∵m⊥n∴ 2m丶n=2sinB ×2sin^2(π/4+B/2)+(2-cos2B)×(-1)=0即2sinB ×2sin^2(π/4+B/2)+(2-cos2B)×(-1)=2sinB×2×[（1-cos（B+π/2）/2]=2sinB+2sin^2B+1-2sin^2-2=2sinB-1=0∴ sinB=1/2∴B=30°
