Question

an areoplane flying horizontally at height 2500√3m above that ground is observed to be at angle of elevation is 60° from the ground.After a flight of 25 secs the angle of elevation is 30°.find the speed of the plane in m/sec.

Answer 1

Given That: Height = 2500√3 m
Therefore, AB=ED= 2500√3 m.
Time = 25 sec.

Solution: (AB/BC) = tan 60°
(2500√3)/BC = √3
Therefore, BC = 2500 m

Similarly, (ED/DC) = tan 30°
(2500√3)/DC = (1/√3)
Therefore, DC = 7500 m

Let x be the distance of the aeroplane flying from point B to point D.

Therefore, x = DC-BC = 7500-2500 =5000 m.

Now, Speed = (5000/25) = 200 m/sec.