In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=60^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$

1. Since $M$ and $N$ are midpoints, $\overrightarrow{MN}=\frac12(U+G)-\frac12(P+A)=\frac12\overrightarrow{PU}+\frac12\overrightarrow{AG}=\dfrac{\mathbf u+\mathbf g}2$, where $\mathbf u,\mathbf g$ are the unit vectors from $P$ toward $T$ and from $A$ toward $T$ (as $PU=AG=1$).
2. With $PA$ along the positive $x$-axis, $\mathbf u$ has direction $36^\circ$ and $\mathbf g$ has direction $180^\circ-60^\circ=120^\circ$. The sum of two unit vectors bisects the angle between them, so $MN$ points in the direction $\frac{36^\circ+120^\circ}2=78^\circ$. (The same method gives $\frac{36^\circ+124^\circ}2=80^\circ$ in the problem.)
3. The acute angle between $MN$ and $PA$ is $\boxed{78^\circ}$.