Geometry

Since we are only interested in angles, we can arbitrarily set the length of the red segments to $1$ .

In parallelogram $FEGC$ this will give us $GC=FE=\frac{1}{2}$ .

From $\triangle ABG$ the full side $AB=\frac{1}{2\sin10^\circ}=2.879$ and the section of it in the middle $DH=2.879-2=0.879$

Projection of this section into the vertical $DF=0.879\times\cos10^\circ=0.866$

From right $\triangle DFE$ the $\angle FDE=\arctan\frac{EF}{DF}=30^\circ$

To get desired angle, we add the $\angle FDH=10^\circ$ and get $30^\circ+10^\circ=\boxed{40^\circ}.$

The really interesting result is obtained when we calculate the length of $DE=\frac{0.866}{\cos30^\circ}=1$ .

A line made of four equal segments attached by flexible joints will fold perfectly into a narrow isosceles triangle with the top angle of $20^\circ$ .