a.
\(\left\{{}\begin{matrix}SH\perp\left(ABC\right)\\BC\in\left(ABC\right)\end{matrix}\right.\) \(\Rightarrow SH\perp BC\)
Lại có \(AB\perp BC\left(gt\right)\)
\(\Rightarrow BC\perp\left(SAB\right)\)
\(AK\in\left(SAB\right)\Rightarrow BC\perp AK\)
Tam giác ABC đều mà K là trung điểm SB
\(\Rightarrow AK\perp SB\)
\(\Rightarrow AK\perp\left(SBC\right)\)
b.
\(\left\{{}\begin{matrix}AK\in\left(AKI\right)\\AK\perp\left(SBC\right)\end{matrix}\right.\) \(\Rightarrow\left(AKI\right)\perp\left(SBC\right)\)
c.
\(\left\{{}\begin{matrix}BC\perp\left(SAB\right)\\BC=\left(SBC\right)\cap\left(ABC\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{SBA}\) là góc giữa (SBC) và (ABC)
Mà tam giác SAB đều \(\Rightarrow\widehat{SBA}=60^0\)
d.
Dựng hình chữ nhật ABCD \(\Rightarrow AB||CD\Rightarrow\left(SC,AB\right)=\left(SC,CD\right)=\widehat{SCD}\)
\(SC=\sqrt{SB^2+BC^2}=2a\)
\(SD=\sqrt{SA^2+AD^2}=\sqrt{SB^2+BC^2}=2a\)
\(CD=AB=a\)
\(\Rightarrow cos\widehat{SCD}=\dfrac{SC^2+CD^2-SD^2}{2SC.CD}=\dfrac{1}{4}\)
\(\Rightarrow\widehat{SCD}\approx75^031'\)
e.
\(AC=\sqrt{AB^2+BC^2}=2a\)
\(cos\widehat{BAC}=\dfrac{AB}{AC}=\dfrac{1}{2}\)
\(\overrightarrow{SB}.\overrightarrow{AC}=\left(\overrightarrow{SH}+\overrightarrow{HB}\right).\overrightarrow{AC}=\overrightarrow{SH}.\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{AC}\)
\(=\dfrac{1}{2}AB.AC.cos\widehat{BAC}=\dfrac{1}{2}.a.2a.\dfrac{1}{2}=\dfrac{a^2}{2}\)
\(\Rightarrow cos\left(AC,SB\right)=\dfrac{\left|\overrightarrow{SB}.\overrightarrow{AC}\right|}{SB.AC}=\dfrac{1}{4}\)
