∵在三角形ABC中,S=(a+b-c)/4√3 且S=absinC ∴(a+b-c)/4√3 =absinC 则a+b-c=2√3absinC 又∵由余弦定理得cosC=(a+b-c)/2ab ∴cosC=2√3absinC÷2ab=√3sinC 解得C=π/6