Combining bcosB=acosA, we get sin2B=sin2A.

∵a>b,∴A>B

∵A, B∈(0, π), ∴2B+2A=π, ∴A+B=π2, that is, C=π2.

∴△ABC is a right triangle;

(2) Remember ∠ACM=θ and get ∠BCN=π2 from (1)? θ

∴AC= 1cosθ,BC=3sinθ

∴f(θ)= 1ac+ 1bc=cosθ+sinθ3=23cos(θ-π6)，

When θ = π 6, the maximum value of f(θ) is 233.