在三角形ABC中,角ABC的对边分别为abc,已知acosB=2bcosA 求证:a2-b2/c2=1/3 怎么算
\uff081\uff09\u7528\u5750\u6807\u6cd5\u8bc1\u660e\u4f59\u5f26\u5b9a\u7406\uff1a\u5df2\u77e5\u5728\u25b3ABC\u4e2d\uff0c\u89d2A\u3001B\u3001C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa\u3001b\u3001c\uff0c\u6c42\u8bc1\uff1aa2=b2+c2-2bccosA\uff1b\uff081\uff09\u4ee5A\u4e3a\u5750\u6807\u539f\u70b9\uff0cAB\u6240\u5728\u76f4\u7ebf\u4e3ax\u8f74\uff0cAB\u7684\u5782\u7ebf\u4e3ay\u8f74\uff0c\u5efa\u7acb\u5e73\u9762\u76f4\u89d2\u5750\u6807\u7cfb\uff0c\u5219C\uff08bcosA\uff0cbsinA\uff09\uff0cB\uff08c\uff0c0\uff09\u2234BC\uff1d(c?bcosA\uff0cbsinA)\u2234a2=\uff08c-bcosA\uff092+\uff08bsinA\uff092=b2+c2-2bccosA\uff1b\uff082\uff09\u75312b=a+c\uff0c\u5f97\u5230b=a+c2\uff0c\u5219cosB=a2+c2?b22ac=a2+c2?(a+c4)22ac=3a2+3c2?2ac8ac\u22654ac8ac=12\uff0c\u7531B\u2208\uff080\uff0c180\u00b0\uff09\uff0ccosB\u4e3a\u51cf\u51fd\u6570\uff0c\u6240\u4ee5\u5185\u89d2B\u7684\u6700\u5927\u503c\u4e3a60\u00b0\uff0e\uff083\uff09\u4e0d\u59a8\u5047\u8bbe\u4e0d\u5b58\u5728\u4ee5a\uff0cb\uff0cc\u4e3a\u4e09\u8fb9\u7684\u4e09\u89d2\u5f62\uff0c\u5373 c+b\uff1ca\u2234c2+b2+2cb\uff1cb2+c2-2bccosA \u2234cosA\uff1c-1\u2235A\u2208\uff080\uff0c\u03c0\uff09\uff0c\u2234\u77db\u76fe\u6545\u5047\u8bbe\u4e0d\u6210\u7acb\uff0c\u5373\u5b58\u5728\u4ee5a\uff0cb\uff0cc\u4e3a\u4e09\u8fb9\u7684\u4e09\u89d2\u5f62
\uff1a\uff081\uff09\u56e0\u4e3ab2+c2-a2=bc\uff0c\u6240\u4ee5cosA=b2+c2-a22bc=12\uff0c
\u53c8\u56e0\u4e3aA\u2208(0\uff0c\u03c02)\uff0c\u6240\u4ee5A=\u03c03\uff1b
（a^2－b^2）/c^2
＝［（sinA）^2－（sinB）^2］/（sinC）^2
＝（sinA＋sinB）（sinB－sinB）/（sinC）^2
＝4［sin（A＋B）/2］［cos（A－B）/2］［sin（A－B）/2］［cos（A＋B）/2］/（sinC）^2
＝sin（A＋B）sin（A－B）/（sinC）^2
＝sin（180°－C）sin（A－B）/（sinC）^2
＝sin（A－B）/sinC。
绛旓細a²+b²-c²>0 鎵浠osC=(a²+b²-c²)/2ab>0 鎵浠鏄攼瑙 鎵浠ユ棤娉曞垽鏂繖鏄粈涔涓夎褰 濡傛灉鍔犱竴涓紝C鏄渶澶х殑瑙 鍒欒繖鏄攼瑙掍笁瑙掑舰
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绛旓細鎵惧埌鍘熼浜嗭紝涓嬮潰鏉ヨˉ鍏呬竴涓:鍘熼涓猴細鍦ㄤ笁瑙掑舰ABC涓紝瑙A,B,C鎵瀵圭殑杈瑰垎鍒负a,b,c锛屽凡鐭(2a锛媌)梅c锛漜os(A锛婥)梅cosC 姹侰鐨勫ぇ灏 鑻 锛2锛屾眰涓夎褰BC闈㈢Н鏈澶ф椂a,b鐨勫.(1)瑙ｏ細鍥犱负 A+B+C=蟺锛涙墍浠 cos(A+C)=-cosB 鎵浠 鍙冲紡=-cosB/cosC (鏆備笉鍙寲绠)鎵浠 宸﹀紡=(2...
