如题所述
ç±CosC=(a^2+b^2-c^2)/2ab CosB=(a^2+c^2-b^2)/2ac CosA=(c^2+b^2-a^2)/2bc
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å°a+b/cosA+cosB=c/cosC ä¸çcos项é½ç¨ä½å¼¦å®çä¸a,b,cæ¿æ¢ï¼åç®å¾ c^2=a^2+b^2-ab,åç»åc^2=a^2+b^2-2*a*b*CosC å¯ç¥2cosC=1,å¨éè§ä¸è§å½¢ä¸ï¼Câï¼0ï¼Ïï¼2ï¼ï¼æ以C=Ïï¼3,æ以A+B=Ï-Ï/3=2Ï/3ï¼å³A+B=2C,æ以è§A,B,Cæçå·®æ°å
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第1个回答 2010-11-17
(a+b)/(cosA+cosB)=(a+b)/[(b²+c²-a²)/2bc+(a²+c²-b²)/2ac]=c/[(a²+b²-c²)/2ab]化解得(a+b)(a²-ab+b²-c²)=0,所以（a²-ab+b²-c²)=0，cosC=(a²+b²-c²)/2ab=1/2,角C=60°
