在△ABC中,tanA=a²,tanB=b²,那么△ABC是什么三角形?
3个回答
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tanA=a²,tanB=b²
a²/sin²A=b²/sin²B
1/sinAcosA=1/sinBcosB
2sinAcosA=2sinBcosB
sin2A=sin2B
2A=2B 或2A=π-2B,A+B=90°
a²/sin²A=b²/sin²B
1/sinAcosA=1/sinBcosB
2sinAcosA=2sinBcosB
sin2A=sin2B
2A=2B 或2A=π-2B,A+B=90°
追问
1/sinAcosA=1/sinBcosB怎么来的
追答
tanA=a²,tanB=b²
a²/sin²A=b²/sin²B
tanA/sin²A=tanB/sin²B
(sinA/cosA)/sin²A=(sinB/cosB)/sin²B
1/sinAcosA=1/sinBcosB
展开全部
tanA=a^2
sinA/cosA = a^2
a/sinA = 1/(acosA)
similarly
tanB=b^2
b/sinB = 1/(bcosB)
a/sinA =b/sinB
1/(acosA) = 1/(bcosB)
a/b = cosB/cosA = sinA/sinB
sinAcosA -sinBcosB =0
(1/2)sin2A - (1/2)sin2B =0
A= B
tanA = a^2
tanB = b^2
tanC = -tan(A+B)
= -(tanA +tanB)/(1-tanAtanB)
=> 1- tanAtanB =0
tanA=tanB =π/4
=>tanC = π/2
=>△ABC是是等腰或直角三角形
sinA/cosA = a^2
a/sinA = 1/(acosA)
similarly
tanB=b^2
b/sinB = 1/(bcosB)
a/sinA =b/sinB
1/(acosA) = 1/(bcosB)
a/b = cosB/cosA = sinA/sinB
sinAcosA -sinBcosB =0
(1/2)sin2A - (1/2)sin2B =0
A= B
tanA = a^2
tanB = b^2
tanC = -tan(A+B)
= -(tanA +tanB)/(1-tanAtanB)
=> 1- tanAtanB =0
tanA=tanB =π/4
=>tanC = π/2
=>△ABC是是等腰或直角三角形
追问
1- tanAtanB作为分母,怎么会 =0
即使1- tanAtanB =0,那又怎么得出tanA=tanB =π/4
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tanA/tanB=a^2/b^2
sinAcosB/sinBcosA=sinA^2/sinB^2
cosB/cosA=sinA/sinB
cosBsinB=cosAsinA
sin2B=sin2A
情况1:2B=2A为等腰
情况2:180-2B=2A
则A+B=90
sinAcosB/sinBcosA=sinA^2/sinB^2
cosB/cosA=sinA/sinB
cosBsinB=cosAsinA
sin2B=sin2A
情况1:2B=2A为等腰
情况2:180-2B=2A
则A+B=90
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