在三角形ABC中,角A,B,C所对的边分别是a.b,c,已知cosC=1/8,cosA=3/4,b=5,则三角 在三角形ABC中，角A，B，C所对的边分别为a，b，c，已知...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c,cosA=3/4 1.cosC=cos2A=2cosA^2 -1=2*9/16 -1=1/8
sinA^2=1-cosA^2=7/16
sinC^2=1-cosC^2=63/64
(a/c)^2=(sinA/sinC)^2=4/9
a/c=2/3
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∵b=5 ，由正弦定理可得a=4
S=½absinC=四分之十五倍的根号七
希望检查后再采纳
把三角形ABC分为2个直角三角形，在边长b上那里分，设边长b上的点为D，CD长度为x，DA长度为y，BD长度为h，则有x+y=b=5, x/a=1/8, y/c=3/4, 所以h^2=(8*x)^2-x^2=(4/3*y)^2-y^2 （勾股定理）.这样就能算出x=0.5 ，y=4.5 ，就是h=根号（0.25*63），面积就是1/2*b*h
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sinBcosC+cosBsinC=2sinAcosB sin(B+C)=2sinAcosB sinA=2sinAcosB cosB=1/2 B=60搴 A+C锛120搴 鍥狅細a/sinA=c/sinC=2R=2鈭3 a+b=2鈭3(sinA+sinC)=4鈭3sin((A+C)/2)cos((A-C)/2)=6cos((A-C)/2)=6 cos((A-C)/2)=1 A=C=60搴 鏁涓夎褰BC鏄瓑杈逛笁瑙掑舰 a=b=...
