Finding angle $x$, a geometry question Here is the question. Find the value of $x$. 
I have solved this question by my own (with 3 different methods). However, all methods of mine are based on the construction of equilateral triangles.I am wondering whether there exist any other possible geometric construction because I don’t think equilateral triangle is the only way out but I couldn’t find any other possible construction apart from equilateral. I would like to understand more about the hidden geometric relation about all given informations so any helps are kindly appreciated. 
The answer for $x$ is $30^o$
 A: 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 $$\angle FKA =  \angle ABX = {1\over 2}\angle ?= 30^{\circ}$$
A: You can also use the law of sin:
$$\triangle AFK:\;\;\; {AF \over \sin x} = {AK\over \sin (168-x)}= {AB\over \sin (12+x)} $$
$$\triangle ABC:\;\;\; {AF \over \sin 48} = {AB\over \sin (84)}$$
So $$ \sin x \sin 84 = \sin (12+x)\sin 48$$
So $$\cot x = {\cos 6 -\cos 12 \cos 42\over \sin 12\cos 42}$$
Solving this (calculator) we get $x= 30$.
A: Clearly $AB = AK$ and $AF = FB$.
Let $ABD$ be equilateral. Then $\angle FAD = \angle FAK = 12^{\circ}$. Since $AD =AK$ we have $\triangle FAD\cong \triangle FAK$ (sas) Also notice that $DF\bot AB$ so $$\angle ADF = 30^{\circ}\implies \angle  FKA = 30^{\circ}$$ 
A: Let $BCF$ be equilateral. Then $F$ is a ceneter of circle through $A,B,C$ so $$\angle ABC = {1\over 2}\angle AFC= 12^{\circ}$$
and $$\angle CAB = {1\over 2}\angle BFC= 30^{\circ}$$
Since $AF = BC$ and $AK = AB$ we have $\triangle ABC\cong \triangle AFK$ (sas) and thus $x= 30^{\circ}$.

