A, B, C, D are four points on a circle. AC and BD intersect at a point E such that `/_BEC = 130^@ and /_ECD = 20^@. /_BAC` is

Options

A. (A) 120
B. (B)`90^@`
C. (C)`100^@`
D. (D)`110^@`

Answer

To solve the problem, we need to find the angle \( \angle BAC \) given the angles \( \angle BEC = 130^\circ \) and \( \angle ECD = 20^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles**: We are given:
- \( \angle BEC = 130^\circ \)
- \( \angle ECD = 20^\circ \)

2. **Use the Property of Angles in a Circle**: We know that the angles formed by two chords intersecting inside a circle can be related. Specifically, the angle formed at the intersection point (in this case, \( \angle BEC \)) is equal to half the sum of the opposite angles.

3. **Calculate \( \angle CEB \)**:
- Since \( \angle ECD = 20^\circ \), we can find \( \angle CEB \) using the property of angles around point E:
\[ \angle CEB = 180^\circ - \angle ECD - \angle BEC \]
\[ \angle CEB = 180^\circ - 20^\circ - 130^\circ = 30^\circ \]

4. **Relate \( \angle BAC \) to \( \angle CEB \)**:
- According to the inscribed angle theorem, \( \angle BAC \) is equal to half of \( \angle CEB \):
\[ \angle BAC = \frac{1}{2} \times \angle CEB = \frac{1}{2} \times 30^\circ = 15^\circ \]

5. **Final Answer**: Thus, the angle \( \angle BAC \) is \( 15^\circ \).

### Summary:
- \( \angle BAC = 15^\circ \)