1.△ABC外接圆半径为根号三且,满足cosC/cosB=2sinA-sinC/sinB 求 B 和 b2.在△ABC中cosA=1/3,求sin2(B+C/2)+cos2A3.在△ABC,tanA:tanB=a2:b2判断形状4.在△ABC中S=3倍根号三/2且c=根号七,tan(C+45°)=-2-根号三⑴
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![1.△ABC外接圆半径为根号三且,满足cosC/cosB=2sinA-sinC/sinB 求 B 和 b2.在△ABC中cosA=1/3,求sin2(B+C/2)+cos2A3.在△ABC,tanA:tanB=a2:b2判断形状4.在△ABC中S=3倍根号三/2且c=根号七,tan(C+45°)=-2-根号三⑴](/uploads/image/z/1170767-47-7.jpg?t=1.%E2%96%B3ABC%E5%A4%96%E6%8E%A5%E5%9C%86%E5%8D%8A%E5%BE%84%E4%B8%BA%E6%A0%B9%E5%8F%B7%E4%B8%89%E4%B8%94%2C%E6%BB%A1%E8%B6%B3cosC%2FcosB%3D2sinA-sinC%2FsinB+%E6%B1%82+B+%E5%92%8C+b2.%E5%9C%A8%E2%96%B3ABC%E4%B8%ADcosA%3D1%2F3%2C%E6%B1%82sin2%EF%BC%88B%2BC%2F2%EF%BC%89%2Bcos2A3.%E5%9C%A8%E2%96%B3ABC%2CtanA%EF%BC%9AtanB%3Da2%EF%BC%9Ab2%E5%88%A4%E6%96%AD%E5%BD%A2%E7%8A%B64.%E5%9C%A8%E2%96%B3ABC%E4%B8%ADS%3D3%E5%80%8D%E6%A0%B9%E5%8F%B7%E4%B8%89%2F2%E4%B8%94c%3D%E6%A0%B9%E5%8F%B7%E4%B8%83%2Ctan%EF%BC%88C%2B45%C2%B0%EF%BC%89%3D-2-%E6%A0%B9%E5%8F%B7%E4%B8%89%E2%91%B4)
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