Giải thích các bước giải:
Ta có $SA\perp (ABCD)\to \widehat{SC, ABCD}=\widehat{SCA}$
$\to \widehat{SCA}=45^o$
Mà $SA\perp AC\to \Delta ASC$ vuông cân tại $A$
$\to SA=AC=a\sqrt{2}$
Kẻ $AE\perp SD$
Ta có $CD\perp AD, SA\perp ABCD\to SA\perp CD$
$\to CD\perp SAD$
$\to CD\perp AE$
Mà $AE\perp SD\to AE\perp (SCD)\to AE\perp EC$
$\to \widehat{AC, SCD}=\widehat{ACE}$
Ta có $SA\perp AD, AE\perp SD$
$\to \dfrac1{AE^2}=\dfrac1{AS^2}+\dfrac1{AD^2}$
$\to AE=\dfrac{a\sqrt{6}}{3}$
$\to \sin\widehat{ACE}=\dfrac{AE}{AC}=\dfrac{\sqrt3}{3}$
$\to \widehat{ACE}=\arcsin(\dfrac{\sqrt3}{3})$
$\to \widehat{AC, SCD}=\arcsin(\dfrac{\sqrt3}{3})$
