如圖，令 \(\overline{CE}=3a, \overline{ED}=a, \overline{AE}=b, \angle CDB=\alpha\Rightarrow\angle ACD=\alpha-45^\circ\)
在 \(\triangle AEC\) 中，由正弦定理，可得 \(\displaystyle \frac{3a}{\sin30^\circ}=\frac{b}{\sin\left(\alpha-45^\circ\right)}\)
\(\displaystyle \Rightarrow \frac{b}{a}=\frac{3\sin\left(\alpha-45^\circ\right)}{\sin30^\circ}\)
在 \(\triangle AED\) 中，由正弦定理，可得 \(\displaystyle \frac{a}{\sin15^\circ}=\frac{b}{\sin\left(180^\circ-\alpha\right)}\)
\(\displaystyle \Rightarrow \frac{b}{a}=\frac{\sin\alpha}{\sin15^\circ}\)
因此，\(\displaystyle \frac{3\sin\left(\alpha-45^\circ\right)}{\sin30^\circ}=\frac{\sin\alpha}{\sin15^\circ}\)
\(\displaystyle \Rightarrow 6\sin\left(\alpha-45^\circ\right)=\frac{4}{\sqrt{6}-\sqrt{2}}\cdot\sin\alpha\)
\(\displaystyle \Rightarrow 3\sqrt{2}\sin\alpha-3\sqrt{2}\cos\alpha=\frac{4}{\sqrt{6}-\sqrt{2}}\cdot\sin\alpha\)
\(\displaystyle \Rightarrow \frac{\sin\alpha}{\cos\alpha}=6+3\sqrt{3}\)
\(\Rightarrow \tan\alpha=6+3\sqrt{3}\)
