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An aircraft is flying at height 3600 m from the ground. If the angle subtended at the ground observation point by aircraft positions 10 s apart is 30º, what is the speed of the aircraft?

(Given Answer= 60*Pi m/s)

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Solution:

The aircraft takes 10s to go from one point (say A) to the other (say B) and AB subtends an angle of 30o at the point of observation (say O) .So, ABO forms a triangle.Now let OC be perpendicular bisector drawn from O to AB(i.e. actually the height).

Then in right trianle, OAC

AC = OC tan 15o = 3600 x 0.2679 = 964.44 m

so, AB = 2 x 964.44 = 1928.88 m

Now, the speed of aircraft , v = AB / t = 1928.88 / 10 = 192.888 m/s.

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