Question

在三角形ABC中,角A,B,C所对的边分别是a,b,c,已知向量m=(cosB,-cosA),向量n=（2c+b,a）且

Answer 1

∵向量m∥向量n,∴cosB、cosA均不为0,且-cosB/cosA = (2c+b)/a = (2sinC+sinB)/sinA

∴-sinAcosB = 2sinCcosA + sinBcosA ,∴2sinCcosA + sin(A+B) = 0 = sinC·(1 + 2cosA)

∵C是内角,∴sinC≠0,∴cosA = -1/2,A = 2π/3,B+C=π/3

而sinB+sinC = 2sin[(B+C)/2]·cos[(B-C)/2] = cos[(B-C)/2],∵0