6. In $\triangle A B C$, $A>90^{\circ}, B=20^{\circ}$, draw $A D \perp A B$ intersecting $B C$ at $D$. Given $A B=1, C D=4$, let $S$ be the area of $\triangle A B C$, then the sum of the numerator and denominator of $S^{2}$ in its simplest form is $\qquad$ .
\begin{array}{l}\text { Solution } \\ \text { As shown in the figure, } B D=\frac{1}{\cos 20^{\circ}}, B C=\frac{1}{\cos 20^{\circ}}+4, \\ \text { then } S=\frac{1}{2}\left(\frac{1}{\cos 20^{\circ}}+4\right) \sin 20^{\circ} \\ =\frac{\sin 20^{\circ}+4 \cos 20^{\circ} \sin 20^{\circ}}{2 \cos 20^{\circ}}=\frac{\sin 20^{\circ}+2 \sin 40^{\circ}}{2 \cos 20^{\circ}}=\frac{2 \sin 30^{\circ} \cos 10^{\circ}+\sin 40^{\circ}}{2 \cos 20^{\circ}}=\frac{\sin 80^{\circ}+\sin 40^{\circ}}{2 \cos 20^{\circ}} \\ =\frac{2 \sin 60^{\circ} \cos 20^{\circ}}{2 \cos 20^{\circ}}=\frac{\sqrt{3}}{2} \Rightarrow S^{2}=\frac{3}{4}, \text { so the sum of the numerator and denominator of the simplest fraction of } S^{2} \text { is } 7 .\end{array}