Question

the angle of elevation of a jet fighter plane from a point lying on the ground is 60 degree. After a flight of 15 second the angle of elevation changes to 30 degree. If jet is flying at a speed of 720km/h .Find the constant height at which it is flying. (use√3=1.732)

Answer 1

Here is the answer to your query.

 



Let P and Q be the two position of jet and the height of the jet be h metres.

Let ABC be the horizontal line through A.

It is given that A is the point of observation and angle of elevation of jet in two position P and Q from point A are 60° and 30° respectively.

∴ ∠PAB = 60°, ∠QAB = 30°

Also speed of jet = 720 km/hr m/sec = 200 m/sec

∴ PQ = 200 × 15 = 3000 m

⇒ BC = PQ = 3000 m

In ΔABP we have



In ΔACQ we have
 



From equation 11 and 22 we have



∴The constant height at which the jet is flying is  metres.