* Xác định góc giữa (SBC) và (ABCD)
Trong mặt phẳng (ABCD), dựng AM vuông góc với BC
$\left\{ \begin{array}{l}
BC \bot AM\\
BC \bot SA
\end{array} \right. \Rightarrow BC \bot \left( {SAM} \right) \Rightarrow BC \bot SM$
$ \Rightarrow \left[ {\widehat {\left( {SBC} \right),\left( {ABCD} \right)}} \right] = \left[ {\widehat {SM,AM}} \right] = \widehat {SMA} = \beta $
* Trong tam giác SAM
$\tan \left( {SMA} \right) = \frac{{SA}}{{AM}} \Leftrightarrow AM = \frac{a}{{\tan \beta }}$
$\sin \left( {SMA} \right) = \frac{{SA}}{{SM}} \Leftrightarrow SM = \frac{a}{{\sin \beta }}$
* Trong tam giác BAO vuông tại O
$\begin{array}{l}
\cos \left( {BAO} \right) = \cos \left( {\frac{\alpha }{2}} \right) = \frac{{AO}}{{AB}} \Leftrightarrow AO = AB.\cos \left( {\frac{\alpha }{2}} \right)\\
\Rightarrow AC = 2AB\cos \left( {\frac{\alpha }{2}} \right)
\end{array}$
* Diện tích tam giác ABC
$\begin{array}{l}
{S_{\Delta ABC}} = \frac{1}{2}BO.AC = \frac{1}{2}AM.BC\\
\Leftrightarrow \frac{1}{2}BO.2AB\cos \frac{\alpha }{2} = \frac{1}{2}\frac{a}{{\tan \beta }}BC\\
\Leftrightarrow BO.\cos \frac{\alpha }{2} = \frac{a}{{2\tan \beta }} \Leftrightarrow BO = \frac{a}{{2\tan \beta .\cos \frac{\alpha }{2}}}\\
\Rightarrow BD = \frac{a}{{\tan \beta .\cos \frac{\alpha }{2}}}
\end{array}$
$\sin \left( {BAO} \right) = \frac{{BO}}{{AB}} \Leftrightarrow AB = \frac{{BO}}{{\sin \left( {BAO} \right)}} = \frac{{\frac{a}{{2\tan \beta .\cos \frac{\alpha }{2}}}}}{{\sin \left( {\frac{\alpha }{2}} \right)}} = \frac{a}{{\tan \beta .\sin \alpha }}$
* Diện tích tam giác SAB, SAD:
${S_{\Delta SAB}} = {S_{\Delta SAD}} = \frac{1}{2}SA.AB = \frac{1}{2}a.\frac{a}{{\tan \beta .\sin \alpha }} = \frac{{{a^2}}}{{2\tan \beta \sin \alpha }}$
* Diện tích tam giác SCB, SCD
\[{S_{\Delta SBC}} = {S_{\Delta SCD}} = \frac{1}{2}SM.BC = \frac{1}{2}.\frac{a}{{\sin \beta }}.\frac{a}{{\tan \beta .\sin \alpha }} = \frac{{{a^2}}}{{2\sin \beta \tan \beta \sin \alpha }}\]
* Diện tích xung quanh:
\[{s_{xq}} = 2\left( {{S_{\Delta SAB}} + {S_{\Delta SBC}}} \right) = 2\left( {\frac{{{a^2}}}{{2\tan \beta \sin \alpha }} + \frac{{{a^2}}}{{2\sin \beta \tan \beta \sin \alpha }}} \right) = \frac{{{a^2}\left( {\sin \beta + 1} \right)}}{{\sin \beta \tan \beta \sin \alpha }}\]
* Thể tích khối chóp:
$AC = 2AB.\cos \left( {\frac{\alpha }{2}} \right) = \frac{{2a\cos \frac{\alpha }{2}}}{{\tan \beta \sin \alpha }}$
${S_{ABCD}} = \frac{1}{2}AC.BD = \frac{1}{2}.\frac{{2a\cos \frac{\alpha }{2}}}{{\tan \beta \sin \alpha }}.\frac{a}{{\tan \beta .\cos \frac{\alpha }{2}}} = \frac{{{a^2}}}{{\sin \alpha .{{\tan }^2}\beta }}$
${V_{S.ABCD}} = \frac{1}{3}SA.{S_{ABCD}} = \frac{{{a^3}}}{{3\sin \alpha .{{\tan }^2}\beta }}$
