[tex]\it ABCD-romb,\ \ \widehat{BAC}=30^o \Rightarrow \widehat{BAD}=2\cdot39^o=60^o \Rightarrow \Delta ABD-echilateral\\ \\ AO-\hat\imath n\breve al\c{\it t}ime\ pentru\ \Delta ABD \Rightarrow AO=\dfrac{\ell \sqrt3}{2}=\dfrac{8\sqrt3}{2}=4\sqrt3\ cm[/tex]
[tex]\it PA\perp (ABC),\ \ AO\subset (ABC) \Rightarrow PA\perp AO \Rightarrow \Delta PAO-dreptunghic,\ \widehat{A}=90^o\\ \\ d(P,\ O) =PO=\sqrt{PA^2+AO^2}=\sqrt{4^2+4^2\cdot3}=\sqrt{16+16\cdot3}=\sqrt{16(1+3)}=\\ \\ =\sqrt{64}=8cm[/tex]
