已知cosc=1 4,求三角形的周长

由正弦定理,b/sinB=c/sinC得b=sinB·c/sinC代入原式得cosC·sinB·c/sinC=c·cosBsinB·cosC=sinC·cosBsinB·cosC-sinC·cosB=

2sinAsinB=cos(A-B)-cos(A+B)=cos(A-B)+cosC=1+cosC所以cos(A-B)=1,A=B,三角形ABC是等腰三角形.

sinA=5/13＜1/2,故可以是A＞150°,或A＜30°1/2＜cosB＜√2/2,故45°＜B＜60°由于A+B＜180°所以A为锐角故cosA=12/13,sinB=4/5cosC=-cos

tanB=√3所以B=60度b=3√6cosC=1/3所以sinC=2√2/3b/sinB=c/sinC所以3√6/(√3/2)=c/(2√2/3)c=8sinA=sin(180-B-C)=sin(B

由cosA=15/17,cosB=4/5求出sinA=根号(1-cos^2A)=根号(1-(15/17)^2)=8/17sinB=3/5cosC=cos(180-(A+B))=cos180cos(A+

sinA=√(1-cos^2A)=√[1-(3/5)^2]=4/5.sinB=√(1-cos^2B)=√[1-(5/13)^2]=12/13.在△ABC中,C=180-(A+B).cosC=cos[1

cosA=(b^2+c^2-a^2)/(2bc)s=a^2-(b-c)^2s=1/2bcsinA得到cosA=15/17sinA=8/17得到直角三角形cosC=0或cosC=8/17

三个内角分别为60-x6060+x,变形整理得cosa^2+cosc^2＝1＋cos(A+C)cos(A-c)=1＋cos120°cos（C-A)=1-(1/2)cos（C-A)而－120°＜C－A＜

2sinAsinB=cos(A-B)-cos(A+B)=cos(A-B)+cosC=1+cosC所以cos(A-B)=1,A=B,三角形ABC是等腰三角形.

因为cosA=五分之四,cosB=十三分之五,所以sinA=3/5,sinB=12/13cosC=cos(π-A-B)=-cos(A+B)=-(cosAcosB-sinAsinB)=-(4/5*5/1

cosB=cosC,∠B=∠C3b=2√3asinB,用正弦定理,两边消去2R,3sinB=2√3sinAsinBsinA=√3/2,A=60°,120°A=60,B=C=60°A=120,B=C=3

三角形ABC中1+cosC=1+cos(180-A-B)=!-cos(A+B)=1-cosAcosB+sinAsinB=2sinAsinB所以sinAsinB+cosAcosB=1即cos(A-B)=

在三角形中sinA,sinB,sinC均大于0sinA=5/13sinB=4/5cosAcosB-sinAsinB=-36/65-20/65=56/65=cos(A+B)=cos(π-C)=-cosC

非常简单,这道题是较为简单的解答题,因此没必要做的太长,适当简洁些即可已知tanB=√3,cosC=1/3则显然BC都为锐角sinB=tanB*cosB=tanB*{1/√[1+(tanB)平方]}=

根据正弦定理sinA:sinB:sinC=a:b:c=3:2:4设a=3t,b=2t,c=4ta+b+c=9t=9t=1a=3,b=2,c=4根据余弦定理cosC=(a^2+b^2-c^2)/2ab=

应该用反cos求出两个角度,再求cosC

因为cosB=4/5,所以sinB=3/5.cosC=-cos(A+B)=sinAsinB-cosAcosB=1/2×3/5-(根号3)/2×4/5=(3-4倍根号3)/10希望有用.

cosB是12/13吧?sinA=3/5,sinB=5/13,cosC=cos[180°-(A+B)]=-cos(A+B)=-(cosAcosB-sinAsinB)=-33/65.

(cosA+2cosC)/(cosA+2cosB)=sinB/sinCcosAsinC+2sinCcosC=cosAsinB+2sinBcosBcosAsinC+sin2C=cosAsinB+sin2

由已知S=0.5ab又有S=0.5*AB*AC*sinA固0.5*AB*AC*sinA=0.5*AB*AC*cosA从而tanA=1固A=45°又有cosC=-cos（A+B)=sinA*sinB-c