Let x be the horizontal (ground) distance from the tower and `theta` the angle of elevation from the tower to the plane.

Then `tan theta=6/x` . Taking the derivative with respect to t we get:

`sec^2theta (d theta)/(dt)=-6/x^2 (dx)/(dt)` where `(dx)/(dt)` is the rate of change of the ground distance from the tower to the plane, or its speed.

We are given that `theta=pi/4,(d theta)/(dt)=-pi/4` and from the right triangle we know x=6. Substituting we get:

`sec^2(pi/4)(-pi/4)=-6/36 (dx)/(dt)`

`-pi/2=-1/6 (dx)/(dt)`

`(dx)/(dt)=3pi`

So the speed of the plane is `3pi "km"/"min"` or approximately 565.5 `"km"/"hr"`

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