Ta có: \(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BD\\BD\perp AC\text{(hai đường chéo hình thoi)}\end{matrix}\right.\) \(\Rightarrow BD\perp\left(SAC\right)\)
\(SA\perp\left(ABCD\right)\Rightarrow\widehat{SCA}\) là góc giữa SC và (ABCD)
\(tan\widehat{SCA}=\dfrac{SA}{AC}=\sqrt{3}\Rightarrow\widehat{SCA}=60^0\)