Kẻ AE vuông góc SC (E thuộc SC)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\AB\perp BC\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp AM\)

\(\Rightarrow AM\perp\left(SBC\right)\Rightarrow AM\perp SC\)

Hoàn toàn tương tự ta có \(AN\perp SC\Rightarrow SC\perp\left(AMN\right)\)

Mà \(AE\perp SC\Rightarrow E\in\left(AMN\right)\)

\(\Rightarrow AE\) là hình chiếu vuông góc của SA lên (AMN)

\(\Rightarrow\widehat{SAE}\) là góc giữa SA và (AMN)

\(AC=a\sqrt{2}\Rightarrow SC=\sqrt{SA^2+AC^2}=2a\)

\(\Delta SAC\) vuông cân tại A \(\Rightarrow AE=SE=\dfrac{1}{2}SC=a\)

\(\Rightarrow\Delta SAE\) vuông cân tại E \(\Rightarrow\widehat{SAE}=45^0\)

Lần lượt kẻ \(AE\perp SB\)  (1) và \(AF\perp SD\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\AB\perp BC\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)

\(\Rightarrow BC\perp AE\) (2)

(1);(2) \(\Rightarrow AE\perp\left(SBC\right)\)

Hoàn toàn tương tự ta có \(AF\perp\left(SCD\right)\)

\(\Rightarrow\) Góc giữa (SBC) và (SCD) là góc giữa AE và AF

Cũng từ \(BC\perp\left(SAB\right)\) mà \(BC=\left(SBC\right)\cap\left(ABCD\right)\Rightarrow\widehat{SBA}\) là góc giữa (SBCD) và đáy

\(\Rightarrow\widehat{SBA}=60^0\Rightarrow SA=AB.tan60^0=a\sqrt{3}\)

Hệ thức lượng: \(\dfrac{1}{AE^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}\Rightarrow AE=\dfrac{a\sqrt{3}}{2}\)

\(\dfrac{1}{AF^2}=\dfrac{1}{SA^2}+\dfrac{1}{AD^2}\Rightarrow AF=\dfrac{a\sqrt{6}}{2}\)

\(SB=\sqrt{SA^2+AB^2}=2a\) ; \(SD=a\sqrt{6}\)

\(BD=\sqrt{AB^2+AD^2}=2a\Rightarrow cos\widehat{BSD}=\dfrac{SB^2+SD^2-BD^2}{2SB.SD}=\dfrac{\sqrt{6}}{4}\)

\(SE=\sqrt{SA^2-AE^2}=\dfrac{3a}{2}\) ; \(SF=\sqrt{SA^2-AF^2}=\dfrac{a\sqrt{6}}{2}\)

\(\Rightarrow EF=\sqrt{SE^2+SF^2-2SE.SF.cos\widehat{BSD}}=\dfrac{a\sqrt{6}}{2}\)

\(\Rightarrow cos\widehat{EAF}=\dfrac{AE^2+AF^2-EF^2}{2AE.AF}=\dfrac{\sqrt{2}}{4}\)

Từ A kẻ \(AE\perp SB\) (\(E\in SB\))

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\AB\perp BC\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)

\(\Rightarrow BC\perp AE\)

\(\Rightarrow AE\perp\left(SBC\right)\)

\(\Rightarrow\widehat{ACE}\) là góc giữa AC và (SBC)

Hệ thức lượng trong tam giác SAB:

\(\dfrac{1}{AE^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}\Rightarrow AE=\dfrac{SA.AB}{\sqrt{SA^2+AB^2}}=\dfrac{a\sqrt{3}}{2}\)

\(AC=AB\sqrt{2}=a\sqrt{2}\)

\(\Rightarrow sin\widehat{ACE}=\dfrac{AE}{AC}=\dfrac{\sqrt{6}}{4}\)

Gọi D là hình chiếu vuông góc của S lên (ABC)

\(SD\perp\left(ABC\right)\Rightarrow SD\perp AB\) , mà \(AB\perp SA\left(gt\right)\Rightarrow AB\perp\left(SAD\right)\Rightarrow AB\perp AD\)

\(\Rightarrow AD||BC\)

Tương tự ta có: \(BC\perp\left(SCD\right)\Rightarrow BC\perp CD\Rightarrow CD||AB\)

\(\Rightarrow\) Tứ giác ABCD là hình vuông

\(\Rightarrow BD=a\sqrt{2}\)

\(SD=\sqrt{SB^2-BD^2}=a\sqrt{2}\)

Gọi P là trung điểm AD \(\Rightarrow MP\) là đường trung bình tam giác SAD

\(\Rightarrow\left\{{}\begin{matrix}MP=\dfrac{1}{2}SD=\dfrac{a\sqrt{2}}{2}\\MP||SD\Rightarrow MP\perp\left(ABC\right)\end{matrix}\right.\)

\(\Rightarrow\alpha=\widehat{MNP}\)

\(cos\alpha=\dfrac{NP}{MN}=\dfrac{NP}{\sqrt{NP^2+MP^2}}=\dfrac{a}{\sqrt{a^2+\dfrac{a^2}{2}}}=\dfrac{\sqrt{6}}{3}\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\BC\perp AB\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)

Kẻ \(AH\perp SB\Rightarrow AH\perp\left(SBC\right)\)

\(\Rightarrow\widehat{ACH}\) là góc giữa AC và (SBC)

\(AC=a\sqrt{2}\) ; \(\dfrac{1}{AH^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}=\dfrac{1}{\dfrac{6a^2}{9}}+\dfrac{1}{a^2}\Rightarrow AH=\dfrac{a\sqrt{10}}{5}\)

\(\Rightarrow sin\widehat{ACH}=\dfrac{AH}{AC}=\dfrac{\sqrt{5}}{5}\Rightarrow\widehat{ACH}\approx26^034'\)

\(\dfrac{V_{SAHKE}}{V_{SABCD}}=\dfrac{2V_{SAHK}}{2V_{SABC}}=\dfrac{V_{SAHK}}{V_{SABC}}\)

\(V_{SABC}=\dfrac{1}{3}SA.\dfrac{1}{2}AB.BC=\dfrac{a^3}{3}\); \(V_{SABCD}=\dfrac{2a^3}{3}\)

\(\dfrac{SH}{SB}=\dfrac{SA^2}{SB}:SB=\left(\dfrac{SA}{SB}\right)^2\); \(\dfrac{SK}{SC}=\dfrac{SA^2}{SC}:SC=\left(\dfrac{SA}{SC}\right)^2\)

\(SB=\sqrt{SA^2+AB^2}=a\sqrt{5}\) ; \(SC=\sqrt{SA^2+AC^2}=a\sqrt{6}\)

\(\dfrac{V_{SAHK}}{V_{SABC}}=\left(\dfrac{SA}{SB}\right)^2.\left(\dfrac{SA}{SC}\right)^2\)

\(\Rightarrow V_{SAHKE}=\left(\dfrac{2a}{a\sqrt{5}}\right)^2.\left(\dfrac{2a}{a\sqrt{6}}\right)^2.\dfrac{2a^3}{3}=\dfrac{16a^3}{45}\)