\(SA\perp\left(ABC\right)\Rightarrow SA\perp BC\)
\(AB\perp BC\Rightarrow BC\perp\left(SAB\right)\)
\(BC\in\left(SBC\right)\Rightarrow\left(SAB\right)\perp\left(SBC\right)\)
b/ \(SA\perp\left(ABC\right)\Rightarrow\widehat{SBA}\) là góc giữa SB và (ABC)
\(tan\widehat{SBA}=\frac{SA}{AB}=1\Rightarrow\widehat{SAB}=45^0\)
c/ \(BC\perp\left(SAC\right)\Rightarrow BC\perp AH\)
Mà \(AH\perp SB\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH\perp SC\)
Lại có \(SC\perp AK\) (giả thiết) \(\Rightarrow SC\perp\left(AHK\right)\)
\(\Rightarrow\widehat{AKH}\) là góc giữa (SBC) và (SAC)
\(AC=\sqrt{AB^2+BC^2}=3a\)
\(\frac{1}{AK^2}=\frac{1}{SA^2}+\frac{1}{AC^2}\Rightarrow AK=\frac{SA.AC}{\sqrt{SA^2+AC^2}}=\frac{6a\sqrt{13}}{13}\)
\(\frac{1}{AH^2}=\frac{1}{SA^2}+\frac{1}{AB^2}=\frac{2}{SA^2}\Rightarrow AH=a\sqrt{2}\)
\(\Rightarrow sin\widehat{AKH}=\frac{AH}{AK}=\frac{\sqrt{26}}{6}\Rightarrow\widehat{AHK}\approx58^0\)
