在三角行ABC中，角A.B.C所对的边分别为a.b.c,设S为三角形ABC的面积,满足S等于4分之根号3括号a的平方加b... 在三角形ABC中，角A,B,C所对的边分别为a,b,c.设S...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A\uff0cB\uff0cc\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa\uff0cb\uff0cc\uff0c\u8bbeS\u4e3a\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef\uff0c\u6ee1\u8db3S\u7b49\u4e8e4\u5206\u4e4bS\uff1d1/2absinC
2abcosC\uff1da^2+b^2-c^2
S\uff1d\u221a3/4(a^2+b^2-c^2)\uff1d\u221a3/4\u00d72abcosC\uff1d\u221a3/2abcosC\uff1d1/2absinC
\u6240\u4ee5tanC\uff1d\u221a3
\u6240\u4ee5C\uff1d\u03c0/3
sinA+sinB\uff1d2sin[(A+B)/2]cos[(A-B)/2]\uff1d\u221a3cos[(A-B)/2]
\u6240\u4ee5\u6700\u5927\u503c\u4e3a\u221a3
\u6b64\u65f6A=B=C\uff1d\u03c0/3
a\u7684\u5e73\u65b9+b\u7684\u5e73\u65b9-c\u7684\u5e73\u65b9=2abcosc
\u5219s=\u4e8c\u5206\u4e4b\u6839\u53f7\u4e09abcosc
\u53c8s=(1/2)*absinc
\u5219tanc=\u6839\u53f7\u4e09 c=60\u5ea6
sinA+sinB=sinA+sin(C+A)=sinA+sinCcosA+cosCsinA
=sinA+\u4e8c\u5206\u4e4b\u6839\u53f7\u4e09cosA+\u4e8c\u5206\u4e4b\u4e00sinA
=\u4e8c\u5206\u4e4b\u4e09sinA+\u4e8c\u5206\u4e4b\u6839\u53f7\u4e09cosA
=\u6839\u53f7\u4e09\uff08\u4e8c\u5206\u4e4b\u6839\u53f7\u4e09sinA+\u4e8c\u5206\u4e4b\u4e00cosA\uff09
=\u6839\u53f7\u4e09sin\uff08A+30\u5ea6\uff09
\u53c8A+B+C=180\u5ea6\uff0cC=60\u5ea6
\u52190\uff1cA\uff1c120
\u521930\uff1cA+30\uff1c150
\u52191/2\uff1csinA\u22641
\u5219\u4e8c\u5206\u4e4b\u6839\u53f7\u4e09<sinA+sinB\u2264\u6839\u53f7\u4e09
\u5219sinA+sinB\u7684\u6700\u5927\u503c\u4e3a\u6839\u53f7\u4e09
1/2absinC=根号3/4*2abcosC，tanC=根号3，所以C=60度
sinA+sinB=sinA+sin(120-A)
=sinA+cosAsin120-sinAcos120
=3/2sinA+根号3/2cosA
=根号3sin(A+30)<=根号3
由余弦定理，a的平方加b的平方减c的平方＝2abcosC，又S＝1/2absinC，由此得tanC＝根号3，C等于60°。和差化积：sinA＋sinB＝sinA＋sin(120°-A)＝2sin60°cos(60°-A)＝根号3×cos(60°-A)，最大值是根号3
绛旓細绠鍗曞垎鏋愪竴涓嬶紝璇︽儏濡傚浘鎵绀
绛旓細.瑙ｏ細b*cosA+(a-2c)*cosB=0 鐢辨寮﹀畾鐞嗗緱sinBcosA+sinAcosB-2sinCcosB=0 鐢卞拰瑙掑叕寮忓緱sin(A+B)-2sinCcosB=0 ; sin(180-C)-2sinCcosB=0 ; sinC-2sinCcosB=0 ; cosB=1/2 ;B=60掳 锛2锛塨=2鏍瑰彿3锛宲=a+b+c,鐢辨寮﹀畾鐞哸/sinA=c/sinC=b/sinB=4,寰梐=4si...
