题目内容
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(1)求角B的大小;
(2)若
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(1)
. ⑵a+c=
.
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试题分析:(1)又A+B+C=π,即C+B=π-A,
∴sin(C+B)=sin(π-A)=sinA,
将(2a-c)cosB=bcosC,利用正弦定理化简得:(2sinA-sinC)cosB=sinBcosC,
∴2sinAcosB=sinCcosB+sinBcosC=sin(C+B)=sinA,
在△ABC中,0<A<π,sinA>0,
∴cosB=
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(2)∵△ABC的面积为
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∴S=
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∴ac=3,又b=
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∴由余弦定理b2=a2+c2-2accosB得:a2+c2-ac=(a+c)2-3ac=(a+c)2-9=3,
∴(a+c)2=12,
则a+c=
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点评:中档题,本题综合考查了正弦、余弦定理的应用,诱导公式,两角和与差的正弦函数公式,三角形的面积公式,以及特殊角的三角函数值。其中(2)将sinB及已知面积代入求出ac的值,利用余弦定理得到b2=a2+c2-2accosB,再利用完全平方公式整理后,按整体思想求出a+c的值。
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