∵2AB*AC=√3|AB|*|AC|
∴AB*AC/(|AB|*|AC|)=√3/2
即 cosA=√3/2
则 角A=π/6
所以 C+B=5π/6
又 √3|AB|*|AC|=3|BC|²
∴ |AB|*|AC|=√3|BC|²
由 正弦定理,有
AB/sinC=AC/sinB=BC/sinA
可得,sinC*sinB=√3sin²A=√3×sin²(π/6)=√3/4
∴ cosC*cosB=cos(C+B)+sinC*sinB
=cos(5π/6)+√3/4
=-√3/4
∴cos(C-B)=cosC*cosB+sinC*sinB
=-√3/4+√3/4=0
∴ C-B=π/2 或C-B=-π/2
由 C+B=5π/6和C-B=π/2 ,得
C=2π/3,B=π/6
同理,由C+B=5π/6和C-B=-π/2,得
C=π/6,B=2π/3
因此,三角形ABC的内角分别为π/6、π/6、2π/3或π/6、2π/3、π/6