zhengkun1981
幼苗
共回答了13个问题采纳率:84.6% 举报
由c=2bcosA可得,c/b=sinC/sinB=2cosA,sinC=sin(A+B)=sinAcosB+sinBcosA,即sinAcosB-sinBcosA=0,sin(A-B)=0,A=B,a=b,则asinA+bsinB=csinC+asinB可化简为:asinA=csinC,a/c=sinA/sinC=sinC/sinA,
(sinA)^2=(sinC)^2,sinA=sinC,A=C,故三角形ABC为等边三角形.
1年前
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