Is there a way to solve for the missing angle?

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 $\angle DCE = 60^\circ - 23^\circ = 37^\circ$.

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 $$\hat{AED}+\hat{AEB}+\hat{BEC}=83^\circ+60^\circ+37^\circ=180^\circ.$$