Ta có: \(SH\perp\left(ABCD\right)\Rightarrow SH\perp AD\) (1)
Mà \(AD\perp AB\) (2)
(1);(2) \(\Rightarrow AD\perp\left(SAB\right)\Rightarrow AD\perp SB\)
b.
Ta có: \(SH=\dfrac{AB\sqrt{3}}{2}=a\sqrt{3}\)
\(OH=\dfrac{1}{2}BC=a\sqrt{2}\Rightarrow SO=\sqrt{SH^2+OH^2}=a\sqrt{5}\)
Do \(OH||AD\) (đường trung bình) \(\Rightarrow OH||\left(SAD\right)\Rightarrow d\left(O;\left(SAD\right)\right)=d\left(H;\left(SAD\right)\right)\)
Từ H kẻ \(HK\perp SA\Rightarrow HK\perp\left(SAD\right)\Rightarrow HK=d\left(H;\left(SAD\right)\right)=d\left(O;\left(SAD\right)\right)\)
\(\dfrac{1}{HK^2}=\dfrac{1}{AH^2}+\dfrac{1}{SH^2}\Rightarrow HK=\dfrac{SH.AH}{\sqrt{SH^2+AH^2}}=\dfrac{a\sqrt{3}}{2}\)
Gọi \(\alpha\) là góc giữa SO và (SAD)
\(\Rightarrow sin\alpha=\dfrac{d\left(O;SAD\right)}{SO}=\dfrac{HK}{SO}=\dfrac{\sqrt{15}}{10}\)
\(\Rightarrow\alpha\approx22^047'\)
