问题
选择题
△ABC中,角A、B、C的对边分别为a、b、c且b2+c2-a2+bc=0,则
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答案
∵△ABC中,b2+c2=a2-bc
∴根据余弦定理,得cosA=
=-b2+c2-a2 2bc 1 2
∵A∈(0,π),∴A=2π 3
由正弦定理,得
=a sinA
=b sinB
=2R,c sinC
∴
=asin(30°-C) b-c
=2RsinAsin(30°-C) 2R(sinB-sinC)
(3 2
cosC-1 2
sinC)3 2 sin(
-C)-sinCπ 3
∵sin(
-C)-sinC=π 3
cosC-3 2
sinC-sinC=1 2
(3
cosC-1 2
sinC)3 2
∴原式=
=
(3 2
cosC-1 2
sinC)3 2
(3
cosC-1 2
sinC)3 2 1 2
故选:A