题目内容
△ABC的内角A、B、C的对边分别是a、b、c,若B=2A,a=1,b=
,则c=( )
A.
B.2 C.
D.1
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A.
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B
∵B=2A,a=1,b=
,
∴由正弦定理
=
得:
=
=
=
,
∴cosA=
,
由余弦定理得:a2=b2+c2﹣2bccosA,即1=3+c2﹣3c,
解得:c=2或c=1(经检验不合题意,舍去),
则c=2.
故选B

∴由正弦定理
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∴cosA=
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由余弦定理得:a2=b2+c2﹣2bccosA,即1=3+c2﹣3c,
解得:c=2或c=1(经检验不合题意,舍去),
则c=2.
故选B

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