Without loss of generality we can assume that
the length of the sides of ABCDE is 1.
Applying xy coordinates to the vertices of
ABCDE we have
B [0,0]
C [1,0]
A [cos(136),sin(136)]
D [1 + cos(76),sin(76)]
Point E must belong to the unit circles
with centers at A and D:
CircleA: [x  cos(136)]^2 +
[y  sin(136)]^2 = 1
CircleD: [x  {1 + cos(76)}]^2 +
[y  sin(76)]^2 = 1
There are two points that satisfy these
conditions. One makes ABCDE convex and
the other does not. The one that does 
E [cos(76),sin(76)]
Clearly it satisfies CircleD. I will let
you confirm that it satisfies CircleA.
Clearly BE = 1. Therefore, triangle ABE
is equilateral and quadrilateral BCDE is
a rhombus. Therefore,
angle(AEB) = 60
angle(BED) = angle(BCD) = 104
angle(AED) = angle(AEB) + angle(BED) = 164.
QED
You can't construct this pentagon with
straightedge and compass.

Posted by Bractals
on 20180412 12:38:48 