What is the polar form of $(9, 12)$ $( 9,12 )$ ?

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Answer 1 · 2017-02-03T04:13:04

$(15, {53.13}^{0})$ $(15,53.13^0)$

polar form is $(r, θ)$ $(r,theta)$

$r = \sqrt{x^{2} + y^{2}} = \sqrt{9^{2} + 12^{2}}$ $r=sqrt(x^2+y^2)=sqrt(9^2+12^2)$

$= \sqrt{81 + 144} = \sqrt{225} = 15$ $=sqrt(81+144)=sqrt(225)=15$

$tan θ = \frac{y}{x} = \frac{12}{9}$ $tan theta = y/x =12/9$
$θ = {tan}^{- 1} (\frac{12}{9})$ $theta = tan^-1(12/9)$

$θ = {53.13}^{0}$ $theta = 53.13^0$

the point of $(9, 12)$ $(9,12)$ is $(15, {53.13}^{0})$ $(15,53.13^0)$ in polar form.

What is the polar form of ( 9,12 )(9,12)( 9,12 )?