题目

在三角形ABC中,已知(a+b)/a= sinB/(sinB -sinA),且cos(A-B)+cosC=1-cos2C
(1)试确定三角形的形状
(2)求(a+ c)/b的取值范围

提问时间：2020-10-31

答案

根据正弦定理,(a+b)/a= sinB/(sinB -sinA) =(sinA+sinB)/sinA
∴sinA·sinB = (sinB+sinA)(sinB-sinA)
= 2sin[(B+A)/2]·cos[(B-A)/2]·2·cos[(B+A)/2]·sin[(B-A)]
=sin(B-A)·sin(B+A)
=sinC·sin(B-A)
cos(A-B)+cosC=1-cos2C 即 2sinA·sinB = 2(sinC)^2,∴(sinC)^2 = sinA·sinB
∴(sinC)^2 = sinC·sin(B-A),∴cosB·sinA = 0,∵sinA≠0,∴cosB = 0,∴B =π/2
∴△ABC是以B为直角的Rt△
又∵sinB/(sinB -sinA) =(sinA+sinB)/sinA,∴1/[1 - sinA] = (1 + sinA)/sinA
∴sinA = 1-(sinA)^2,解得sinA = (√5 - 1)/2 ,∴sinC = cosA = [(2√5 - 2)^(1/2)]/2
∴(a+ c)/b = (sinA + sinC)/sinB = sinA + sinC = (√5 - 1)/2 + {[(2√5 - 2)^(1/2)]/2}