we have the point
(5, -pi/6)
where
r=5
θ=-pi/6
see the figure below to better understand the problem
The given point lies in the IV quadrant
so
the x-coordinate is positive
the y-coordinate is negative
Remember that
x=rcosθ
y=rsinθ
so
x=5*cos(pi/6)=(5√3)/2
y=5*sin(pi/6)=-5/2
therefore
the coordinates are
((5√3)/2,-5/2)
