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

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Answer 1 · 2016-01-29T09:59:37

The polar coodinates : $(r, theta)=(2sqrt3, 120^@)$

Given: the rectangular coordinates $(-sqrt3, 3)$

this means $x=-sqrt3$ and $y=3$

Convert now to Polar corrdinate system:

Compute $r$ :

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

$r=sqrt(4(3))$

$r=2sqrt3$

Compute the angle $theta$ :

$theta=tan^-1(y/x)=tan^-1(3/-sqrt3)=120^@$

The polar coodinates : $(r, theta)=(2sqrt3, 120^@)$

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

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