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The inscribed angle ABC should be half of the intercepted arc.
If angle AOC = 100, then intercepted arc is 360 - 100 = 260
Angle ABC is 1/2 * 260 = 130.

Answer D
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We are given that O is the center of the circle, and A , B , C , and O form a quadrilateral inside the circle. We are also given \angle ACO = 40^\circ , and we are trying to find \angle ABC .

Reanalysis of the problem:

1. Angle ACO :
Given \angle ACO = 40^\circ .
2. Central angle relationship:
O is the center of the circle, and OA and OC are radii of the circle. So, triangle OAC is isosceles. But in this problem, the focus is on the angles related to the quadrilateral ABCO .
3. Cyclic Quadrilateral Property:
The quadrilateral ABCO is inscribed in the circle. One key property of cyclic quadrilaterals is that opposite angles are supplementary (their sum is 180^\circ ).
4. Finding \angle ABC :
You are given that \angle ACO = 40^\circ . Since \angle ABC and \angle ACO are opposite angles in the cyclic quadrilateral, they must add up to 180^\circ :

\angle ABC + \angle ACO = 180^\circ

Therefore:

\angle ABC = 180^\circ - 40^\circ = 140^\circ


Conclusion:

So, the correct measure of \angle ABC is 140 degrees, based on the cyclic quadrilateral property, not 130 degrees.
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