Question 337098
{{{r^2=(-2)^2+(2sqrt(3))^2}}}
{{{r^2=4+12}}}
{{{r^2=16}}}
{{{r=4}}}
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{{{tan(theta)=(2sqrt(3))/(-2)}}}
{{{tan(theta)=-sqrt(3)}}}
{{{theta=(2*pi)/3}}}
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b) is the correct answer.