在锐角三角形ABC中,已知A=2B,求a/c的取值范围
c^2=a^2+b^2-2ab*cosC 将A=2B即B=A/2带入,得 C^2=（5/4-cosC）a^2(a/c)^2=1/(5/4-cosC) 锐角三角形中,0
绛旓細A=2B 閿愯涓夎褰 浠绘剰涓や釜瑙掍箣鍜屽簲璇ュぇ浜90掳锛屽皬浜180掳 90掳< A+B = 3B < 180掳 A = 2B 锛 90掳 30掳锛 B < 45掳 鈭2/2 < cosB < 鈭3/2 鈭2 < 2cosB < 鈭3 鈭2 < a/b< 鈭3
绛旓細A=2C,鎵浠+3C=180搴︼紝B=180-3A,鍥犱负浠讳竴瑙掑皬浜庡叾浣欎袱瑙掑拰锛屽ぇ浜庝袱瑙掑樊锛屾墍浠<B<3C,瑙ｅ緱30搴<C<45搴︺俛/c=sinA/sinC=sin2C/sinC=2sinCcosC/sinA=2cosC,鎵浠ヨ寖鍥翠负 鏍瑰彿2<a/c<鏍瑰彿3
绛旓細鍦ㄩ攼瑙 ABC 涓紝宸茬煡 A = 蟺/3銆傛垜浠笇鏈涙眰瑙 (a-c)/b 鐨勫彇鍊艰寖鍥淬傛牴鎹笁瑙掑嚱鏁扮殑瀹氫箟锛屾垜浠煡閬鍦ㄩ攼瑙掍笁瑙掑舰涓紝瀵逛簬瑙 ABC锛歴in(A) = a/b cos(A) = c/b 鎴戜滑鍙互浠ｅ叆宸茬煡鏉′欢 A = 蟺/3锛屽緱鍒帮細sin(蟺/3) = a/b cos(蟺/3) = c/b sin(蟺/3) = 鈭3/2 cos(蟺/3) =...
绛旓細A=2B 0锛淎+B锛180掳 0锛2B+B锛180掳 0锛3B锛180掳 0锛淏锛60掳 C=180掳-A-B=180掳-3B 鏍规嵁姝ｅ鸡瀹氱悊锛歝/sinC=b/sinB c/b=sinC/sinB=sin(180-3B)/sinB=sin3B/sinB=sin(2B+B)/sinB =(sinBcos2B+cosBsin2B)/sinB =(sinBcos2B+2sinBcosBcosB)/sinB =cos2B+2cos^2B =2cos^...
绛旓細鍦ㄩ攼瑙掍笁瑙掑舰ABC涓,瑙扐,B,C鎵瀵圭殑杈瑰垎鍒槸a,b,c,宸茬煡a=鏍瑰彿7,b=3,鏍瑰彿7鍊峴inB+sinA=2鍊嶆牴鍙蜂笁姹傝A鐨勫ぇ灏忔眰涓夎褰㈤潰绉ユユ!!... 鍦ㄩ攼瑙掍笁瑙掑舰ABC涓,瑙扐,B,C鎵瀵圭殑杈瑰垎鍒槸a,b,c,宸茬煡a=鏍瑰彿7,b=3,鏍瑰彿7鍊峴inB+sinA=2鍊嶆牴鍙蜂笁姹傝A鐨勫ぇ灏忔眰涓夎褰㈤潰绉ユユ!! 灞曞紑 鎴戞潵绛 1...
绛旓細/2bc=1/2 b²+c²-1=bc 锛坆+c锛²-1=3bc锛屸埖bc鈮1/4锛坆+c锛²鈭达紙b+c锛²-1鈮3/4锛坆+c锛²锛屸埓锛坆+c锛²鈮4 鈭碽+c鈮2锛屸埓a+b+c鈮3锛屸埖b+c锛瀉锛涓夎褰涓よ竟涔嬪拰澶т簬绗笁杈癸級锛屸埓a+b+c锛2锛屸埓鈻ABC鐨鍛ㄩ暱鍙栧艰寖鍥达紙2锛3]...
绛旓細鍥犱负涓夎褰BC鐨勯潰绉疭=鈭3/4锛坆^2+c^2-a^2锛夊張鐢辨寮﹀畾鐞嗭紝寰 S=1/2bcsinA 鎵浠 b^2+c^2-a^2=2/鈭3 bcsinA a^2=b^2+c^2-2 1/鈭3 bcsinA 鎵浠 cosA=1/鈭3 sinA 鍗 tanA=鈭3 鎵浠 A=60掳銆
绛旓細鐢辨寮﹀畾鐞嗙煡a/sinA=b/sinB,鍒2bsinA/sinA=b/sinB,sinB=1/2 鐢涓夎褰闈㈢Н鍏紡鐭ワ紝S=(1/2)ac*sinB=(1/4)ac=(1/4)a(6-a)<=(1/4)[(a+6-a)/2]^2=9/4,褰撲笖浠呭綋a=6-a ,a=3鏃剁瓑鍙锋垚绔嬨傛墍浠ラ潰绉渶澶у间负9/4
绛旓細涓夎褰负閿愯涓夎褰紝鎵浠ヨB灏忎簬90搴︼紝鍙堣C涔熷繀椤诲皬浜90搴︼紝鍗宠A+瑙払 > 90搴︼紝寰楀埌瑙払 > 60搴 鍗 60搴 < 瑙払 < 90搴 鐢辨寮︽鐞嗙煡锛歛 / sinA = b / sinB锛屽張宸茬煡 a = 2 瑙扐 = 30搴︼紝鎵浠 b = 4 x sinB 鍙堝凡鐭ヤ簡B鐨勫彇鍊艰寖鍥达紝鐩稿簲鍙互姹傚緱b鐨勫彇鍊艰寖鍥 鐢辩鎰忔暀...
绛旓細瑙ｏ細涓夎褰闈㈢НS=1/2AB脳AC脳sinA 鈭礎B=8锛孉C=10锛孲=25 鈭磗inA=5/8 鍙堚埖A涓閿愯 鈭碿osA=鈭氾紙1-sin²A锛=鈭39/8 tanA=sinA/cosA=5鈭39/39 cotA=1/tanA=鈭39/5
