Từ A kẻ \(AH\perp CD\Rightarrow\widehat{ASH}\) là góc giữa SA và (SCD)
\(DH=\dfrac{AB-DC}{2}=\dfrac{a}{2}\)
\(\Rightarrow AH=\sqrt{AD^2-DH^2}=\sqrt{a^2-\left(\dfrac{a}{2}\right)^2}=\dfrac{a\sqrt{3}}{2}\)
\(\Rightarrow tan\widehat{ASH}=\dfrac{AH}{SA}=\dfrac{1}{2}\)
\(\Rightarrow\widehat{ASH}\approx26^034'\)



