In the given figure, we are tasked with calculating two angles: $\angle ABC$ and $\angle ADC$, given that $\angle AOC = 130^\circ$.

Solution:

1. Angle at the center and angle at the circumference subtended by the same arc:

   • In a circle, the angle subtended at the center is twice the angle subtended at the circumference by the same arc. This is known as the angle at the center theorem.
   • Arc $AC$ subtends the central angle $\angle AOC = 130^\circ$.
   • Therefore, the angle subtended by the same arc $AC$ at point $B$ on the circumference (i.e., $\angle ABC$) is: $\angle ABC = \frac{1}{2} \times \angle AOC = \frac{1}{2} \times 130^\circ = 65^\circ$

2. Opposite angles in a cyclic quadrilateral:

   • Quadrilateral $ABCD$ is a cyclic quadrilateral (all vertices lie on the circumference of the circle).
   • The opposite angles in a cyclic quadrilateral sum to $180^\circ$.
   • Therefore, $\angle ABC + \angle ADC = 180^\circ$.
   • We already found that $\angle ABC = 65^\circ$, so: $\angle ADC = 180^\circ - 65^\circ = 115^\circ$

Final Answer:

• $\angle ABC = 65^\circ$
• $\angle ADC = 115^\circ$

These results follow directly from the properties of circles and cyclic quadrilaterals.

Would you like further clarification on any of the steps or explanations?

Here are 5 related questions to further explore this topic:

1. What is the angle subtended at the circumference by a semicircle?
2. Can all quadrilaterals be cyclic? Why or why not?
3. How do you prove that opposite angles in a cyclic quadrilateral are supplementary?
4. What happens if the central angle is a right angle? What can we say about the arc?
5. How can the inscribed angle theorem be applied to real-world problems involving circles?

Tip: Always remember that in a cyclic quadrilateral, the sum of opposite angles is 180°, a crucial property that simplifies solving many geometric problems in circles.