6. In $\triangle A B C$, it is known that $b c=b^{2}-a^{2}$, and $\angle B-$ $\angle A=80^{\circ}$. Then $\angle C=$ $\qquad$ (answer in degrees).
6. $60^{\circ}$.

From $b c=b^{2}-a^{2}$
$$
\begin{aligned}
\Rightarrow & \sin B \cdot \sin C=\sin ^{2} B-\sin ^{2} A \\
& =\frac{1}{2}(\cos 2 A-\cos 2 B) \\
& =\sin (B+A) \cdot \sin (B-A) \\
\Rightarrow & \sin B=\sin (B-A)=\sin 80^{\circ} \\
\Rightarrow & \angle B=100^{\circ}, \angle A=20^{\circ} \\
\Rightarrow & \angle C=60^{\circ} .
\end{aligned}
$$