已知三角形的三边为a,b,c,a＋b＝3,ab=2,且C=120°.求：⑴边C.⑵△ABC的面积

(1)用余弦定理：
c^2=a^2+b^2-2abcosC
=(a+b)^2-2ab-2abcosC
=9-4-4cos120°
=5+4*(1/2)
=7
c=√7
(2)S=(1/2)absinC
=(1/2)*2*(√3/2)
=√3/2