \; m$ from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are $30^\circ$ and $60^\circ$ respectively. Find the distance between the two planes at that instant.
$P_1, P_2$: Positions of the two planes
$P_1P_2$: Distance between the two planes
$A$: Point of observation
Height of plane $P_1$ above ground $= OP_1 = 3125 \; m$
In $\triangle OP_1A$, $\;$ $\dfrac{OP_1}{OA} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
$\implies$ $OA = \sqrt{3} \times OP_1$
In $\triangle OP_2A$, $\;$ $\dfrac{OP_2}{OA} = \tan 60^\circ = \sqrt{3}$
$\implies$ $OP_2 = \sqrt{3} \times OA = \sqrt{3} \times \sqrt{3} \times OP_1 = 3 OP_1 = 3 \times 3125 = 9375 \; m$
$\therefore \;$ $P_1 P_2 = OP_2 - OP_1 = 9375 - 3125 = 6250$
$\therefore \;$ Distance between the two planes $= 6250 \; m$