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已知在△ABC中,(1)若sinc+sin(B—A)=sin2A,则三角形的的形状 (2)若sinA=sinB+sinC/cosB+cosC,
已知在△ABC中,(1)若sinc+sin(B—A)=sin2A,则三角形的的形状
(2)若sinA=sinB+sinC/cosB+cosC,则三角形的的形状
人气:342 ℃ 时间:2019-11-02 21:27:48
解答
sinC+sin(B-A)=2sin2Asin[π-(A+B)]+sin(B-A)=2sin2Asin(A+B)+sin(B-A)=2sin2AsinAcosB+cosAsinB+sinBcosA-cosBsinA=2sin2A2sinBcosA=2sinAcosAcosA(sinA-sinB)=0直角或等腰三角形.sinA=(sinB+sinc)/(cosB+cosC)...
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