Kẻ $SH\perp(ABC)$ Từ H kẻ $HE//AC; HI//AB\Rightarrow \widehat{HES}=\widehat{HIS}=60^{o}$
nhận thấy rằng $HE=HI$
Ta có : $a=BC=BH+CH=\frac{EH}{\sin 30}+\frac{IH}{\sin 60}(với \widehat{ABC}=30^{o};\widehat{ACB}=60^{o}$
$\Rightarrow IH=\frac{3-\sqrt{3}}{4}\Rightarrow SH=\frac{-3+3\sqrt{3}}{4}$
$\Rightarrow V_{S.ABC}=\frac{3-\sqrt{3}}{32}=\frac{1}{3}d_{(B;(SAC))}S_{SAC}(*)$
Có: $SI=\frac{3-\sqrt{3}}{2}\Rightarrow S_{SAC}=\frac{1}{2}AC.SI=...$
$\Rightarrow thay vào (*)\Rightarrow ...$