How do you convert $(11, - 9)$ $(11,-9)$ into polar coordinates?

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How do you convert $(11, - 9)$ $(11,-9)$ into polar coordinates?

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Answer 1 · 2018-05-06T07:33:54

$(\sqrt{202}, {tan}^{- 1} (- \frac{9}{11}) + 2 π) or (14.2, {5.60}^{c})$ $(sqrt202,tan^-1(-9/11)+2pi) or (14.2,5.60^c)$

Explanation:

$(x, y) \to (r, θ); (r, θ) = (\sqrt{x^{2} + y^{2}}, {tan}^{- 1} (\frac{y}{x}))$ $(x,y)->(r,theta);(r,theta)=(sqrt(x^2+y^2),tan^-1(y/x))$

$r = \sqrt{x^{2} + y^{2}} = \sqrt{11^{2} + {(- 9)}^{2}} = \sqrt{121 + 81} = \sqrt{202} \approx 14.2$ $r=sqrt(x^2+y^2)=sqrt(11^2+(-9)^2)=sqrt(121+81)=sqrt202~~14.2$

$θ = {tan}^{- 1} (- \frac{9}{11})$ $theta=tan^-1(-9/11)$

However, $(11, - 9)$ $(11,-9)$ is in quadrant 4, and so we must add $2 π$ $2pi$ to our answer.

$θ = {tan}^{- 1} (- \frac{9}{11}) + 2 π \approx {5.60}^{c}$ $theta=tan^-1(-9/11)+2pi ~~5.60^c$

$(\sqrt{202}, {tan}^{- 1} (- \frac{9}{11}) + 2 π) or (14.2, {5.60}^{c})$ $(sqrt202,tan^-1(-9/11)+2pi) or (14.2,5.60^c)$

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How do you convert $(11, - 9)$ $(11,-9)$ into polar coordinates?

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