a) Notam {O}=AC Π BD si {O`}=A`C`Π B`D`. In ΔBB`O prin Teorema lui Pitagora⇒B`O=√[(6√2)²+(3√2)²]=√(72+18)=√90=3√10
sin∡((B`AC)(ACD`))=sin∡(B`OD`)=... .In ΔB`OD`⇒Aria ΔB`OD`= (OB`·OD`sin∡(B`OD`))/2=(3√10·3√10sin∡(B`OD`))/2=45·sin∡(B`OD`) (1)
AriaΔB`OD`=(B`D`·OO`)/2=(6√2·6√2)/2=36 (2). Din (1) si (2) ⇒ 45·sin∡(B`OD`) =36⇒sin∡(B`OD`)=4/5=0,8
b) DucemDE⊥AD`(3); E∈AD`. Din CD⊥(ADD`) si (3) prin Teorema celor 3 perpendiculare⇒CE⊥AD`⇒tg∡((CAD`)(A`AD))=tg∡(CED)=CD/ED=6/ED=6/(2√6)=√6/2 deoarece ED=(6·6√2)/(6√3)= 2√6(s-a aplicat teorema a doua a inaltimii in ΔADD`)
c) Din C`C⊥(ABC) si CO⊥BD prin Teorema celor 3 perpendiculare⇒C`O⊥BD d(C;(C`BD))=d(C;C`O)=CF=(3√2·6√2)/(3√10)=...=6√10/5 (s-a aplicat teorema a doua a inaltimii in ΔOCC`) iar {F}∈OC` unde CF⊥OC`.
