How do you convert $r=25/( 5-3cos(theta-30))$ into rectangular form?

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Answer 1 · 2016-10-26T10:42:29

$5sqrt(x^2+y^2)-(3sqrt3)/2x -1/2y=25$

Polar coordinates $(r,theta)$ and Cartesian coordinates $(x,y)$ are related as

$x=rcostheta$ and $y=rsintheta$ , i.e. $r^2=x^2+y^2$ and $tantheta=y/x$

Hence, $r=25/(5-3cos(theta-30^o))$ can be written as

$r(5-3cos(theta-30^o))=25$

or $r(5-3(costhetacos30^o +sinthetasin30^o))=25$

or $r(5-3(costhetaxxsqrt3/2 +sinthetaxx1/2))=25$

or $5r-(3sqrt3)/2rcostheta -1/2rsintheta=25$

or $5sqrt(x^2+y^2)-(3sqrt3)/2x -1/2y=25$

How do you convert r=25/( 5-3cos(theta-30))r=25/( 5-3cos(theta-30)) into rectangular form?