Answer to Question #347026 in Analytic Geometry for Busi

Question

Convert (3–√,−1) into polar coordinates (r,θ) so that r≥0 and 0≤θ<2π

Answer 1

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

$r\ge0=>r=\sqrt{4}=2$

Quadrant III

$\tan \theta=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

$\theta=\pi+\tan^{-1}(\dfrac{1}{\sqrt{3}})=\dfrac{7\pi}{6}$

$(2, \dfrac{7\pi}{6})$