绛旓細瑙ｇ瓟锛氬埄鐢ㄦ寮﹀畾鐞哸/sinA=b/sinB=c/sinC 鈭礱(cosC+鈭3sinC)=b 鈭磗inA(cosC+鈭3sinC)=sinB=sin(A+C)鈭磗inAcosC+鈭3sinAsinC=sinAcosC+cosAsinC 鈭粹垰3sinAsinC=cosAsinC 鈭 tanA=鈭3/3 锛1锛夆埓 A=30掳 锛2锛塖=(1/2)bcsinA=鈭3/2锛屸埓bc=2鈭3 鈶 鍒╃敤浣欏鸡瀹氱悊 a²=b&...
绛旓細1.s=1/2脳5脳asin60掳=10鏍3锛宎=8 鐢ㄤ綑寮﹀畾鐞哻^2=a^2+b^2-2abcos60掳=49 c=7 2.鐢辨寮﹀畾鐞嗭細a/sinA=c/sinC sinA=4鏍3/7,鐢变綑寮﹀畾鐞嗗彲寰梒osA=1/7 sin(A+60掳锛=5鏍3/14
绛旓細a=ksina b=ksinb c=ksinc 鎵浠ュ彧瑕佹眰鍑簁灏卞彲浠ヤ簡 s=1/2 b*c*sina =1/2*2*c*sin60=2鏍瑰彿3 c=4 a^2=b^2+c^2-2bccosa=4+16-2*2*4*cos60=20-8=12 a=2鏍瑰彿3 鎵浠=a/sina=4 鎵浠ュ師寮=4
绛旓細a(cosC+鏍瑰彿3sinC)=b 鐢辨寮﹀畾鐞嗗緱鍒帮細sinAcosC锛嬧垰3sinAsinC锛峴inB=0 sinAcosC锛嬧垰3sinAsinC锛峴in(A锛婥)=0 sinAcosC锛嬧垰3sinAsinC锛峴inAcosC锛峜osAsinC=0 鈭3sinAsinC锛峜osAsinC=0 鈭3sinA=cosA 鍥爐an(A)=鈭3/3 寰楋細A=30掳 (2)S=1/2bcsinA=1/2bc*1/2=鏍瑰彿3/2,寰楀埌bc=2鏍瑰彿3 ...
绛旓細瑙ｏ細锛1锛夌敱姝ｅ鸡瀹氱悊 b/sinB=c/sinC 鍙緱锛 sinC=csinB/b=1Xsin45搴/鏍瑰彿2 =锛1/2鏍瑰彿2)/(鏍瑰彿2锛=1/2锛屾墍浠 瑙扖=30搴︼紝鐢变綑寮﹀畾鐞 b^2=a^2+c^2--2accosB 鍙緱锛 2=a^2+1--2acos45搴 2=a^2+1--(鏍瑰彿2)a a^2--(鏍瑰彿2)a--1=0 a= (鏍瑰彿6--鏍瑰彿...
绛旓細鍒╃敤姝ｅ鸡瀹氱悊.a/sinA=b/sinB=c/sinC c/b=sinC/sinB=sin2B/sinB=2sinBcosB/sinB=2cosB
绛旓細鍦ㄤ笁瑙掑舰ABC涓,瑙扐BC鎵瀵圭殑杈瑰垎鍒负abc,cosA B 鍦ㄤ笁瑙掑舰ABC涓,瑙扐BC鎵瀵圭殑杈瑰垎鍒负abc,cosA+C/2=鈭3/3[1]姹俢osB鐨勫糩2]鑻=3,b=2鈭2,姹俢鐨勫... 鍦ㄤ笁瑙掑舰ABC涓,瑙扐BC鎵瀵圭殑杈瑰垎鍒负abc,cosA+C/2=鈭3/3[1]姹俢osB鐨勫糩2]鑻=3,b=2鈭2,姹俢鐨勫. 灞曞紑 鎴戞潵绛 1...
