The diagonals Ac and BD of a parallelogram intersects at O. If `angleOAD=40^(@),angleOAB=20^(@)andangleCOD=75^(@)`, then evaluate the following,
`angleADC`

Answer

To solve the problem, we need to find the angle \( \angle ADC \) in the parallelogram \( ABCD \) where the diagonals \( AC \) and \( BD \) intersect at point \( O \). We are given:

- \( \angle OAD = 40^\circ \)
- \( \angle OAB = 20^\circ \)
- \( \angle COD = 75^\circ \)

### Step-by-Step Solution:

1. **Identify the Angles**:
- We know that \( \angle OAD = 40^\circ \) and \( \angle OAB = 20^\circ \).
- Therefore, we can find \( \angle BAD \):
\[ \angle BAD = \angle OAD + \angle OAB = 40^\circ + 20^\circ = 60^\circ \]

2. **Using the Property of Parallel Lines**:
- Since \( AB \parallel CD \) and \( AD \) is a transversal, we can use the property that the sum of the interior angles on the same side of the transversal is \( 180^\circ \).
- Thus, we have:
\[ \angle BAD + \angle ADC = 180^\circ \]

3. **Substituting the Known Value**:
- We already calculated \( \angle BAD = 60^\circ \). Now we can substitute this value into the equation:
\[ 60^\circ + \angle ADC = 180^\circ \]

4. **Solving for \( \angle ADC \)**:
- Rearranging the equation gives:
\[ \angle ADC = 180^\circ - 60^\circ = 120^\circ \]

### Final Answer:
\[ \angle ADC = 120^\circ \]