In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. 
If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ, \measuredangle ABC = 100^\circ, \measuredangle ACB = 20^\circ$ and 
$\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals
$\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad \textbf{(B) }\frac{\sqrt{3}}{8}\qquad \textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad \textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad \textbf{(E) }\frac{1}{8}$
Let $F$ be the point on the extension of side $AB$ past $B$ for which $AF=1$. Since $AF=AC$ and $\measuredangle FAC = 60^\circ$,$\triangle ACF$ is equilateral. Let $G$ be the point on line segment $BF$ for which $\measuredangle BCG=20^\circ$. Then $\triangle BCG$ is similar to $\triangle DCE$ and $BC=2(EC)$. Also $\triangle FGC$ is congruent to $\triangle ABC$. Therefore, $[\triangle ACF]  = ([\triangle ABC] + [\triangle GCF]) + [\triangle BCG]$. Plugging in the values that we know and then dividing by 2 results in an answer of $\boxed{B) \frac{\sqrt{3}}{8}.}$
This solution is from the solution manual but was typed here by alpha_2.