**Solution:**

The aircraft takes 10s to go from one point (say A) to the other (say B) and AB subtends an angle of 30^{o} at the point of observation (say O) .So, ABO forms a triangle.Now let OC be perpendicular bisector drawn from O to AB(i.e. actually the height).

Then in right trianle, OAC

AC = OC tan 15^{o} = 3600 x 0.2679 = 964.44 m

so, AB = 2 x 964.44 = 1928.88 m

**Now, the speed of aircraft , v = AB / t = 1928.88 / 10 = 192.888 m/s.**