\(\Delta SAB\) đều \(\Rightarrow SH\perp AB\)
Mà \(\left\{{}\begin{matrix}\left(SAB\right)\perp\left(ABCD\right)\\\left(SAB\right)\cap\left(ABCD\right)=AB\end{matrix}\right.\) \(\Rightarrow SH\perp\left(ABCD\right)\)
\(\Rightarrow SH\perp AD\) (1)
\(AD\perp AB\) (2) (đáy là hv)
(1);(2) \(\Rightarrow AD\perp\left(SAB\right)\)
b/ \(SH\perp\left(ABCD\right)\Rightarrow\widehat{SCH}\) là góc giữa SC và (ABCD)
\(SH=\frac{a\sqrt{3}}{2}\) (đường cao tam giác đều), \(CH=\sqrt{BC^2+BH^2}=\frac{a\sqrt{5}}{2}\)
\(tan\widehat{SCH}=\frac{SH}{CH}=\frac{\sqrt{15}}{5}\Rightarrow\widehat{SCH}\approx37^045'\)
\(\left\{{}\begin{matrix}AD//BC\\AD\perp\left(SAB\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)
\(\Rightarrow\widehat{BSC}\) là góc giữa SC và (SAB)
\(SB=BC=a\Rightarrow\Delta SBC\) vuông cân tại B \(\Rightarrow\widehat{BSC}=45^0\)
Gọi K là trung điểm CD \(\Rightarrow HK\perp CD\Rightarrow CD\perp\left(SHK\right)\)
\(\Rightarrow\widehat{SKH}\) là góc giữa (SCD) và (ABCD)
\(HK=AD=a\Rightarrow tan\widehat{SKH}=\frac{SA}{HK}=\frac{\sqrt{3}}{2}\Rightarrow\widehat{SKH}\approx40^053'\)
c/ Trong tam giác SHK, từ H kẻ \(HP\perp SK\Rightarrow HP\perp\left(SCD\right)\)
\(\Rightarrow HP=d\left(H;\left(SCD\right)\right)\)
\(\frac{1}{HP^2}=\frac{1}{SH^2}+\frac{1}{HK^2}\Rightarrow HP=\frac{SH.HK}{\sqrt{SH^2+HK^2}}=\frac{a\sqrt{21}}{7}\)
\(AB=2HB\Rightarrow d\left(A;\left(SBC\right)\right)=2d\left(H;\left(SBC\right)\right)\)
Từ H kẻ \(HQ\perp SB\Rightarrow HQ\perp\left(SBC\right)\Rightarrow HQ=d\left(H;\left(SBC\right)\right)\)
\(\frac{1}{HQ^2}=\frac{1}{SH^2}+\frac{1}{HB^2}\Rightarrow HQ=\frac{SH.HB}{\sqrt{SH^2+HB^2}}=\frac{a\sqrt{3}}{4}\)
\(\Rightarrow d\left(A;\left(SBC\right)\right)=\frac{a\sqrt{3}}{2}\)
