Bài 1:
- a) Chứng minh $\left( SBC \right)\bot \left( SAB \right)$
Ta có:
$\begin{cases}BC\bot AB\\BC\bot SA\\AB\cap SA=A\end{cases}$
$\to BC\bot \left( SAB \right)$
$\to \left( SBC \right)\bot \left( SAB \right)$
b)
$\left( SBC \right)\cap \left( ABC \right)=BC$
Trong $\left( SBC \right)$ có $BC\bot SB$ ( vì $BC\bot \left( SAB \right)$
Trong $\left( ABC \right)$ có $BC\bot AB$
$\to \left( \widehat{\left( SBC \right);\left( ABC \right)} \right)=\left( \widehat{SB;AB} \right)=\widehat{SBA}$
$AB=\sqrt{A{{C}^{2}}-B{{C}^{2}}}=a\sqrt{2}$
$\tan \left( \widehat{SBA} \right)=\frac{SA}{AB}=\frac{a\sqrt{6}}{a\sqrt{2}}=\sqrt{3}$
$\to \widehat{SBA}=60{}^\circ $
Bài 2:
a)
Ta có:
$\begin{cases}BD\bot AC\\BD\bot SA\\AC\cap SA=A\end{cases}$
$\to \left(BD\right)\bot \left(SAC\right)$
$\to\left(SBD\right)\bot\left(SAC\right)$
b)
Gọi $O$ là tâm hình vuông $ABCD$
$\left( SBD \right)\cap \left( ABCD \right)=BD$
Trong $\left( SBD \right)$ có $BD\bot SO$ ( Vì $BD\bot \left( SAC \right)\,\,,\,\,SO\subset \left( SAC \right)$ )
Trong $\left( ABCD \right)$ có $BD\bot AO$
$\to \left( \widehat{\left( SBD \right);\left( ABCD \right)} \right)=\left( \widehat{\left( SO;AO \right)} \right)=\widehat{SOA}$
$AO=\frac{1}{2}AC=\frac{1}{2}\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\frac{a\sqrt{2}}{2}$
$AO=SA=\frac{a\sqrt{2}}{2}$
$\to $$\Delta SAO$ là tam giác vuông cân
$\to \widehat{SOA}=45{}^\circ $
