∵cosA=
,cosB=
b2+c2?a2
2bc
,
a2+c2?b2
2ac
∴c(acosB-bcosA)=b2,变形得:ac?
-bc?
a2+c2?b2
2ac
=b2,
b2+c2?a2
2bc
整理得:a=
b,
2
∵△ABC的面积为
b2,1 2
∴S△ABC=
absinC=1 2
?1 2
b?b?sinC=
2
b2,1 2
整理得:sinC=
,
2
2
则∠C=45°或135°.
故答案为:45°或135°