I've been trying this for days I can't get my head round it any help would be appreciated.

The one line of symmetry is vertical, so we could fold the hexagon in half in such a way that the vertices A and B would meet at the same point, and the same goes for the pairs C,F and D,E. Because of this symmetry, we know angle AFE is congruent to BCD, and angle FED is congruent ot CDE.

Let x be the measure of angle CDE.

In any convex polygon with n sides, the interior angles sum to (n - 2)*180º in measure. ABCDEF is a hexagon, so n = 6.

We have 2 angles of measure 123º, 2 of measure x, and 2 of measure 2x. So

2(123º + x + 2x ) = (6 - 2)*180º

246º + 2x + 4x = 720º

6x = 474º

x = 79º

Angle AFE is congruent to angle BCD, which is twice the measure of CDE, so angle AFE has measure 2*79º = 158º.