Ta có: \(\left(SBC\right)\cap\left(ABC\right)=BC\)
Mà lại có: \(SA\perp BC\left(SA\perp\left(ABCD\right)\right);AB\perp BC\)
Do đó \(BC\perp\left(SAB\right)\)
Mặt khác \(\left(SAB\right)\cap\left(ABCD\right)=AB;\left(SAB\right)\perp\left(SBC\right)=SB\)
Vậy \(\left(\left(SBC\right),\left(ABC\right)\right)=\left(SB,AB\right)=\widehat{SBA}\)