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 $10.0\; s$ apart is $30^o$, what is the speed in $m/s$ of the aircraft ?
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 $10.0\; s$ apart is $30^o$, what is the speed in $m/s$ of the aircraft ?
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, $\angle POQ =30^{\circ}$ Time $=10\, s$
In $\Delta PRO:$
$\tan 15^{\circ}=\frac{ PR }{ OR }$
$PR = OR \tan 15^{\circ}$
$=3400 \times \tan 15^{\circ}$
$\triangle PRO$ is similar to $\Delta RQO$
$\therefore PR = RQ$
$PQ = PR + RQ$
$=2 PR =2 \times 3400 \tan 15^{\circ}$
$=6800 \times 0.268=1822.4 \,m$
$\therefore$ Speed of the aircraft $=\frac{1822.4}{10}=182.24 \,m / s$
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