解:(Ⅰ)∵2asinB-
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=0
∴由正弦定理,得:2sinAsinB-
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=0,
∵B是三角形內(nèi)角,可得sinB>0…(3分)
∴等式的兩邊約去sinB,得2sinA-
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=0,即sinA=
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…(5分)
因此,A=
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或A=
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…(7分)
(Ⅱ)∵A為銳角,∴結(jié)合(I)得A=
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結(jié)合三角形內(nèi)角和,得B+C=
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…(9分)
∵y=
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sinB+sin(C-
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)=
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sinB+sin(
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-B)
=
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sinB+cosB=2sin(B+
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) …(12分)
∵B∈(0,
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),得B+
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∈(
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,
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)
∴sin(B+
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)∈
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,可得2sin(B+

)∈(1,2]
因此,函數(shù)y=
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sinB+sin(C-

)的值域域為(1,2]…(14分)
分析:(I)根據(jù)正弦定理,化簡2asinB-
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=0得2sinAsinB-
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=0,結(jié)合sinB>0算出sinA=
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,由A∈(0,π)即可得到A=
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或A=
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;
(II)因為A為銳角,可得A=
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,從而得到B+C=
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,將函數(shù)y=
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sinB+sin(C-
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)化簡為y=
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sinB+sin(
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-B),再由兩角差的正弦公式和輔助角公式化簡整理,得y=2sin(B+

),最后根據(jù)三角函數(shù)的圖象與性質(zhì),結(jié)合角B的取值范圍,即可求出函數(shù)y=
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sinB+sin(C-

)的值域.
點評:本題給出三角形中的邊角關(guān)系,求角A的大小并依此求一個三角函數(shù)式的值域,著重考查了用正余弦定理解三角形、三角函數(shù)的圖象與性質(zhì)和三角恒等變換等知識,屬于中檔題.