已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值请教步骤
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![已知A,B,C是△ABC的三个内角,向量m=(1,-√3),n=(cosA,sinA),且m·n=-1, (1)求角A(2)若(sinB+cosB)/(sinB-cosB)=3,求tanC的值请教步骤](/uploads/image/z/3684837-21-7.jpg?t=%E5%B7%B2%E7%9F%A5A%2CB%2CC%E6%98%AF%E2%96%B3ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E5%90%91%E9%87%8Fm%3D%EF%BC%881%2C-%E2%88%9A3%EF%BC%89%2Cn%3D%EF%BC%88cosA%2CsinA%EF%BC%89%2C%E4%B8%94m%C2%B7n%3D-1%2C+%281%29%E6%B1%82%E8%A7%92A%EF%BC%882%29%E8%8B%A5%28sinB%2BcosB%29%2F%28sinB-cosB%29%3D3%2C%E6%B1%82tanC%E7%9A%84%E5%80%BC%E8%AF%B7%E6%95%99%E6%AD%A5%E9%AA%A4)
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