题目内容
在△ABC中,角A,B,C所对的边长分别为a,b,c.若C=120°,c=
a,则( )
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A.a>b |
B.a<b |
C.a=b |
D.a与b的大小关系不能确定 |
A
方法一:由余弦定理得2a2=a2+b2-2abcos120°,
∴b2+ab-a2=0,
即(
)2+
-1=0
=
<1,故b<a.
方法二:由余弦定理得2a2=a2+b2-2abcos120°,
∴b2+ab-a2=0,
即b2=a2-ab=a(a-b)>0,∴a>b.
∴b2+ab-a2=0,
即(
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方法二:由余弦定理得2a2=a2+b2-2abcos120°,
∴b2+ab-a2=0,
即b2=a2-ab=a(a-b)>0,∴a>b.

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