Chọn B
Gọi I là trung điểm AB, ta có:\( IH\bot AB\Rightarrow AB\bot \left(SIH\right)\Rightarrow AB\bot SI.\)
\(\left(\left(SAB\right),\left(ABC\right)\right)=\widehat{SIH}. AH=\frac{BC}{2} =\frac{a}{2} , SH=\sqrt{SA^{2} -AH^{2} } =\frac{a}{\sqrt{2} } ;\)
\(IH=\frac{AC}{2} =\frac{a\sqrt{6} }{6} . \tan \widehat{SIH}=\frac{SH}{IH} =\frac{\frac{a}{\sqrt{2} } }{\frac{a\sqrt{6} }{6} } =\sqrt{3} .\)
Vậy \(\left(\left(SAB\right),\left(ABC\right)\right)=\widehat{SIH}=\frac{\pi }{3} .\)