Question

在三角形ABC中,角A,B,C所对的边分别是a,b,c,且1+tanA/tanB=2c/b,

若三角形ABC是锐角三角形，求sinB+sinC的取值范围

Answer 1

解：∵1+tanA/tanB=2c/b
∴tanB+tanA=2tanB*c/b,
∵c/b=sinC/sinB
∴tanB+tanA
=2tanB*sinC/sinB
=2sinC/cosB
即tanB+tanA=2sinC/cosB
sinB*cosA+sinA*cosB=2sinC*cosA
sin(A+B)=2sinC*cosA,
∵sinC=sin(A+B), ∴sinC=2sinC*cosA,
∵sinC≠0
∴cosA=1/2＞0(∴A是锐角）
∴A=π/3
sinB+sinC=sinB+sin(2π/3-B)
=(3/2)sinB+(√3/2)cosB