设â AOBä¸åä½åç交ç¹æ¯Bï¼Bç¹å¨xè½´çæå½±ç¹æ¯c
å 为â AOB=5Ï/3=-Ï/3=-60ºï¼OBæ¯åä½å çåå¾OB=1
æ以ä¸è§å½¢BOCåè¾¹çæ¯ä¾å ³ç³»ï¼
OCï¼OB=xï¼1
x=OC/OB
å 为OCãOBæ¯å«60ºç´è§ä¸è§å½¢ç两个边
æ以
x=OC/OB=1ï¼2=1/2
åç
BC=y=-â3/2
æ以ç¹Bçåæ æ¯ï¼1/2ï¼-â3/2ï¼ã
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第1个回答 2019-07-26
记 B 到 Ox 轴的垂线（图中虚线）垂足为 A，
则 ∠AOB = π/3, OB = 1, 则 OA = cos (π/3) = 1/2, AB = -cos (π/3) = - √3/2
则 ∠AOB = π/3, OB = 1, 则 OA = cos (π/3) = 1/2, AB = -cos (π/3) = - √3/2
第2个回答 2019-07-26
如图可知，满意请采纳，谢谢
