Question

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.7

Answer 1

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=0

c=90

更简单算法：

c=m2+n2,c^2=(m^2+n^2)^2=m^4+2m^2*n^2+n^4

a=m2-n2,a^2=(m^2-n^2)^2=m^4-2m^2*n^2+n^4

b=2mn,b^2=2mn*2mn=4m^2*n^2

a^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^2

a^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