2. In convex pentagon $A B C D E, A B=B C, C D=D E, \angle A B C=100^{\circ}, \angle C D E=80^{\circ}$, and $B D^{2}=$ $\frac{100}{\sin 100^{\circ}}$. Find the area of the pentagon.
Answer: 50
Solution: Let $A B=B C=p, A C=r, C D=D E=q, C E=s$, and $\theta=\angle A C E$. Then $r=2 p \cos 40^{\circ}, s=2 q \cos 50^{\circ}$, and
$$
\begin{aligned}
\frac{100}{\sin 100^{\circ}} & =B D^{2}=p^{2}+q^{2}-2 p q \cos \left(90^{\circ}+\theta\right) \\
& =p^{2}+q^{2}+2 p q \sin \theta
\end{aligned}
$$

Thus, the required area is given by
$$
\begin{aligned}
\frac{1}{2} p^{2} \sin 100^{\circ}+\frac{1}{2} q^{2} \sin 80^{\circ}+\frac{1}{2} r s \sin \theta & =\frac{1}{2} p^{2} \sin 100^{\circ}+\frac{1}{2} q^{2} \sin 80^{\circ}+\frac{1}{2}\left(4 p q \cos 40^{\circ} \sin 40^{\circ}\right) \sin \theta \\
& =\frac{1}{2} p^{2} \sin 100^{\circ}+\frac{1}{2} q^{2} \sin 100^{\circ}+p q \sin 100^{\circ} \sin \theta \\
& =\frac{1}{2} \sin 100^{\circ}\left(p^{2}+q^{2}+2 p q \sin \theta\right) \\
& =\frac{1}{2} \sin 100^{\circ}\left(\frac{100}{\sin 100^{\circ}}\right) \\
& =50
\end{aligned}
$$