How do you write polar coordinates of #(-sqrt 3, 3)#?

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Answer 1 · 2016-01-29T09:59:37

The polar coodinates : #(r, theta)=(2sqrt3, 120^@)#

Given: the rectangular coordinates #(-sqrt3, 3)#

this means #x=-sqrt3# and #y=3#

Convert now to Polar corrdinate system:

Compute #r#:

#r=sqrt(x^2+y^2)=sqrt((-sqrt3)^2+3^2)=sqrt(3+9)=sqrt(12)#

#r=sqrt(4(3))#

#r=2sqrt3#

Compute the angle #theta#:

#theta=tan^-1(y/x)=tan^-1(3/-sqrt3)=120^@#

The polar coodinates : #(r, theta)=(2sqrt3, 120^@)#

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