在锐角三角形ABC中,角A=2角B,则c/b的取值范围

先确定∠B的范围
∠A=2∠B<π/2, ∴ ∠B<π/4
∠C=π-∠A-∠B=π-3∠B<π/2,∴ ∠B>π/6
即π/6 < ∠B <π/4
利用正弦定理a/sinA=b/sinB=c/sinC
∴ c/b
=sinC/sinB
=sin(π-3B)/sinB
=sin(3B)/sinB
=sin(2B+B)/sinB
=(sin2BcosB+cos2BsinB)/sinB
=2cos²B+cos2B
=2cos²B+2cos²B-1
=4cos²B-1
∵ π/6 < ∠B <π/4
∴cosB∈(√2/2,√3/2)
cos²B∈(1/2,3/4)
∴ 4cos²B-1∈(1,2)