What is the polar form of $(10, 10)$ $( 10,10 )$ ?

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What is the polar form of $(10, 10)$ $( 10,10 )$ ?

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Answer 1 · 2016-02-20T01:15:54

Use the following formulas:

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

$θ = {tan}^{- 1} (\frac{y}{x})$ $theta=tan^-1(y/x)$

Explanation:

The rectangular coordinate $(10, 10)$ $(10,10)$ lies in quadrant I. So, $θ$ $theta$ must fall between $0$ $0$ and $90$ $90$ degrees

First, solve for r:

$r = \sqrt{x^{2} + y^{2}} = \sqrt{10^{2} + 10^{2}} = \sqrt{2 \times 100} = 10 \sqrt{2}$ $r=sqrt(x^2+y^2)=sqrt(10^2+10^2)=sqrt(2xx100)=10sqrt2$

Now, find $θ$ $theta$ ...

$θ = {tan}^{- 1} (\frac{y}{x}) = {tan}^{- 1} (\frac{10}{10}) = {tan}^{- 1} (1) = 45^{o}$ $theta=tan^-1(y/x)=tan^-1(10/10)=tan^-1(1)=45^o$ or $\frac{π}{4}$ $(pi)/4$

Polar Form : $(r, θ) = (10 \sqrt{2}, \frac{π}{4})$ $(r,theta)=(10sqrt2,(pi)/4)$

hope that helped

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What is the polar form of $(10, 10)$ $( 10,10 )$ ?

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