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