Let B & D be the two positions of the plane and let A be the point of observation. Let ABE the horizontal line through A. It is given that angles of elevation of the plane into positions B and D from a point A are 60 °and 30°.
Angle BAC = 60° , angle DAB = 30°
In ∆ ABC
tan 60 °= BC/AC = 2500/AC
√3 = 2500/AC
AC = 2500/√3
IN ∆ AED
tan 30° = ED/AE
1/√3= 2500/AE
AE= 2500√3
BD = CE
CE= AE-AC
BD = 2500√3 - 2500 /√3
BD = 2500 ( √3 - 1/√3)
BD = 2500 ( √3×√3 - 1)/√3
BD = 2500 (3-1)/√3
BD =( 2500 ×2)/√3
BD = 5000 /√3 m
BD =( 5000/ √3 ) × 1/1000 km
BD = 5/√3 km ( Distance)
Plane travels 5/√3 km in 15 sec
Time = 15 sec (given )
Time = 15/3600= 1/240 hr
Speed= Distance/time
Speed=( 5/√3) / (1/240)
Speed = 5/√3 × 240
Speed = (5 × 240)/1.732 [ √3= 1.732]
Speed= 1200 /1.732
Speed = 692.84 km/h
Hence, the speed of the aeroplane in km/h is 692.84 km/h.