a)DO⊥(ABC),BC⊂(ABC),OM⊥BC⇒folosind teo. 3⊥⇒DM⊥BC
m(∡(DMC),(ABC))=m(∡DMO)=60°
OM apotema in ΔABC echilateral⇒OM=l√3/6=6√3*√3/6=3 cm.
Folosind teorema unghiului de 30° in ΔDOM dreptunghic in O⇒DM=2OM=6 cm
Folosind teorema lui Pitagora=>h=DO=3√3 cm.
b)d(O,(DBC))=d(O,DM)=c1*c2/ip=DO*OM/DM=3√3*3/6=3√3/2 cm
c)tg∡DCO=DO/OC=3√3/6=√3/2, unde
l=R√3=>R=l/√3=6√3/√3=6 cm.
