设三角形三边abc且a=b+1=c+2
a/sinA=c/sinC =＞ ( c+2)/(2sinCcosC)=c/sinC =＞ 2c cosC=c+2 .①
c²=a²+b²-2abcosC =＞ c²=(c+2)²+(c+1)²-2(c+2)(c+1)cosC .②
① ②f方程组可解c值
则三角形的周长为3c+3接下去呢 ？① ②方程组可解c值(将cosC 去掉)则三角形的周长为3c+3 (把c值代入3c+3 )能不能把 2c cosC=c+2c²=(c+2)²+(c+1)²-2(c+2)(c+1)cosC 这两个的具体解题过程帮我算下c²=(c+2)²+(c+1)²-2(c+2)(c+1)cosC=＞2(c+2)(c+1)cosC =c²+6c+5=(c+1)(c+5)=＞ cosC = (c+5)/[2(c+2)] 代入2c cosC=c+2得：2c(c+5)/[2(c+2)]=c+2 =＞ c(c+5)=(c+2)(c+2) =＞ c=4故三角形的周长为3c+3 =15