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°#