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The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the plane b flying at a constant height of 1500√3 m. Find the speed of the plane.

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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

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