In $△ABC$, $AB$ and $AC$ are produced to points $D$ and $E$, respectively. If the bisectors of $\angle CBD$ and $\angle BCE$ meet at point $O$, so that $\angle BOC=76^{\circ }$, then find the measure of $\angle A$.

Point $O$ is the ex-center of $△ABC$ formed by the intersection of the external angle bisectors at vertices $B$ and $C$.According to the external angle bisector theorem:
$$\angle BOC=90^{\circ }-\frac{\angle A}{2}$$
Substitute the given value $\angle BOC=76^{\circ }$ into the formula:
$$76^{\circ }=90^{\circ }-\frac{\angle A}{2}$$
$$\frac{\angle A}{2}=90^{\circ }-76^{\circ }=14^{\circ }$$
$$\angle A=14^{\circ }\times 2=28^{\circ }$$