jmc

The sum of the interior angles of an $n$-sided polygon is $180(n-2)$. For a regular hexagon, the interior angles sum to $180(4)$, so each interior angle has a measure of $\frac{180(4)}{6}=30\cdot 4=120^{\circ }$. Since $\overline{PO}$ and $\overline{PQ}$ are congruent sides of a regular hexagon, $△POQ$ is an isosceles triangle. The two base angles are congruent and sum to a degree measure of $180-120=60^{\circ }$, so each base angle has a measure of $30^{\circ }$. There are now a couple approaches to finishing the problem.

$\text{Approach 1}$: We use the fact that trapezoid $PQLO$ is an isosceles trapezoid to solve for $x$ and $y$. Since $\overline{PO}$ and $\overline{QL}$ are congruent sides of a regular hexagon, trapezoid $PQLO$ is an isosceles trapezoid and the base angles are equal. So we know that $x+30=y$. Since the interior angle of a hexagon is $120^{\circ }$ and $m\angle PQO=30^{\circ }$, we know that $\angle OQL$ is a right angle. The acute angles of a right triangle sum to $90^{\circ }$, so $x+y=90$. Now we can solve for $x$ with $x+(x+30)=90$, which yields $x=30$. The degree measure of $\angle LOQ$ is $\boxed{30^{\circ }}$.

$\text{Approach 2}$: We use the fact that trapezoid $LMNO$ is an isosceles trapezoid to solve for $x$. Since $\overline{NO}$ and $\overline{ML}$ are congruent sides of a regular hexagon, trapezoid $LMNO$ is an isosceles trapezoid and the base angles are equal. The interior angles of a trapezoid sum to $360^{\circ }$, so we have $2z+120+120=360$, which yields $z=60$. Angle $O$ is an interior angle of a hexagon that measure $120^{\circ }$, so $z+x+30=120$. We found that $z=60$, so $x=30$. The degree measure of $\angle LOQ$ is $\boxed{30^{\circ }}$.