Finding angle $x$, a geometry question

Let $X$ be such that $X,B$ are on the same side of line $AF$ and $AXF$ is equilateral.

Then $F$ is a center of a circle through $A,X,B$ and thus $$\angle XAB = {1\over 2}\angle XFB = 12^{\circ}$$ Now $AX =AF$ and $AK=AB$ so $\triangle AXB\cong \triangle AFK$ (sas) which means $$\angle FKA = \angle ABX = {1\over 2}\angle ?= 30^{\circ}$$