在三角形ABC中,内角A,B,C的对边分别是a,b,c

\u5728\u4e09\u89d2\u5f62ABC\u4e2d,\u4e09\u4e2a\u5185\u89d2\u6ee1\u8db3\u2220B-\u2220A=\u2220C-\u2220B,\u5219\u2220B\u7b49\u4e8e

\u2220B-\u2220A=\u2220C-\u2220B \u5f97\u52302B=A+C\uff0c\u53c8\u56e0\u4e3aA+B+C=180\u5ea6 \u6240\u4ee53B=180\uff0cB=60\u5ea6

\u8bbe\u2220B=X\uff0c\u5219\u2220A=x-10,\u2220C=x+10
\u2220A+\u2220B+\u2220C
=x-10+x+x+10
\u89e3\u5f97x=60
\u2220B=60\u00b0
\u2220A=60-10=50\u00b0
\u2220C=60+10=70\u00b0





  • 鍦ㄤ笁瑙掑舰ABC涓,鍐呰A,B,C瀵硅竟鐨勮竟闀垮垎鍒槸a,b,c銆傚凡鐭=2C=蟺/3,
    绛旓細鎵浠 sinA=鈭3/4a锛宻inB=鈭3/4b 鎵浠 b=2a CosC=锛坅^2+b^2-c^2锛/2ab 1/2 =(a^2+4a^2-4)/4a^2 a^2=4/3 涓夎褰鐨勯潰绉疭=1/2absinC=鈭3/ 2a^2 =2鈭3/3
  • 鍦ㄤ笁瑙掑舰ABC涓,鍐呰A,B,C鎵瀵圭殑杈瑰垎鍒负a,b,c,宸茬煡asin2B=鏍瑰彿3b...
    绛旓細asin2B=鈭3bsinA sinA路2sinBcosB=鈭3sinBsinA A銆B鍧囦负涓夎褰㈠唴瑙掞紝sinA>0锛宻inB>0 cosB=鈭3/2 B=蟺/6 (2)sinB=sin(蟺/6)=½sinA=鈭(1-cos²A)=鈭(1-⅓²)=2鈭2/3 sinC=sin(A+B)=sinAcosB+cosAsinB =(2鈭2/3)路(鈭3/2)+⅓路½=(1...
  • 鍦ㄤ笁瑙掑舰ABC涓,鍐呰A,B,C,瀵硅竟鍒嗗埆涓篴,b,c,宸茬煡b/a+c=a+b-c (1)姹...
    绛旓細b(a+b)=(a+c)(a+b-c)ab+b^2=(a+c)(a-c)+(a+c)b ab+b^2=a^2-c^2+ab+bc 鈭碽^2+c^2-a^2=bc 鈶 cosA=(b^2+c^2-a^2)/(2bc)=1/2 鈭碅=60º(2)鍚戦噺AC涓嶤B澶硅涓180º-C 鈭礲=5锛屽悜閲廇C路鍚戦噺CB=5 鈭磡AC||CB|cos(180º-C)=5 鍗-...
  • 鍦ㄤ笁瑙掑舰ABC涓,鍐呰A,B,C鎵瀵圭殑杈瑰垎鍒槸a,b,c,宸茬煡涓夎褰BC鐨勯潰绉疭=...
    绛旓細1.鏍规嵁涓夎褰闈㈢Н鍏紡s=1/2acsinb=1/2bcsina=1/2absinc 宸茬煡s=a²-锛坆-c锛²鍒嗗埆鍙互姹傚嚭sina鍜宑osb鐨勫 2.cosc=4/5 sinc骞虫柟+cosc骞虫柟=1锛 绠楀嚭sinc=3/5 1/2bcsina=1/2absinc c=asinc/sina 1/2absinc=a²-锛坆-c锛²,鍒嗗埆浠e叆灏卞彲浠ユ眰鍑轰簡锛屽氨鏄...
  • 鍦ㄢ柍ABC涓,鍐呰A,B,C鎵瀵圭殑杈瑰垎鍒负a,b,c,.宸茬煡asinA=4bsinB,ac=鏍瑰彿5...
