Qua A kẻ đường thẳng vuông góc AC cắt BC kéo dài tại D
\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp AD\\AD\perp AC\end{matrix}\right.\) \(\Rightarrow AD\perp\left(SAC\right)\)
Từ A kẻ \(AH\perp SB\)
\(\left\{{}\begin{matrix}BC\perp AB\left(gt\right)\\SA\perp\left(ABC\right)\Rightarrow SA\perp BC\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp AH\)
\(\Rightarrow AH\perp\left(SBC\right)\)
\(\Rightarrow\widehat{HAD}\) là góc giữa (AC) và (SBC) hay \(\widehat{HAD}=60^0\)
\(AC=\sqrt{AB^2+BC^2}=2a\)
\(AD=AC.tanC=\dfrac{AC.AB}{BC}=\dfrac{2a\sqrt{3}}{3}\)
\(\Rightarrow AH=AD.cos60^0=\dfrac{a\sqrt{3}}{3}\)
\(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}\Rightarrow\dfrac{1}{SA^2}=\dfrac{1}{AH^2}-\dfrac{1}{AB^2}=\dfrac{2}{a^2}\Rightarrow SA=\dfrac{a\sqrt{2}}{2}\)
\(V=\dfrac{1}{6}.SA.AB.BC=\dfrac{a^3\sqrt{6}}{12}\)
