Trong `(ABCD), AC∩BD=O`

`=>SO⊥(ABCD)`

Trong `(ABCD),` kẻ `OE⊥CD`

`OE⊥CD;SO⊥CD=>CD⊥(SOE)=>CD⊥SE`

`=>[S,CD,B]=hat(SEO)`

`OE = 1/2 AD = a/2; OA = sqrt2/2 a`

`SO=sqrt(SA^2-OA^2)=sqrt6/6 a`

`SE=sqrt(SO^2+OE^2)= sqrt15/6`

`cos hat(SEO) = (OE)/(SE) = sqrt15/5`

`=>B`