The angle of elevation of a jet plane 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 plane is flying at a constant height of  find the speed of the jet plane.

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 1500 


[Since jet plane is flying at constant height, therefore, CB = ED = 
In right triangle ABC, we have


In right triangle ADE, we have

Putting the value of (i) in (ii), we get
1500 + BD = 4500
⇒    BD = 3000
∵ Distance travelled in 15 sec
=  CE = BD = 3000 metres,

Now, speed of plane (m/s)     =    

and speed of plane (km/h)     =   
                                           =       720 km/hr