What is the polar form of #( -1,16 )#?

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Answer 1 · 2018-05-06T17:50:54

the polar cordinates #(r,theta)=(sqrt257,-1.5084)#

the point #( -1,16 )#

#r=sqrt(x^2+y^2)=sqrt[(-1)^2+(16)^2]=sqrt257#

#tantheta=y/x=16/-1=-16#

we will take arctan to each side

#arctan(tantheta)=arctan(-16)#

#theta=arctan(-16)#

#theta=-1.5084#

the polar cordinates #(r,theta)=(sqrt257,-1.5084)#