10. (20 points) In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If
$$
\angle A=39^{\circ},\left(a^{2}-b^{2}\right)\left(a^{2}+a c-b^{2}\right)=b^{2} c^{2} \text {, }
$$

find the degree measure of $\angle C$.
10. Note that,
$$
\begin{array}{l}
\sin ^{2} A-\sin ^{2} B=\sin (A+B) \cdot \sin (A-B) \\
=\sin C \cdot \sin (A-B) .
\end{array}
$$

By the Law of Sines, the original equation becomes
$$
\begin{array}{l}
\left(\sin ^{2} A-\sin ^{2} B\right)\left(\sin ^{2} A+\sin A \cdot \sin C-\sin ^{2} B\right) \\
=\sin ^{2} B \cdot \sin ^{2} C \\
\Leftrightarrow \sin (A-B)(\sin (A-B)+\sin A)=\sin ^{2} B \\
\Leftrightarrow \sin ^{2}(A-B)-\sin ^{2} B+\sin (A-B) \cdot \sin A=0 \\
\Leftrightarrow \sin (A-2 B) \cdot \sin A+\sin (A-B) \cdot \sin A=0 \\
\Leftrightarrow \sin (A-2 B)+\sin (A-B)=0 \\
\Leftrightarrow \sin \frac{2 A-3 B}{2} \cdot \cos \frac{B}{2}=0 . \\
\text { Given } \angle A, \angle B, \angle C \in(0, \pi) \\
\Rightarrow \sin \frac{2 A-3 B}{2}=0 \\
\Rightarrow \angle B=26^{\circ} \Rightarrow \angle C=115^{\circ} .
\end{array}
$$