在三角形ABC中，已知tanB等于根号三，cosC等于1/3，AC等于三倍的根号六，则三角形ABC的面积。 三角形ABC中,tanB=根号3,cosC=1/3,AC=3...
\u6b63\u5f26\u5b9a\u7406\u7684\u9898\u3002\u4e09\u89d2\u5f62ABC\u4e2dtanB=\u6839\u53f7\u4e09\u3001cosC=1/3,AC=\u4e09\u500d\u6839\u53f7\u516d\u3002\u6c42\u4e09\u89d2\u5f62\u7684\u9762\u79eftanB=\u6839\u53f73\uff0c\u5f97sinB=\u4e8c\u5206\u4e4b\u6839\u53f7\u4e09\uff0ccosC=1/3,\u5f97sinC=\u4e09\u5206\u4e4b\u4e8c\u6839\u53f7\u4e8c\uff0c
\u7531\u6b63\u5f26\u5b9a\u7406\u5f97AB=\u56db\u6839\u53f7\u516d\uff0c
sinA\u7531sin\uff08180-B-C\uff09\u8ba1\u7b97\uff0c\u5f97\u516d\u5206\u4e4b\uff08\u6839\u53f73+2\u6839\u53f72\uff09\uff0c
\u9762\u79ef=\u4e8c\u5206\u4e4bAB*AC*sinA=\u516d\u6839\u53f7\u4e09+\u5341\u4e8c\u6839\u53f7\u4e8c
\u89e3\uff1a\u5df2\u77e5tanB=\u221a3,cosC=1/3 \u5219\u663e\u7136B C\u90fd\u4e3a\u9510\u89d2 sinB=tanB*cosB=tanB*{1/\u221a[1+(tanB)\u5e73\u65b9]}=\u221a3/2 cosB=\u221a[1-(sinB)\u5e73\u65b9]=1/2 sinC=\u221a[1-(cosC)\u5e73\u65b9]=2\u221a2/3\u53c8\u77e5\u9053 AC=3\u221a6\u6839\u636e\u6b63\u7384\u5b9a\u7406 AC/sinB=AB/sinC \u5219AB=8\u6839\u636e\u4e24\u89d2\u548c\u7684\u6b63\u7384\u5c55\u5f00\u5f0fsinA=sin[\u03c0-(B+C)]=sin(B+C)=sinB*cosC+cosB*sinC=(2\u221a2+\u221a3)/6\u6545\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef=(1/2)*AC*AB*sinA=8\u221a3+6\u221a2
tanB=根号3, B=60度 sinB=根号3/2sinC=(2倍根号2)/3 AC=b=3倍根号6
由c/sinC=b/sinB,可知c=8
由cosC=1/3=(a2+b2-c2)/2ab可知a=根号6+4
S=sinB*ac/2=6*根号2+8*根号3
绛旓細2sinAcosB=sinBcosC+cosBcosC 2sinAcosB=sin(B+C)鍙堝洜涓篈+B+C=180搴 鎵浠in(B+C)=sinA,鑰孉鏄涓夎褰鐨 鍐呰 鎵浠inA涓嶇瓑浜0鐨 鎵浠2cosB=1 cosB=1/2 B=60掳
绛旓細a/cosB=b/cosA acosA=bcosB sinA/cosB=sinB/cosA sinAcosA=sinBcosB sin2A=sin2B 绗竴绉嶆儏鍐碉細2A=2B 鍗矨=B 涓虹瓑鑵涓夎褰 绗簩绉嶆儏鍐碉細2A+2B=180鍗矨+B=90 涓虹洿瑙掍笁瑙掑舰
绛旓細tanB=鈭3 鎵浠=60搴 b=3鈭6 cosC=1/3 鎵浠inC=2鈭2/3 b/sinB=c/sinC 鎵浠3鈭6/(鈭3/2)=c/(2鈭2/3)c=8 sinA=sin(180-B-C)=sin(B+C)=sinBcosC+cosBsinC=鈭3/6+鈭2/3=(鈭3+2鈭2)/6 鎵浠=(bcsinA)/2=6鈭2+4鈭3 ...
