jmc

Since the square and the equilateral triangle share a side, all sides of the square are the same length as all sides of the equilateral triangle. Specifically, we have $CD=CB$, and so $△BCD$ is isosceles with equal angles at $B$ and $D$. Let $x$ denote the number of degrees in the measure of each of these two angles (that is, $x^{\circ }=m\angle CDB=m\angle CBD$).

Angle $\angle BCD$ is equal to $90^{\circ }+60^{\circ }=150^{\circ }$ (since a square and an equilateral triangle have interior angles of $90^{\circ }$ and $60^{\circ }$, respectively). Since the sum of angles in a triangle is $180^{\circ }$, we have $$x+x+150=180,$$ giving $x=\boxed{15}$.