(1)由已知当a0a=110mm时,可得ac=300+80-110=270(mm),又ab=l=300mm,bc=r=80mm,∴cosθ=ac2+bc2?ab2 2ac?bc =2702+802?3002 2×270×80 =107 432 ;(2)设ac=xmm,若θ=0,则a0a=0mm;若θ=π,则a0a=2rmm,若0<θ<π时,在△abc中,由余弦定理得:ab2=ac2+bc2-2ac?bccosc,即x2-2(rcosθ)x-(l2-r2)=0,解得:x1=rcosθ+ (rcosθ)2+l2?r2 =rcosθ+ l2?r2sin2θ (mm),x2=rcosθ- (rcosθ)2+l2?r2 <0(不合题意,舍去),∴a0a=a0c-ac=l+r-rcosθ- l2?r2sin2θ (mm),∴当θ为任意角时,有a0a=l+r-rcosθ- l2?r2sin2θ (mm).