75 degrees

• Draw a ray $\vec{OA}$.
• Place the compass at point $O$ (the vertex) and draw a large arc that intersects $\vec{OA}$ at point $P$.
• With the same compass width, place the compass at $P$ and draw an arc that intersects the first arc at point $Q$.
• Draw a ray from $O$ through $Q$. This ray, $\vec{OQ}$, forms a $60^\circ$ angle with $\vec{OA}$ (i.e., $\angle QOA = 60^\circ$).

• With the same compass width, place the compass at $Q$ and draw an arc that intersects the initial large arc at point $R$. (This creates a $120^\circ$ angle, $\angle ROA$).
• Place the compass at $Q$ and draw an arc above the initial large arc.
• Place the compass at $R$ and draw another arc that intersects the arc drawn from $Q$. Let this intersection point be $S$.
• Draw a ray from $O$ through $S$. This ray, $\vec{OS}$, forms a $90^\circ$ angle with $\vec{OA}$ (i.e., $\angle SOA = 90^\circ$).

The angle between $\vec{OQ}$ ($60^\circ$) and $\vec{OS}$ ($90^\circ$) is $\angle SOQ = 90^\circ - 60^\circ = 30^\circ$.

• Place the compass at $Q$ (where the initial arc intersects $\vec{OQ}$) and draw an arc inside $\angle SOQ$.
• Let $T$ be the point where the initial large arc intersects $\vec{OS}$. Place the compass at $T$ and draw another arc that intersects the arc drawn from $Q$. Let this intersection point be $U$.
• Draw a ray from $O$ through $U$. This ray, $\vec{OU}$, bisects $\angle SOQ$. Therefore, $\angle UOQ = \frac{1}{2} \times 30^\circ = 15^\circ$.

The angle $\angle UOA$ is the sum of $\angle UOQ$ and $\angle QOA$.

The final answer is $\text{The angle } \angle UOA = 75^\circ$.