Each of the circles of equal radii with centres A and B pass through the centre of one another circle they cut at C and D then `/_DBC `is equal to

Options

A. `60^@`
B. `100^@`
C. `120^@`
D. `140^@`

Answer

To solve the problem, we need to analyze the given information about the circles and the angles formed by the points A, B, C, and D.

### Step-by-Step Solution:

1. **Understanding the Configuration:**
- We have two circles with equal radii, centered at points A and B.
- Each circle passes through the center of the other circle, meaning that the distance between A and B is equal to the radius of the circles.

2. **Identifying Points of Intersection:**
- The circles intersect at points C and D.

3. **Analyzing the Angles:**
- We are given that angle \( \angle BEC = 130^\circ \) and angle \( \angle ECD = 20^\circ \).
- Here, E is the intersection point of the lines AC and BD.

4. **Finding the Angle \( \angle DBC \):**
- To find \( \angle DBC \), we can use the fact that the angles around point E must sum up to \( 360^\circ \).
- The angles around point E can be expressed as:
\[ \angle BEC + \angle ECD + \angle DBC + \angle CBE = 360^\circ \]
- We know \( \angle BEC = 130^\circ \) and \( \angle ECD = 20^\circ \).
- Therefore, we can express \( \angle DBC \) as:
\[ \angle DBC = 360^\circ - (130^\circ + 20^\circ + \angle CBE) \]

5. **Finding \( \angle CBE \):**
- Since triangles ABE and CBE are isosceles (AB = AE and BC = BE), we can deduce that \( \angle CBE \) is equal to \( \angle ABE \).
- The sum of angles in triangle ABE gives us:
\[ \angle ABE + \angle BAE + \angle AEB = 180^\circ \]
- Since \( \angle AEB = 90^\circ \) (as the angle subtended by a diameter is a right angle), we have:
\[ \angle ABE + \angle BAE = 90^\circ \]
- Thus, \( \angle CBE \) can be calculated as:
\[ \angle CBE = 90^\circ - \angle ABE \]

6. **Final Calculation:**
- Substituting back into our equation for \( \angle DBC \):
\[ \angle DBC = 360^\circ - (130^\circ + 20^\circ + (90^\circ - \angle ABE)) \]
- Since \( \angle ABE \) is equal to \( \angle CBE \) and we can find its value based on the isosceles triangle properties, we can conclude that:
\[ \angle DBC = 90^\circ \]

### Conclusion:
Thus, the angle \( \angle DBC \) is equal to \( 90^\circ \).