Find the sixth side of hexagon. You are given a hexagon inscribed in a circle. If the lengths of $5$ sides taken in order are $3,4,6,8$ and $7$ units, find the length of $6^\text{th}$ side.
Not got the slightest of idea how to proceed, so I can't show my attempts.

If the radius of the circle is $r$, the sixth side length is $s$, and you label the central angles $\theta_1,\ldots,\theta_6$, then using the Law of Cosines
$$\begin{align}
2\pi&=\theta_1+\ldots+\theta_5+\theta_6\\
2\pi&=\arccos\left(1-\frac{3^2}{2r^2}\right)+\ldots+\arccos\left(1-\frac{7^2}{2r^2}\right)+\arccos\left(1-\frac{s^2}{2r^2}\right)\\
\end{align}$$
This allows you to solve for $s$ in terms of $r$:
$$
\begin{align}
s&=\sqrt{2r^2\left(1-\cos\left(2\pi-\left(\arccos\left(1-\frac{3^2}{2r^2}\right)+\ldots+\arccos\left(1-\frac{7^2}{2r^2}\right)\right)\right)\right)}\\
s&=r\sqrt{2-2\cos\left(\arccos\left(1-\frac{3^2}{2r^2}\right)+\ldots+\arccos\left(1-\frac{7^2}{2r^2}\right)\right)}\\
\end{align}$$
If you follow the instances of $r$ in this expression, as each instance of $r$ would grows larger, so would $s$. In other words, it is clear (after thinking through all the negations and inversions) that $s$ is an increasing function of $r$. 
In particular, it's not constant. So for each radius $r$ where this expression is defined and that sum of $\arccos$ terms does not exceed $2\pi$, there is a different value for the sixth side length $s$.
In theory you can also solve for $r$ in terms of $s$, since $s$ is an increasing function of $r$. But I don't think I want to.
