How do you write polar coordinates of $(- \sqrt{3}, 3)$ $(-sqrt 3, 3)$ ?

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Answer 1 · 2016-01-29T09:59:37

The polar coodinates : $(r, θ) = (2 \sqrt{3}, 120^{\circ})$ $(r, theta)=(2sqrt3, 120^@)$

Given: the rectangular coordinates $(- \sqrt{3}, 3)$ $(-sqrt3, 3)$

this means $x = - \sqrt{3}$ $x=-sqrt3$ and $y = 3$ $y=3$

Convert now to Polar corrdinate system:

Compute $r$ $r$ :

$r = \sqrt{x^{2} + y^{2}} = \sqrt{{(- \sqrt{3})}^{2} + 3^{2}} = \sqrt{3 + 9} = \sqrt{12}$ $r=sqrt(x^2+y^2)=sqrt((-sqrt3)^2+3^2)=sqrt(3+9)=sqrt(12)$

$r = \sqrt{4 (3)}$ $r=sqrt(4(3))$

$r = 2 \sqrt{3}$ $r=2sqrt3$

Compute the angle $θ$ $theta$ :

$θ = {tan}^{- 1} (\frac{y}{x}) = {tan}^{- 1} (\frac{3}{- \sqrt{3}}) = 120^{\circ}$ $theta=tan^-1(y/x)=tan^-1(3/-sqrt3)=120^@$

The polar coodinates : $(r, θ) = (2 \sqrt{3}, 120^{\circ})$ $(r, theta)=(2sqrt3, 120^@)$

Have a nice day!!! from the Philippines...

How do you write polar coordinates of (-sqrt 3, 3)(−√3,3)(-sqrt 3, 3)?