How does the law of cosines help with this problem? I've gotten stuck on this problem from Stewart Calculus, 7th edition:

A runner sprints around a circular track of radius 100 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m?

From the hints provided on the website, I know that the law of cosines is somehow involved in finding the solution.
So far I think I've figured out that the angle of his position on the track is changing by 7/200pi radians/second, as he's moving at 7 meters and the circumference of the track is 200pi meters. I recognize that two sides of the triangle are not changing; the runner's position is a constant 100 m from the track's center and his friend's position is a constant 200 m from the track's center. I cannot figure out how to combine this information to get any closer to the answer. Help, please?

Edit: I'm now not even sure I've set up this for correct application of the law of cosines. Given side lengths of 100, 200, and 200, I tried this:

200^2 = 200^2 + 100^2 - 2*200*100*cos(x), where x is presumably the angle formed between the friend, the center of the track, and the runner.

This gives 40,000 = 40,000 + 10,000 - 40000*cos(x). For this to work, cos(x) would have to be 1/4, but that seems impossible. 

Let $C$ be the distance between the runner and his friend.
Let $A$ be the distance between the runner and the center of the circle.
Let $B$ be the distance between the friend and the center of the circle.
Now you have a triangle.  Let $\theta$ be the angle opposite the side of length $C$.  You can see that the change in $\theta$ is a constant.
We know $C^2 = A^2 + B^2 - 2ABcos(\theta)$.
So $2C\frac{dC}{dt} = 0 + 0 - 2ABcos(\theta + \frac \pi 2)\frac{d\theta}{dt}$.
You are trying to solve for $\frac{dC}{dt}$, and you have already figured out $\frac{d\theta}{dt}$.
Edit: oops I forgot to take the derivative of the cosine, added that in.
