\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\AB\perp BC\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp AM\)
Mà \(AM\perp SB\) (gt)
\(\Rightarrow AM\perp\left(SBC\right)\Rightarrow AM\perp SC\)
Hay góc giữa 2 đường thẳng là \(\dfrac{\pi}{2}\)