Khẳng định nào sau đây đúng?

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{SA \bot (ABCD)}\\{SA \subset (SAB)}\end{array} \Rightarrow (SAB) \bot (ABCD)} \right.\) hay $\left(\right(SAB),(ABCD\left)\right)={90}^{°}$

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{BC \bot AB}\\{BC \bot SA({\rm{ do }}SA \bot (ABCD))}\end{array} \Rightarrow BC \bot (SAB) \Rightarrow BC \bot SB} \right.\).

Khi đó: \(\left\{ {\begin{array}{*{20}{l}}{BC = (SBC) \cap (ABCD)}\\{AB \bot BC,SB \bot BC}\\{AB \subset (ABCD),SB \subset (SBC)}\end{array}} \right.\)

\( \Rightarrow ((SBC),(ABCD)) = (SB,AB) = \widehat {SBA}\) (góc \(\widehat {SBA}\) nhọn vì $\widehat{SAB}={90}^{°}$

Tam giác \(SAB\) vuông tại \(A\) có: $\tan \widehat{SBA}=\frac{SA}{AB}=\frac{a\sqrt{3}}{a}=\sqrt{3}\Rightarrow \widehat{SBA}={60}^{°}$

Vậy $\left(\right(SBC),(ABCD\left)\right)=\widehat{SBA}={60}^{°}$

Kẻ đường cao \(AK\) của tam giác \(ABD\).

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{BD \bot AK}\\{BD \bot SA}\end{array} \Rightarrow BD \bot (SAK) \Rightarrow BD \bot SK} \right.\).

Khi đó: \(\left\{ {\begin{array}{*{20}{l}}{(SBD) \cap (ABCD) = BD}\\{AK \bot BD,SK \bot BD}\\{AK \subset (ABCD),SK \subset (SBD)}\end{array}} \right.\)

\( \Rightarrow ((SBD),(ABCD)) = (SK,AK) = \widehat {SKA}\) (góc \(\widehat {SKA}\) nhọn vì $\widehat{SAK}={90}^{°}$

Tam giác \(ABD\) vuông tại \(A\) có đường cao \(AK\) nên

$\tan \widehat{SKA}=\frac{SA}{AK}=\frac{a\sqrt{3}}{\frac{a\sqrt{3}}{2}}=2\Rightarrow \widehat{SKA}\approx 63,{43}^{°}$

Tam giác \(SAK\) vuông tại \(A\) có: \(\tan \widehat {SKA} = \frac{{SA}}{{AK}} = \frac{{a\sqrt 3 }}{{\frac{{a\sqrt 3 }}{2}}} = 2 \Rightarrow \widehat {SKA} \approx 63,{43^^\circ }\).

Vậy $\left(\right(SBD),(ABCD\left)\right)=\widehat{SKA}\simeq 63,{43}^{°}$