在三角形abc中,角a,b,c分别为三角形的三个内角,且sin b/ sin a=2√3sin c
sin3C=sin(2C+C)=sin2CcosC+cos2CsinC∴2sin2CcosC-sin3C=sin2CcosC-cos2CsinC=sin(2C-C)=sinC
即sinC=√3(1-cosC)
∴1/2sinC+√3/2cosC=√3/2
即sin(C+60°)=√3/2
C=60°
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osB=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3sinAcosA=sinCcosB+sinBcosC=sin(B+C)=sinA 鍗冲緱 3cosA=1 鎵浠 cosA=1/3 锛2锛夆埖cosA= 1/3 鈭磗inA= 2鈭2/3 cosB=-cos锛圓+C锛=-cosAcosC+sinAsinC=- 1/3cosC+ 2鈭2/3sinC 鍙 cosB+cosC=2鈭3/3 鈭2鈭3/3-cosC=- 1/3cosC+ 2鈭2/3sinC 鍖栫畝锛屽緱 鈭3-cosC=鈭...
绛旓細瑙ｏ細鍒╃敤姝ｅ鸡瀹氱悊鍖栫畝宸茬煡绛夊紡寰楋細锛坅+c锛/b=(a−b)/(a−c)锛屽寲绠寰梐^2+b^2-ab=c^2锛屽嵆a^2+b^2-c^2=ab锛鈭碿osC=(a^2+b^2−c^2)/2ab=1/2锛屸埖C涓涓夎褰鐨勫唴瑙掞紝鈭碈=蟺/3 (a+b)/c =(sinA+sinB)/sinC =2/鈭3[sinA+sin锛2蟺/3-A锛塢=2sin锛...
绛旓細璇侊細bccosA=1;accosB=1 鍥犳涓ゅ紡鐩搁櫎寰楋細acosB/bcosA=1 涓夎褰腑鐢辨寮﹀畾鐞嗗緱锛歴inAcosB/sinBcosA=1 绉婚」锛歴inAcosB-sinBcosA=0 鍥犳锛歴in(A-B)=0,A-B=0 A=B 瑙:bccosA=1 鍥犱负璇ヤ笁瑙掑舰涓虹瓑鑵锛宐cosA=AD(D涓篶杈逛腑鐐)=c/2;鎵浠ワ細(c骞虫柟)/2=1 c=鏍瑰彿2 ...
绛旓細asinA = b锛坅²+b²-c²锛/ 2ab + c 锛坅²+c²-b²锛/ 2ac = 锛坅²+b²-c²锛/ 2a + 锛坅²+c²-b²锛/ 2a = 2a²/(2a)=a 鈭 sinA =1 鈭 A =90掳 鏁涓夎褰鏄洿瑙掍笁瑙掑舰銆傦紙2锛bcosB/a+c...
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b�0�5=a�0�5+c �0�5-2accos60�0�2=a�0�5+c �0�5-ac 鈶,灏嗏憼浠ｅ叆鈶,骞跺寲绠寰:(a-c)�0�5=0,鈭碼=c ,鏍规嵁鈶犲彲寰:a=b=c,鎵浠涓夎褰BC鏄...
绛旓細锛1锛夊垽鏂涓夎褰BC鐨勫舰鐘 鐢ㄤ綑寮﹀畾鐞嗙殑鍏紡锛屾妸瑙掑叧绯昏浆鍖栦负杈圭殑鍏崇郴锛屽彲浠ヨВ鍐虫闂 鍥犱负c*(cosA+cosB)=c[(b²+c²-a²)/2bc+(a²+c²-b²)/2ac]=(b²+c²-a²)/b+(a²+c²-b²)/a=a+b 鎵浠 a(b²...
绛旓細1锛岃В锛氬洜涓A锛孊锛孋鎴愮瓑宸暟鍒楋紝鎵浠ヨB=60搴 鎵浠osB=1/2 2锛岃В锛氱敱姝ｅ鸡瀹氱悊锛宻inA:sinB:sinC=a:b:c sinA=asinB/b sinC=csinB/b 鎵浠inAsinC=acsinBsinB/(b^2)宸茬煡sinB=浜屽垎涔嬫牴鍙蜂笅涓锛宎c=b^2 鎵浠ワ紝sinAsinC=3/4
绛旓細锛1锛鍦ㄤ笁瑙掑舰ABC涓紝瑙扐,B,C鎵瀵圭殑杈瑰垎鍒负a,b,c,宸茬煡鏍瑰彿3sinAsinB+cosAsinB-sinC=0 sinC=sin(A+B)=sinAcosB+cosAsinB 鏍瑰彿3sinAsinB+cosAsinB-sinC =鏍瑰彿3sinAsinB+cosAsinB-sinAcosB-cosAsinB =(鏍瑰彿3-1)sinAcosB=0 鎵浠inAcosB=0 鍥犱负 0<A<蟺 sinA>0 鎵浠 cosB=0 ...
