Question

An aircraft is flying at a height of $3400m$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10s$ apart is ${30}^o$, then the speed of the aircraft is:

• A. 10.8 $m{s}^{-1}$
• B. 1963 $m{s}^{-1}$
• C. 108 $m{s}^{-1}$
• D. 196.3 $m{s}^{-1}$

Answer 1

Answer: (D)

196.3 $m{s}^{-1}$

O is the observation point at the ground. A and B are the positions of aircraft for which $\angle$AOB = 30${}^{\circ }$.Time taken by aircraft from A to B is 10 s.

In $\Delta$ AOB,

$tan{30}^{\circ }=\frac{AB}{3400};AB=3400tan{30}^{\circ }=\frac{3400}{\sqrt{3}}$m

$\therefore$ Speed of aircraft, $v=\frac{AB}{10}=\frac{3400}{10\sqrt{3}}=196.3m{s}^{-1}$