在三角形ABC中,若sinA=（sinB+sinC）/(cosB+cosC),则△ABC的形状为

sin(B+C)=(sinB+sinC)/(cosB+cosC)
(cosB+cosC)(sinBcosC+cosBsinC)=sinB+sinC
展开得 sinBcosBcosC+sinC(cosB)^2+sinB(cosC)^2+sinCcosCcosB=sinB+sinC
sinBcosBcosC+sinCcosCcosB=sinB[1-(cosC)^2]+sinC[1-(cosB)^2]
cosBcosC(sinB+sinC)=sinB(sinC)^2+sinC(sinB)^2
cosBcosC(sinB+sinC)=sinBsinC(sinB+sinC)
cosBcosC=sinBsinC
cosBcosC-sinBsinC=0
cos(B+C)=0
B+C=90
直角三角形.