[tex]ArcAB= 90^{0} ; unghiAOB=90^{0}; AOB.triunghi.dreptunghic.isoscel \\ OA=BO=12 \sqrt{3} \\ AB^{2} = 2OA^{2};AB=OA \sqrt{2}=12 \sqrt{3}* \sqrt{2} ; AB=12 \sqrt{6} \\ ArcAB= 120^{0} ; unghiAOB=120^{0}; OA=BO=12 \sqrt{3}[/tex]
ducem OM perpendicular pe AB si avem AM≡MB ; <AOM≡<BOM=60* ; sin<AOM=AM/AO⇒sin60*=AM/AO⇒√3/2=AM/12√3⇒AM=18⇒AM=2*AM⇒AB=36