How do you find the polar coordinates given $(- 2, - 2 \sqrt{3})$ $(-2, -2sqrt3)$ ?

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Answer 1 · 2016-11-17T22:30:41

Please see the explanation.

Polar coordinates are an order pair of $(r, θ)$ $(r, theta)$

The conversion from $(x, y)$ $(x, y)$ to r is:

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

The conversion from $(x, y)$ $(x, y)$ to $θ$ $theta$ is more complicated:

If $x > 0 and y \geq 0$ $x > 0 and y >= 0$ , then $θ = {tan}^{- 1} (\frac{y}{x}) [1]$ $theta = tan^-1(y/x)" [1]"$

If $x = 0 and y > 0$ $x = 0 and y > 0$ , then $θ = \frac{π}{2} [2]$ $theta = pi/2" [2]"$

If $x = 0 and y < 0$ $x = 0 and y < 0$ , then $θ = \frac{3 π}{2} [3]$ $theta = (3pi)/2" [3]"$

If $x < 0$ $x < 0$ , then $θ = π + {tan}^{- 1} (\frac{y}{x}) [4]$ $theta = pi + tan^-1(y/x)" [4]"$

If $x > 0 and y < 0$ $x > 0 and y < 0$ , then $θ = 2 π + {tan}^{- 1} (\frac{y}{x}) [5]$ $theta = 2pi + tan^-1(y/x)" [5]"$

For the given point $(- 2, - 2 \sqrt{3})$ $(-2, -2sqrt(3))$

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

$r = \sqrt{4 + 12}$ $r = sqrt(4 + 12)$

$r = \sqrt{16}$ $r = sqrt(16)$

$r = 4$ $r = 4$

Because the x coordinate is less than zero, we use equation [4]:

$θ = π + {tan}^{- 1} (\frac{- 2 \sqrt{3}}{- 2})$ $theta = pi + tan^-1((-2sqrt(3))/(-2))$

$θ = π + \frac{π}{3}$ $theta = pi + pi/3$

$θ = \frac{4 π}{3}$ $theta = (4pi)/3$

The polar point is $(4, \frac{4 π}{3})$ $(4, (4pi)/3)$

How do you find the polar coordinates given (-2, -2sqrt3)(−2,−2√3)(-2, -2sqrt3)?