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.
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