绛旓細鏍规嵁tanB姹傚唴瑙払鐨勫ぇ灏忥紝鍐嶇畻鍑築鐨剆in鍊硷紝鏍规嵁cosC绠楀嚭sinC,涔嬪悗鐢ㄦ寮﹀畾鐞嗭細b/sinB=c/sinC,姹傚嚭c鍗充负AB鐨勫笺傛牴鎹綑寮﹀畾鐞嗭細cosC = (a^2 + b^2 - c^2) / (2路a路b) ,绠楀嚭a,鍐嶆牴鎹叕寮:S=(absinC)/2绠楀嚭闈㈢Н銆(鍏朵腑鈥/鈥濅负闄わ紝鈥淾2鈥濅负骞虫柟锛夈
绛旓細tanC=1/3锛tanB=1/2锛宼anA=tan(蟺-B-C)=-tan锛圔+C锛夛紱tan锛圔+C锛夊彲浠ラ氳繃tanC=1/3鍜宼anB=1/2姹傚嚭鏉ワ紝閫氳繃a/sinA=c/sinC鍙互姹傚嚭a
绛旓細璁剧孩绾块暱 X锛宼anB=1/2,tanC=1/3,鎵浠=5X锛岃绠楀嚭X=5鐨1/2娆℃柟锛屾墍浠=90搴︼紝鍒檛anA=鏃犵┓澶
绛旓細宸茬煡tanb=-2锛屽垯sin b=锛2*鏍瑰彿5锛/5锛宑os b =-鏍瑰彿5/5锛屽張sin a =鏍瑰彿10/10锛屽垯cosa=3鏍瑰彿10/10锛屽張瑙抌涓洪挐瑙掞紝sinc=sin(a+b锛=sinacosb+sinbcosa=-鏍瑰彿2/10+3鏍瑰彿2/5=鏍瑰彿 2/2锛屾墍浠ヨc=45搴︺傚竻鍝ョ湅鍦ㄦ垜鐢ㄦ墜鏈烘墦瀛楋紝搴婁笂鍙ｇ畻鐨勫垎涓婇噰绾冲惂锛岃阿璋簡銆
绛旓細鍦ㄤ笁瑙掑舰 ABC 涓紝瑙 B 绛変簬 60 搴︼紝涓$b^2=ab$銆傜敱$b^2=ab$锛屽彲寰$b=\frac{a}{b}$锛屽嵆$\tan B=\frac{1}{b}=\frac{1}{a}$銆傚洜涓哄湪涓夎褰腑锛屾鍒囧嚱鏁板湪 0 鍒 90 搴︿箣闂存槸鍗曡皟閫掑鐨勶紝鎵浠$\frac{1}{a}=\tan B=\tan60^{\circ}$锛岃В寰$a=1$銆傚洜涓$b^2=ab=a$...
绛旓細1.cosB=sinA =0.5 a=tanAb 2.b^2=c^2-a^2 绠楀嚭b tanB=b/a 3.a=btanA c鐢ㄥ嬀鑲″畾鐞 sinB=b/c 4.a=3 b=6 a*b=2*9 a^2+b^2=45 瑙ｅ嚭鏉ョ殑 5涓鐭ラ亾 鍏朵粬鐨勮嚜宸辩畻鍑烘潵灏辫浜
绛旓細sin 瑙扖AD=鏍瑰彿13/13 浠鐐逛负璧风偣浣滅嚎娈礏A寤堕暱绾跨殑鍨傜嚎 浜ょ偣涓篍鐐, 宸茬煡 tan B=AD/AB=1/3 鎵浠 AB=3AD AE=AB=3AD CE=2AD 涓夎褰鍏崇郴寰楀嚭AC=鏍瑰彿13*AD 浠ヤ笅涓嶉毦瑙ｅ嚭
