What is the polar form of $(5, 14)$ $( 5,14 )$ ?

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What is the polar form of $(5, 14)$ $( 5,14 )$ ?

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Answer 1 · 2016-01-27T20:50:51

$(\sqrt{221}, 1.23)$ $(sqrt221 , 1.23 )$

Explanation:

Using the formulae that link Cartesian to Polar coordinates.

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

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

The point (5 , 14 ) is in the 1st quadrant and care must be taken

to ensure that $θ is in the 1st quadrant$ $theta color(black)( " is in the 1st quadrant")$

(here x = 5 and y = 14)

$r^{2} = 5^{2} + 14^{2} = 25 + 196 = 221 \Rightarrow r = \sqrt{221}$ $r^2 = 5^2 + 14^2 = 25 + 196 = 221 rArr r = sqrt221$

and $θ = {tan}^{- 1} (\frac{14}{5}) = 1.23 radians$ $theta = tan^-1 (14/5 ) =1.23 color(black)(" radians ")$

and $θ is in the 1st quadrant$ $theta color(black)(" is in the 1st quadrant")$

Polar coordinates are therefore ( $\sqrt{221}, 1.23)$ $sqrt221 , 1.23 )$

Answer 2 · 2016-01-27T20:51:07

$\sqrt{221} ∠ 70, 3^{\circ}$ $sqrt221/_70,3^@$

Explanation:

Rectangular form $(x, y)$ $(x,y)$ can be converted into polar form $(r, θ)$ $(r,theta)$ as follows :

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

So in this particular case we get

$r = \sqrt{5^{2} + 14^{2}} = \sqrt{221}$ $r=sqrt(5^2+14^2)=sqrt221$

$θ = {tan}^{- 1} (\frac{14}{5}) = 70, 3^{\circ}$ $theta=tan^(-1)(14/5)=70,3^@$

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What is the polar form of $(5, 14)$ $( 5,14 )$ ?

2 Answers

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What is the polar form of (5,14)( 5,14 )?

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What is the polar form of $(5, 14)$ $( 5,14 )$ ?