10. (14 points) Given $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, and point $P$ is on the curve $C$. If $S_{\triangle P F_{1} F_{2}}=\sqrt{3}$, find $\angle F_{1} P F_{2}$.
10. Suppose point \( P\left(x_{0}, y_{0}\right) \) is on the right branch of the hyperbola.
From the given conditions, it is easy to get \( F_{1} F_{2}=2 \sqrt{2} \). Then
\[
\left.S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2} \|\right| y_{0}\left|=\frac{1}{2} \times 2 \sqrt{2}\right| y_{0} \right\rvert\,=\sqrt{3} \text {. }
\]

Solving for \( \left|y_{0}\right| \) gives \( \left|y_{0}\right|=\frac{\sqrt{6}}{2} \).
Also, from \( x_{0}^{2}-y_{0}^{2}=1 \), we get
\[
x_{0}^{2}=1+y_{0}^{2}=1+\frac{6}{4}=\frac{5}{2} \text {. }
\]

By the second definition of the hyperbola, we have
\[
\begin{array}{l}
\left|P F_{1}\right|=e\left[x_{0}-\left(-\frac{a^{2}}{c}\right)\right] \\
=a+e x_{0}=1+\sqrt{2} x_{0},
\end{array}
\]

and \( \left|P F_{2}\right|=e\left(x_{0}-\frac{a^{2}}{c}\right)=e x_{0}-a=\sqrt{2} x_{0}-1 \).
Using the cosine rule, we get
\[
\begin{aligned}
& \cos \angle F_{1} P F_{2} \\
= & \frac{\left(1+\sqrt{2} x_{0}\right)^{2}+\left(\sqrt{2} x_{0}-1\right)^{2}-(2 \sqrt{2})^{2}}{2\left(1+\sqrt{2} x_{0}\right)\left(\sqrt{2} x_{0}-1\right)} \\
= & \frac{2\left(2 x_{0}^{2}+1\right)-8}{2\left(2 x_{0}^{2}-1\right)} \\
= & \frac{2 \times(5+1)-8}{2 \times(5-1)}=\frac{1}{2} .
\end{aligned}
\]

Thus, \( \angle F_{1} P F_{2}=60^{\circ} \).