The answer is 112.5.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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 $180^\circ$ , $\angle A_7A_9A_{13} = 180^\circ - 67.5^\circ = \boxed{112.5}^\circ$ .