The plane travels horizontally over the station.
If we consider a horizontal distance of x km the velocity means the the rate at which the distance change. So if we denote time t in hours;
For horizontal direction;
`(dx)/dt = 800 (km)/h`
Also if we consider the angle between the line joint the radar with the plane and the ground to be alpha then;
`tanalpha = 2/x`
`xtanalpha = 2`
Now lets us differentiate the above equation with respect to t.
`x*sec^2alpha*(dalpha)/dt+tanalpha*dx/dt = 0`
`x(1+tan^2alpha)*(dalpha)/dt+tanalpha*dx/dt = 0`
`x*(1+4/x^2)*(dalpha)/dt+2/x*800 = 0`
So in the question it is asked to find `(dalpha)/dt` at x = 3 km
`3*(1+4/3^2)*(dalpha)/dt+2/3*800 = 0`
`(dalpha)/dt = -123.076 `
The (-) is due to decreasing the angle.
So the rate of change of angle is 123.076 deg/h