Do \(SA\perp\left(ABCD\right)\Rightarrow\) góc giữa \(SO\) và \(\left(ABCD\right)\) là \(\widehat{SOA}\)
\(AO=\frac{AC}{2}=\frac{a\sqrt{2}}{2}\)
\(\Rightarrow tan\widehat{SOA}=\frac{SA}{OA}=\frac{a\sqrt{2}}{\frac{a\sqrt{2}}{2}}=2\)
\(\Rightarrow\widehat{SOA}=arctan2\approx63^026'\)