An airplane at a height of $6 \; km$ passes vertically above another plane at an instant when their angles of elevation at the same observing point are $60^\circ$ and $45^\circ$ respectively. Find the difference between the heights of the planes. Take $\sqrt{3} = 1.732$
$P_1, \; P_2 =$ Airplanes
$O =$ Point of observation
$AP_1 =$ Height of airplane $P_1 = 6 \; km$
$P_1 P_2 =$ Difference between the heights of the planes
In $\triangle OP_1A$, $\;$ $\dfrac{AP_1}{OA} = \tan 60^\circ$
$\implies$ $OA = \dfrac{AP_1}{\tan 60^\circ} = \dfrac{6}{\sqrt{3}} = 2\sqrt{3} \; km$
In $\triangle OP_2A$, $\;$ $\dfrac{AP_2}{OA} = \tan 45^\circ$
$\implies$ $AP_2 = OA \times \tan 45^\circ = 2 \sqrt{3} \times 1 = 2 \sqrt{3} \; km$
Difference between the heights of the planes $= P_1P_2$
$P_1 P_2 = AP_1 - AP_2 = 6 - 2 \sqrt{3} = 2.536 \; km$