问题描述： 在△ABC满足,sinA=(sinB+sinC)/(cosB+cosC),此三角形的形状是? 1个回答 分类：数学 2014-11-19 问题解答： 我来补答 sinA=(sinB+sinC)/(cosB+cosC) sin(B+C)=(sinB+sinC)/(cosB+cosC) sinBcosC+cosBsinC=(sinB+sinC)/(cosB+cosC) sinBcosBcosC+sinB(cosC)^2+(cosB)^2sinC+cosBsinCcosC=sinB+sinC sinBcosBcosC+cosBsinCcosC=sinB-sinB(cosC)^2+sinC-(cosB)^2sinC sinBcosBcosC+cosBsinCcosC=sinB(sinC)^2+(sinB)^2sinC cosBcosC(sinB+sinC)=sinBsinC(sinB+sinC) (cosBcosC-sinBsinC)(sinB+sinC)=0 cos(B+C)(sinB+sinC)=0 sinB+sinC≠0 所以cos(B+C)=0 B+C=90度,直角三角形 展开全文阅读
