在△ABC中,已知内角A=3 π
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(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
角A=π/3,角B=x,则角C=π-x-π/3=2π/3-x,边BC=2根号3,由正弦定理得BC/sinA=AB/sinC=AC/sinB,AB=BCsinC/sinA=(2根号3)sin(2π/3-
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
(1)∵c=2,C=π/3由余弦定理得:c^2=a^2+b^2-2abcosC∴4=a^2+b^2-ab又S△ABC=√3∴1/2absinC=√3=>√3/4*ab=√3=>ab=4联立方程组:{a
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/
△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
第一个问题:∵∠A=π/3,∴sin∠A=√3/2.显然有:∠C=π-∠A-∠B=π-π/3-x,∴sin∠C=sin(π/3+x).由正弦定理,有:AC/sin∠B=BC/sin∠A=2√3/(√3
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
三角型为等边三角形时候面积最大,为3倍根号3.方法,做内角A角平分线,当线垂直于BC时,面积最大,用勾股定理算出另外两边相等.三角形为等边三角形.且高为3.再问:为什么等边三角形时面积最大?再答:证明
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
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
△ABC的面积等于根号3,1/2absinC=√3,sinC=√3/2,ab=4,由余弦定理有,c^2=a^2+b^2-2abcosC,cosC=1/2所以a^2+b^2=8,a^2+b+^2+2ab
面积S=1/2*ab*SINC=√3得到ab=4余弦定理COSC=1/2=(a*a+b*b-c*c)/2ab,代入ab=4,c=2得到a*a+b*b=8于是a^2+b^2-2ab=0...(a-b)^
∵△ABC的面积等于3,c=2,C=π3,∴S=12absinC=34ab=3,即ab=4,∴由余弦定理c2=a2+b2-2abcosC得:4=a2+b2-ab=(a+b)2-3ab=(a+b)2-1