Ncert Exercises & Solutions

4.25 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°, what is the speed of the aircraft?

NEETprep Answer:

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

Height of the aircraft from the ground, OR = 3400 m

The angle subtended between the positions, ∠POQ = 30°

Time = 10 s

In ΔPRO:

\begin{array}{l} \tan 15^{\circ} = \frac{P R}{O R} \\ P R = O R \times \tan 15^{\circ} \\ = 3400 \times \tan 15^{\circ} \end{array}

ΔPRO is similar to ΔRQO.

So,PR = RQ

PQ = PR + RQ

= 2PR = 2 × 3400 tan 15°

= 6800 × 0.268 = 1822.4 m

\Rightarrow S p e e d o g t h e a i r c r a f t = \frac{1822.4}{10} = 182.24 m / s

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