How to find the coordinates of the vertices of a pentagon centered at the origin

Let us start from vertex $b=(0 ; 8.1)$.

Turning clockwise by $72°$ (recall: $72°=360°/5$), one meets vertex $a$. It means that if one materiallizes the $x$ axis, the angle between $Oa$ and the horizontal axis is $90°-72°=18°$. Thus the coordinates of $a$ are $(8.1 \cos(18°) ; 8.1 \sin(18°))$.

Doing once again the same thing (turning clockwise by $72°$), one meets vertex $C$, with angle cOa $= 72°$, thus line segment makes an angle $72°-18°=54°$ below the horizontal axis, thus considered as $-54°$.

Therefore, the coordinates of $c$ are $(8.1 \cos(-54°) ; 8.1 \sin(-54°))$.

The coordinates of $e$ and $d$ are easily deduced from the coordinates of $a$ and $c$ by symmetry.

To find the coordinates of the vertices of a pentagon centered at the origin:

Assuming starting with b on $y$ axis,

$$(x,y)= ( 8.1 \cos ( t + k \, 2 \pi/5) , 8.1 \sin( t + k\, 2 \pi/5)) $$

where $t$ is polar coordinate angle at (0,0) and $k$ is varied between 0 to 5 to arrive at all vertex coordinates of the regular pentagon.