The point $ O$ is the center of the circle circumscribed about $ \triangle ABC$, with $ \angle BOC \equal{} 120^\circ$ and $ \angle AOB \equal{} 140^\circ$, as shown. What is the degree measure of $ \angle ABC$?
[asy]unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair B=dir(80), A=dir(220), C=dir(320), O=(0,0);

draw(unitcircle);
draw(A--B--C--O--A--C);
draw(O--B);

draw(anglemark(C,O,A,2));

label("$A$",A,SW);
label("$B$",B,NNE);
label("$C$",C,SE);
label("$O$",O,S);
label("$140^{\circ}$",O,NW,fontsize(8pt));
label("$120^{\circ}$",O,ENE,fontsize(8pt));[/asy]$ \textbf{(A)}\ 35 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$
1. Given that \( O \) is the center of the circle circumscribed about \( \triangle ABC \), we know that \( \angle BOC = 120^\circ \) and \( \angle AOB = 140^\circ \).
2. Since \( O \) is the center of the circle, \( \angle BOC \) and \( \angle AOB \) are central angles. The corresponding arcs subtended by these angles are \( \overset{\frown}{BC} \) and \( \overset{\frown}{AB} \) respectively.
3. The measure of the arc \( \overset{\frown}{BC} \) is equal to the measure of the central angle \( \angle BOC \), which is \( 120^\circ \).
4. The measure of the arc \( \overset{\frown}{AB} \) is equal to the measure of the central angle \( \angle AOB \), which is \( 140^\circ \).
5. The measure of the arc \( \overset{\frown}{AC} \) can be found by subtracting the measures of arcs \( \overset{\frown}{BC} \) and \( \overset{\frown}{AB} \) from \( 360^\circ \):
   \[
   \overset{\frown}{AC} = 360^\circ - 120^\circ - 140^\circ = 100^\circ
   \]
6. The inscribed angle \( \angle ABC \) subtends the arc \( \overset{\frown}{AC} \). By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the arc it subtends:
   \[
   \angle ABC = \frac{1}{2} \times \overset{\frown}{AC} = \frac{1}{2} \times 100^\circ = 50^\circ
   \]

The final answer is \(\boxed{50^\circ}\)