Is there a way to solve for the missing angle? 
I was working on this problem. I tried to draw ${AC}$, ${BD}$ as isosceles triangle and divide into cases to find the missing angle $??$, but I got stuck. Can someone help me please or give me a clue?
 


$$AB=AD=BC=x,AC=t\\t^2=x^2+x^2-2x^2\cos166\\\to t^2=2x^2(1-\cos166)=2x^2(2\sin^2(83))=4x^2\cos^2(7)\\t=2x\cos 7$$so
$$\frac{t}{\sin(120-c)}=\frac{x}{sin(c-7)}\\\frac{2x\cos 7}{\sin(120-c)}=\frac{x}{sin(c-7)}\\2\sin(c-7)\cos 7=\cos(30-c)$$using product to sum formula
$$\sin(c-14)+\sin c=\cos(30-c)\\
\sin 30 \cos c+ \cos 30 \sin c+\sin c=\cos 30 \cos c + \sin30 \sin c\\\sin c(1-\frac 12 +\cos 14)=\cos c(\cos 30+\sin14)\\\tan c=\frac{\cos 30+\sin 14}{\frac12+\cos 14}=\frac{\sin 60+\sin 14}{\cos 60+\cos 14}=\frac{2\sin37 \cos 23}{2\cos37 \cos 23}=\tan(37)$$finally
$$\tan(c)=\tan(37) \to c=37$$
