As shown in the figure, two angle bisectors of \(\triangle ABC\),\(\overline {BE}\) and \(\overline {CF}\), intersect at P. If \(\angle EPF= 111^{\circ}\), what is \(\angle A\) in degrees?
\(\text{Let }\angle A = x, \angle PBC = \angle PBF = y, \angle PCB = \angle PCE = z\\ \begin{cases} x + y +z = 111^{\circ}\\ x+2y+2z = 180^{\circ} \end{cases} \implies y + z = 69^{\circ}\\ x + 69^{\circ} = 111^{\circ}\\ x = 42^{\circ}\\ \angle A = 42^{\circ}\)
+586 
