如图所示,在四边形ADC1C2中,DC1=DC2=290,AD=280,∠C1DC2=32°,∠C1AC2=20°,求AC1和AC2的长度。 如图所示,在四边形ABCD中AB=AD,BC=DC,连接AC...
\u5982\u56fe\u6240\u793a,\u5728\u56db\u8fb9\u5f62ABCD\u4e2d,\u5df2\u77e5\u2220A=\u2220C=90\u00b0\u5e73\u884c
\u2235\u2220A=\u2220C=90º
\u2234\u2220A+\u2220C=180º
\u2234\u2220ABC+\u2220ADC=360º-(\u2220A+\u2220C)
=360º-180º=180º
\u2235BE\u5e73\u5206\u2220ABC
\u2234\u2220ABC=2\u22201
\u540c\u7406\uff1a\u2220ADC=2\u22203
\u2234\u2220ABC+\u2220ADC=2\u22201+2\u22203=180º
\u5219\u22201+\u22203=90º
\u5728Rt\u25b3DAF\u4e2d\uff1a\u22203+\u22205=90º
\u2234\u22201+\u22203=\u22203+\u22205
\u5373\uff1a\u22201=\u22205
\u2234DF\u2225BE
\u89e3\uff1a\u5f53\u2220CAB=\u2220ACD\u65f6\uff0c\u56db\u8fb9\u5f62ABCD\u662f\u83f1\u5f62\uff0c
\u8bc1\u660e\uff1a\u2235AB=AD\uff0cBC=DC\uff0cAC=AC\uff0c
\u2234\u0394ACB\u224c\u0394ACD\uff0c\u2234\u2220CAB=\u2220CAD\uff0c
\u2235\u2220CAB=\u2220ACD\uff0c\uff0c\u2234\u2220CAD=\u2220ACD\uff0c
\u2234AD=CD\uff0c
\u2234AB=AD=BC=CD\uff0c
\u2234\u56db\u8fb9\u5f62ABCD\u662f\u83f1\u5f62\u3002
△DC1C2是等腰△,作其底边C1C2上的高DE,E为C1C2的中点。
C1E=290sin16°=79.93483319,DE=290cos16°=278.7658918,
∠C1AC2=20°,看成点A对C1C2的张角,A点位于以C1C2为弦的一个圆周上,是这个圆对C1C2的圆周角,这个圆对C1C2的圆心角是40°,圆心在C1C2的中垂线DE上,设圆心为O,则∠C1OE=∠C2OE=20°,半径OC1=OC2=OA=C1E/sin20°=290sin16°/sin20°=233.713817
OE=C1E/tan20°=290sin16°/tan20°=219.6191492
OD=DE-OE=290cos16°-290sin16°/tan20°=59.14674261
△OAD三边已知,可以求出∠AOD,余弦定理:
cos∠AOD=(OA²+OD²-AD²)/(2OA.OD)
=(233.713817²+59.14674261²-280²)/(2×233.713817×59.14674261)
=-0.7335206417
∠AOD=137.1823596°
∠AOC1=180°-∠C1OE-∠AOD=180°-20°-137.1823596°=22.81764037°
AC1=2OAsin(∠AOC1/2)=2×233.713817sin(22.81764037°/2)
=92.46103865
∠AOC2=∠C1OC2+∠AOC1=40°+22.81764037°=62.81764037°
AC2=2OAsin(∠AOC2/2)=2×233.713817sin(62.81764037°/2)
=243.5957151
本想用余弦定理计算的,三元方程组也列出了,但用计算器结算的结果不对,没办法只好图解了,如下图
绛旓細鈻矰C1C2鏄瓑鑵扳柍锛屼綔鍏跺簳杈笴1C2涓婄殑楂楧E锛孍涓篊1C2鐨勪腑鐐广侰1E=290sin16掳=79.93483319锛孌E=290cos16掳=278.7658918锛屸垹C1AC2=20掳锛岀湅鎴愮偣A瀵笴1C2鐨勫紶瑙掞紝A鐐逛綅浜庝互C1C2涓哄鸡鐨勪竴涓渾鍛ㄤ笂锛屾槸杩欎釜鍦嗗C1C2鐨勫渾鍛ㄨ锛岃繖涓渾瀵笴1C2鐨勫渾蹇冭鏄40掳锛屽渾蹇冨湪C1C2鐨勪腑鍨傜嚎DE涓婏紝璁惧渾蹇冧负O锛屽垯...
绛旓細1銆佽繛鎺D锛屸埖BD=BD锛孉B=DC锛孉D=BC 鈭粹柍ABD鈮屸柍CDB(SSS)鈭粹垹A=鈭C 2銆佽繛鎺C锛屸埖AC=AC锛孉B=CD锛孉D=BC 鈭粹柍ADC鈮屸柍CBA(SSS)鈭粹垹B=鈭燚
绛旓細鍋囧畾鍥涜竟褰ABCD瑙掕竟鐩稿浣嶇疆濡備笅鍥撅細鏍规嵁浣欏鸡瀹氱悊 a2=b2+c2- 2bc cos鈭燚AB ?BD2 = a2 + d2 - ad cos鈭燚AB 灏卞彲浠ユ眰寰桞D闀垮害 鍐嶆牴鎹寮﹀畾鐞 BD / sin鈭燚AB = a / sin鈭1 = d / sin鈭2 渚垮彲姹傚緱鈭1銆佲垹2 鍚岀悊涔熷彲浠ユ眰鍑 鈭3銆佲垹4 ?鈭燗BC = 鈭2 +鈭4 鈭ADC = 鈭1 +鈭...
