The angle of elevation of an aircraft from a point on horizontal ground is found to be 30°. The angle of elevation of same aircraft after 24 sec which is moving horizontally to the ground is found to be 60°. If the height of the aircraft from the ground is 3600√3 m , find the velocity of the aircraft.
Answers
Let P and Q be the two positions of the aircraft and A be the point of observation.
Let ABC be the horizontal line through A. It is given that angles of elevation of the aircraft in two positions P and Q from a point A are 60° and 30° respectively.
So, ∠PAB = 60°, ∠QAC = 30°
PB = 3600√3 m
In ΔABP,
tan 60° = BP/AB
√3 = 3600√3/AB
AB = 3600 m
In ΔACQ,
tan 30° = CQ/AC
1/√3 = 3600√3/AC
AC = 3600√3 * √3
AC = 3600 * 3
AC = 10800 m
Distance = BC = AC – AB
BC = (10800 – 3600) m
BC = 7200 m
Speed of aircraft = 7200m/24 sec
= 300 m/sec
OR
=(300/1000)*60*60 km/hr
= (3/10)x3600 km/hr
Speed of aircraft = 1080 km/hr
[1km=1000 m, 1hr=60 min=60 * 60 sec]
