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

Answer

To solve the problem, we will follow these steps:

1. **Identify the angles in the parallelogram**: We are given the angles \( \angle OAD = 40^\circ \), \( \angle OAB = 20^\circ \), and \( \angle COD = 75^\circ \).

2. **Find \( \angle BAD \)**:
- Since \( \angle BAD = \angle OAD + \angle OAB \):
\[ \angle BAD = 40^\circ + 20^\circ = 60^\circ \]

3. **Find \( \angle AOD \)**:
- We know that the sum of angles around point O is \( 360^\circ \). Therefore, \( \angle AOD \) can be calculated as:
\[ \angle AOD = 180^\circ - \angle COD = 180^\circ - 75^\circ = 105^\circ \]

4. **Find \( \angle ADO \)**:
- In triangle \( AOD \), the sum of angles is \( 180^\circ \). Thus:
\[ \angle ADO + \angle OAD + \angle AOD = 180^\circ \]
\[ \angle ADO + 40^\circ + 105^\circ = 180^\circ \]
\[ \angle ADO = 180^\circ - 145^\circ = 35^\circ \]

5. **Find \( \angle ADC \)**:
- Since \( \angle ADC = \angle ADO \) (because opposite angles in a parallelogram are equal):
\[ \angle ADC = 35^\circ \]

6. **Find \( \angle BDC \)**:
- In parallelogram \( ABCD \), the sum of adjacent angles is \( 180^\circ \):
\[ \angle BAD + \angle ADC + \angle BDC = 180^\circ \]
\[ 60^\circ + 35^\circ + \angle BDC = 180^\circ \]
\[ \angle BDC = 180^\circ - 95^\circ = 85^\circ \]

Thus, the final answer is:
\[ \angle BDC = 85^\circ \]