2cosC＝2b-√3c

根据余弦定理,得：2abcosC=a^2+b^2-c^22bccosA=b^2+c^2-a^2所以2b×cosA－c×cosA=(2b-c)×cosA=(b^2+c^2-a^2)(2b-c)/(2bc

cosC/cosB=（2a-c）/b=(2sinA-sinC)/sinBcoscsinB=2sinAcosB-sinCcosBcoscsinB+sinCcosB-2sinAcosB=0sin(B+C)

用正弦定理化等式右边为角,得到：cosB/cosC=－sinB/(2sinA＋sinC),去分母后有cosB(2sinA＋sinC)＋sinBcosC=0,2cosBsinA＋(cosBsinC＋si

因为a/sinA=b/sinB=c/sinC所以-b/（2a+c）=-sinB/（2sinA+sinC）再问：麻烦写一下中间转化过程和约掉的东西。。3Q再答：a=ksinAb=ksinBc=ksinC

m⊥n=>m.n=0(2cosc/2,-sinc).(cosc/2,2sinc)=02(cosc/2)^2-2(sinc)^2=0cosC+1-2(1-(cosC)^2)=02(cosC)^2+cos

cosC/cosA=3c/[3a+(2√3)b=3sinC/[3sinA+2√3sinB]3sinAcosC+2√3sinBcosC=3cosAsinC等式的一边加一个“-”号则,3sin(A+C)=

题目应该是这样的：(2b-[根号3]c)cosA)=[根号3]acosC求角A.利用正弦定理,有：(2sinB－√3sinC)×cosA=√3sinAcosC,展开后得到：2sinBcosA=√3si

sinC-√3/2+√3cosC-2sinCcosC=0(sinC-√3/2)-2cosC(sinC-√3/2)=0(sinC-√3/2)(cosC-1/2)=0∴sinC=√3/2或cosC=1/2

90,30

√2/2A+C=2[180-(A+C)]=>A+C=1201/cosA+1/cosC=-√2/cosB=>(cosA+cosC)/cosAcosC=√2cos(A+C)带入A+C=120=>(cosC

1.A2.B3.D

由正弦定理a/sinA=b/sinB=c/sinC=2R得：a=2RsinA,b=2RsinB,c=2RsinC,将上式代入已知cosB/cosC=-(b/2a+c),得cosB/cosC=-sinB

A+C=2B,B=60,A+C=1201/cosA+1/cosC=-√2/cosB=-2√2cosA+cosC=-2√2cosAcosC2cos[(A+C)/2]cos[(A-C)/2]=-√2[co

2a=√3c,a=√3/2cc^2=a^2+b^2-2abcosC3/4c^2+b^2-√3/2*√3/2cb-c^2=0b^2-3/4cb+c^2/4=0(b-c)(b+c/4)=0得b=c,另一个

[[1]]∵cosC=3/4.0＜C＜180º∴sinc=(√7)/4再由正弦定理可得:a/sinA=c/sinC∴sinA=(a/c)sinC=(1/√2)×(√7/4)=(√14)/8∴

再答：亲，我已经帮你解决问题了，说好的好评呢再问：那些划掉的是什么意思再答：两边约掉的

根据正弦定理,a=2RsinA,b=2RsinB,c=2RsinC,代入已知的式子,整理有,2sinB*cosA=√3sin（A+C）=√3sinB,即cosA=√3/2,所以A=π/6设AC=2x,

因为：a/sinA=b/sinB=c/sinC=2R所以：a^2=4R^2*sinAb^2=4R^2*sinBc^2=4R^2*sinC所以:(a^2-b^2)/(cosA+cosB)=4R^2*(s

1.将a,b,c用sinA,sinB,sinC替换即√3*sinB*cosA-sinC*cosA=sinA*cosC移项：√3*sinB*cosA=sinC*cosA+sinA*cosC将右边合并：√

1.cosC=√3/4sinC=√13/42a=√3c正弦定理2sinA=√3sinCsinA=√39/8cosA=5/8sinB=sin(A+C)=sinAcosC+cosAsinC=√39/8*√