已知三角形ABC中的三边分别为a b c,且3a²+3b²-3c²+2ab=0,则sinc=
如上,最好有过程
人气：440 ℃ 时间：2019-12-20 19:27:55
解答
3a²+3b²-3c²+2ab=0
3c²=3a²+3b²+2ab
c²=a²＋b²-2ab×（-1/3）
所以
cosC=-1/3
sinC=√1-cos²C=2√2/3
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