Kẻ $SH \perp AC$
$SH \perp AC, AC=(SAC) \cap (ABC)\\ \Rightarrow SH \perp (ABC)$
Kẻ $HD \perp AB$
$SH \perp AB(SH \perp (ABC))\\ \Rightarrow AB \perp (SHD)\\ \Leftrightarrow (SAB) \perp (SHD)$
Kẻ $HE \perp SD$
$HE \perp SD; HE=((SAB) \cap (SHD)\\ \Rightarrow HE \perp (SAB)\\ \Rightarrow ((SAB);(ABC))=(SH;HE)=\widehat{SHE}$
$\Delta SAC$ vuông cân tại $A$ có $SH$ vừa là đường cao vừa là trung tuyến
$\Rightarrow AH=\dfrac{1}{2}AC\\ HD//BC(HD \perp AB,BC \perp AB) \Rightarrow AD=\dfrac{1}{2}AB$
$\Rightarrow HD$ là đường trung bình $\Delta ABC$
$\Rightarrow HD=\dfrac{1}{2}BC=\dfrac{1}{2}\\ SH =\dfrac{\sqrt{2}}{2}\\ \Rightarrow HE=\sqrt{\dfrac{1}{\dfrac{1}{SH^2}+\dfrac{1}{DH^2}}}=\dfrac{\sqrt{6}}{6}\\ \Rightarrow SE =\dfrac{\sqrt{3}}{3}\\ \Rightarrow \tan \widehat{SHE}=\dfrac{SE}{HE}=\sqrt{2}\\ \Rightarrow B$
