# The angle of elev

Let P and Q be the two positions of the aeroplane and A be the point of observation. Let ABC be the horizontal line through A. It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.

PAB = 60° and QAB = 30°

Also given PB = 3000√3 m

In Δ ABP,

tan 60° = BP/AP

√3

AB = 3000 m

In Δ ACQ,

tan 30° = CQ/AC

AC = 3000 (3)

AC = 9000 m

Distance, BC = AC – AB = 9000 – 3000 = 6000 m

Thus, the plane travels 6000m in 30 seconds.

We know that speed = distance/time

Speed of plane = 200 m/sec

Speed of the aeroplane is 200 m/sec.

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