Construct the following angle with your compass and pencil only

• Draw a ray $\vec{OA}$.
• Place the compass at point $O$ 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 $Q$.
• With the same compass width, place the compass at $Q$ and draw an arc that intersects the first arc at $R$.
• With the compass at $Q$ and then at $R$, draw two arcs that intersect each other above the first arc at point $S$.
• Draw a ray from $O$ through $S$. This ray, $\vec{OS}$, forms a $90^\circ$ angle with $\vec{OA}$ (i.e., $\angle SOA = 90^\circ$).

• Extend the ray $\vec{OA}$ backwards through $O$ to form a straight line. Let $P'$ be a point on this extended line to the left of $O$.
• Place the compass at $O$ and draw an arc that intersects the extended line at $P'$. (This might be the same arc as in Step 1 if it was drawn large enough).
• With the same compass width, place the compass at $P'$ and draw an arc that intersects the initial large arc (centered at $O$) at point $U$.
• Draw a ray from $O$ through $U$. This ray, $\vec{OU}$, forms a $60^\circ$ angle with $\vec{OP'}$ (i.e., $\angle UOP' = 60^\circ$).

• Place the compass at $P'$ and draw an arc inside $\angle UOP'$.
• With the same compass width, place the compass at $U$ and draw another arc that intersects the arc drawn from $P'$. Let this intersection point be $V$.
• Draw a ray from $O$ through $V$. This ray, $\vec{OV}$, bisects $\angle UOP'$. So $\angle VOP' = 30^\circ$.

• Place the compass at $P'$ and draw an arc inside $\angle VOP'$.
• Let $W$ be the point where $\vec{OV}$ intersects the initial large arc (centered at $O$). With the same compass width, place the compass at $W$ and draw another arc that intersects the arc drawn from $P'$. Let this intersection point be $X$.
• Draw a ray from $O$ through $X$. This ray, $\vec{OX}$, bisects $\angle VOP'$. So $\angle XOP' = 15^\circ$.

The angle $\angle XOS$ is the sum of $\angle XOP'$ and $\angle P'OS$. Since $\vec{OP'}$ is the extension of $\vec{OA}$, $\angle P'OS$ is the $90^\circ$ angle constructed in Step 1.

The final answer is $\text{The angle } \angle XOS = 105^\circ$.