In an isosceles triangle, angle “B” = 120 degrees. The radius of the circumscribed circle is 2cm. Find side AB.

О – cent of the circumscribed circle, ОВ = ОА = OC = 2 AB = BC trace triangles AOB and COB are equal Consequently, angles ABO and CBO are equal, Therefore, both are 60 degrees, ОВ = ОА = OC angles BAO and BCO are also 60 each (equal as the angles of isosceles triangles, that is, two equilateral triangles are obtained (all angles are 60 each) and therefore AB = 2