Problem - 2997
Let $R=\frac{7\sqrt{3}}{3}$ be the circumradius of $\triangle{ABC}$. If $\angle{B} = 60^\circ$ and its area $S_{\triangle{ABC}}=10\sqrt{3}$, find the lengths of $a$, $b$, and $c$.
Solving the following system:
$$
\left\{
\begin{array}{rl}
a^2 + c^2 - 2ac\cos{60^\circ} &= b^2 \\
\\
\frac{1}{2}ac\sin{60^\circ} &= 10\sqrt{3}\\
\\
2\times\frac{7\sqrt{3}}{3}\sin{60^\circ} &= b
\end{array}
\right.
$$
leads to $\boxed{a=5}, \boxed{b=7}$, and $\boxed{c=8}$.