Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta < 60^{\circ}$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan\theta$? $\textbf{(A)}~\frac{3}{4}\qquad$ $\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad$ $\textbf{(C)}~\frac{4}{5}\qquad$ $\textbf{(D)}~\frac{11}{13}\qquad$ $\textbf{(E)}~\frac{7\sqrt{3}}{13}$
Solution
The hexagon $ADBECF$ consists of $\triangle ABC$ and three congruent triangles $ADB$, $BEC$, $CFA$. Since $[ABC]=\frac{\sqrt3}4\cdot14^2=49\sqrt3$, we get $[ADB]=\frac13\left(91\sqrt3-49\sqrt3\right)=14\sqrt3$.
Let $O$ be the center and $R=OA=\dfrac{14}{\sqrt3}$. Then $\angle AOD=\theta$ and $\angle DOB=120^\circ-\theta$. Points $D$ and $O$ lie on opposite sides of $AB$, so $$\begin{gathered}[ADB]=[AOD]+[DOB]-[AOB]\\=\tfrac12R^2\left(\sin\theta+\sin(120^\circ-\theta)\right)-\tfrac13\cdot49\sqrt3 .\end{gathered}$$
Put $\varphi=60^\circ-\theta$. Then $\sin\theta+\sin(120^\circ-\theta)=2\sin60^\circ\cos\varphi=\sqrt3\cos\varphi$, and $\frac12R^2\sqrt3=\frac{98\sqrt3}3$. So $$\begin{gathered}14\sqrt3=\frac{98\sqrt3}{3}\cos\varphi-\frac{49\sqrt3}{3}\\ \Longrightarrow\quad\cos\varphi=\frac{13}{14},\ \ \sin\varphi=\frac{3\sqrt3}{14},\ \ \tan\varphi=\frac{3\sqrt3}{13}.\end{gathered}$$
Equilateral $\triangle ABC$ with side length $14$ (its vertices labeled counterclockwise) is rotated counterclockwise about its center by an angle $\theta$, where $0<\theta<60^\circ$, taking $A,B,C$ to $D,E,F$, respectively. The six points $A,D,B,E,C,F$ (in this order) form a hexagon of area $77\sqrt3$. What is $\tan\theta$?
$\textbf{(A)}~\frac{3}{4}\qquad$ $\textbf{(B)}~\frac{5\sqrt{3}}{11}\qquad$ $\textbf{(C)}~\frac{4}{5}\qquad$ $\textbf{(D)}~\frac{11}{13}\qquad$ $\textbf{(E)}~\frac{7\sqrt{3}}{13}$ 