sinα=-4√3/7cosβ=cos(α+β-α)=cos(α+β)cosα+sin(α+β)sinα①设π/2<α+β<π则sin(α+β)=5√3/14cosβ=-11/14*1/7-4√3/7*5√3/14=-71/98,β=arc(-71/98)②设π<α+β<3π/2则sin(α+β)=-5√3/14cosβ=-11/14*1/7+4√3/7*5√3/7=49/98=1/2∵β∈(-π,0)∴β=-60°