Question

13£®ÒÑÖªÏòÁ¿$\overrightarrow{m}$=£¨2cosx+2$\sqrt{3}$sinx£¬1£©£¬ÏòÁ¿$\overrightarrow{n}$=£¨cosx£¬-y£©£¬ÇÒ$\overrightarrow{m}$¡Í$\overrightarrow{n}$£®
£¨¢ñ£©½«y±íʾΪxµÄº¯Êýf£¨x£©£¬²¢Çóf£¨x£©µÄµ¥µ÷µÝÔöÇø¼ä£»
£¨¢ò£©ÒÑÖªA£¬B£¬C·Ö±ðΪ¡÷ABCµÄÈý¸öÄڽǣ¬Èôf£¨$\frac{A}{2}$£©=3£¬ÇÒsinBsinC=$\frac{3}{4}$£¬ÊÔÅжϡ÷ABCµÄÐÎ×´£®

Answer 1

·ÖÎö £¨¢ñ£©ÓÉƽÃæÏòÁ¿ÊýÁ¿»ýµÄÔËËã¼°Èý½Çº¯ÊýÖеĺãµÈ±ä»»Ó¦Óÿɵú¯Êý½âÎöʽΪ£ºf£¨x£©=2sin£¨2x+$\frac{¦Ð}{6}$£©+1£¬ÓÉ2k¦Ð-$\frac{¦Ð}{2}$¡Ü2x+$\frac{¦Ð}{6}$¡Ü2k¦Ð+$\frac{¦Ð}{2}$£¬k¡ÊZ£¬¿É½âµÃf£¨x£©µÄµ¥µ÷µÝÔöÇø¼ä£®
£¨¢ò£©ÓÉf£¨$\frac{A}{2}$£©=3¿ÉµÃsin£¨A+$\frac{¦Ð}{6}$£©=1£¬½áºÏA·¶Î§¿ÉÇóA£¬C=$\frac{2¦Ð}{3}$-B£¬ÓÉsinBsinC=$\frac{3}{4}$ÀûÓÃÈý½Çº¯ÊýÖеĺãµÈ±ä»»Ó¦ÓÿɵÃsin£¨2B-$\frac{¦Ð}{6}$£©=1£¬ÓÖÓÉ-$\frac{¦Ð}{6}$£¼2B-$\frac{¦Ð}{6}$$£¼\frac{7¦Ð}{6}$£¬¿ÉµÃ2B-$\frac{¦Ð}{6}$=$\frac{¦Ð}{2}$£¬´Ó¶øµÃ½âB£¬CµÄÖµ£¬¿ÉµÃ¡÷ABCΪµÈ±ßÈý½ÇÐΣ®

½â´ð £¨±¾ÌâÂú·ÖΪ12·Ö£©
½â£º£¨¢ñ£©ÓÉ$\overrightarrow{m}$¡Í$\overrightarrow{n}$£¬¿ÉµÃ$\overrightarrow{m}$•$\overrightarrow{n}$=0£¬¿ÉµÃ2cos²x+2$\sqrt{3}$sinxcosx-y=0£¬¼´ÓУºf£¨x£©=2sin£¨2x+$\frac{¦Ð}{6}$£©+1£¬
ÓÉ2k¦Ð-$\frac{¦Ð}{2}$¡Ü2x+$\frac{¦Ð}{6}$¡Ü2k¦Ð+$\frac{¦Ð}{2}$£¬k¡ÊZ£¬¿ÉµÃf£¨x£©µÄµ¥µ÷µÝÔöÇø¼äΪ£º[k$¦Ð-\frac{¦Ð}{3}$£¬k$¦Ð+\frac{¦Ð}{6}$]£¬k¡ÊZ¡­6·Ö
£¨¢ò£©¡ßf£¨$\frac{A}{2}$£©=3£¬
¡à2sin£¨A+$\frac{¦Ð}{6}$£©+1=3£¬¼´sin£¨A+$\frac{¦Ð}{6}$£©=1£¬
ÓÖ¡ß0£¼A£¼¦Ð£¬¡à¿ÉÇóA=$\frac{¦Ð}{3}$£¬
¡ßA+B+C=¦Ð£¬
¡àC=$\frac{2¦Ð}{3}$-B£¬
ÓÉsinBsinC=$\frac{3}{4}$£¬
⇒sinBsin£¨$\frac{2¦Ð}{3}$-B£©=$\frac{3}{4}$£¬
⇒sinB£¨sin$\frac{2¦Ð}{3}$cosB-cos$\frac{2¦Ð}{3}$sinB£©=$\frac{3}{4}$£¬
⇒$\frac{\sqrt{3}}{2}sinBcos\\;B+\frac{1}{2}si{n}^{2}B=\frac{3}{4}$B$+\frac{1}{2}si{n}^{2}B=\frac{3}{4}$£¬
⇒$\frac{\sqrt{3}}{4}$sin2B+$\frac{1}{4}£¨1-cos2B£©=\frac{3}{4}$£¬
⇒$\frac{\sqrt{3}}{2}$sin2B-$\frac{1}{2}$cos2B=1
⇒sin£¨2B-$\frac{¦Ð}{6}$£©=1
ÓÖ¡ß-$\frac{¦Ð}{6}$£¼2B-$\frac{¦Ð}{6}$$£¼\frac{7¦Ð}{6}$£¬
¡à2B-$\frac{¦Ð}{6}$=$\frac{¦Ð}{2}$£¬
¡àB=$\frac{¦Ð}{3}$£¬C=$\frac{¦Ð}{3}$£¬¡÷ABCΪµÈ±ßÈý½ÇÐΡ­12·Ö

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËƽÃæÏòÁ¿ÊýÁ¿»ýµÄÔËË㣬Èý½Çº¯ÊýÖеĺãµÈ±ä»»Ó¦Óã¬ÕýÏÒº¯ÊýµÄͼÏóºÍÐÔÖÊ£¬ÊôÓÚ»ù±¾ÖªÊ¶µÄ¿¼²é£®