(10) Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C: x^{2}-y^{2}=1$, and point $P$ is on $C$. If the area of $\triangle P F_{1} F_{2}$ is $\sqrt{3}$, 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 see that \( F_{1} F_{2} = 2 \sqrt{2} \). Note that
\[
\begin{aligned}
S_{\triangle P F_{1} F_{2}} & =\frac{1}{2}\left|F_{1} F_{2}\right| \cdot\left|y_{0}\right| \\
& =\frac{1}{2} \times 2 \sqrt{2}\left|y_{0}\right|=\sqrt{3},
\end{aligned}
\]
Solving for \( \left|y_{0}\right| \) gives \( \left|y_{0}\right|=\frac{\sqrt{6}}{2} \). From \( x^{2}-y^{2}=1 \), we have \( x_{0}^{2}-y_{0}^{2}=1 \), solving for \( x_{0} \) gives
\[
x_{0}^{2}=1+y_{0}^{2}=1+\frac{6}{4}=\frac{5}{2}.
\]

By the second definition of the hyperbola,
\[
\left|P F_{1}\right|=e\left[x_{0}-\left(-\frac{a^{2}}{c}\right)\right]=a+e x_{0}=1+\sqrt{2} x_{0}
\]
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 have
\[
\cos \angle F_{1} P F_{2}=\frac{\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-\left|F_{1} F_{2}\right|^{2}}{2\left|P F_{1}\right|\left|P F_{2}\right|},
\]
Thus,
\[
\begin{aligned}
\cos \angle F_{1} P F_{2} & =\frac{\left(1+\sqrt{2} x_{0}\right)^{2}+\left(\sqrt{2} x_{0}-1\right)^{2}-(2 \sqrt{2})^{2}}{2\left(1+\sqrt{2} x_{0}\right)\left(\sqrt{2} x_{0}-1\right)} \\
& =\frac{2\left(2 x_{0}^{2}+1\right)-8}{2\left(2 x_{0}^{2}-1\right)} \\
& =\frac{2\left(2 \times \frac{5}{2}+1\right)-8}{2\left(2 \times \frac{5}{2}-1\right)} \\
& =\frac{2 \times(5+1)-8}{2 \times(5-1)} \\
& =\frac{2 \times 6-8}{8}=\frac{1}{2}.
\end{aligned}
\]

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