绛旓細cosA=(b^2+c^2-a^2)/2bc=2/鈭5 sinA=1/鈭5 sin2A=4/5 cos2A=1-2(sinA)^2=3/5 sin(2A-蟺/3)=(4/5)*1/2-(3/5)鈭3/2=(4-3鈭3)/10
绛旓細鐢变簬 a,b,c鎴愮瓑宸暟鍒楋紝鎵浠2b=a+c 鐢辨寮﹀畾鐞嗭紝鏄撳緱 2sinB=sinA+sinC 鍗 sinA+sinC=鈭7/2 (1)璁綾osA-cosC=x 锛2锛(1)²+(2)²锛屽緱 2+2(sinAsinC-cosAcosC)=7/4 +x²鎵浠 x²=-2cos(A+C) +1/4 鍗 x²=2cosB+1/4 鍥犱负鍏樊涓烘锛屼粠...
绛旓細a^2+b^2-c^2=2abcosC锛屼唬鍏,S=鏍瑰彿3/4*2abcosC 1/2absinC=鏍瑰彿3/4*2abcosC锛宼anC=鏍瑰彿3锛屾墍浠=60搴 sinA+sinB=sinA+sin(120-A)=sinA+cosAsin120-sinAcos120 =3/2sinA+鏍瑰彿3/2cosA =鏍瑰彿3sin(A+30)<=鏍瑰彿3
绛旓細鍥犱负AC=b=2 瑕佷娇涓夎褰鏈変袱瑙,灏辨槸瑕佷娇浠涓哄渾蹇,鍗婂緞涓2鐨勫渾涓嶣A鏈変袱涓氦鐐,褰瑙扐绛変簬90鏃剁浉鍒,褰撹A绛変簬45鏃朵氦浜B鐐,涔熷氨鏄彧鏈変竴瑙.鎵浠ヨA澶т簬45灏忎簬90.鏍瑰彿2/2<sinA<1 鐢辨寮﹀畾鐞嗭細a*sinB=b*sinA.浠ｅ叆寰楀埌锛歛=x=b*sinA/sinB=2鍊嶇殑鏍瑰彿2*sinA 2<a<2鍊嶇殑鏍瑰彿2 ...
绛旓細C=45.鐢盿^2+c^2=b^2+ac寰梐^2+c^2-b^2=ac,鐢变綑寮﹀畾鐞嗗緱cosB=(a^2+c^2-b^2)/(2ac)=ac/(2ac)=1/2 鏁呮湁B=60,A+C=180-B=120.A=120-C.鍐嶇敱姝ｅ鸡瀹氱悊寰梥inA/sinC=a/c=(鈭3+1)/2 2sinA=(鈭3+1)sinC,2sin(120-C)=(鈭3+1)sinC 2sin120cosC-2sinCcos120=(鈭...
绛旓細/2=sinB*sinC,sin(A+B)*sin(A-B)=sinB*sinC,鑰,A+B+C=180,A+B=180-C,sin(A+B)=sinC,鍗虫湁,sin(A-B)=sinB,A-B=B,A=2B,寰楄瘉.2)鈭礱=鈭3b sinA =鈭3sinB =sin2B =2sinBcosB cosB=鈭3/2 B=30掳 A=2B=60掳 C=180掳-30掳-60掳=90掳 涓夎褰BC鏄洿瑙掍笁瑙掑舰銆
绛旓細瑙ｏ細锛1锛夊凡鐭inC=鈭/3锛屽垯C=60掳 鐢遍潰绉叕寮忥細 s=1/2*ab*sinC=1/2*a*5*sin60掳锛屽緱a=8 锛2锛夋牴鎹綑寮﹀畾鐞嗭細c*c=a*a+b*b-2a*b*cosC=64+25-2*8*5*cos60掳=49锛屽緱 c=7 鍒欑敱锛歝osA=(b*b+c*c-a*a)/(2bc)sinA*sinA+cosA*cosA=1 鍙緱锛歝osA=1/7锛宻inA=7鍒嗕箣4鍊...
绛旓細2cosB+1)sinA =0 鍦ㄤ笁瑙掑舰ABC涓,sinA>0 鎵浠ュ彧鏈:cosB=-1/2 閭ｄ箞:B=120 (2).b=鏍瑰彿13,a+c=4 cosB=-1/2=(a^2+c^2-b^2)/2ac=[(a+c)^2-2ac-b^2]/2ac =(16-2ac-13)/2ac =(3-2ac)/2ac 鎵浠:3-2ac=-ac ac=3 鎵浠ョ敱a+c=4,ac=3鍙互瑙ｅ緱 a=3鎴栬卆=1 ...
