How do you convert $(- 6, - 3)$ $(-6,-3)$ into polar coordinates?

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How do you convert $(- 6, - 3)$ $(-6,-3)$ into polar coordinates?

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Answer 1 · 2016-02-13T17:17:23

Polar coordinates of $(- 6, - 3)$ $(−6,−3)$ are $(3 \sqrt{5}, {206.565}^{o})$ $(3sqrt5, 206.565^o)$

Explanation:

If $(x, y)$ $(x, y)$ are converted into polar coordinates $(r, θ)$ $(r, theta)$ ,

while $(x, y)$ $(x, y)$ in terms of $r$ $r$ and $θ$ $theta$ are $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ , $(r, θ)$ $(r, theta)$ in terms of $x$ $x$ and $y$ $y$ are $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ and $θ = {tan}^{- 1} (\frac{y}{x})$ $theta=tan^-1(y/x)$

Hence, converting $(- 6, - 3)$ $(−6,−3)$ into polar coordinates

$r = \sqrt{{(- 6)}^{2} + (- 3^{2})}$ $r=sqrt((-6)^2+(-3^2)$ = $\sqrt{45}$ $sqrt45$ = $3 \sqrt{5}$ $3sqrt5$

$θ = {tan}^{- 1} (\frac{- 3}{- 6})$ $theta=tan^-1((-3)/-6)$ = ${tan}^{- 1} (\frac{1}{2})$ $tan^-1(1/2)$ . Further, as $(- 6, - 3)$ $(−6,−3)$ is in third quadrant $θ = (180 + 26.565) = {206.565}^{o}$ $theta=(180+26.565)=206.565^o$

Hence polar coordinates of $(- 6, - 3)$ $(−6,−3)$ are $(3 \sqrt{5}, {206.565}^{o})$ $(3sqrt5, 206.565^o)$

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How do you convert $(- 6, - 3)$ $(-6,-3)$ into polar coordinates?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

How do you convert (−6,−3) (-6,-3) into polar coordinates?

1 Answer

Explanation:

Related questions

Impact of this question

How do you convert $(- 6, - 3)$ $(-6,-3)$ into polar coordinates?