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*

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