Question

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, what is the speed of the aircraft ?

Answer 1

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, $OR$ = $3400 \,m$
Angle subtended between the positions, $\angle POQ$ = $30\degree$
Time = $10 \,s$
In $\triangle PRO$:
$\tan 15\degree$= $\frac{PR }{OR}$
$PR$ = $OR \tan15\degree$
= $3400 \times tan 15\degree$
$\triangle PRO$ is similar to $\triangle RQO$.
$\therefore$ $PR$ = $RQ$
$PQ$ = $PR + RQ$
= $2PR$
= $2 \times 3400 \,\tan 15\degree$
= $6800 × 0.268$
= $1822.4 \,m$
$\therefore$ Speed of the aircraft = $\frac{1822.4 }{19}$
= $182.24 \,m/s$