Question

10£®ÔÚ¡÷ABCÖУ¬½ÇA£¬B£¬CËù¶ÔµÄ±ß·Ö±ðΪa£¬b£¬c£®ÒÑÖª$\overrightarrow m=£¨{sinC£¬{b^2}-{a^2}-{c^2}}£©£¬\overrightarrow n=£¨{2sinA-sinC£¬{c^2}-{a^2}-{b^2}}£©$ÇÒ$\overrightarrow m¡Î\overrightarrow n$£»
£¨¢ñ£©Çó½ÇBµÄ´óС£»
£¨¢ò£©ÉèT=sin²A+sin²B+sin²C£¬ÇóTµÄÈ¡Öµ·¶Î§£®

Answer 1

·ÖÎö £¨1£©ÓÉÒÑÖªÀûÓÃÏòÁ¿¹²ÏßµÄÐÔÖÊ£¬ÓàÏÒ¶¨Àí£¬ÕýÏÒ¶¨Àí¿ÉµÃ$\frac{sinC}{2sinA-sinC}$=$\frac{sinCcosB}{sinBcosC}$£¬ÀûÓÃsinC¡Ù0£¬sinA¡Ù0£¬½áºÏÈý½Çº¯ÊýºãµÈ±ä»»µÄÓ¦Óã¬Èý½ÇÐÎÄڽǺͶ¨Àí¿ÉÇó$cosB=\frac{1}{2}$£¬½áºÏBµÄ·¶Î§¼´¿ÉµÃ½âBµÄÖµ£®
£¨¢ò£©ÀûÓÃÈý½Çº¯ÊýºãµÈ±ä»»µÄÓ¦Óû¯¼ò¿ÉµÃT=$\frac{7}{4}$-$\frac{1}{2}$cos£¨2A+$\frac{¦Ð}{3}$£©£¬ÓÉ·¶Î§$0£¼A£¼\frac{2¦Ð}{3}$£¬¿ÉÇó$\frac{¦Ð}{3}£¼2A+\frac{¦Ð}{3}£¼\frac{5¦Ð}{3}$£¬ÀûÓÃÓàÏÒº¯ÊýµÄÐÔÖʼ´¿ÉµÃ½âTµÄÈ¡Öµ·¶Î§£®

½â´ð £¨±¾ÌâÂú·ÖΪ12·Ö£©
½â£º£¨1£©ÒòΪ£º$\overrightarrow m=£¨{sinC£¬{b^2}-{a^2}-{c^2}}£©£¬\overrightarrow n=£¨{2sinA-sinC£¬{c^2}-{a^2}-{b^2}}£©$ÇÒ$\overrightarrow m¡Î\overrightarrow n$£»
ËùÒÔ£º$\frac{sinC}{2sinA-sinC}=\frac{{{b^2}-{a^2}-{c^2}}}{{{c^2}-{a^2}-{b^2}}}=\frac{-2accosB}{-2abcosC}=\frac{{c{cosB}}}{bcosC}=\frac{sinCcosB}{sinBcosC}$£¬¡­£¨1·Ö£©
ÒòΪ£ºsinC¡Ù0£¬
ËùÒÔ£ºsinBcosC=2sinAcosB-sinCcosB£¬¡­£¨2·Ö£©
ËùÒÔ£º2sinAcosB=sinBcosC+sinCcosB=sin£¨B+C£©=sinA£¬¡­£¨4·Ö£©
ÒòΪ£ºsinA¡Ù0£¬
ËùÒÔ£º$cosB=\frac{1}{2}$£¬
ÒòΪ£º0£¼B£¼¦Ð£¬
ËùÒÔ£º$B=\frac{¦Ð}{3}$£»¡­£¨6·Ö£©
£¨¢ò£©ÒòΪ£º$T={sin^2}A+{sin^2}B+{sin^2}C=\frac{1}{2}£¨1-cos2A£©+\frac{3}{4}+\frac{1}{2}£¨1-cos2C£©$¡­£¨7·Ö£©
=$\frac{7}{4}-\frac{1}{2}£¨cos2A+cos2C£©=\frac{7}{4}-\frac{1}{2}[{cos2A+cos£¨{\frac{4¦Ð}{3}-2A}£©}]$¡­£¨8·Ö£©
=$\frac{7}{4}-\frac{1}{2}£¨{\frac{1}{2}cos2A-\frac{{\sqrt{3}}}{2}sin2A}£©=\frac{7}{4}-\frac{1}{2}cos£¨{2A+\frac{¦Ð}{3}}£©$¡­£¨9·Ö£©
ÒòΪ£º$0£¼A£¼\frac{2¦Ð}{3}$£¬
ËùÒÔ£º$0£¼2A£¼\frac{4¦Ð}{3}$£¬
¹Ê£º$\frac{¦Ð}{3}£¼2A+\frac{¦Ð}{3}£¼\frac{5¦Ð}{3}$£¬¡­£¨10·Ö£©
Òò´Ë£º-1¡Ücos£¨2A+$\frac{¦Ð}{3}$£©$£¼\frac{1}{2}$£¬
ËùÒÔ£º$\frac{3}{2}$£¼T¡Ü$\frac{9}{4}$£®¡­£¨12·Ö£©

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÏòÁ¿¹²ÏßµÄÐÔÖÊ£¬ÓàÏÒ¶¨Àí£¬ÕýÏÒ¶¨Àí£¬Èý½Çº¯ÊýºãµÈ±ä»»µÄÓ¦Óã¬Èý½ÇÐÎÄڽǺͶ¨Àí£¬ÓàÏÒº¯ÊýµÄÐÔÖÊÔÚ½âÈý½ÇÐÎÖеÄ×ÛºÏÓ¦Ó㬿¼²éÁËת»¯Ë¼Ï룬ÊôÓÚÖеµÌ⣮