a,β∈(0,π/2)
则a+β∈(0,π)
又 cos(a+β)=-√10/10
所以,sin(a+β)=3/10
则tan(a+β)=-3
tanβ=tan[(a+β)-a]
=[tan(a+β)-tana]/[1+tan(a+β)tana]
=(-3-2)/(1-6)
=1
又β∈(0,π/2),
所以,β=π/4