If $xcos alpha+ysin alpha=xcos beta+ysin beta=k$ then show that $x/cos ((alpha+beta)/2)=y/sin ((alpha+beta)/2)=k/cos ((alpha-beta)/2)$ ?

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Answer 1 · 2017-10-22T08:12:50

Given
$xcos alpha+ysin alpha-k=0....[1]$

$xcos beta+ysin beta-k=0....[2]$

To show that $x/cos ((alpha+beta)/2)=y/sin ((alpha+beta)/2)=k/cos ((alpha-beta)/2)$

On cross multiplication of [1] an[2] we get

$x/(sinbeta -sin alpha)=y/(cosbeta-cosalpha)=k/(sinbetacosalpha-cosbetasinalpha)$

$=>x/(-2cos((alpha +beta)/2)sin((alpha-beta)/2))=y/(-2sin((alpha+beta)/2)sin((alpha-beta)/2))=k/(-sin(alpha-beta))$

$=>x/(-2cos((alpha +beta)/2)sin((alpha-beta)/2))=y/(-2sin((alpha+beta)/2)sin((alpha-beta)/2))=k/(-2sin((alpha-beta)/2)cos((alpha-beta)/2))$

$=>x/(cos((alpha +beta)/2))=y/(sin((alpha+beta)/2))=k/(cos((alpha-beta)/2))$

If xcos alpha+ysin alpha=xcos beta+ysin beta=kxcos alpha+ysin alpha=xcos beta+ysin beta=k then show that x/cos ((alpha+beta)/2)=y/sin ((alpha+beta)/2)=k/cos ((alpha-beta)/2)x/cos ((alpha+beta)/2)=y/sin ((alpha+beta)/2)=k/cos ((alpha-beta)/2)?