a,
Gọi $O=AC\bot BD$
$SA\bot (ABCD)\to BD\bot SA$
Suy ra $BD\bot (SAC)$
Trong $(SAC)$, kẻ $OH\bot SC$
$BD\bot (SAC)\to BD\bot OH$
$\to d(BD; SC)=OH$
$AC=AB\sqrt2=a\sqrt2$
$\to OC=\dfrac{AC}{2}=\dfrac{a\sqrt2}{2}$
$SC=\sqrt{SA^2+AC^2}=3a$
Ta có $\Delta HOC\backsim\Delta ASC$ (g.g)
$\to \dfrac{OC}{OH}=\dfrac{SC}{SA}$
$\to OH=\dfrac{\dfrac{a\sqrt2}{2}.a\sqrt7}{3a}=\dfrac{a\sqrt{14}}{6}$
Vậy $d(BD;SC)=\dfrac{a\sqrt{14}}{6}$
b,
Có $BO\bot (SAC)$
$\to (SB,(SAC))=(SB,SO)$
$BO=\dfrac{a\sqrt2}{2}$
$SB=\sqrt{SA^2+AB^2}=a\sqrt8$
$\Delta SBO$ vuông tại $O$ có:
$\sin\widehat{BSO}=\dfrac{BO}{SB}=\dfrac{\sqrt2}{2\sqrt8}=\dfrac{1}{4}$
Vậy $\sin(SB,(SAC))=\dfrac{1}{4}$
$CD\bot AD, CD\bot SA\to CD\bot SD$
Có: $AD\bot CD, SD\bot CD, (SCD)\cap(ABCD)=CD$
$\to ((SCD),(ABCD))=(SD,AD)$
$\Delta SAD$ vuông tại $A$ có:
$\tan\widehat{SDA}=\dfrac{SA}{AD}=\sqrt7$
Vậy $\tan((SCD),(ABCD))=\sqrt7$
b,
$DO\bot (SAC)$
$\to d(D;(SAC))=DO=\dfrac{a\sqrt2}{2}$
