ΔABC 中，E 是 AC 中点，BC=1，∠A=100°，∠B=60°，∠CED=80°，求 SΔABC+2SΔCDE

王守恩

\(由CB=1,\ \ 得CA=\frac{\sin60^\circ}{\sin80^\circ}\ \ \ \ CE=CD=\frac{\sin60^\circ}{2\sin80^\circ}\)

\(S_{ΔABC}+2S_{ΔCDE}=\frac{CB*CA\sin20^\circ}{2}+\frac{2*CE*CD\sin20^\circ}{2}=\frac{\sin60^\circ\sin20^\circ}{2\sin80^\circ}+\frac{\sin60^\circ\sin60^\circ\sin20^\circ}{4\sin80^\circ\sin80^\circ}\)