题目内容
(本小题满分12分)
在△ABC中,角A、B、C的对边分别为
、
、
,且满足
.
(1)求角B的大小;
(2)设
,求
的最小值.
在△ABC中,角A、B、C的对边分别为
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(1)求角B的大小;
|
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(1)
(2) 当
时,
取得最小值0.
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试题分析:解:(1)由正弦定理
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代入(2a-c)cosB=bcosC,得(2sinA-sinC)cosB="sinBcosC."
即2sinAcosB=sinBcosC+sinCcosB=sin(B+C)
∵A+B+C=π,∴2sinAcosB="sinA."
∵0<A<π,∴sinA≠0.
∴cosB=
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∵0<B<π,∴B=
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(2)
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由B=
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所以,当
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点评:解决的关键是根据已知的边角关系化简变形,结合正弦定理和来得到结论,同时结合向量的数量积来求解最值,属于基础题。
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