Two lines AB and CD intersect at a point O such that `angleBOC+angleAOD`=`280^(@)`, as shown in the figure. Find all the four angles.

Explanation

To solve the problem, we need to find the four angles formed by the intersection of lines AB and CD at point O, given that \( \angle BOC + \angle AOD = 280^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles**:
Let \( \angle BOC = x \) and \( \angle AOD = y \). According to the problem, we have:
\[ x + y = 280^\circ \]

2. **Use the Property of Vertically Opposite Angles**:
Since AB and CD intersect at point O, the angles \( \angle BOC \) and \( \angle AOD \) are vertically opposite angles. Therefore, we have:
\[ x = y \]

3. **Substitute and Solve**:
Substitute \( y \) with \( x \) in the equation from step 1:
\[ x + x = 280^\circ \]
This simplifies to:
\[ 2x = 280^\circ \]
Dividing both sides by 2 gives:
\[ x = 140^\circ \]
Thus, we find:
\[ \angle BOC = 140^\circ \quad \text{and} \quad \angle AOD = 140^\circ \]

4. **Find the Remaining Angles**:
The other two angles \( \angle AOC \) and \( \angle BOD \) are also vertically opposite angles. Therefore:
\[ \angle AOC = \angle BOD \]
Since \( \angle AOC + \angle BOC = 180^\circ \) (as they are on a straight line), we can write:
\[ \angle AOC + 140^\circ = 180^\circ \]
Thus:
\[ \angle AOC = 180^\circ - 140^\circ = 40^\circ \]
Therefore, we also have:
\[ \angle BOD = 40^\circ \]

### Summary of Angles:
- \( \angle BOC = 140^\circ \)
- \( \angle AOD = 140^\circ \)
- \( \angle AOC = 40^\circ \)
- \( \angle BOD = 40^\circ \)