2acosC=2b-根号3c

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

=2acosC,sinB=2sinAcosCsin（180-A-C)=2sinAcosCsin(A+C)=2sinAcosCsinAcosC+cosAsinC=2sinAcosCcosAsinC=si

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

∠A=60° 我用的是几何方法,画出图.作BD⊥AC,设AD=x那么cosA=AD/AB=x/ccosC=CD/CB=(b-x)/a代入(2b-c)cosA-acosC=0得(2b-c)x/

利用正弦定理：a/sinA=b/sinB=c/sinC,2bcosA=ccosA＋acosC>>>>>A=60°===>>>cosA=[b²＋c²－a²]/(2bc)=[

(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,

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

由正弦定理知：cosc=sinc所以C=45*又已知a+b=2+2根号2所以由余弦定理：a平方+b平方=c平方+2abCOSC=（a+b）平方-2ab=12+8根号2=4+2ab*根号2除2即12+8