Question

△ABC中,若asinA+bsinB=csinC+asinB,且c=2bcosA,试判断△ABC的形状

Answer 1

因为：a/sinA=b/sinB=c/sinC

sinC=sin(A+B)=sinAcosB+sinBcosA,

化简c=2bcosA

sinC=2sinBcosA,

sinAcosB+sinBcosA=2sinBcosA

sinAcosB-sinBcosA=0,

sin(A-B)=0,

A=B,a=b,

asinA+bsinB=csinC+asinB

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.