Q. 284.4( 26 Votes )

The angle of elevation of a jet plane from point A on the ground is 60˚. After A flight of 15 seconds, the angle of elevation changes to 30˚. If the jet plane is flying at a constant height of 1500 √ 3 m, find the speed of the jet plane.

Answer :


Let the jet plane goes from point P to point C and we have given,


Initially angle of elevation from point A, PAQ = θ1 = 60°


After 15 seconds,


Angle of elevation from point A, CAB = θ2 = 30°


As the plane is flying at a constant height,


BC = PQ = 1500√3 m


Now,


In ΔABC





In ΔAPQ





So, we have


QB = AB – AQ = 4500 – 1500 = 3000 m


And


QB = PC


So, jet plane travels 3000 m in 15 seconds


And we know,




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