在三角形ABC中，a.b.c分别是三个内角A,B,C的对边。若a=2,C=派/4,cosB/2= 在三角形ABC中a,b,c分别为三个内角A,B,C的对边，a...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d,a,b,c\u5206\u522b\u662f\u4e09\u4e2a\u5185\u89d2,A,B,C\u7684\u5bf9\u8fb9,\u82e5a=2,C=\u03c0/4,cosB/2=2\u6839\u53f75/5.\u6c42\u89d2B\u7684\u4f59\u5f26\u503c.1\u3001cosB=2cos²(B/2)\uff0d1=3/5\uff1b
2\u3001sinB=4/5\uff0c\u6240\u4ee5sinA=sin(C\uff0bB)=sinCcosB\uff0bcosCsinB=7\u221a2/10\uff0c\u5229\u7528\u6b63\u5f26\u5b9a\u7406a/sinA=c/sinC\uff0c\u5f97c=10/7\uff0c\u4ece\u800cS=(1/2)acsinB=8/7\u3002
\u89e3\uff1a\u6839\u636e\u4e09\u89d2\u51fd\u6570\u534a\u89d2\u516c\u5f0f\uff0ccosB/2=\u00b1\u221a(1+cosB)/2\uff0c2cosB/2\uff1d1+cosB\uff0c\u5373\uff1a2(2\u221a5/5)=1+cosB\uff0c cosB=3/5\uff0c\u2235sinB=1-cosB=1-9/25=16/25\uff0c\u2234sinB=\u00b14/5\uff0c\u2235C=\u03c0/4=45\u00b0\uff0c\u2234sinC=cosC=\u221a2/2\uff0c sinA=sin(180\u00b0-(B+C))=sin(B+C)=sinB*cosC+cosB*sinC=3/5*\u221a2/2\u00b14/5*\u221a2/2=7\u221a2/10\uff0c\u6216=-\u221a2/10(\u2235sinC\u3001cosB\u5747>0\uff0c\u820d\u53bb)\uff0c \u7531\u6b63\u5f26\u5b9a\u7406\u5f97\uff1aa/sinA=c/sinC\uff0cc=a*sinC/sinA=2*\u221a2/2\u00f77\u221a10/10=2\u221a5/7
cosB=cosB/2的平方减去1，用反余弦可得角B大小；角B得到后，角C大小已知，因此可得角A。而后用a/sinA=b/sinB(正弦定理),可得b;
a*sinB即可得b边上的高，并且b已知，因此可以的面积
简单分析一下，详情如图所示
请问 什么是COS
S=5
不知
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...
绛旓細鍦ㄤ笁瑙掑舰abc涓,a,b,c鍒嗗埆涓鸿a,b,c鐨勫杈广傚鏋渁,b,c鎴愮瓑宸暟鍒,瑙抌=30搴,涓夎褰bc闈㈢Н涓3/2,姹俠鐨勫約=acsinb/2=3/2,ac=6,ac=2b,a^22acc^2=4b^2,a^2c^2-b^2=3b^2-12.cosb=(a^2c^2-b^2)/(2ac)=鈭3/2,(3b^2-12)/12=鈭3/2,b^2=2鈭34,b=鈭31. 宸茶禐杩 ...
绛旓細浣犵殑棰樼洰娌℃湁闂鍝!渚濋鐢诲嚭涓夎褰BC 瑙:(1)杩炴帴AD,浜C浜嶥,AD鍨傜洿BC銆俢osB=2c鍒涔媋=BD/BC 鍗 2c鍒嗕箣a=BD/c 寰:BD=1/2a 鎵浠ヤ笁瑙掑舰鏄瓑鑵颁笁瑙掑舰 (2)鍥犱负sinB=3鍒嗕箣鏍瑰彿3=AD/AB,涓夎褰BC鏄瓑鑵颁笁瑙掑舰 鎵浠inC=sinB=3鍒嗕箣鏍瑰彿3=AD/AC 鍥犱负b=AC=3 鎵浠D=鏍瑰彿3 鎵浠C=2...
绛旓細1銆乧os(B-C)=1-cosA鈫抍osBcosC+sinBsinC=1+cos(B+C)鈫抍osBcosC+sinBsinC=1+cosBcosC-sinBsinC鈫2sinBsinC=1鈫抯inBsinC=1/2銆2銆乥ac鎴愮瓑姣旀暟鍒楋紝鏍规嵁姝ｅ鸡瀹氱悊鍙緱sinB銆乻inA銆乻inC鎴愮瓑姣旀暟鍒楋紝鏈塻in²A=sinBsinC=1/2锛屾墍浠inA=鈭2/2锛屾晠鈭燗=45掳銆3銆乼anB+tanC=sinB/cosB + ...
绛旓細sinA/a=sinB/b=sinC/c=k 2ka^2=(2b-c)bk+(2c-b)ck a^2=b^2+c^2-bc=b^2+c^2-2bccosA cosA=1/2 A=60
绛旓細1+cosa=(b+c)/c cosa=b/c b=ccosa 鎵浠鍒癮c鐨勫瀭瓒虫槸c 鎵浠モ垹c=90掳 c=5 寰楀埌b=4 鎵浠a=3 鍐呭垏鍦嗗崐寰剅=锛坅+b-c)/2=1 鎵浠ラ潰绉负蟺
绛旓細2cosC=1銆傚緱cosC=1/2寰楄C绛変簬60搴︺傝B灏辩瓑浜75搴︺傜敱椤剁偣B鍚慉C浣滃瀭绾緽D锛屼娇BD鍨傜洿AC锛岀偣D鍦ˋC涓娿傚氨鐭ヨABD绛変簬45搴︼紝瑙扗BC绛変簬30搴︺傚湪鐩磋涓夎褰BDC涓紝瑙扗BC绛変簬30搴︼紝鍒2DC=BC锛孉D=BD銆傞鐩篃璁稿皯浜嗕竴涓潯浠讹紝鏄竴鏉¤竟鐨勯暱搴︽槸宸茬煡鐨勩傚簲璇ユ槸AC鐨勯暱銆傚氨鍐嶅皢DC璁句负X銆傚垯BC=2X=...
绛旓細鍦ㄤ笁瑙掑舰ABC涓紝瑙扖=90搴︼紝瑙扐銆佽B銆佽C鎵瀵圭殑杈瑰垎鍒负a銆b銆c锛岃嫢a:b=2:3,c=2鏍瑰彿13,鍒檃=___b=___锛熺敱浜嶤=90掳 鎵浠锛宎^2+b^2=c^2=52 鍙堢敱浜巃/b=2/3 涓や釜寮忓瓙鍙互寰楀埌a=4 b=6
