Question 1177649
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Hi
(8 sqrt3, −8) 4th quadrant
 r = sqrt(x^2 + y^2) = sqrt(4*64) = ± 16
for the point for 0 ≤ 𝜃 < 2𝜋
tan θ = x/y = - √3 look below: θ is 11π/6 (4th quadrant)  0r 11π/6 - π = 5π/6 if r<0
two sets of polar coordinates:
(16, 11π/6)  and (-16, 5π/6) 
Wish You the Best in your Studies
(cos, sin)  Unit Circle:
     *[illustration wheel].
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