Question
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 a speed of 720 km/hour, find the constant height at which the jet is flying.
Solution
Let A be the point of observation, C and E be the two points of the plane. It is given that after 15 seconds angle of elevation changes from 60° to 30°.
i.e., ∠BAC = 60° and ∠DAE = 30°. It is also given that height of the jet plane is
[Sincc jet plane is flying at constant height, Let, CB = ED = h km]
In right triangle ABC, we have
In right triangle ADL, we have
Comparing (i) and (ii) we get,
[Hence, constant height at which the jet is flying = 2.6 km (app)]
i.e., ∠BAC = 60° and ∠DAE = 30°. It is also given that height of the jet plane is
[Sincc jet plane is flying at constant height, Let, CB = ED = h km]
In right triangle ABC, we have
In right triangle ADL, we have
Comparing (i) and (ii) we get,
[Hence, constant height at which the jet is flying = 2.6 km (app)]
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