在三角形ABC中,角A、B、C所对的边分别为a、b、c.已知A=兀/3,cosB=(根号6)/3,且c^2=a^2+[(

2个回答

  • cosB=根号6/3,则有sinB=根号3/3.

    那么有sinC=sin(180-(A+B))=sin(A+B)=sinAcosB+cosAsinB=根号3/2*根号6/3+1/2*根号3/3=(3根号2+根号3)/6

    a/sinA=c/sinC=b/sinB

    b/a=sinB/sinA=(根号3/3)/(根号3/2)=2/3

    a/c=sinA/sinC=(根号3/2)*6/(3根号2+根号3)=3根号3/(3根号2+根号3)=3/(根号6+1)

    b^2=a^2+c^2-2accosB=a^2+a^2+(根号6-1)b-2a*(根号6+1)/3*a*根号6/3

    b^2=2a^2+(根号6-1)b-a^2*(4/3+2根号6/9)

    4/9a^2=2a^2+(根号6-1)*2/3a-a^2*(4/3+2根号6/9)

    (2根号6/9-2/9)a^2=(2根号6-2)/3*a

    a=3

    故有b=2/3a=2.