jmc

The sum of the angle measures in a pentagon is $180(5-2)=540$ degrees, so each of the interior angles of regular pentagon $FGHIJ$ has measure $540^{\circ }/5=108^{\circ }$. Specifically, $\angle JFG=108^{\circ }$, so $$\angle AFJ=180^{\circ }-\angle JFG=180^{\circ }-108^{\circ }=72^{\circ }.$$ Similarly, we have $\angle AJF=180^{\circ }-108^{\circ }=72^{\circ }$. Finally, the angles of $△AFJ$ sum to $180^{\circ }$, so $$\angle FAJ=180^{\circ }-\angle AFJ-\angle AJF=180^{\circ }-72^{\circ }-72^{\circ }=\boxed{36^{\circ }}.$$