由韦达定理得
tanα+tanβ=-5/3
tanαtanβ=-2/3
α∈(0°,90°) tanα>0
β∈(90°,180°) tanβ0
( tanα-tanβ)²=(tanα+tanβ)²-4tanαtanβ
=(-5/3)²-4(-2/3)
=49/9
tanα-tanβ=7/3
tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)
=(7/3)/[1+(-2/3)]
=7
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)
=(-5/3)/[1-(-2/3)]
=-1
α∈(0°,90°) β∈(90°,180°)
90°