4. $S=\frac{1}{2} a b \sin C$ so $\sin C=\frac{\sqrt{3}}{2}$ so $\angle C=60^{\circ}$ or $120^{\circ}$

Also $c^{2}=a^{2}+b^{2}-2 a b \cos C \quad$ we know: so $\angle C=60^{\circ} \quad c=\sqrt{21} ; \angle C=120^{\circ} \quad c=\sqrt{61}$.
4. Given $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ in $\triangle A B C$, and $S$ is the area of $\triangle A B C$. If $a=4, b=$ $5, S=5 \sqrt{3}$, find the length of $c$.