假设外接圆半径rsinA=a/(2r),sinB=b/(2r),sinC=c/(2r)代入2asinA=(2b+c)sinB+(2c+b)sinC化简转换得:b^2+c^2+bc-a^2=0用余弦定理(b^2+c^2-a^2)/(2bc)=-1/2=cosA得A=120,B+C=60即A=2π/3,则B+C=π/3sinB+sinC=sinB+sin(π/3-B)=sinB+√3/2cosB-1/2sinB=1/2sinB+√3/2cosB=sin(B+π/3)因为0