Given m( ABC) = 60^ and m( ACB) = 40^:

B. $160^\circ$

Step 1: Find the measure of angle BAC. The sum of angles in a triangle is $180^\circ$. In $\triangle ABC$: $m(\angle BAC) + m(\angle ABC) + m(\angle ACB) = 180^\circ$ Given $m(\angle ABC) = 60^\circ$ and $m(\angle ACB) = 40^\circ$: $m(\angle BAC) + 60^\circ + 40^\circ = 180^\circ$ $m(\angle BAC) + 100^\circ = 180^\circ$ $m(\angle BAC) = 180^\circ - 100^\circ$ $m(\angle BAC) = 80^\circ$

Step 2: Relate the inscribed angle BAC to the central angle BOC. The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle. In this case, $\angle BOC$ is the central angle subtended by arc BC, and $\angle BAC$ is the inscribed angle subtended by the same arc BC. Therefore: $m(\angle BOC) = 2 \times m(\angle BAC)$ $m(\angle BOC) = 2 \times 80^\circ$ $m(\angle BOC) = 160^\circ$

The final answer is $\boxed{160^\circ}$.

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