Áp dụng định lý hàm cos:
\(AH=\sqrt{AC^2+CH^2-2AC.CH.cosC}=\frac{a\sqrt{7}}{3}\)
\(\widehat{SAH}=45^0\Rightarrow\Delta SAH\) vuông cân tại H \(\Rightarrow SH=AH=\frac{a\sqrt{7}}{3}\)
Qua C kẻ đường thẳng song song AB cắt AH kéo dài tại D
\(\Rightarrow AB//\left(SCD\right)\Rightarrow d\left(AB;SC\right)=d\left(AB;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)
Áp dụng talet: \(\frac{BH}{HC}=\frac{AH}{HD}=\frac{1}{2}\Rightarrow\frac{AD}{HD}=\frac{3}{2}\)
\(\Rightarrow d\left(A;\left(SCD\right)\right)=\frac{3}{2}d\left(H;\left(SCD\right)\right)\)
Từ H kẻ \(HP\perp CD\) , từ H kẻ \(HQ\perp SP\)
\(\Rightarrow HQ\perp\left(SCD\right)\Rightarrow HQ=d\left(H;\left(SCD\right)\right)\)
\(\widehat{PCH}=B=60^0\) (so le trong) \(\Rightarrow HP=HC.sin60^0=\frac{a\sqrt{3}}{3}\)
\(\frac{1}{HQ^2}=\frac{1}{HP^2}+\frac{1}{SH^2}\Rightarrow HQ=\frac{HP.SH}{\sqrt{HP^2+SH^2}}=\frac{a\sqrt{210}}{30}\)
\(\Rightarrow d\left(AB;SC\right)=\frac{3}{2}HQ=\frac{a\sqrt{210}}{20}\)