\(\left\{{}\begin{matrix}SA\perp BC\\AB\perp BC\end{matrix}\right.\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp SB\)
\(\left\{{}\begin{matrix}BC\perp SB\\BC\perp AB\\\left(SAB\right)\cap\left(SBC\right)=BC\end{matrix}\right.\Rightarrow\left(\left(SAB\right),\left(SBC\right)\right)=\left(SB,AB\right)=\widehat{SBA}\)