Let P and Q are the two position of plane and A is a observation point.
Let ABC is a horizontal line from A. The angle of elevation of P and Q from A are 60° and 30° respectively. Now
∠PAC = 60°
∠QAC = 30°
and PB = QC = 1500√3 m
From right angled ∆PBA,
tan 60° = PB/AB
⇒ √3 = 1500√3/AB
AB = 1500 m
From right angled ∆ACQ,
tan 30° = QC/AC
⇒ 1/√3 = 1500√3/AC
⇒ AC = 1500 × √3 × √3
= 4500 m
Now, PQ = BC
= AC – AB
= 4500 – 1500
= 3000 m
Hence, plane covers 3000 m distance in 15 seconds.
So, speed of plane = 3000/15
= 200 m/sec
= (200 x 60 x 60)/1000
= 720 km/h