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17.解:(1)∵△ABC的面积为c(asin bsin B-csin-)=absin∴由正弦定理可得:c(a2 b2-c2)abc,即:2 2abcos=ab,可得cosC=21∵C∈(0,)∴C=3二2)∵C=3D为AB中点,且c=2,CD=√3,π
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