绛旓細鍏崇郴锛欰C1脳AC2锛滱B脳AD 鐞嗙敱濡備笅锛氳繛鎺C1 鈭鍥涜竟褰BC1C2鏄姍O鐨勫唴鎺鍥涜竟褰 鈭粹垹B锛嬧垹AC2C1锛180掳 鍙堚垹AC2D锛嬧垹AC2C1锛180掳 鈭粹垹B锛濃垹AC2D 鍚(1)鏈夛細鈭燗DC2锛濃垹AC1B 鈭粹柍ADC2鈭解柍AC1B 鈭碅B/AC2锛滱C1/AD 鍗筹細AC1脳AC2锛滱B脳AD ...
绛旓細1銆佷笉鐩哥瓑锛屽湪BC涓婂彇BE=AB锛岃繛鎺E AB=BE锛孊D鍏辩敤锛孊D骞冲垎鈭燗BC锛屸柍ABD鈮屸柍EBD锛屸垹A=鈭燘ED 鑰屸垹BED=鈭燙ED+鈭燙锛屽洜姝も垹A > 鈭C 2銆佲垹A澶 3銆佲垹A+鈭燙=180搴 鈻矨BD鈮屸柍EBD锛孉D=ED, AD=CD, ED=CD, 鈭燙=鈭燙ED 鑰屸垹A + 鈭燙=鈭燘ED + 鈭燙ED =骞宠=180搴 ...
绛旓細鈭鍥涜竟褰ABCD鏄钩琛屽洓杈瑰舰 鈭碅D鈥朆C,AB鈥朇D(骞宠鍥涜竟褰㈢殑瀵硅鐩哥瓑) 鈭燗=鈭燙,鈭燘=鈭燚(骞宠鍥涜竟褰㈢殑瀵硅竟鐩哥瓑) AO=CO,BO=DO(骞宠鍥涜竟褰㈢殑瀵硅绾夸簰鐩稿钩鍒) 骞宠鍥涜竟褰㈢殑鍒ゅ畾 鍒ゅ畾瀹氱悊1 涓ょ粍瀵硅竟鍒嗗埆骞宠鐨勫洓杈瑰舰鏄钩琛屽洓杈瑰舰 鍑犱綍璇█: 鈭礎D鈥朆C,AB鈥朇D 鈭村洓杈瑰舰ABCD鏄钩琛屽洓杈瑰舰 (涓ょ粍瀵硅竟鍒嗗埆骞宠鐨勫洓杈...
绛旓細鍥犱负AD骞宠鍙堢瓑浜嶣1C1 鎵浠鍥涜竟褰DC1B1涓哄钩琛屽洓杈瑰舰. 鎵浠C1骞宠鍙堢瓑浜嶢B1 鎵浠D1涓嶥C1鎵鎴愮殑瑙掕浆鍖栦负姹傗垹D1AB1 鍥犱负AD1,AB1,B1D1閮芥槸姝f柟褰㈢殑瀵硅绾.涓旀鏂逛綋鐨勫叚涓潰閮芥槸鐩哥瓑鐨. 鎵浠D1=AB1=B1D1 鎵浠ヤ笁瑙掑舰AB1D1鏄涓夎褰 鎵浠モ垹D1AB1=60掳 鍗矨D1涓嶥C1鎵鎴愮殑瑙掓槸60掳 宸茶禐杩 宸茶俯杩...
绛旓細瑙o細杩炴帴AC锛屸埖C1D=DC锛孉D=AD1锛鈭碨鈻矨DC=12S鈻矰1C1D锛孲鈻矨BC=12S鈻矨1B1B锛屽悓鐞嗗緱鍑猴細S鍥涜竟褰DCB=12S鈻矯1B1C+12S鈻矪B1A1锛屸埓姣忕粡杩囦竴娆℃搷浣滐紝閮藉皢闈㈢Н鎵╁ぇ5鍊嶏紝濡傛灉lg5n锛瀕g20092锛宯锛9.45锛岄偅涔堟渶灏戠粡杩10娆℃搷浣滐紝鍥涜竟褰㈢殑闈㈢Н瓒呰繃20092锛
绛旓細鍥犱负AD=DC BD=BD 鈭燗BD=鈭燙BD 鈭粹柍ABD≅鈻矯1BD(SSA)鍥犱负SSA锛屼笖鈭燗BD=鈭燙BD<90掳鏄攼瑙掞紝鈭存湰棰樻湁涓よВ锛(1)S鍥涜竟褰ABCD=2S鈻矨BD=2•3鈭(3)=6鈭(3)(2)鍥犱负鈭燚C1B=鈭燚AB=120掳 鈭粹垹DC1C2=180-120=60掳 鍥犱负DC1=DC2=4 鈭粹垹DC2C1=60掳 鈭粹垹C1DC2=30掳 鈭碨...
绛旓細1,鐢遍鎰,ABCD-A1B1C1D1鏄钩琛屽叚闈綋,涓斿簳闈㈡槸鑿卞舰鎵浠1D1//BD,AD1//BC1 鎵浠ラ潰C1BD//闈B1D1 杩欎釜鏄叕鐞,鎴戜笉鎯冲仛杩囧瑙i噴,鎴戜笅闈㈤兘鐢ㄧ殑鏄悜閲忓拰鏁伴噺绉 2,BD=BC+CD(鍚戦噺),鍋囪妫遍暱涓篴 BD*C1C=(BC+CD)C1C=BC*C1C+CD*C1C =a²cos60`+a²cos120`=0 鎵浠1C鈯...