What is the polar form of #(5,-3)#?

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Answer 1 · 2015-12-06T21:22:17

#r=sqrt(x^2+y^2)#

#theta=tan^-1(y/x)#

First note, the coordinate #(5.-3)# is in Quadrant IV so #270< theta<360#

#r=sqrt(5^2+(-3)^2)=sqrt34#

Reference Angle in Quad IV #=tan^-1(3/5)~~31^o#

#theta =(270^o)+# (reference angle) #=270+31=301^o#

polar form #=(sqrt34, 301^o)#

hope that helped