\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BD\\BD\perp AC\end{matrix}\right.\) \(\Rightarrow BD\perp\left(SAC\right)\)
Mà \(BD\in\left(SBD\right)\Rightarrow\left(SAC\right)\perp\left(SBD\right)\)
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp AB\\AB\perp AD\end{matrix}\right.\) \(\Rightarrow AB\perp\left(SAD\right)\)
\(\Rightarrow\) SA là hình chiếu vuông góc của SB lên (SAD)
\(\Rightarrow\widehat{BSA}\) là góc giữa SB và (SAD)
\(tan\widehat{BSA}=\frac{AB}{SA}=\frac{1}{2}\Rightarrow\widehat{BSA}\approx26^034'\)
Kẻ \(AH\perp SD\) (1)
\(\left\{{}\begin{matrix}AB\perp\left(SAD\right)\\CD//AB\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\Rightarrow CD\perp AH\) (2)
(1);(2) \(\Rightarrow AH\perp\left(SCD\right)\Rightarrow AH=d\left(A;\left(SCD\right)\right)\)
\(\frac{1}{AH^2}=\frac{1}{SA^2}+\frac{1}{AD^2}\Rightarrow AH=\frac{SA.AD}{\sqrt{SA^2+AD^2}}=\frac{2a\sqrt{5}}{5}\)
