设C = 60,三角形三边分别为a,b,c,a+b+c = 20,
则面积S = 1/2 ab*sinC = 10√3,解得ab = 40,
根据余弦定理,cosC = (a^2 + b^2 - c^2)/2ab = [(a+b)^2 - 2ab -c^2]/2ab = 1/2,将ab = 40代入得(a+b)^2 = c^2 + 120,
因为a+b+c = 20,所以(a+b)^2 = (20-c)^2 = c^2 + 120,接方程得c = 7,
所以a+b = 20 - c = 13,由a+b = 13与ab = 40可求出a = 5,b = 8或a = 8,b = 5.
综上:三角形的各边长分别为5cm,7cm,8cm.
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