在三角形ABC中,a,b,c分别为A,B,C所对的三边,已知（a+b-c）（a-b+c）=bc,求A 在三角形ABC中，角A,B,C所对的边分别为a,b,c,已知...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c\uff0c\u5df2\u77e5(2a\uff0bb)\u00f7c\uff1dcos(A\uff0bC)\u00f7c\u627e\u5230\u539f\u9898\u4e86\uff0c\u4e0b\u9762\u6765\u8865\u5145\u4e00\u4e0b:
\u539f\u9898\u4e3a\uff1a\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c\uff0c \u5df2\u77e5(2a\uff0bb)\u00f7c\uff1dcos(A\uff0bC)\u00f7cosC \u6c42C\u7684\u5927\u5c0f \u82e5c \uff1d2\uff0c\u6c42\u4e09\u89d2\u5f62ABC\u9762\u79ef\u6700\u5927\u65f6a,b\u7684\u503c.
(1)\u89e3\uff1a
\u56e0\u4e3a A+B+C=\u03c0\uff1b
\u6240\u4ee5 cos(A+C)=-cosB
\u6240\u4ee5 \u53f3\u5f0f=-cosB/cosC (\u6682\u4e0d\u53ef\u5316\u7b80)
\u6240\u4ee5 \u5de6\u5f0f=(2sinA+sinB)/sinC (\u7528\u6b63\u5f26\u5b9a\u7406\u5316\u7b80)
2sinAcosC+sinBcosC=-cosBsinC ( \u4e24\u8fb9\u540c\u4e58cosCsinC \u53bb\u5206\u6bcd)
\u7531\u4f59\u5f26\u5b9a\u7406 \u5f97\uff1a2sinAcosC=-sin(B+C)
2sinAcosC=-sinA
\u56e0\u4e3a sinA\u22600
\u6240\u4ee5 cosC=-1/2
\u53c8 0<C<\u03c0
\u6240\u4ee5 C=2\u03c0/3
(2)\u89e3\uff1a
\u7531\u4f59\u5f26\u5b9a\u7406\u5f97\uff1a cosC=-1/2=\uff08a^2+b^2-c^2\uff09/(2ab)
4-ab=a^2+b^2\u300b2ab \uff08\u57fa\u672c\u4e0d\u7b49\u5f0f\uff09
ab\u300a4/3
\u5f53\u4e14\u4ec5\u5f53 a=b=\u4e09\u5206\u4e4b\u4e8c\u6839\u53f7\u4e09 \u65f6 \u53d6\u7b49
\u53c8 S=1/2 absinC
S\u6700\u5927\u65f6\uff0cab\u53d6\u6700\u5927\u503c4/3
\u6240\u4ee5 S\u6700\u5927\u65f6\uff0ca=b=\u4e09\u5206\u4e4b\u4e8c\u6839\u53f7\u4e09
\u5982\u56fe
[ a＋（b-c）] [ a-（b-c）]＝bc
a²-（b-c）²＝bc
a²-（b²-2bc＋c²）＝bc
a²-b²＋2bc-c²＝bc
a²-b²-c²＝-bc
-a²＋b²＋c²＝bc
cosA＝（b²＋c²-a²）/2bc＝1/2
所以A＝60°
绛旓細鐢变綑寮﹀畾鐞嗗彲寰锛宑osB=锛坅鏂+c鏂-b鏂癸級/ 2ac = -ac/2ac= -1/2 鎵浠ワ紝B=120掳 锛2锛夛紝鍥犱负瑙払鏈澶э紝鏁呮渶澶ц竟闀夸负b=鏍瑰彿7, 鍙坰inC=2sinA,鎵浠澶т簬A锛屽垯鏈灏忚竟涓篴 .鐢辨寮﹀畾鐞嗭紝鏈塩=2a.鍒欐湁涓変釜鏉′欢锛歝=2a, b=鏍瑰彿7锛宎(骞虫柟)+c骞虫柟-b骞虫柟 = -ac 鐢辫繖涓変釜鏉′欢瑙ｆ柟绋...
绛旓細鍦ㄤ笁瑙掑舰abc涓紝a,b,c鍒嗗埆涓鸿a锛宐锛宑鐨勫杈广傚鏋渁,b,c鎴愮瓑宸暟鍒楋紝瑙抌=30搴︼紝涓夎褰bc闈㈢Н涓3/2锛屾眰b鐨勫 s=acsinb/2=3/2,ac=6,a c=2b,a^2 2ac c^2=4b^2,a^2 c^2-b^2=3b^2-12.cosb=(a^2 c^2-b^2)/(2ac)=鈭3/2锛(3b^2-12)/12=鈭3/2锛宐^2=2鈭3...
