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?

Asked by Abhisek | 1 year ago |  91

##### Solution :-

Height at which the aircraft is flying = 3400 m

Let A and B be the positions of the aircraft making an angle ∠AOB = 30°.

The perpendicular OC is drawn on AB. Here OC is the height of the aircraft which

is equal to 3400 m and ∠AOC =  ∠COB = 15°.

In the ΔAOC, AC = OC tan 15°

= 3400 x 0.267 = 910.86 m

AB = AC + CB = AC + AC = 2 AC = 2 x 910.86 m

Speed of the aircraft = distance AB/time

= $$\dfrac{ (2 \times 910.86)}{10}$$= 182.17 m/s =182.2 m/s

Answered by Pragya Singh | 1 year ago

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