Lời giải:
a) Ta có:
$\begin{cases}SA\perp BC\quad (SA\perp (ABC))\\BC\perp AB\quad (gt)\\SA\cap AB = \{A\}\end{cases}$
$\Rightarrow BC\perp (SAB)$
$\Rightarrow BC\perp SB$
$\Rightarrow \triangle SBC$ vuông tại $B$
b) Ta có:
$\begin{cases}(SBC)\cap (ABC) = BC\\SB\perp BC\quad \text{(câu a)}\\SB\subset (SBC)\\AB\perp BC\quad (gt)\\AB\subset (ABC)\end{cases}$
$\Rightarrow \widehat{((SBC);(ABC))} = \widehat{SBA}$
