解:(1)∵accosC+ c=b, 由正弦定理得2RsinAcosC+ 2RsinC=2RsinB, 即sinAcosC+ sinC=sinB, 又∵sinB=sin(A+C)=sinAcosC+cosAsinC, ∴ sinC=cosAsinC, ∵sinC≠0, ∴ , 又∵0<A<π, ∴ . (2)由正弦定理得:b= = ,c= , ∴l=a+b+c =1+ (sinB+sinC) =1+ (sinB+sin(A+B)) =1+2( sinB+ cosB) =1+2sin(B+ ), ∵A= ,∴B ,∴B+ ,∴ , 故△ABC的周长l的取值范围为(2,3]. |