The angle of elevation of a jet fighter from a point A on the ground is 60° . after a flight of 15 seconds the angle of elevation changes to 30° .if the jet is flying at the speed 720 km/h find the constant height at which the jet is flying.
The figure is given below:

Let jet fly at a constant height of y km
and let the distance from a point to the starting point of jet bet x km.
Thus, BD = EC = y km
And, AB = x km
As distance = speed × time.
Distance took by jet to travel from D to E = ![]()
DE = 3km
Using trigonometry;
∠ DAB = 60°
tan 60° = √3
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Also,∠ EAC = 30°
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y = 2.59 m
So, the constant height of the jet is 2.59 km
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