How do you convert $(\sqrt{3}, 1)$ $(sqrt(3), 1)$ into polar forms?

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How do you convert $(\sqrt{3}, 1)$ $(sqrt(3), 1)$ into polar forms?

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Answer 1 · 2016-01-09T17:04:22

If $(a, b)$ $(a,b)$ is a are the coordinates of a point in Cartesian Plane, $u$ $u$ is its magnitude and $α$ $alpha$ is its angle then $(a, b)$ $(a,b)$ in Polar Form is written as $(u, α)$ $(u,alpha)$ .
Magnitude of a cartesian coordinates $(a, b)$ $(a,b)$ is given by $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ and its angle is given by ${tan}^{- 1} (\frac{b}{a})$ $tan^-1(b/a)$

Let $r$ $r$ be the magnitude of $(\sqrt{3}, 1)$ $(sqrt3,1)$ and $θ$ $theta$ be its angle.
Magnitude of $(\sqrt{3}, 1) = \sqrt{{(\sqrt{3})}^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2 = r$ $(sqrt3,1)=sqrt((sqrt3)^2+1^2)=sqrt(3+1)=sqrt4=2=r$
Angle of $(\sqrt{3}, 1) = {tan}^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$ $(sqrt3,1)=Tan^-1(1/sqrt3)=pi/6$

$\Rightarrow$ $implies$ Angle of $(\sqrt{3}, 1) = \frac{π}{6} = θ$ $(sqrt3,1)=pi/6=theta$

$\Rightarrow (\sqrt{3}, 1) = (r, θ) = (2, \frac{π}{6})$ $implies (sqrt3,1)=(r,theta)=(2,pi/6)$
$\Rightarrow (\sqrt{3}, 1) = (2, \frac{π}{6})$ $implies (sqrt3,1)=(2,pi/6)$
Note that the angle is given in radian measure.

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