ÌâÄ¿ÄÚÈÝ

20£®ÒÑÖªOÊÇÈñ½ÇÈý½ÇÐÎABCµÄÍâ½ÓÔ²Ô²ÐÄ£¬¡ÏA=60¡ã£¬$\frac{cosB}{sinC}$•$\overrightarrow{AB}$+$\frac{cosC}{sinB}$•$\overrightarrow{AC}$=2m•$\overrightarrow{AO}$£¬ÔòmµÄֵΪ£¨¡¡¡¡£©
A£®$\frac{\sqrt{3}}{2}$B£®$\sqrt{2}$C£®1D£®$\frac{1}{2}$

·ÖÎö ¸ù¾ÝOÊÇ¡÷ABCµÄÍâ½ÓÔ²Ô²Ðıã¿ÉµÃ³ö$\overrightarrow{AB}•\overrightarrow{AO}=\frac{|\overrightarrow{AB}{|}^{2}}{2}£¬\overrightarrow{AC}•\overrightarrow{AO}=\frac{|\overrightarrow{AC}{|}^{2}}{2}$£¬ÕâÑùÔÚ$\frac{cosB}{sinC}•\overrightarrow{AB}+\frac{cosC}{sinB}•\overrightarrow{AC}=2m•\overrightarrow{AO}$Á½±ßͬ³ËÒÔ$\overrightarrow{AO}$£¬±ã¿ÉµÃµ½$\frac{cosB}{sinC}•\frac{|\overrightarrow{AB}|}{2}+\frac{cosC}{sinB}•\frac{|\overrightarrow{AC}{|}^{2}}{2}=2m•|\overrightarrow{AO}{|}^{2}$£¬¶ø¿ÉÉè¡÷ABCµÄÍâ½ÓÔ²°ë¾¶ÎªR£¬´Ó¶ø$|\overrightarrow{AO}|=R$£¬²¢ÇÒÓÉÕýÏÒ¶¨ÀíÓÐ$|\overrightarrow{AB}|=2RsinC£¬|\overrightarrow{AC}|=2RsinB$£¬´øÈëÉÏʽ±ã¿ÉµÃµ½£¬sinCcosB+cosCsinB=m£¬¸ù¾Ý¡ÏA=60¡ã¼°Á½½ÇºÍµÄÕýÏÒ¹«Ê½±ã¿ÉÇó³ömµÄÖµ£®

½â´ð ½â£ºÈçͼ£¬È¡ABµÄÖеãD£¬ACµÄÖеãE£¬Á¬½ÓOD£¬OE£¬Ôò£º
OD¡ÍAB£¬OE¡ÍAC£»
¡à$\overrightarrow{AB}•\overrightarrow{AO}=|\overrightarrow{AB}||\overrightarrow{AO}|cos¡ÏBAO$=$\frac{|\overrightarrow{AB}{|}^{2}}{2}$£¬$\overrightarrow{AC}•\overrightarrow{AO}=\frac{|\overrightarrow{AC}{|}^{2}}{2}$£»
¡àÓÉ$\frac{cosB}{sinC}•\overrightarrow{AB}+\frac{cosC}{sinB}•\overrightarrow{AC}=2m•\overrightarrow{AO}$µÃ£¬$\frac{cosB}{sinC}•\overrightarrow{AB}•\overrightarrow{AO}+\frac{cosC}{sinB}•\overrightarrow{AC}•\overrightarrow{AO}=2m•{\overrightarrow{AO}}^{2}$£»
¡à$\frac{cosB}{sinC}•\frac{|\overrightarrow{AB}{|}^{2}}{2}+\frac{cosC}{sinB}•\frac{|\overrightarrow{AC}{|}^{2}}{2}=2m•|\overrightarrow{AO}{|}^{2}$£¨1£©£»
Éè¡÷ABCµÄÍâ½ÓÔ²°ë¾¶ÎªR£¬Ôò$|\overrightarrow{AO}|=R$£»
ÓÉÕýÏÒ¶¨ÀíµÃ£¬$\frac{|\overrightarrow{AB}|}{sinC}=\frac{|\overrightarrow{AC}|}{sinB}=2R$£»
¡à$|\overrightarrow{AB}|=2RsinC£¬|\overrightarrow{AC}|=2RsinB$£¬ÇÒ$|\overrightarrow{AO}|=R$£¬´úÈ루1£©µÃ£º
2cosBsinC•R2+2cosCsinB•R2=2mR2£»
¡àsinCcosB+cosCsinB=sin£¨B+C£©=sinA=m£»
ÓÖ¡ÏA=60¡ã£»
¡à$m=\frac{\sqrt{3}}{2}$£®
¹ÊÑ¡£ºA£®

µãÆÀ ¿¼²éÈý½ÇÐεÄÍâ½ÓÔ²Ô²ÐĵĸÅÄÏòÁ¿ÊýÁ¿»ýµÄ¼ÆË㹫ʽ£¬Èý½Çº¯ÊýµÄ¶¨Ò壬ÒÔ¼°ÕýÏÒ¶¨Àí£¬Á½½ÇºÍµÄÕýÏÒ¹«Ê½£¬²¢Çå³þÈý½ÇÐεÄÄڽǺͣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø