问题描述: 已知a、b、c分别为△ABC的三内角A、B、C的对边,acosc+3asinc-b-c=0 1个回答 分类:数学 2014-12-15 问题解答: 我来补答 已知等式利用正弦定理化简得:sinAcosC+3sinAsinC-sinB-sinC=0,∴sinAcosC+3sinAsinC-sin(A+C)-sinC=0,即sinAcosC+3sinAsinC-sinAcosC-cosAsinC-sinC=0,∴3sinAsinC-cosAsinC-sinC=0,∵sinC≠0,∴3sinA=cosA+1,即sinA1+cosA=33,∴tanA2=sinA1+cosA=33,∴A2=π6,即A=π3.故选:B. 展开全文阅读