acosc根号3asincbc 0

根号3－c）cosA＝acosC这个条件应该是（根号3b－c）cosA＝acosC否则无解利用正弦定理sqr(3)*2RsinBcosA-2RsinCcosA=2RsinAcosC两边除掉2R并移向s

acosC+√3asinB-b-c=0利用正弦定理a/sinA=b/sinB=c/sinCsinAcosC+√3sinAsinC-sinB-sinC=0∵sinB=sin(A+C),sinAcosC+

(√3b-c)cosA=acosC(√3sinB-sinC)cosA=sinAcosC√3sinBcosA=sinAcosC+sinCcosA√3sinBcosA=sin(A+C)√3sinBcosA

（√3b-c）cosA=acosC（√3sinB-sinC）cosA=sinAcosC,∴√3sinBcosA=sinAcosC+sinCcosA=sin（A+C）=sinB,∴cosA=√3／3．再

csinA=acosC=>a/c=sinA/cosC由正弦定理a/c=sinA/sinC∴sinC=cosC=>∠C=π/4∴∠A+∠B=3π/4==>∠B=3π/4-∠A3sinA-cos(B+π/

一问：sinAcosC＋√3sinAsinC－sinB－sinC=0sinAcosC＋√3sinAsinC－sin(A＋C)－sinC=0sinAcosC＋√3sinAsinC－sinAcosC－co

(√3b-c)cosA=acosC(√3sinB-sinC)cosA=sinAcosC√3sinBcosA=sinAcosC+sinCcosA√3sinBcosA=sin(A+C)√3sinBcosA

(√3b-c)cosA=acosC(√3sinB-sinC)cosA=sinAcosC√3sinBcosA=sinAcosC+sinCcosA√3sinBcosA=sin(A+C)√3sinBcosA

将（2b-根号3c）cosA=根号3acosC代入正弦定理得：（2sinB-根号3sinC）cosA=根号3sinAcosC,A为30°选12ABC为钝角三角形,用正弦定理得b为2根号2,C为105°

(√3×b-c)cosA=acosC根据正弦定理（√3sinB-sinC)cosA=sinAcosC∴√3sinBcosA=sinAcosC+cosAsinC=sin(A+C)=sinB∵sinB>0

(1)2bcosA=√3ccosA+√3acosC=√3(ccosA+acosC)=√3b∴cosA=√3/2∴A=30°(2)若a=2B=45°则：2/sin30°=b/sin45°,∴b=2√2,

acosC＋√3asinC-b-c＝0根据正弦定理a＝2RsinA,b＝2RsinB,c＝2RsinC∴sinAcosC＋√3sinAsinC－sinB－sinC＝0（＊）∵sinB＝sin［180&

1.sinAcosC+根号3/2sinC=sinB又∵sinB=sinAcosC+cosAsinC∴cosA=根号3/2∴A=π/62.a=1,根号3c=1+2b代入原式得cosC+(1+2b)/2=

（1）acosC+√3asinB-b-c=0利用正弦定理a/sinA=b/sinB=c/sinCsinAcosC+√3sinAsinC-sinB-sinC=0∵sinB=sin(A+C),sinAco

acosC+√3asinB-b-c=0利用正弦定理a/sinA=b/sinB=c/sinCsinAcosC+√3sinAsinC-sinB-sinC=0∵sinB=sin(A+C),sinAcosC+

①过B作BE垂直AC交AC于E,(2b-根号3c)cosA=根号3acosC,所以2b•cosA-根号3c•cosA=根号3acosC推出2b•cosA=根号3&#

题目条件有错误,应该是acosC+√3asinC-b-c=0,算死我了.答：（1）三角形ABC中,acosC+√3asinC-b-c=0acosC+√3asinC=b+c结合正弦定理a/sinA=b/

前面我发了封私信你,作废,我用另外个号,就是这个号,帮你答了再问：第二行怎么得出来的？O(∩_∩)O谢谢再答：用了正弦定理，a/sinA=2R左右同时乘2R啦

s代表sin正弦定理a/sA=b/sB=c/sC得b=asB/sA,c=asC/sA代入得(2asinB/sinA-根3asinC/sinA)cosA=根3acosC2cosAsinB=根3cosAs

2bcosA=ccosA+acosC利用正弦定理,a/sinA=b/sinB=c/sinC=2R∴2*2RsinBcosA=2RsinCcosA+2RsinAcosC即2sinBcosA=sinCco