$1)(SAB) \perp (ABCD)\\ \Rightarrow ((SAB);(ABCD))=90^o$
$2)\Delta SAB$ đều, $SH$ vừa là trung tuyến vừa là đường cao
$\Rightarrow SH \perp AB\\ (SAB) \perp (ABCD);SH \subset (SAB);(SAB) \cap (ABCD)=AB; SH \perp AB\\ \Rightarrow SH \perp (ABCD)\\ (SC;(ABCD))=(SC;CH)$
\Delta SAB đều, $SH=\dfrac{3a\sqrt{3}}{2}$
$\Delta CBH,\widehat{CBH}=90^o\\ \Rightarrow CH=\sqrt{CB^2+BH^2}=\sqrt{CB^2+\left(\dfrac{AB}{2}\right)^2}=\dfrac{\sqrt{73}a}{2}\\ (SC;CH)=\widehat{SCH}=\arctan\dfrac{SH}{CH}\approx 31,3^o$
$3)E$ là trung điểm $DC$
$EH$ là đường trung bình hình thang $ABCD$
$\Rightarrow EH//AD//BC,EH=\dfrac{AD+BC}{2}=AD=BC\\ \Rightarrow EH\perp AB,EH\perp DC\\ SH \perp (ABCD)\\ DC \subset (ABCD)\\ \Rightarrow SH \perp DC\\ EH\perp DC\\ \Rightarrow DC \perp (SHE)\\ \Rightarrow DC \perp SE\\ ((SCD);(ABCD))=(SE;EH)=\widehat{SEH}=\arctan\dfrac{SH}{EH}\approx 33^o$$4)$Kẻ $AF\perp SB;BG\perp SA$
$AF=BG=SH=\dfrac{3a\sqrt{3}}{2}$
$ABCD$ là hình chữ nhật
$\Rightarrow CB \perp AB\\ CB \perp SH(SH\perp(ABCD);CB \subset(ABCD))\\ \Rightarrow CB\perp(SAB)\\ AF \subset (SAB)\\ \Rightarrow CB \perp AF\\ AF\perp SB\\ \Rightarrow AF \perp (SBC)$
Tương tự
$\Rightarrow BG \perp (SAD)\\ ((SBC);(SAD))=(AF;BG)=60^o$
$5)$Trong $(SHE)$, kẻ $HI \perp SE(1)$
$HI=\sqrt{\dfrac{1}{\dfrac{1}{SH^2}+\dfrac{1}{EH^2}}}=\dfrac{12\sqrt{273}}{91}\\ DC \perp (SHE)\\ DC \subset(SCD)\\ \Rightarrow (SCD)\perp(SHE)(2)\\ (SCD)\cap(SHE)=HE\\ (1)(2)HI\perp(SCD)\\ EH \perp AB\\ EH \perp SH(SH\perp(ABCD);EH \subset(ABCD))\\ \Rightarrow EH\perp(SAB)\\ ((SAB);(SCD))=(EH;HI)=\widehat{EHI}=\arccos\dfrac{HI}{EH}\approx 57^o$
$6)\Delta SAC, GO$ là đường trung bình
$\Rightarrow GO//SC,GO=\dfrac{1}{2}SC$
$J$ là trung điểm $DG$
$\Delta DGB, JO$ là đường trung bình
$\Rightarrow JO//BG;JO=\dfrac{1}{2}BG=\dfrac{3a\sqrt{3}}{4}\\ BG \perp (SAD)\\ \Rightarrow JO\perp (SAD)\\ (SC;(SAD))=(GO;(SAD))=(GO;JG)=\widehat{JGO}$
$\Delta SBC$ vuông tại $B$
$\Rightarrow SC=\sqrt{BC^2+SB^2}=5a\\ \Rightarrow GO=2,5a\\ \widehat{JGO}=\arcsin\dfrac{JO}{GO}\approx 31,3^o$
