How do you convert the cartesian coordinate (3√3, 3) into polar coordinates?

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Answer 1 · 2015-10-19T01:07:29

$P = 6 ∠ 30^{o}$ $P=6/_30^(o)$

The module is:

$m^{2} = {(3 \sqrt{3})}^{2} + 3^{2}$ $m^2=(3sqrt(3))^2+3^2$

$m^{2} = 27 + 9$ $m^2=27+9$

$m = \sqrt{36} = 6$ $m=sqrt(36)=6$

The angle is:

$ϕ = {tan}^{- 1} (\frac{3}{3 \sqrt{3}})$ $phi=tan^-1(3/(3sqrt(3)))$

$ϕ = 30^{o}$ $phi=30^(o)$

$P = 6 ∠ 30^{o}$ $P=6/_30^(o)$