5. Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{19}+\frac{y^{2}}{3}=1$, and point $P$ is on the ellipse $C$. If $S_{\triangle P F_{1} F_{2}}=\sqrt{3}$, then $\angle F_{1} P F_{2}=$ $\qquad$
5. $60^{\circ}$.

Let $\angle F_{1} P F_{2}=\theta$. Then
$$
\left\{\begin{array}{l}
P F_{1}+P F_{2}=2 \sqrt{19}, \\
P F_{1}^{2}+P F_{2}^{2}-2 P F_{1} \cdot P F_{2} \cos \theta=64 .
\end{array}\right.
$$

Thus, $P F_{1} \cdot P F_{2}=\frac{6}{1+\cos \theta}$.
And $S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2} \sin \theta=3 \tan \frac{\theta}{2}=\sqrt{3}$, so $\theta=60^{\circ}$.