In $\triangle ABC$, $\angle B=75^\circ$ and $\angle C=45^\circ$. Let $O$ be the center of the circle through $A$, $B$, and $C$, and let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. What is the degree measure of $\angle OAH$?

1. $\angle A=180^\circ-75^\circ-45^\circ=60^\circ$. The inscribed angle $\angle ABC=75^\circ$ intercepts arc $AC$, so $\angle AOC=150^\circ$. Triangle $AOC$ is isosceles, so $\angle OAC=\dfrac{180^\circ-150^\circ}2=15^\circ$.
2. In the right triangle $AHC$, $\angle HAC=90^\circ-\angle C=45^\circ$.
3. The rays $AO$ and $AH$ both lie inside $\angle BAC$, so $\angle OAH=\angle HAC-\angle OAC=45^\circ-15^\circ=\boxed{30^\circ}$.