在三角形ABC中·角ABC所对应的边分别为abc且满足bSINA=根号3aCOSB
2012-01-09 · 知道合伙人教育行家
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bsinA = √3a cosB
b/a = √3 cosB/sinA
根据正弦定理:b/a = sinB/sinA
∴sinB/sinA = √3 cosB/sinA
sinB = √3 cosB
tanB = √3
B=π/6
第二问楼主写错了cosA/2=2√5/2?
如果cosA/2 = 2/√5
则cosA = 2(cosA/2)^2-1 = 2*(2/√5)^2-1 = 3/5
A为锐角
sinA = √{1-(cosA)^2} = 4/5
sinC = sin(π-A-B) = sin(A+B) = sinAcos(π/3)+cosAsin(π/3) = 4/5*1/2+3/5*√3/2 = (4+3√3)/10
b/a = √3 cosB/sinA
根据正弦定理:b/a = sinB/sinA
∴sinB/sinA = √3 cosB/sinA
sinB = √3 cosB
tanB = √3
B=π/6
第二问楼主写错了cosA/2=2√5/2?
如果cosA/2 = 2/√5
则cosA = 2(cosA/2)^2-1 = 2*(2/√5)^2-1 = 3/5
A为锐角
sinA = √{1-(cosA)^2} = 4/5
sinC = sin(π-A-B) = sin(A+B) = sinAcos(π/3)+cosAsin(π/3) = 4/5*1/2+3/5*√3/2 = (4+3√3)/10
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(1)根据正弦定理a/sinA=b/sinB可知
a/sinA=b/sinB (A)
而bsinA = √3a cosB (B)
方程(A)*(B)得
sinB/cosB =√3
即tanB=√3
所以B=π/3
(2)因为cosA/2 =2 /√5
则cosA = 2(cosA/2)*(cosA/2)-1 = 2*(2/√5)*((2/√5))-1 = 3/5
sinA=√(1-cosA *cosA )=4/5
因为在三角形ABC中,A+B+C=π
所以sinC = sin(π-A-B) = sin(A+B)
= sinAcos(π/3)+cosAsin(π/3)
= 4/5*1/2+3/5*√3/2
= (4+3√3)/10
a/sinA=b/sinB (A)
而bsinA = √3a cosB (B)
方程(A)*(B)得
sinB/cosB =√3
即tanB=√3
所以B=π/3
(2)因为cosA/2 =2 /√5
则cosA = 2(cosA/2)*(cosA/2)-1 = 2*(2/√5)*((2/√5))-1 = 3/5
sinA=√(1-cosA *cosA )=4/5
因为在三角形ABC中,A+B+C=π
所以sinC = sin(π-A-B) = sin(A+B)
= sinAcos(π/3)+cosAsin(π/3)
= 4/5*1/2+3/5*√3/2
= (4+3√3)/10
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1.∵ bsinA=√3acosB.
sinBsinA= √3sinAcosB,
∵sinA≠0,即tanB=√ 3,
0<B<π,
B= π/3.
2.cosA= 2cos2A/2-1=3/5,
sinA>0,
sinA= 4/5,
sinC=sin(A+ π/3)= 1/2sinA+√3/2cosA= (4+3√3)/10.
sinBsinA= √3sinAcosB,
∵sinA≠0,即tanB=√ 3,
0<B<π,
B= π/3.
2.cosA= 2cos2A/2-1=3/5,
sinA>0,
sinA= 4/5,
sinC=sin(A+ π/3)= 1/2sinA+√3/2cosA= (4+3√3)/10.
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因为bsinA=根号3acosB,所以(a/sinA)*根号3=b/cosB,由正弦定理得;::a/sinA=b/sinB,所以tanB=sinB/coB=根号3,所以角B=60度
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