cosA=1/2,sinA=√3/2,cosB=√6/3,sinB=√3/3,
由正弦定理知:a/sinA=b/sinB,a=b*sinA/sinB=3/2*b
cosC=cos(兀-A-B)=-cos(A+B)=sinAsinB-cosAcosB=1/2-√6/6
由余弦定理知:c^2=a^2+b^2-2ab*cosC,
a^2+(√6-1)b=a^2+b^2-2ab(1/2-√6/6),
(√6-1)b=b^2-2ab(1/2-√6/6)
将a=3/2*b代入,
(√6-1)b=b^2-2*3/2*b*b(1/2-√6/6)=b^2*(1-3/2+√6/2),
b^2*(1-3/2+√6/2)-(√6-1)b=0,
b(1-3/2+√6/2)-(√6-1)=0
解得:b=(√6-1)/(1-3/2+√6/2)=(√6-1)/(-1/2+√6/2)=2(√6-1)/(-1+√6)=2
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