# How do you express the Cartesian coordinates (3, - 4) as polar coordinates?

Polar Coordinates $\left(5 , {306.86989765}^{\circ}\right)$
OR
Polar Coordinates $\left(5 , - {53.13010235}^{\circ}\right)$

#### Explanation:

Given: $x = 3$ and $y = - 4$

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{3}^{2} + {\left(- 4\right)}^{2}} = \sqrt{25} = 5$

$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right) = {\tan}^{-} 1 \left(- \frac{4}{3}\right) = {306.86989765}^{\circ}$

God bless....I hope the explanation is useful.

May 15, 2016

$\left(r , \theta\right) = 5 \angle - 0.927295 \left[r a \mathrm{di} a n\right]$

#### Explanation:

The pass equations are
$x = r \cos \left(\theta\right)$
$y = r \sin \left(\theta\right)$
and dividing each equation side
$\frac{y}{x} = \tan \left(\theta\right)$ and also its inverse function
$\theta = \arctan \left(\frac{y}{x}\right)$
also x^2+y^2=r^2cos(theta)^2+r^2sin(theta)^2 = r^2(cos(theta)^2+sin(theta)^2)=r²
Applying all those formulas we get
${3}^{2} + {\left(- 4\right)}^{2} = {r}^{2} = 25 = {5}^{2}$
$\theta = \arctan \left(- \frac{4}{3}\right) = - 0.927295 \left[r a \mathrm{di} a n\right]$