Question #8a89a

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Question #8a89a

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Answer 1 · 2016-12-21T22:27:22

Use the equations, $r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2 + y^2)$ and $θ = π + {tan}^{- 1} (\frac{y}{x})$ $theta = pi + tan^-1(y/x)$
$(- 3 \sqrt{3}, - 3) \to (6, \frac{7 π}{6})$ $(-3sqrt(3), -3) to (6, (7pi)/6)$

Explanation:

Given the point:

Use the equation:

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

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

$r = \sqrt{27 + 9}$ $r = sqrt(27 + 9)$

$r = 6$ $r = 6$

There are many equations for the angle, $θ$ $theta$ ; which one you use depends on x and y:

$[1] θ = {tan}^{- 1} (\frac{y}{x}); x > 0 and y \geq 0$ $"[1] "theta = tan^-1(y/x); x > 0 and y ge 0$
$[2] θ = \frac{π}{2}; x = 0 and y > 0$ $"[2] "theta = pi/2; x = 0 and y > 0$
$[3] θ = π + {tan}^{- 1} (\frac{y}{x}); x < 0$ $"[3] " theta = pi + tan^-1(y/x); x < 0$
$[4] θ = \frac{3 π}{2}; x = 0 and y < 0$ $"[4] "theta = (3pi)/2; x = 0 and y < 0$
$[5] θ = 2 π + {tan}^{- 1} (\frac{y}{x}); x > 0 and y < 0$ $"[5] "theta = 2pi + tan^-1(y/x); x > 0 and y < 0$

Because $x = - 3 \sqrt{3}$ $x = -3sqrt(3)$ , then we do not care about the value of y and we use equation [3}:

$θ = π + {tan}^{- 1} (\frac{- 3}{- 3 \sqrt{3}}); x < 0$ $theta = pi + tan^-1((-3)/(-3sqrt(3))); x < 0$

$θ = \frac{7 π}{6}$ $theta = (7pi)/6$

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Question #8a89a

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