解由m⊥n
知mn=0
即√3cos(π/2-A)+[cos(π-A)-1]×1=0
即√3sinA-cosA-1=0
即√3sinA-cosA=1
即2(√3/2sinA-1/2cosA)=1
即2sin（A-π/6)=1
即sin（A-π/6)=1/2
即A-π/6=π/6或A-π/6=5π/6
即A=π/3或A=π(舍去)
故A=π/3