在三角形ABC中,B=六分之派,c=根号3
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![在三角形ABC中,B=六分之派,c=根号3](/uploads/image/f/3207412-28-2.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CB%3D%E5%85%AD%E5%88%86%E4%B9%8B%E6%B4%BE%2Cc%3D%E6%A0%B9%E5%8F%B73)
a=2c,则:sinA=2sinC又:A=π/2+C,则:sinA=sin(π/2+C)=cosC则:2sinC=cosC4sin²C=cos²C4(1-cos²C)=c
COS(B/2)=2√5/5.SIN(B/2)=√5/5SINB=2SIN(B/2)COS(B/2)=4/5S△ABC=acSINB/2=π/5
将第一个式子化成:sinA=√2sinB,第二个式子化为:√3cosA=√2cosB,将一二两式相乘得:√(3/2)sin(2A)=sin(2B),再因为A+B
A=π/4,tan(A+B)=7,∴tanC=-7,cosC=-1/√[(-7)^+1]=-√2/10,∴sinC=7√2/10.sinB=sin(A+C)=3/5,∴S△ABC=(1/2)acsin
B=π/3,cosA=4/5,b=√3C=π-B-A=2π/3-AsinC=sin(2π/3)cosA-cos(2π/3)sinA=(√3/2)(4/5)-(1/2)(3/5)=2√3/5-3/10
(1):由题意得:因为cosA=4/5又因为A、B、C是三角形ABC的内角.所以sinA=[根号下(5^2-4^2)]/5=3/5又因为角B=60度所以sinB=(根号3)/2,B=1/2所以可得si
不妨设sinB/2=x,cosB/2=y,列方程:1式:x平方+y平方=1;2式:2xy=1/3,sinA=(根号2)/2*[y-x],用1式-2式可凑出y-x的平方,可求出sinA=三分之根号3;2
余弦定理:a^2=b^2+c^2-2*b*c*CosA代入数据c=√3,b=1,A=30°所以a^2=1,a=1又由正弦定理a/sinA=b/sinB,sinB=sinA=√3/2,B=30°(150
sinA+sinB=2sin(A+B)sinA+1/2=√3sinA+cosAS△ABC=ac/4=1/2ac=2b=1
由题可得,A=3分之派
解(1):∵cosA=4/5∴sinA=√(1-cos²A)=√[1-(4/5)²]=√(9/25)=3/5sinC=sin[180°-(A+B)]=sin(A+B)=sinAco
1、正弦定理:a/sinA=b/sinB=c/sinC=ta=tsinAb=tsinBc=tsinCbsin(π/4+C)-csin(π/4+B)=a所以sinAsin(π/4+C)-sinCsin(
1、cosB=√6/3,sinB=√3/3,sinA=√3/2,cosA=1/2sinC=sin(π-A-B)=sin(A+B)=sinAcosB+cosAsinB=√3/2*√6/3+1/2*√3/
sinC-cosC=√2sin(C-π/4)=√2cos(3π/4-C)=-√2cos(π/4+C)显然C>=π/4,否则cosC>=√2/2,sinC+sinB=√3+cosC>=2,不可能成立因此
sin(B-π/6)cos(B-π/3)=1/2(√3/2sinB-1/2cosB)(1/2cosB+√3/2sinB)=1/23/4sin^2B-1/4cos^2B=1/23sin^2B-(1-si
1)根据余弦公式:c^2=a^2+b^2-2abcosA故:a^2+b^2-2abcos(π/3)=2^2=4a^2+b^2-ab=4……(1)又面积S=absinC/2=√3absin(π/3)=2
由正弦定理得:a/sinA=b/sinBsinB=bsinA/a=√3*(√3/2)/3=1/2则:B=π/6或5π/6(舍去)所以,C=π-π/3-π/6=π/2
三角形中A+B+C=πC=π-A-B=5π/6=150度cosC=-√3/2
c=2分之3倍的派?是2π/3吧余弦定理c²=a²+b²-2abcosC3=a²+1-2*a*1*(-1/2)a²+a-2=0(a+2)(a-1)=0
由正弦定理a/SinA=b/SinB得a=bSinA/SinB=5根号2/3.