16 Sides

Choose any point from $A_{14}$ to $A_6$ (I have chosen $A_2$ for convenience) to make $A_2A_7A_9A_{13}$ a cyclic quadrilateral , so that we can use its properties. Let the center of the circle be $O$ . Then $\angle A_7OA_{13} = \dfrac 6{16} \times 360^\circ = 135^\circ$ . For the same chord the angle it extends at the circumference is half that of the angle at the center. Therefore $\angle A_7A_2A_{13} = \dfrac {135^\circ}2 = 67.5^\circ$ . Since the opposite angles of a cyclic quadrilateral add to $180^\circ$ , $\angle A_7A_9A_{13} = 180^\circ - 67.5^\circ = \boxed{112.5}^\circ$ .

Let $O$ be the center of the circle and draw $A_{13}O$ and $A_{7}O$ .

Then $\angle A_{13}OA_{7}$ (the angle that subtends the major arc) is $\frac{10}{16} \cdot 360° = 225°$ , and since an inscribed angle subtended by two points is half the central angle subtended by the same two points, $\theta = \frac{1}{2} \cdot 225° = \boxed{112.5°}$ .