The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If `angleDAC= 32^@ and angleAOB= 70^@, " then "angleDBC` is equal to

Options

A. `24^@`
B. `86^@`
C. `38^@`
D. `32^@`

Answer

To find the angle \( \angle DBC \) in the parallelogram ABCD where the diagonals AC and BD intersect at point O, we will follow these steps:

### Step 1: Understand the given information
We know:
- \( \angle DAC = 32^\circ \)
- \( \angle AOB = 70^\circ \)

### Step 2: Find \( \angle AOD \)
Since \( AOB \) and \( AOD \) are angles around point O, we can use the property that the sum of angles around a point is \( 360^\circ \). However, since we are interested in the linear pair formed by \( \angle AOB \) and \( \angle AOD \):
\[ \angle AOD + \angle AOB = 180^\circ \]
Substituting the known value:
\[ \angle AOD + 70^\circ = 180^\circ \]
\[ \angle AOD = 180^\circ - 70^\circ = 110^\circ \]

### Step 3: Find \( \angle ADO \)
In triangle AOD, we can find \( \angle ADO \) using the sum of angles in a triangle:
\[ \angle AOD + \angle DAC + \angle ADO = 180^\circ \]
Substituting the known values:
\[ 110^\circ + 32^\circ + \angle ADO = 180^\circ \]
\[ \angle ADO = 180^\circ - 110^\circ - 32^\circ = 38^\circ \]

### Step 4: Find \( \angle DAB \)
Since ABCD is a parallelogram, opposite angles are equal. Therefore:
\[ \angle DAB = \angle ADO = 38^\circ \]

### Step 5: Find \( \angle DBC \)
Since AD is parallel to BC, we can use the property of alternate interior angles:
\[ \angle DAB = \angle DBC \]
Thus:
\[ \angle DBC = 38^\circ \]

### Final Answer
Therefore, \( \angle DBC \) is equal to \( 38^\circ \).

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