Question

△ABC中,向量m=(b,3a),向量n=(c,6),且m平行n,∠C-∠A=π/2,求角B

Answer 1

由向量m//向量n,知b^2=3ac.又有∠C-∠A=π/2,故sinC=cosA

由正弦定理知,b/a=sinB/sinA, c/b=sinC/sinB,故b/a=3c/b可知sinB^2=3sinAsinC

又有B=π/2-2A,故sinB=cos2A

代入得cos2A^2=3sinAcosA=3/2sin2A

由三角函数性质可化为sin2A^2+3/2sin2A-1=0,解之得sin2A=1/2,

故A=π/12,

B=π/3,