题目

已知△ABC的面积为(a2+b2-c2),则角C的度数为(    )A.135°              B.120°                C.60°                   D.45° 答案：解析:∵S△ABC=absinC,∴absinC=(a2+b2-c2),即c2=a2+b2-2absinC.    又根据余弦定理得c2=a2+b2-2abcosC,∴-2absinC=-2abcosC,即sinC=cosC.∴C=45°.答案:D

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