A fun follow up problem about a convex polygon

Question

Find the remaining two angles of this polygon:

No trigonometry or Pytagoras' theorem allowed :D

Answer 1

Oh, that was really fun. You can do this, by introducing equilateral triangles at the angles of 60° namely the one at 150°-90° and the one at 60° which both split your side of length 2 in two parts. Since 180°-2*60°=60° you get there a third equilateral triangle (the green sides represent length 1 sides and the red angles are all 60°). I think you can take it from here :-) (using the sum of the angles)

Answer 2

Consider the following decomposition, placing in evidence 3 equilateral triangles:

Angle in D in isosceles triangle GDF is clearly $360^{\circ}-90^{\circ}-2 \times 60^{\circ}=150^{\circ}$.

Thus angles in $G$ and $F$ in the same triangle GDF are equal to $15^{\circ}.$

Thus the looked for angles in G and F are resp. $90^{\circ}+15^{\circ}=105^{\circ}$ and $2 \times 60^{\circ}+15^{\circ}=135^{\circ}.$