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?
 
 A: Only thing to realise here was that $\sin(\pi - x) = \sin (x)$. So $\sin DAC = \sin (\pi - DBC) = \sin DBC$. Also $AD = AB = BC$
Then we apply sine rule on $DAC$ and $DBC$ triangles.
$$\frac{\sin DCA}{ AD} = \frac{\sin DAC}{DC}$$ and $$\frac{\sin CDB}{ BC} = \frac{\sin DBC}{DC}$$ 
So we get angle $\angle DCA = \angle CDB = 30$ (seen from triangle $DOC$ where centre point is $O$) or the required angle as $37 ^\circ$.
A: This is somehow an ugly approach. Nevertheless, you can calculate $BD$ using the law of cosines. Then apply the law of cosines in $BDC$ to calculate $DC$ and use the law of sines to get the unknown angle.
A: 
Though Robert gave a great solution, I post another solution with a picture. Draw a line $\overline{DE}$ parallel to $\overline{AB}$, toward $\overline{BC}$. Then $\triangle DEC$ becomes an equilateral triangle and $\square ABED$ becomes a parallelogram, and in turn rhombus which makes $\triangle BEC$ be isosceles triangle. Thus we have $\angle BED = 74^\circ$,$\angle DEC=60^\circ$, and $\angle BCE = 23^\circ$. Therefore, we have $\angle DCE = 60^\circ - 23^\circ = 37^\circ$. 
A: There is a point $E$  along $DC$ such that $\triangle ABE$ is equilateral (why? See my P.S.). Therefore $$\hat{ADC}=\frac{1}{2}(180^\circ-(74^\circ-60^\circ))=83^\circ\quad\text{and}\quad \hat{BCD}=\frac{1}{2}(180^\circ-(166^\circ-60^\circ))=37^\circ.$$

Note that your picture is misleading because the angle at $D$ should be acute!
P.S. Go backwards. Start form the equilateral triangle $\triangle ABE$. Then draw externally the isosceles triangle $\triangle AED$ with an angle at $A$ of $14^\circ$, and the isosceles triangle $\triangle BEC$ with an angle at $B$ of $106^\circ$. Then $D$, $E$ and $C$ are on the same line because
$$\hat{AED}+\hat{AEB}+\hat{BEC}=83^\circ+60^\circ+37^\circ=180^\circ.$$
A: 
$$AB=AD=BC=x,AC=t\\t^2=x^2+x^2-2x^2\cos166\\\to t^2=2x^2(1-\cos166)=2x^2(2\sin^2(83))=4x^2\cos^2(7)\\t=2x\cos 7$$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
$$\tan(c)=\tan(37) \to c=37$$
