Question #064f8

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Question #064f8

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Answer 1 · 2016-05-23T11:03:31

$(5 \sqrt{2}, - 1.429)$ $(5sqrt2,-1.429)$

Explanation:

Using the formulae that links Cartesian to Polar coordinates.

$∙ r = \sqrt{x^{2} + y^{2}} and θ = {tan}^{- 1} (\frac{y}{x})$ $•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)$

here x = 1 and y = -7

$\Rightarrow r = \sqrt{1^{2} + {(- 7)}^{2}} = \sqrt{50} = 5 \sqrt{2}$ $rArrr=sqrt(1^2+(-7)^2)=sqrt50=5sqrt2$

and $θ = {tan}^{- 1} (- 7) \approx - 1.429 radians or \approx - {81.87}^{\circ}$ $theta=tan^-1(-7)≈-1.429" radians"" or" ≈-81.87^@$

$\Rightarrow (1, - 7) = (5 \sqrt{2}, - 1.429) = (5 \sqrt{2}, - {81.87}^{\circ})$ $rArr(1,-7)=(5sqrt2,-1.429)=(5sqrt2,-81.87^@)$

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Question #064f8

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