在三角形ABC中,A=4分之派,cosB=10分之根号10,1>求cosC,2
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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√3
a=180/8b=540/8c=180/2=90直角三角形
sinB=3/5,cosC=cos[180°-(A+B)]=-cos(A+B)=-(cosAcosB-sinAsinB)=-[(√2/2)*4/5-(√2/2)*3/5]∴cosC=-√2/10,2、
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
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
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
tan(A+π/4)=-2-√3;tanA=√3;A=π/3Y=2sin^2B+sin(2B+π/6)=1+sin(2B-π/6)≤2当sin(2B-π/6)=1亦即B=π/3时Y取得最大值考虑到三角
1、∵A、B、C是三角形的内角∴sin(A+B)=sinC∴√2asin(B+π/4)=c√2sinAsin(B+π/4)=sinC(根据正弦定理)√2sinA[(√2/2)sinB+(√2/2)co
由题可得,A=3分之派
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(
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
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
a/sinA=c/sinCsinA=a×sin(π/4)/c=a×(√2/2)/2a=√2/4cos2A=1-2sin²A=1-2×(√2/4)²=3/4sin2A=√7/4(由△
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
AB/sinC=BC/sinAsinC=AB*sinA/BC=√6*√3/2/3=√2/2BC>AB∴C
由正弦定理a/SinA=b/SinB得a=bSinA/SinB=5根号2/3.