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

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

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Answer 1 · 2016-05-15T13:42:27

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

Explanation:

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

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

$θ = {tan}^{- 1} (\frac{y}{x}) = {tan}^{- 1} (- \frac{4}{3}) = {306.86989765}^{\circ}$ $theta=tan^-1 (y/x)=tan^-1 (-4/3)=306.86989765^@$

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

Answer 2 · 2016-05-15T13:52:42

$(r, θ) = 5 ∠ - 0.927295 [r a d i a n]$ $(r,theta) = 5 angle -0.927295[radian]$

Explanation:

The pass equations are
$x = r cos (θ)$ $x = r cos(theta)$
$y = r sin (θ)$ $y = r sin(theta)$
and dividing each equation side
$\frac{y}{x} = tan (θ)$ $y/x=tan(theta)$ and also its inverse function
$θ = arctan (\frac{y}{x})$ $theta = arctan(y/x)$
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
$3^2+(-4)^2=r^2=25 = 5^2$
$theta = arctan(-4/3) =-0.927295[radian]$

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

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