问题描述: 如果三角形的三条边分别为a=m2-n2,b=2mn,c=m2+n2(m>n大于0),则角C的度数为.选项A.12 B.72 C.25 1个回答 分类:数学 2014-11-18 问题解答: 我来补答 由余弦定理:cosC=(a^2+b^2-c^2)/(2ab)cosC=[(m^2-n^2)^2+(2mn)^2-(m^2+n^2)^2]/[2*2mn*(m^2-n^2)]cosC=(m^4-2m^2n^2+n^4+4m^2n^2-m^4-2m^2n^2-n^4)/[4mn(m^2-n^2)]cosC=0c=90更简单算法:c=m2+n2,c^2=(m^2+n^2)^2=m^4+2m^2*n^2+n^4a=m2-n2,a^2=(m^2-n^2)^2=m^4-2m^2*n^2+n^4b=2mn,b^2=2mn*2mn=4m^2*n^2a^2+b^2=m^4-2m^2*n^2+n^4+4m^2*n^2=m^4+2m^2*n^2+n^4=(m^2+n^2)^2=c^2a^2+b^2=c^2(勾股定理)三角形为直角三角形,角C=90,选项好像没有正确答案 展开全文阅读