在△ABC中角A,B,C所对的角分别是a,b,c已知角C=60度c=根号六求a+b的取值范围

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  • 由余弦定理得:c^2=a^2+b^2-2abcosC.

    a^2+b^2-2abcos60=(√6)^2.

    a^2+b^2-ab=6.

    a^2+b^2+2ab-3ab=6.

    (a+b)^2=6-3ab (1).

    ∵a/sinA=c/sinC,b/sinB=c/sinC.

    ( a/sinA)*(b/sinB)=(c/sinC)^2

    ab/(sinAsinB=(c/sinC)^2=[√6/(√3/2)]^2=8.

    ab=8sinAsinB=4*(2sinAsinB)=4[c0s(A-B)+cos(A+B)]

    =4[cos(A-B)-cosC].

    =4cos(A-B)-2.

    (a+b)^2=6-3[4cos(A-B)-2].

    =6-12cos(A-B)+6.

    =12-12cos(A-B).

    =12{1-cos(A-B)]

    =12*2sin^2[(A-B)/2].

    =24sin^2[(A-B)/2].

    a+b=2√6sin[(A-B)/2]

    ∵0<sin(A-B)/2≤1.

    0< a+b≤2√6.

    ∵a+b>c=√6

    ∴a+b的取值范围为:√6<a+b≤2√6.