题目内容
已知角A、B、C为△ABC的三个内角,其对边分别为a、b、c,若
=(-cos
,sin
),
=(cos
,sin
),a=2
,且
·
=
.
(1)若△ABC的面积S=
,求b+c的值.
(2)求b+c的取值范围.
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(1)若△ABC的面积S=
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(2)求b+c的取值范围.
(1)4;(2)(2
,4]
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试题分析:(1)由
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(2)由(1)可知角A,以及边长
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(1)∵
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∴-cos2
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又A∈(0,π),∴A=
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又由S△ABC=
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由余弦定理得:a2=b2+c2-2bc·cos
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∴16=(b+c)2,故b+c=4.………7分
(2)由正弦定理得:
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∴b+c=4sinB+4sinC=4sinB+4sin(
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∵0<B<
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