已知在锐角三角形ABC中,a,b,c分别为角A,B,C所对的边,且（2c-b）cosA＝acosB 1.求角A的值 2.

(2c-b)cosA = a.cosB => (2sinC-sinB)cosA = sinAcosB .正弦定理=> 2sinCcosA = sinBcosA+sinAcosB => 2sinCcosA = sin(A+B) = sin (π-C) = sinC=> 2cosA=1 => cosA=1/2 => A=π/3由1,得 (2c-b)cosA = acosB => (2c-b)/2=acosB=> (2c-b)/2 * 2c = 2ac * cosB = a²+c²-b² .余弦定理=> 2c²-bc = a²+c²-b² => c²-bc+b²=3 .代入a=> (b+c)²=3+3bc <= 3+3*[(b+c)/2]² . 均值不等式=> (b+c)² <= 12 => b+c <=2√3b+c>a=√3 综上所述, b+c∈(√3,2√3 ]

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