b,
$BC\bot AB, BC\bot SA$ nên $BC\bot (SAB)$
$\to AM\bot BC$
Mà $AM\bot SB$ nên $AM\bot (SBC)$
$\to AM\bot SC$
CMTT, ta có $AN\bot (SCD)$ nên $AN\bot SC$
$\to SC\bot (AMN)$
$\to SC\bot AP\quad\forall P\in (AMN)$
Trong $(SAC)$, kẻ $AI\bot SC$
$\to I\in (AMN)$
$\to SC\bot(AMIN)$
Có $CI\bot (AMIN)\to d(C,(AMN))=CI$
$AC=AB\sqrt2=a\sqrt2$
$\to SC=\sqrt{SA^2+AC^2}=a\sqrt6$
Ta có $\Delta AIC\backsim\Delta SAC$ (g.g)
$\to \dfrac{AC}{IC}=\dfrac{SC}{AC}$
$\to IC=\dfrac{a\sqrt2.a\sqrt2}{a\sqrt6}=\dfrac{a\sqrt6}{3}$
$\to AI=\sqrt{AC^2-IC^2}=\dfrac{2a\sqrt3}{3}$
$\Delta SAB=\Delta SAD$ (c.g.c)
$\to AM=AN$
$\widehat{IMA}=\widehat{INA}=90^o$
$\to \Delta IMA=\Delta INA$ (ch-cgv)
$\to \widehat{IAM}=\widehat{IAN}$
$\dfrac{1}{SA^2}+\dfrac{1}{AB^2}=\dfrac{1}{AM^2}\to AM=\dfrac{2a\sqrt5}{5}=AN$
$\to \cos\widehat{IAM}=\dfrac{AM}{IA}=\dfrac{\sqrt{15}}{5}$
$\to \sin\widehat{IAM}=\dfrac{\sqrt{10}}{5}$
$\to \sin\widehat{MAN}=\sin(2\widehat{IAM})=2\sin\widehat{IAM}\cos\widehat{IAM}= \dfrac{2\sqrt6}{5}$
$\to S_{AMN}=\dfrac{1}{2}MA.NA\sin\widehat{MAN}= \dfrac{4a^2\sqrt6}{25}$
Kẻ $MR//SA//NQ$
$\to \Delta ARQ$ là hình chiếu $\Delta AMN$ trên mặt đáy
$AM^2=AR.AB\to AR=AQ=\dfrac{4a}{5}$
$\to S_{ARQ}= \dfrac{1}{2}AR.AQ= \dfrac{8a^2}{25}$
Đặt $\varphi=((AMN),(ABCD))$
Có: $S_{ARQ}=S_{AMN}\cos\varphi$
Vậy $\cos\varphi= \dfrac{\sqrt6}{3}$