绛旓細)/2 (鍑犱綍骞冲潎鏁板皬浜庣瓑浜庣畻鏈钩鍧囨暟)鈭4=b²+c²-bc鈮²+c²-(c²+b²)/2 鍗砨²+c²鈮8 bc鈮4 b²+c²+2bc鈮16 (b+c)²鈮4²b+c鈮4 鈭礲+c>a=2 锛涓夎褰浠绘剰涓よ竟涔嬪拰澶т簬绗笁杈癸級鈭2<b+c鈮4 ...
绛旓細m骞宠浜巒 鈭磗inB=2sinA 鈭b=2a=2鈭3 cosC=(a^2+b^2-c^2)/(2ab)鈭碼b=a^2+b^2-c^2 6=3+12-c^2 c=3 (2)cosA=(b^2+c^2-a^2)/(2bc)=鈭3/2 鈭碅=蟺/6 涓夎褰BC鐨勯潰绉=1/2ab*sinC=3鈭3/2 濡傛灉鎮ㄨ鍙垜鐨勫洖绛旓紝璇峰強鏃剁偣鍑诲彸涓嬭鐨勩愭弧鎰忋戞寜閽垨鐐瑰嚮鈥滈噰绾充负...
绛旓細A=B 涓夎褰负绛夎叞涓夎褰=b 鏍规嵁浣欏鸡瀹氱悊锛 cosC=(a2+b2-c2)/(2ab) = (b2+b2-c2)/(2b2) = (2b2-c2)/(2b2) = 1-2(c/2b)2 = 1 - 2脳(1/鈭3)2 = 1/3 sinC=鈭(1-cos2C) = 2鈭2/3 cos2C=2cos2C-1=-7/9 sin2C=2sinCcosC=4鈭2/9 cos(2C+蟺/4) = cos...
绛旓細1.cosB=3/5.sinB=鈭(1-cos^2B)=4/5.a/sinA=b/sinB,sinA=a*sinB/b=2/5.2.S涓夎褰BC鐨勯潰绉=4=1/2*sinB*ac,c=8/sinB*a=5.b^2=a^2+c^2-2cosB*ac=17,b=鈭17.
绛旓細鍦ㄤ笁瑙掑舰abc涓紝a,b,c鍒嗗埆涓鸿a锛宐锛宑鐨勫杈广傚鏋渁,b,c鎴愮瓑宸暟鍒楋紝瑙抌=30搴︼紝涓夎褰bc闈㈢Н涓3/2锛屾眰b鐨勫 s=acsinb/2=3/2,ac=6,a c=2b,a^2 2ac c^2=4b^2,a^2 c^2-b^2=3b^2-12.cosb=(a^2 c^2-b^2)/(2ac)=鈭3/2锛(3b^2-12)/12=鈭3/2锛宐^2=2鈭3...
绛旓細瑙ｏ細鍒╃敤姝ｅ鸡瀹氱悊鍖栫畝宸茬煡绛夊紡寰楋細锛坅+c锛/b=(a−b)/(a−c)锛屽寲绠寰梐^2+b^2-ab=c^2锛屽嵆a^2+b^2-c^2=ab锛鈭碿osC=(a^2+b^2−c^2)/2ab=1/2锛屸埖C涓涓夎褰鐨勫唴瑙掞紝鈭碈=蟺/3 (a+b)/c =(sinA+sinB)/sinC =2/鈭3[sinA+sin锛2蟺/3-A锛塢=2sin锛...
绛旓細鐢辨寮﹀畾鐞嗗緱,(3a-c)/b=(3sinA-sinC)/sinB=cosC/cosB 鎵浠,3sinAcosB-sinCcosB=sinBcosC 3sinAcosB=sin(B+C)=sinA 鎵浠 cosB=1/3 鎵浠 sinB=鏍瑰彿(1-1/9)=(2鏍瑰彿2)/3
绛旓細鐢憋紙1锛夊拰锛2锛夎В寰楋細a=2锛宐=2 2锛塻inC+sin锛圔-A锛夛紳2sin2A sin(蟺/3)+sin(2蟺/3-2A)=2sin2A 鏁寸悊寰楋細sin2A=鈭3(1+cos2A)/3鈥︹︼紙3锛(sin2A)^2+(cos2A)^2=1鈥︹︼紙4锛夎仈绔嬶紙3锛夊拰锛4)瑙ｅ緱锛歝os2A=1/2 2A=蟺/3锛屾墍浠=30掳锛孊=120掳-A=90掳 RT鈻ABC涓AB=c=2...
