(1)
∵a/sinA=b/sinB=c/sinC=2R
∴a=2RsinA b=2RsinB
∵acosB+bcosA=R
∴2RsinAcosB+2RsinBcosA=R
∴sin(A+B)=1/2
∴sinC=1/2
∴C=π/6或C=5π/6
∵是锐角△ABC中
∴C=5π/6舍去,从而C=π/6
(2)
S=½absinC
=½ax1x1/2
=a/4=√3/4
a=√3
根据余弦定理,得
c²=a²+b²-2abcosC
=3+1-2√3cosπ/6
=4-2√3x√3/2
=4-3
=1
所以c=1