在三角形abc中，cosB=1/2，cosC=1/3,AC=3根号6，求三角形ABC的面积。 在三角形ABC中 tanB= 根号3 cos=1/3 A...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0cTANB=\u6839\u53f73\uff0cCOSC=\u4e09\u5206\u4e4b\u4e00\uff0cAC=3\u6839\u53f76\uff0c\u6c42\u4e09\u89d2\u5f62ABC\u7684\u9762\u79efS\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\u5f0f
sinA=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
B=60
sinB=\u221a3/2\uff0ccosB=1/2
b=AC=3\u221a6
\u53c8cosC=1/3 \u5219sinC=2\u221a2/3
\u7531\u6b63\u5f26\u5b9a\u7406\uff0cb/sinB=c/sinC
\u5219c=8
\u5219\u9762\u79ef=1/2bcsinA=1/2bcsin(180-\uff08B+C\uff09\uff09=1/2bcsin\uff08B+C\uff09
=1/2bc(sinBcosC+cosBsinC)=2\u221a6*\uff08\u221a3+2\u221a2\uff09
绛旓細鍦ㄤ笁瑙掑舰ABC涓瑙扐BC鎵瀵圭殑杈瑰垎鍒负abc宸茬煡cosB=-1/2(1)鑻=2b=2鈭3姹備笁瑙掑舰ABC鐨勯潰绉(2)姹俿inA涔榮inC鐨勫彇鍊艰寖鍥... 鍦ㄤ笁瑙掑舰ABC涓ABC鎵瀵圭殑杈瑰垎鍒负abc宸茬煡cosB=-1/2(1)鑻=2 b=2鈭3 姹備笁瑙掑舰ABC鐨勯潰绉(2)姹俿inA涔榮inC鐨勫彇鍊艰寖鍥 灞曞紑 鎴戞潵绛 1...
绛旓細鐢辨寮﹀畾鐞嗘湁b/sinB=a/sinA 鑰宎=3 鍒檅sinA=3sinB锛圛锛夊洜A+B+C=180掳 鍒檚inC=sin(A+B)=sinAcosB+cosAsinB 鑰cosB=1/3 鍗3sinC=sinA+3cosAsinB锛圛I锛夌敱姝ｅ鸡瀹氱悊鏈塩/sinC=a/sinA 鑰宎=3 鍗3sinC=csinA锛圛II锛夌敱锛圛锛夛紙II锛夛紙III锛夊緱 csinA=sinA+bsinAcosA 鍥爏inA鈮0 鍒檆=1+b...
绛旓細鍦≧T螖ABC锛屸垹C=90掳锛屸埓cosB=sinA,鈭(cosA+sinA)^2=1,2sinA*cosA=0,sin2A=0,A=0.鈭村凡鐭ユ潯浠跺瓨鍦ㄩ敊璇備簨瀹炰笂锛氬湪RT螖ABC涓紝sinA+cosA=a/c+b/c=(a+b)/c>1.鈭存湰棰樻棤瑙ｃ
绛旓細锛1锛夊洜涓篵=1,c=鈭3,鈭燙=2/3蟺,鎵浠ョ敱姝ｅ鸡瀹氱悊寰楋細sinB=1/2锛孊=30掳锛屾墍浠cosB=鈭3/2,(2)鍥犱负C=120掳锛孊=30掳锛屾墍浠=30掳锛屾墍浠=b=1
绛旓細鏍规嵁浣欏鸡瀹氱悊 b²= a²+ c² -2acCosB = 2² + 3² - 2脳2脳3脳(1/4)= 13 - 3 = 10 鎵浠 b = 鈭10
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绛旓細杩欎笉灏辨槸姝ｅ鸡瀹氱悊璺熶綑寮﹀畾鐞嗕箞锛熸瘡涓畾鐞嗙敤涓娆★紝閮藉嚭鏉ヤ簡銆 b=a+c-2accosB 寰楀嚭b=10.鎵浠=鏍瑰彿10 cosB=1/4 寰楀嚭 SINB=鏍瑰彿15/4 鐢辨寮﹀畾鐞嗗緱鍑猴細b/SINB=C/SINC 鎵浠inc=锛3鍊嶆牴鍙6锛/8 楹荤儲閲囩撼锛岃阿璋!
绛旓細b鏂=a鏂+c鏂-2acCosB=4+9-12*1/4=10 b=鏍瑰彿10 SinB鏂=1-CosB鏂=1-1/16=15/16 SinB=鏍瑰彿15 / 4 c/sinC = b/sinB sinC=c*sinB/b=3*锛堟牴鍙15 / 4锛/鏍瑰彿10=3*鏍瑰彿6 / 8
绛旓細寰堥珮鍏翠负鎮ㄨВ绛旓紝绁濅綘瀛︿範杩涙锛併愬涔犲疂鍏搞戝洟闃熶负鎮ㄧ瓟棰樸傛湁涓嶆槑鐧界殑鍙互杩介棶锛佸鏋滄偍璁ゅ彲鎴戠殑鍥炵瓟銆傝鐐瑰嚮涓嬮潰鐨勩愰変负婊℃剰鍥炵瓟銆戞寜閽紝绛旈涓嶆槗锛岃阿璋紒濡傛灉鏈夊叾浠栭渶瑕佸府鍔╃殑棰樼洰锛屾偍鍙互姹傚姪鎴戙傝阿璋紒锛
绛旓細蟺-(B+C)]+sinC =sin(B+C)+sinC =sinBcosC+cosBsinC+sinC =鈭3/2cosC+3/2sinC =鈭3sin(C+蟺/6)sinA+sinC鈮も垰3 (3)cosB=(a^2+c^2-b^2)/2ac=(a^2+c^2-ac)/2ac=1/2 a^2+c^2-ac=ac a^2+c^2-2ac=0 (a-c)^2=0 a=c锛孊=蟺/3 鈭粹柍ABC涓虹瓑杈涓夎褰 ...
