How do you convert (15, -3) into polar coordinates?

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How do you convert (15, -3) into polar coordinates?

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Answer 1 · 2016-01-20T19:19:29

$(\sqrt{234}, - {11.3}^{\circ})$ $(sqrt234 , - 11.3^@ )$

Explanation:

To convert from Cartesian coordinates to Polar . Use the
following formulae.

$∙ r^{2} = x^{2} + y^{2}$ $• r^2 = x^2 + y^2$

$∙ tan θ = (\frac{y}{x})$ $• tantheta = (y/x)$

In this question x = 15 and y = - 3

so $r^{2} = 15^{2} + {(- 3)}^{2} = 225 + 9 = 234 \Rightarrow r = \sqrt{234}$ $r^2 = 15^2 + (- 3 )^2 = 225 + 9 = 234 rArr r = sqrt234$

and $tan θ = (- \frac{3}{15}) \Rightarrow θ = {tan}^{- 1} (- \frac{3}{15}) = - {11.3}^{\circ}$ $tantheta = (-3/15) rArr theta = tan^-1 (-3/15) = - 11.3^@$

Note : ( 15 , - 3 ) is a point in the 4th quadrant and so $θ$ $theta$ must
be in this quadrant

Since negative angles are measured from the x-axis in a
clockwise direction then $θ = - {11.3}^{\circ}$ $theta = -11.3^@$ is in this quadrant.

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How do you convert (15, -3) into polar coordinates?

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