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＝acosB1.求角A的值2.若a=√3,则求b+c的取值范围

Question

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.若a=√3,则求b+c的取值范围

Answer 1

1.  (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

2. 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√3

   b+c>a=√3 

   综上所述, b+c∈(√3,2√3 ]