How do you find the polar coordinates given $(\sqrt{3}, 1)$ $(sqrt3, 1)$ ?

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Answer 1 · 2017-02-09T15:38:39

The answer is $= (2, \frac{π}{6})$ $=(2,pi/6)$

To convert from cartesian coordinates $(x, y)$ $(x,y)$ to polar coordinates $(r, θ)$ $(r,theta)$ , we use the following equations

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

$θ = arctan (\frac{y}{x})$ $theta=arctan(y/x)$

Here, we have

$(x, y) = (\sqrt{3}, 1)$ $(x,y)=(sqrt3,1)$

$r^{2} = {(\sqrt{3})}^{2} + 1 = 4$ $r^2=(sqrt3)^2+1=4$

$r = 2$ $r=2$

$θ = arctan (\frac{1}{\sqrt{3}}) = arctan (\frac{\sqrt{3}}{3})$ $theta=arctan(1/sqrt3)=arctan(sqrt3/3)$

$θ = \frac{π}{6}$ $theta=pi/6$

So,

the polar coordinates are $(2, \frac{π}{6})$ $(2,pi/6)$

How do you find the polar coordinates given (√3,1)(sqrt3, 1)?