sinA= (sinB+sinC)/ (cosB+cosC )
sinA(cosB+cosC)=sinB+sinC
sinA2cos(B+C)/2cos(B-C)/2=2sin(B+C)/2cos(B-C)/2
cos(B-C)/2*[sinAcos(B+C)/2-sin(B+C)/2]=0
cos(B-C)/2[sinAcos(180-A)/2-sin(180-A)/2]=0
cos(B-C)/2[sinAcos(90-A/2)-sin(90-A/2)]=0
cos(B-C)/2(sinAsinA/2-cosA/2)=0
cos(B-C)/2(2sinA/2cosA/2sinA/2-cosA/2)=0
cosA/2cos(B-C)/2(2sin^2A/2-1)=0
cosA/2cos(B-C)/2*(-cosA)=0
cosAcosA/2cos(B-C)/2=0
A=π/2 A/2=π/2 -->A=π(舍去) (B-C)/2=π/2 B-C=π(舍去)
所以是直角三角形