在△abc中,已知内角A=丌3,边Bc二2根号3

（1）利用正弦定理a/sinA=b/sinB=c/sinC∴√3/sin(π/3)=b/sinx=c/sin(2π/3-x)即2=b/sinx=c/sin(2π/3-x)∴b=2sinx,c=2sin

根据正弦定理,b/sinB=a/sinA,a=2√3,A=π/3,B=x,b=4sinx,c/sinC=a/sinA,c=2√3/(√3/2)*sinC=4sinC=4sin(A+B)=4sin(π/

1)AC=BC*sinx/sinA=4sinxAB=BC*sin(120°-x)/sinA=4sin(120°-x)y=2√3+4[sinx+sin(120°-x)]=2√3+4√3[√3/2*sin

因为√37>4>3所以角C最大CosC=(a²+b²-c²)/2ab=（3²+4²-37)/(2×3×4)=-12/24=-1/2所以C=2π/3,也

1.根据正弦定理：b/sinB=c/sinC∵B=C∴b=c∵2b=√3a∴a=2b/√3余弦定理：cosA=(b^2+c^2-a^2)/2bc=[b^2+b^2-(2b/√3)^2]/2b*b=1/

(1).∵a,b,c成等比数列,∴b²=ac∵b²=a²+c²-2accosB∴ac=a²+c²-3ac/2即a=2c或者c=2a不妨设a=

因为cosB=3/4,0

△ABC中，∵b2=ac，a+c=3，cosB=34，∴b2=a2+c2-2ac•cosB=（a+c）2-72ac=9-72b2，∴b2=2．则AB•BC=ca•cos（π-B）=b2 （-

（1）由余弦定理c2=a2+b2-2abcosC，的a2+b2-ab=4，又∵△ABC的面积等于3，∴12absinC=12ab•32=3，∴ab=4，得a=b=2．（2）sin（A+C）=2sinA

1.只要想办法将AB边与AC边表示出来就行了,根据正弦定理,不难得出AB=[2根3/sin(π/3)]*sinXAC=[2根3/sin(π/3)]*sin（X+π/3）∴F（x）=2根3+[2根3/s

利用正弦定理BC/sinA=AC/sinB=AB/sinCBC/sinA=4=AC/sinx=AB/sin(2/3π-x)f(x)=AB+BC+AC=2根号3+4sinx+4sin(2/3π-x)定义

由于正弦定理.BC/SINA=AB/SINC=AC/SINB所以,AC/SINB=AB/SINC=4,AC=4SINX,AB=4SIN(pai-pai/3-x)即AB=4SIN(2pai/3-x)所以

角A=60度角B=X角C=180-60-X=120-XSIN角A:BC=SIN角B:AC=SIN角C:AB=根号3/2:2根号3=1:4AC=4*SINXAB=SIN(120-X)*4Y=2根号3+4

(1)因为内角C=π-(π/3+x)>0所以0

2√3/sin60°=AC/sinxAC=（2√3/sin60°）sinx2√3/sin60°=AB/sin（180°-60°x）AB=（2√3/sin60°）sin（180°-60°-x）AB=（2

a/SinA=(2√3)/(√3/2)=4=b/SinB=c/SinCb=4SinX,C=180-60-X=120-XSinC=(√3/2)CosX+0.5*SinXc==2√3*CosX+2*Sin

sinC=sin(A+B)=sinAcosB+sinBcosA=2cosAsinB+sinBcosA=3cosAsinB∴cosA=sinC/3sinB=c/3b(正弦定理)余弦定理cosA=(c&s

sinC+sin(B-A)=2sin2Asin(B+A)+sin(B-A)=2*2sinAcosA2sinBcosA=4sinAcosA2cosA(sinB-2sinA)=0cosA=0或sinB=2

1.三角形面积(1/2)*b*sin60*a=根号3可得a*b=4根据余弦公式,a^2+b^2-c^2=2abcosC可得a^2+b^2=8所以a=2,b=22.根据正弦公式,b/sinB=a/sin

由sinAcosC=3cosAsinC得a×（a^2+b^2-c^2）/2ab=3c×(b^2+c^2-a^2)/2bca^2+b^2-c^2=3×(b^2+c^2-a^2)2a^2=b^2+2c^2