How do you convert $(−1, 5)$ to polar form?

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Answer 1 · 2018-05-04T11:47:55

$(sqrt26, 101.31°)$

Given (-1, 5) -> (r, $theta)$

$r= sqrt(x^2+y^2)$
$theta= arctan(y/x)$

$r= sqrt((-1)^2+(5)^2)=sqrt26$
$theta= arctan(5/-1)=-78.69°+180^@$ as the angle has to be in the second quadrant

Answer 2 · 2018-05-04T11:53:02

$(x,y) -> (r, theta)$
$(-1,5) -> (\sqrt(26), 3.38)$

$(x,y)$
$r^2=x^2+y^2$
$tan(theta)=tan(y/x)$

$r=\sqrt(x^2+y^2)=\sqrt((-1)^2+5^2)=\sqrt(26)$
$theta=tan^(-1)(y/x)=tan^(-1)(5/-1)=1.37 ["rad"]=78.7°$

How do you convert (−1, 5) (−1, 5) to polar form?