Kẻ AD vuông góc BC
\(\Rightarrow BC\perp\left(SAD\right)\Rightarrow\widehat{SDA}\) là góc giữa (SBC) và (ABC)
\(\Rightarrow\widehat{SDA}=60^0\)
\(\Rightarrow AD=\dfrac{SA}{tan60^0}=\dfrac{a\sqrt{3}}{3}\)
\(\Rightarrow AB=\dfrac{AD}{sin\widehat{ABD}}=\dfrac{AD}{sin60^0}=\dfrac{2a}{3}\)
\(V=\dfrac{1}{3}SA.S_{ABC}=\dfrac{1}{3}.a.\dfrac{1}{2}.\left(\dfrac{2a}{3}\right)^2.sin120^0=\dfrac{a^3\sqrt{3}}{27}\)