在三角形ABC中,若a平方＝b(b c),求证A＝2B
因为 a^2=b(b+c),(sinA)^2=(sinB)^2+sinBsinC,(sinA)^2=(sinB)^2+sinBsin(A+B) 所以 (sinA+sinB)(sinA-sinB)=sinBsin(A+B) 所以 4sin[(A+B)/2]*cos[(A-B)/2]*cos[(A+B)/2]*sin[(A-B)/2]=sinBsin(A+B) (此处用到了和...绛旓細鍥犱负 a^2=b(b+c),(sinA)^2=(sinB)^2+sinBsinC,(sinA)^2=(sinB)^2+sinBsin(A+B) 鎵浠 (sinA+sinB)(sinA-sinB)=sinBsin(A+B) 鎵浠 4sin[(A+B)/2]*cos[(A-B)/2]*cos[(A+B)/2]*sin[(A-B)/2]=sinBsin(A+B) (姝ゅ鐢ㄥ埌浜嗗拰...
绛旓細a/(b+c)=b/a 鍗筹細CB鈭禖D=CA鈭禖B. 鍙堚垹ACB=鈭燘CD.鈭粹娍CAB鈭解娍CBD 锛堝搴旇竟鎴愭瘮渚嬶紝澶硅鐩哥瓑鐨勪袱涓涓夎褰鐩镐技锛岋級鈭粹垹CBA=鈭燚=(1/2)鈭燘AC 鍗筹細A=2B
绛旓細a^2=b^2+c^2-2bcCosA 鑱旂珛涓婂紡鍙緱锛歝osA=-1/2 鎵浠A=120搴
绛旓細璇佹槑锛氬洜涓篴^2=b^2+c^2-2bccosA,鍙堢敱棰樻剰鐭锛宎^2=b^2+bc鎵浠^2-2bccosA=bc鍒檆=b(1+2cosA)鎵浠ョ敱姝ｅ鸡瀹氱悊c/sinC=b/sinB寰梥inB+2cosAsinB=sinC=sin(A+B)=sinAcosB+sinBcosA鍒檚inB=sinAcosB-sinBcosA=sin(A-B)鍙A,B,C閮芥槸涓夎褰鐨勫唴瑙掞紝鎵浠B=A-B鍗矨=2B ...
绛旓細=(b^2-bc+1/4*c^2+3/4*c^2)/(2bc)=[(b-c/2)^2+3/4*c^2]/(2bc)鈭(b-c/2)^2+3/4*c^2>0 鈭碿osA>0 鍙 鍦ㄧ涓,鍥涜薄闄愬唴鏃讹紝鎵嶆湁 cosA>0 鈶 A鏄涓夎褰鐨勪竴涓唴瑙掞紝瀹冨睘浜庣涓鎴栫浜岃薄闄愬唴鐨勮 鈶 鐢扁憼鈶″緱 A灞炰簬绗竴璞￠檺鍐呯殑瑙 鈭碅<90搴 鈭磋A涓90...
绛旓細a²=b²+c²+bc 涓ゅ紡姣旇緝鍙煡 -2bcCosA =bc -2CosA=1 CosA=-1/2 A=2蟺/3,5,鏍规嵁浣欏鸡瀹氱悊 cosA=锛坆²+c²-a²锛/锛2bc锛夎宎²=b²+c²+bc锛屼唬鍏ヤ笂寮忥紝鍗冲緱 cosA=锛坆²+c²-锛坆²+c²+bc锛夛級/锛2bc...
绛旓細a²=b²-c²a²+c² =b²涓夎褰BC涓虹洿瑙掍笁瑙掑舰
绛旓細1銆鍦ㄤ笁瑙掑舰ABC涓紝鑻鐨骞虫柟=b鐨勫钩鏂+c鐨勫钩鏂广傞偅涔堝畠鏄洿瑙掍笁瑙掑舰锛屽苟涓攁鏄枩杈广傛墍浠モ垹A=90搴 2銆侊紙1锛塧=5,b=12,c=13 c鏂圭瓑浜巄鏂瑰姞a鏂广傛墍浠ョ洿瑙掍笁瑙掑舰 (2)a=2锛宐=鏍瑰彿2锛宑=鏍瑰彿3+1 c鏂瑰ぇ浜巄鏂瑰姞a鏂 銆傛墍浠ラ挐瑙掍笁瑙掑舰锛堟敞鎰忓洜涓篶姣攁銆乥閮藉ぇ锛屾墍浠ユ墠閫塩鏂规斁涓杈广佸彟...
绛旓細sin²A=(1-cos2A)/2, sin²B=(1-cos2B)/2 鈭磗in²A-sin²B=(1-cos2A)/2-(1-cos2B)/2=(cos2B-cos2A)/2=3/8 鈭碿os2B-cos2A=3/4 cos2B-cos2A=cos[(A+B)-(A-B)]-cos[(A+B)+(A-B)]=cos(A+B)cos(A-B)+sin(A+B)sin(A-B)-[cos(A+B...
绛旓細鈭礱^2=b(b+c)鈭磗in^2A=sin^2B+sinB*sinC 鈭(sinA+sinB)(sinA-sinB)=sinB*sin(A+B)鈭2sin[(A+B)/2]cos[(A-B)/2]*2cos[(A+B)/2]sin[(A-B)/2]=sinB*sin(A+B)鈭磗in)(A+B)*sin(A-B)=sinB*sin(A+B)鈭磗in(A-B)=sinB 鈭碅-B=B 鈭A=2B 濡傛灉鏈鏈変粈涔堜笉鏄庣櫧...
