Từ O kẻ \(OH\perp BC\Rightarrow OH\perp\left(SBC\right)\)
\(\Rightarrow\widehat{OSH}\) là góc giữa SO và (SBC)
\(OH=\frac{1}{2}AB=\frac{a}{2}\)
\(BD=a\sqrt{2}\Rightarrow OB=\frac{a\sqrt{2}}{2}\) \(\Rightarrow SO=\sqrt{SB^2+OB^2}=\frac{a\sqrt{14}}{2}\)
\(sin\widehat{OSH}=\frac{OH}{SO}=\frac{\sqrt{14}}{14}\Rightarrow\widehat{OSH}\approx15^030'\)