請教一題三角函數測量的題目~

令\(\overline{BC}=1,\overline{AB}=\overline{FB}=x\)
則\(\overline{BD}=\tan 15{}^\circ x=\left( 2-\sqrt{3} \right)x,\overline{AD}=\left( \sqrt{3}-1 \right)x\)
由孟氏定理
\(\begin{align}
  & \frac{\sqrt{3}-1}{2-\sqrt{3}}\times \frac{x}{x-1}\times \frac{1}{3}=1 \\
& \tan \alpha =x=6+3\sqrt{3} \\
\end{align}\)