解ï¼
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ï¼a+cï¼/b=(a−b)/(a−c)ï¼
åç®å¾a^2+b^2-ab=c^2ï¼
å³a^2+b^2-c^2=abï¼
â´cosC=(a^2+b^2−c^2)/2ab=1/2ï¼
âµC为ä¸è§å½¢çå
è§ï¼
â´C=Ï/3
(a+b)/c
=(sinA+sinB)/sinC
=2/â3[sinA+sinï¼2Ï/3-Aï¼]
=2sinï¼A+Ï/6ï¼ï¼
âµAâï¼0ï¼2Ï/3ï¼ï¼
â´A+Ï/6âï¼Ï/6ï¼5Ï/6ï¼ï¼
â´sinï¼A+Ï/6ï¼âï¼1/2ï¼1]ï¼
å(a+b)/cçåå¼èå´æ¯ï¼1ï¼2]ï¼
追é®2/â3[sinA+sinï¼2Ï/3-Aï¼]
=2sinï¼A+Ï/6ï¼ä¸ç详ç»è¿ç¨
追ç2/â3[sinA+sinï¼2Ï/3-Aï¼]
=2/â3(sinA+sin2Ï/3cosA-cos2Ï/3sinA)
=2/â3(sinA+â3/2cosA+1/2*sinAï¼
=2/â3(3/2sinA+â3/2cosA)
=2/â3*â3(â3/2sinA+1/2cosA)
=2sinï¼A+Ï/6ï¼