The math model that represents this situation is a triangle. The plane has already flown beyond the radar station which is on the ground. A line segment from the radar station to the plane is the hypotenuse of the triangle. The vertical distance from the plane to ground is constant at 5 km. The horizontal distance (x) changes along the ground.

A relationship between the angle and the sides of the triangle is:

`tan(theta)=y/x`

Since the vertical distance is constant, `tan(theta)=5/x`

Solving for x, `x=5/(tan(theta))`

Take the derivative of both sideswith respect to t:

`dx/dt=(-5*sec^2(theta)*(d theta)/dt)/(tan^2 (theta))`

Substiting in values when theta = pi/3 and d(theta)/dt = -pi/6.

`dx/dt=(-5*sec^2(pi/3)*(-pi/6))/(tan^2(pi/3))`

`dx/dt=(-5*2^2*(-pi/6))/(sqrt(3)^2)`

`dx/dt=((20pi)/6)/3=((10pi)/3)*(1/3)`

`dx/dt=((10pi)/9)=3.491`

The plane is traveling at **10pi/9 or 3.491 km/min.**

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