What is the polar form of $(5, - 3)$ $(5,-3)$ ?

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Answer 1 · 2015-12-06T21:22:17

$r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$

$θ = {tan}^{- 1} (\frac{y}{x})$ $theta=tan^-1(y/x)$

First note, the coordinate $(5 . - 3)$ $(5.-3)$ is in Quadrant IV so $270 < θ < 360$ $270< theta<360$

$r = \sqrt{5^{2} + {(- 3)}^{2}} = \sqrt{34}$ $r=sqrt(5^2+(-3)^2)=sqrt34$

Reference Angle in Quad IV $= {tan}^{- 1} (\frac{3}{5}) \approx 31^{o}$ $=tan^-1(3/5)~~31^o$

$θ = (270^{o}) +$ $theta =(270^o)+$ (reference angle) $= 270 + 31 = 301^{o}$ $=270+31=301^o$

polar form $= (\sqrt{34}, 301^{o})$ $=(sqrt34, 301^o)$

hope that helped

What is the polar form of (5,-3)(5,−3)(5,-3)?