https://file.aiquickdraw.com/mathbot/file/17261594811324w8cnknf.jpg

The image depicts a geometric configuration involving a circle, angles, and labeled points \( A \), \( B \), \( C \), \( D \), and \( E \). Here's a breakdown of the problem from what I can observe:

- Angle \( \angle OAB = 56^\circ \) (where \( O \) is the center of the circle, and \( AB \) is a chord).
- Angle \( \angle EDC = 21^\circ \) (where \( D \) and \( E \) are points on the circle and line respectively).
- \( \theta \) is the angle at \( C \) outside the circle.
- The circle appears to be inscribed in the configuration.

To solve for \( \theta \), we can use properties of circles, particularly the *angles in a semicircle* theorem, *exterior angles*, and possibly *cyclic quadrilateral* properties.

### Possible Approach
- Use the fact that angle subtended by a chord at the center is twice the angle subtended at the circumference.
- Use the fact that the exterior angle \( \theta \) of a triangle is equal to the sum of the two opposite interior angles.

Would you like me to proceed with a detailed calculation of \( \theta \)? Let me know if you have any other specific questions!

### Related Questions:
1. How can we use the angle at the center of a circle to solve for other angles?
2. What is the exterior angle theorem, and how does it apply to this scenario?
3. How do the angles in a cyclic quadrilateral relate to each other?
4. Can we use any other properties of chords and tangents to simplify this problem?
5. How would the solution change if the points \( D \) and \( E \) were positioned differently?

**Tip:** The angle subtended by a diameter at any point on the circle is always \( 90^\circ \).

In the given geometric diagram, angle OAB is 56°, angle EDC is 21°, and angle θ is at point C. How can we solve for θ using geometric properties?

Learn how to solve for angle θ in a circle geometry problem involving circle theorems and exterior angle properties. Given angles OAB = 56° and EDC = 21°, find θ using step-by-step reasoning. Suitable for students in Grades 9-11.

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