在三角形abc中，a2-（b-c）2=bc，求角A 在△ABC中，a，b，c分别是角A，B，C的对边，已知b2=...
\u5728\u4e09\u89d2\u5f62abc\u4e2d a b c\u5206\u522b\u662f\u89d2abc\u7684\u5bf9\u8fb9,a2-(b-c)2=bc,\u6c42\u89d2aa2-(b-c)2=bc
a²-b²-2bc-c²=bc
a²=b²+bc+c²=b²+c²-2bccos120\u00b0
\u6240\u4ee5\uff0c\u2220A=120\u00b0
\uff081\uff09\u5728\u25b3ABC\u4e2d\uff0c\u2235b2=ac\uff0c\u4e14a2-c2=ac-bc\uff0c\u2234b2+c2-a2=bc\uff0c\u2234b2+c2?a2 2bc\uff1d12\uff0c\u2234cosA=12\uff0c\u53c8A\u662f\u4e09\u89d2\u5f62\u7684\u5185\u89d2\uff0c\u6545A=\u03c03\uff0e\uff082\uff09\u56e0\u4e3af(x)\uff1dcos(\u03c9x?A2)+sin(\u03c9x)=cos(\u03c9x?\u03c06)+sin(\u03c9x)=32cos\u03c9x+12sin\u03c9x+sin\u03c9x=32cos\u03c9x+32sin\u03c9x=3sin\uff08\u03c9x+\u03c06\uff09\uff0c\u56e0\u4e3af\uff08x\uff09\u7684\u6700\u5c0f\u6b63\u5468\u671f\u4e3a\u03c0\uff0c\u6240\u4ee5\u03c9=2\uff0c\u51fd\u6570\u89e3\u6790\u5f0f\u4e3a\uff1af\uff08x\uff09=3sin\uff082x+\u03c06\uff09\uff0cx\u2208[0\uff0c\u03c02]\uff0c2x+\u03c06\u2208[\u03c06\uff0c2\u03c03]\uff0c\u5f53x=\u03c06\u65f6\uff0c\u51fd\u6570\u7684\u6700\u5927\u503c\u4e3a3\uff0e
在三角形abc中，a2-（b-c）2=bc，求角A根据余弦定理有：
a^2=b^2+c^2-2bc*cosA
因为a^2-(b-c)^2=bc
即a^2=b^2-bc+c^2
所以2cosA=1
cosA=0.5
A=60度
a²-（b-c）²=bc
a²-（b²-2bc+c²）=bc
a²-b²+2bc-c²=bc
a²=b²+c²-bc
a²=b²+c²-2bc(1/2)
sinA=1/2
A=π/6
绛旓細a2=b2+c2+bc b2+c2-a2=-bc (b2+c2-a2)/bc=-1 cosA=(b2+c2-a2)/2bc =-1/2 A=2蟺/3
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細绠鍗曡绠椾竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細浣欏鸡瀹氱悊锛歛²=b²+c²-2bc脳cosA 宸茬煡锛歛²=b²+bc+c²锛屾墍浠²+c²-2bccosA =b²+bc+c² 瑙ｅ緱锛歝osA=-1/2 A=120
绛旓細sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B))sin^A*2cosAsinB=sin^B*2sinAcosB sin^A*2cosAsinB-sin^B*2sinAcosB=0 sinAsinB(sin2A-sin2B)=0 sin2A=sin2B 2A=2B 鎴2A+2B=180搴 A=B鎴朅+B=90搴 鏁呪柍ABC鏄瓑鑵涓夎褰鎴栫洿瑙掍笁瑙掑舰銆傛湁涓嶆槑鐧界殑鍦版柟鍐嶉棶鍝燂紝...
绛旓細a/sinA=b/sinB=c/sinC sin(A+B)=sinC (sin^2A-sin^2B)/sin^2C=(sinA+sinB)(sinA-sinB)/sin^2C=2sin(A+B)/2*cos(A-B)/2*2cos(A+B)/2sin(A-B)/2 /sin^2C=sin(A+B)sin(A-B)/sin^2C=sin(A-B)/sinC
绛旓細璇佹槑 锛氱敱姝ｅ鸡瀹氱悊鐭ワ紝a/sinA=b/sinB,sinB^2/b^2=sinA^2/a^2 鍦ㄤ笁瑙掑舰abc涓,cos2A/a2-cos2B/b2=1/a2-1/b2 宸﹁竟锛濓紙1锛2sinA^2锛/a^2-锛1锛2sinB^2锛/b^2 =(1/a^2-1/b^2)+2sinB^2/b^2-2sinA^2/a^2 =(1/a^2-1/b^2)+2锛坰inB^2/b^2-sinA^2/a^2锛=1/a...
绛旓細B 鐢辨寮﹀畾鐞嗗彲浠ョ煡閬揳/sinA=b/sinB 鈭碼²/sin²A=b²/sin²B 鈭 sinAcosA=sinBcosB 鈭 2sinAcosA=2sinBcosB 鈭 sin2A=sin2B 鈭 2A=2B 鎴栬 2A=180掳-2B 鈭 A=B鎴栬匒+B=90掳 鈭 鈻ABC鏄瓑鑵涓夎褰鎴栬呯洿瑙掍笁瑙掑舰 ...
绛旓細鍥犱负a锛宐锛c鎴愮瓑姣旀暟鍒 鎵浠2=ac 甯﹀叆鍘熷紡寰a2-c2=b2-bc鍗砤2=b2+c2-bc 鏍规嵁浣欏鸡瀹氱悊a2=b2+c2-2bcCosA 鎵浠2cosA=1 cosA=1/2 鍥犱负a鍦(0,TT)涓 鎵浠=60`鎴朅=120`bsinB/c=b*bsinB/bc=acsinB/bc=asinB/b 鏍规嵁姝ｅ鸡瀹氱悊a/sinA=b/sinB 鎵浠/b=sinA/sinB 鎵浠sinB/b=sinAsinB...
绛旓細鍥炵瓟锛氳В:鐢变綑寮﹀畾鐞嗗緱: cosC=(a^2+b^2-c^2)/2ab 鍥犱负a^2+b^2-c^2=鏍瑰彿2*ab 鎵浠osC=鏍瑰彿2/2 鎵浠ヨC=45搴
