在△ABC中,(a²+b²-c²)²-(ab)²=0,则C=

问题描述:

在△ABC中,(a²+b²-c²)²-(ab)²=0,则C=

(a²+b²-c²)²-(ab)²=0
[(a²+b²-c²)+(ab)][(a²+b²-c²)-(ab)]=0
(a²+b²-c²)+(ab)=0或(a²+b²-c²)-(ab)=0
cosC=(a²+b²-c²)/2ab=-1/2 或cosC=(a²+b²-c²)/(2ab)=1/2
C=60°或C=120°