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 $\tan\theta$?

1. The hexagon is $\triangle ABC$ together with three congruent triangles $ADB$, $BEC$, $CFA$. Since $[ABC]=\frac{\sqrt3}4\cdot14^2=49\sqrt3$, we get $[ADB]=\frac13\left(77\sqrt3-49\sqrt3\right)=\frac{28\sqrt3}3$.
2. Let $O$ be the center, $R=OA=\dfrac{14}{\sqrt3}$, so $\angle AOD=\theta$ and $\angle DOB=120^\circ-\theta$. As $D$ and $O$ are on opposite sides of $AB$, $[ADB]=[AOD]+[DOB]-[AOB]$.
3. With $\varphi=60^\circ-\theta$ we have $\sin\theta+\sin(120^\circ-\theta)=\sqrt3\cos\varphi$, hence
   $$\begin{gathered}\frac{28\sqrt3}{3}=\frac{98\sqrt3}{3}\cos\varphi-\frac{49\sqrt3}{3}\\ \Longrightarrow\quad\cos\varphi=\frac{11}{14},\ \ \sin\varphi=\frac{5\sqrt3}{14},\ \ \tan\varphi=\frac{5\sqrt3}{11}.\end{gathered}$$
4. Therefore
   $$\tan\theta=\tan(60^\circ-\varphi)=\frac{\sqrt3-\frac{5\sqrt3}{11}}{1+\sqrt3\cdot\frac{5\sqrt3}{11}}=\frac{6\sqrt3/11}{26/11}=\boxed{\frac{3\sqrt3}{13}}.$$