The diagonals `AC` and `BD` of a parallelogram `ABCD` intersect each other at the point O. If `angle DAC = 32^(@) and angle AOB =70^(@), " find " angle DBC.`

Options

A. `35^(@)`
B. `38^(@)`
C. `40^(@)`
D. `45^(@)`

Explanation

To solve the problem, we will use the properties of a parallelogram and the relationships between the angles formed by the diagonals.

### Step-by-Step Solution:

1. **Identify the Given Angles:**
- We are given that \( \angle DAC = 32^\circ \) and \( \angle AOB = 70^\circ \).

2. **Use the Property of Parallelogram:**
- In a parallelogram, the diagonals bisect each other. Therefore, \( AO = OC \) and \( BO = OD \).

3. **Find \( \angle AOD \):**
- Since \( \angle AOB = 70^\circ \), and \( \angle AOB \) and \( \angle AOD \) are vertically opposite angles, we have:
\[ \angle AOD = \angle AOB = 70^\circ \]

4. **Find \( \angle DAC + \angle DAB \):**
- In triangle \( AOD \), the sum of angles is \( 180^\circ \):
\[ \angle DAC + \angle DAB + \angle AOD = 180^\circ \]
- Let \( \angle DAB = x \). Therefore, we can write:
\[ 32^\circ + x + 70^\circ = 180^\circ \]
- Simplifying this gives:
\[ x + 102^\circ = 180^\circ \]
- Thus,
\[ x = 180^\circ - 102^\circ = 78^\circ \]
- Therefore, \( \angle DAB = 78^\circ \).

5. **Find \( \angle DBC \):**
- Since \( AB \parallel CD \) (as it is a property of parallelograms), we have:
\[ \angle DAB = \angle DBC \]
- Therefore, \( \angle DBC = 78^\circ \).

### Final Answer:
\[ \angle DBC = 78^\circ \]