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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