    绛旓細寰梥inB=asinA/4b=5/鈭5.鐢憋紙1锛夌煡锛孉涓洪挐瑙掞紝鍒橞涓洪攼瑙掋傗埓cosB=鈭1-sinB鐨勫钩鏂=2鈭5/5.浜庢槸sin2B=2sinBcosB=4/5 cos2B=1−2sinB鐨勫钩鏂=3/5 鏁卻in锛2B−A锛=sin2BcosA−cos2BsinA=-2鈭5/5.涓夎鍑芥暟鏄暟瀛︿腑灞炰簬鍒濈瓑鍑芥暟涓殑瓒呰秺鍑芥暟鐨勫嚱鏁般傚畠浠殑鏈川鏄换浣曡...
  • 鍦ㄢ柍ABC涓,鍐呰A,B,C鎵瀵圭殑杈瑰垎鍒负a,b,c.宸茬煡a鈮燽,c=鏍瑰彿3,cos²A...
    绛旓細鎮ㄥソ锛鍦ㄤ笁瑙掑舰ABC涓紝鍐呰A,B,C鎵瀵圭殑杈瑰垎鍒负a锛宐锛宑锛屽凡鐭鈮燽锛宑=鏍瑰彿3锛宑osA^2-cosB^2=鏍瑰彿3sinAcosA-鏍瑰彿3sinBcosB.1.姹傝C鐨勫ぇ灏忋俢osA^2-cosB^2=鏍瑰彿3sinAcosA-鏍瑰彿3sinBcosB cosA^2-鏍瑰彿3sinAcosA=cosB^2-鏍瑰彿3sinBcosB cosA(cosAcos蟺/3-sinAsin蟺/3)=cosB(cosBcos蟺/3-...
  • 鍦ㄢ柍ABC涓,鍐呰A,B,C鎵瀵圭殑杈瑰垎鍒负a,b,c,bsinA=-鈭3acos...
    绛旓細瑙o細锛1锛夆埖鍦ㄢ柍ABC涓紝鏍规嵁姝e鸡瀹氱悊寰梐sinA=bsinB锛鈭碽sinA=asinB锛庡張鈭电敱宸茬煡寰梑sinA=-鈭3acosB锛屸埓sinB=-鈭3cosB锛屽彲寰梩anB=-鈭3锛屸埖鍦ㄢ柍ABC涓紝0锛淏锛溝锛屸埓B=2蟺3锛涳紙2锛夛紙鈪帮級鈭礏D涓衡垹ABC鐨勫钩鍒嗙嚎锛屸埓鈭燗BD=鈭燙BD=蟺3锛庘埖S鈻矨BC=S鈻矪CD+S鈻矨BD锛孊D=1銆丅C=x涓擝A=y锛庘埓...
  • 鍦ㄤ笁瑙掑舰ABC涓,鍐呰A,B,C鐨勫杈逛负a,b,c,涓攁sinB=鏍瑰彿3bcosA(1)姹傝A...
    绛旓細绛旓細1锛涓夎褰BC涓锛歛sinB=鈭3bcosA 鏍规嵁姝e鸡瀹氱悊锛歛/sinA=b/sinB=c/sinC=2R 鍒欐湁锛歴inAsinB=鈭3sinBcosA 鍥犱负锛歴inB>0 鎵浠ワ細sinA=鈭3cosA 鎵浠ワ細tanA=鈭3 鎵浠ワ細A=60掳 2锛b=2锛宑=鈭3+1 鏍规嵁浣欏鸡瀹氱悊鏈夛細a^2=b^2+c^2-2bccosA a^2=4+3+2鈭3+1-4(鈭3+1)*cos60掳 a...
  • 鍦ㄢ柍ABC涓,鍐呰A,B,C鐨勫杈瑰垎鍒槸a,b,c,涓攃路sinA+鈭3a路cosC=0,姹傝...
    绛旓細鍙堝洜涓恒CD鏄涓夎褰BC鐨涓嚎锛屾墍浠ャ鏄撶煡锛氫笁瑙掑舰BCD鍏ㄧ瓑浜庝笁瑙掑舰AED,鎵浠ャAE=BC=A=8, 瑙扐ED=瑙払CD,鎵浠ャ瑙扐ED+瑙扐CD=瑙払CD+瑙扐CD =瑙扐CB =120搴︼紝鎵浠ャ瑙扖AE=180搴--锛堣AED+瑙扐CD锛=180搴--120搴 锛60搴︺傛墍浠ャ鍦ㄤ笁瑙掑舰ACE涓紝鐢变綑寮﹀畾鐞嗗彲寰楋細CE^2=AC^2+BC^2--2ACxBCxcos...
  • 鍦ㄢ柍ABC涓,鍐呰A,B,C鎵瀵圭殑杈逛緷娆′负a,b,c,濡傛灉婊¤冻B=30掳,b=4鐨勨柍ABC...
    绛旓細鍦ㄢ柍ABC涓,鍐呰A,B,C鎵瀵圭殑杈逛緷娆′负a,b,c,濡傛灉婊¤冻B=30掳,b=4鐨勨柍ABC鎭版湁涓涓紝鍒檃鐨勫彇鍊艰寖鍥存槸锛氳В锛氬洓涓瓟妗堥兘涓嶅锛佹纭瓟妗堟槸4<a鈮4鈭3.杩欎釜闂鏈濂界敤浣滃浘娉曟眰瑙c傛敞鎰忔潯浠讹細B=30掳锛宐=4鐨勨柍ABC鎭版湁涓涓傜壒鍒敞鎰忊滄伆鏈変竴涓濄備綔鈭燲BY=30掳锛涢《鐐笰鍦ㄥ皠绾緽Y涓婄Щ鍔紝浠 ...
  • 本站交流只代表网友个人观点,与本站立场无关
    欢迎反馈与建议,请联系电邮
    2024© 车视网