We are given a circle with center $O$. Points $L$, $M$, and $N$ are on the circumference. We are given that $\angle NLM = 74^\circ$ and $\angle LMN = 39^\circ$. We need to find the measure of $\angle LOM$, which is denoted by $x$.

1. Problem Description

We are given a circle with center $O$. Points $L$, $M$, and $N$ are on the circumference. We are given that $\angle NLM = 74^\circ$ and $\angle LMN = 39^\circ$. We need to find the measure of $\angle LOM$, which is denoted by $x$.

2. Solution Steps

First, we find the measure of $\angle LNM$ in triangle $LMN$. The sum of angles in a triangle is $180^\circ$, so

$$\angle LNM = 180^\circ - \angle NLM - \angle LMN = 180^\circ - 74^\circ - 39^\circ = 180^\circ - 113^\circ = 67^\circ.$$

The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference.

In this case, $\angle LOM$ is the angle at the center subtended by arc $LM$, and $\angle LNM$ is the angle subtended by the same arc at point $N$ on the remaining circumference.

Therefore, we have

$$\angle LOM = 2 \cdot \angle LNM.$$

Substituting the value of $\angle LNM$ we found earlier:

$$\angle LOM = 2 \cdot 67^\circ = 134^\circ.$$

Thus, $x = 134^\circ$.

3. Final Answer

The value of $x$ is $134^\circ$.

The answer is A. $134^\circ$.