In a quadrilateral ABCD, OA and OB are the angle bisectors of `angleDAB" and " angleCBA`. If `angleADC=70^(@)" and "angleBCD=80^(@)` then find `angleAOB` ?

Options

A. `70^(@)`
B. `80^(@)`
C. `90^(@)`
D. `75^(@)`

Answer

To find the angle \( \angle AOB \) in the quadrilateral \( ABCD \) where \( OA \) and \( OB \) are the angle bisectors of \( \angle DAB \) and \( \angle CBA \) respectively, we can follow these steps:

### Step 1: Understand the angles in the quadrilateral
In any quadrilateral, the sum of the interior angles is \( 360^\circ \). We know:
- \( \angle ADC = 70^\circ \)
- \( \angle BCD = 80^\circ \)

### Step 2: Calculate the sum of the remaining angles
Let \( \angle DAB = x \) and \( \angle CBA = y \). According to the properties of quadrilaterals, we have:
\[ x + y + 70^\circ + 80^\circ = 360^\circ \]
This simplifies to:
\[ x + y + 150^\circ = 360^\circ \]
Thus,
\[ x + y = 360^\circ - 150^\circ = 210^\circ \]

### Step 3: Use the angle bisector property
Since \( OA \) and \( OB \) are the angle bisectors, we have:
\[ \angle DAB = x \quad \text{and} \quad \angle CBA = y \]
The angle bisectors divide these angles into two equal parts:
\[ \angle OAD = \frac{x}{2} \quad \text{and} \quad \angle OBA = \frac{y}{2} \]

### Step 4: Relate the angles to triangle \( AOB \)
In triangle \( AOB \), we know that:
\[ \angle AOB + \angle OAD + \angle OBA = 180^\circ \]
Substituting the values:
\[ \angle AOB + \frac{x}{2} + \frac{y}{2} = 180^\circ \]
Since \( x + y = 210^\circ \), we can substitute:
\[ \angle AOB + \frac{210^\circ}{2} = 180^\circ \]
This simplifies to:
\[ \angle AOB + 105^\circ = 180^\circ \]
Thus,
\[ \angle AOB = 180^\circ - 105^\circ = 75^\circ \]

### Final Answer
Therefore, the angle \( \angle AOB \) is \( 75^\circ \).

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