已知 $F_{1}, F_{2}$ 分别为双曲线 $x^{2}-y^{2}=1$ 的左, 右焦点, 点 $P$ 在双曲线上. 若 $\angle F_{1} P F_{2}=60^{\circ}$, 则 $\triangle P F_{1} F_{2}$的面积为 ( )
由题意可得双曲线 $x^{2}-y^{2}=1$ 的, 即 $a=1, b=1, c=\sqrt{2}$, 则 $F_{2}(\sqrt{2}, 0), F_{1}(-\sqrt{2}, 0)$,又 $\left|F_{1} F_{2}
ight|^{2}=8,|| P F_{1}|-| P F_{2}||=2 a=2$,由余弦定理可得: $\left|F_{1} F_{2}
ight|^{2}=\left|P F_{1}
ight|^{2}+\left|P F_{2}
ight|^{2}-2\left|P F_{1}
ight| \cdot\left|P F_{2}
ight| \cos 60^{\circ}=\left(\left|P F_{1}
ight|-\left|P F_{2}
ight|
ight)$ ${ }^{2}+\left|P F_{1}
ight| \cdot\left|P F_{2}
ight|=24+\left|P F_{1}
ight| \cdot\left|P F_{2}
ight|, 	herefore\left|P F_{1}
ight| \cdot\left|P F_{2}
ight|=4, 	riangle F_{1} P F_{2}$ 的面积 $S, \left.S=rac{1}{2} 
ight
vert\, P F_{1}$ $|\cdot| P F_{2} \left\lvert\, \sin 60^{\circ}=rac{1}{2} 	imes 4 	imes rac{\sqrt{3}}{2}=\sqrt{3}
ight.$,