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

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Answer 1 · 2016-10-26T10:42:29

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

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

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ , i.e. $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$

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

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

or $r (5 - 3 (cos θ {cos 30}^{o} + sin θ {sin 30}^{o})) = 25$ $r(5-3(costhetacos30^o +sinthetasin30^o))=25$

or $r (5 - 3 (cos θ \times \frac{\sqrt{3}}{2} + sin θ \times \frac{1}{2})) = 25$ $r(5-3(costhetaxxsqrt3/2 +sinthetaxx1/2))=25$

or $5 r - \frac{3 \sqrt{3}}{2} r cos θ - \frac{1}{2} r sin θ = 25$ $5r-(3sqrt3)/2rcostheta -1/2rsintheta=25$

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

How do you convert r=25/( 5-3cos(theta-30))r=255−3cos(θ−30)r=25/( 5-3cos(theta-30)) into rectangular form?