For example, let $F_{1}, F_{2}$ be the two foci of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{45}=1$, and $P$ be a point on the hyperbola. It is known that $\left|P F_{2}\right|$, $\left|P F_{1}\right|,\left|F_{1} F_{2}\right|$ form an arithmetic sequence (or $2\left|P F_{1}\right|=\left|P F_{2}\right|+\left|F_{1} F_{2}\right|$), and the common difference is greater than 0. Try to find $\angle F_{1} P F_{2}$.
From the given, we know that $a^{2}=4, b^{2}=45$, then $c=7$.
Also, $2\left|P F_{1}\right|=\left|P F_{2}\right|+2 c$, so $2\left|P F_{1}\right|-\left|P F_{2}\right|=14$.
And $\left|P F_{1}\right|-\left|P F_{2}\right|=2 a=4$, thus we find $\left|P F_{1}\right|=10,\left|P F_{2}\right|=6$. Therefore, by property 3(II), we know $60=\left|P F_{1}\right| \cdot\left|P F_{2}\right|=\frac{b^{2}}{\sin ^{2} \theta}=\frac{2 b^{2}}{1-\cos 2 \theta}$, which gives $\cos \theta=-\frac{1}{2}$.
Thus, $\theta=120^{\circ}$, i.e., $\angle F_{1} P F_{2}=120^{\circ}$.