Question

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+[(根号6)-1]b.〈...

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+[(根号6)-1]b.〈l〉求sinC的值.〈2〉求边b的长.

Answer 1

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.

Answer 2

l〉cosB=(根号6)/3
sinB=(根号3)/3
sinC=sin（B+A)=sinBcosA+cosBsinA=(√3+3√2)/6