Gọi H là hình chiếu của S lên đáy, từ H kẻ \(HM\perp CD\)
\(\Rightarrow CD\perp\left(SHM\right)\Rightarrow\widehat{SMH}\) là góc giữa (SCD) và (ABCD)
\(\Rightarrow\widehat{SMH}=60^0\)
Talet: \(\dfrac{HM}{AD}=\dfrac{CH}{AC}=\dfrac{3}{4}\Rightarrow HM=\dfrac{3a}{4}\)
\(\Rightarrow SH=HM.tan60^0=\dfrac{3a\sqrt{3}}{4}\)
Do \(AB||\left(SCD\right)\Rightarrow d\left(B;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)
Mà \(AC=\dfrac{4}{3}HC\Rightarrow d\left(A;\left(SCD\right)\right)=\dfrac{4}{3}d\left(H;\left(SCD\right)\right)\)
Từ H kẻ \(HK\perp SM\Rightarrow HK\perp\left(SCD\right)\Rightarrow HK=d\left(H;\left(SCD\right)\right)\)
\(\dfrac{1}{HK^2}=\dfrac{1}{SH^2}+\dfrac{1}{HM^2}=\dfrac{64}{27a^2}\Rightarrow HK=\dfrac{3a\sqrt{3}}{8}\)
\(\Rightarrow d\left(B;\left(SCD\right)\right)=\dfrac{4}{3}HK=\dfrac{a\sqrt{3}}{2}\)
b.
Ta có: \(\left\{{}\begin{matrix}SH\perp\left(ABCD\right)\Rightarrow SH\perp BD\\BD\perp AC\end{matrix}\right.\) \(\Rightarrow BD\perp\left(SAC\right)\)
Trong mp (SAC), từ H kẻ \(HI\perp SO\)
\(\Rightarrow HI\perp\left(SBD\right)\Rightarrow HI=d\left(H;\left(SBD\right)\right)\)
\(OH=\dfrac{1}{2}AO=\dfrac{1}{4}AC=\dfrac{a\sqrt{2}}{4}\)
\(\dfrac{1}{HI^2}=\dfrac{1}{OH^2}+\dfrac{1}{SH^2}=\dfrac{232}{27a^2}\Rightarrow HI=a\sqrt{\dfrac{27}{232}}\)
AH cắt (SBD) tại O, mà \(AO=2HO\Rightarrow d\left(A;\left(SBD\right)\right)=2d\left(H;\left(SBD\right)\right)=\dfrac{3a\sqrt{174}}{58}\)
