In a `Delta ABC, angle A=65^(@), angle C=75^(@)`, where O is circumcentre , find `angle OAC`.

Answer

To find the angle \( OAC \) in triangle \( ABC \) where \( \angle A = 65^\circ \) and \( \angle C = 75^\circ \), we will follow these steps:

### Step 1: Calculate \( \angle B \)
The sum of the angles in a triangle is always \( 180^\circ \). Therefore, we can find \( \angle B \) using the formula:
\[ \angle B = 180^\circ - \angle A - \angle C \]
Substituting the known values:
\[ \angle B = 180^\circ - 65^\circ - 75^\circ = 40^\circ \]

### Step 2: Use the Angle at the Center Theorem
The angle at the center of a circle is twice the angle at the circumference subtended by the same arc. In our case, the angle \( AOC \) at the center \( O \) is twice the angle \( ABC \):
\[ \angle AOC = 2 \times \angle B = 2 \times 40^\circ = 80^\circ \]

### Step 3: Determine \( OAC \) and \( OCA \)
Since \( O \) is the circumcenter, the segments \( OA \) and \( OC \) are equal (they are both radii of the circumcircle). In triangle \( AOC \), we have:
- \( \angle AOC = 80^\circ \)
- \( OA = OC \)

### Step 4: Apply the Isosceles Triangle Property
In triangle \( AOC \), since \( OA = OC \), the angles opposite these sides are equal:
\[ \angle OAC = \angle OCA \]
Let \( \angle OAC = \angle OCA = x \).

### Step 5: Set Up the Equation
The sum of angles in triangle \( AOC \) gives us:
\[ \angle OAC + \angle OCA + \angle AOC = 180^\circ \]
Substituting the known values:
\[ x + x + 80^\circ = 180^\circ \]
This simplifies to:
\[ 2x + 80^\circ = 180^\circ \]

### Step 6: Solve for \( x \)
Subtract \( 80^\circ \) from both sides:
\[ 2x = 180^\circ - 80^\circ = 100^\circ \]
Now divide by 2:
\[ x = \frac{100^\circ}{2} = 50^\circ \]

### Conclusion
Thus, the angle \( OAC \) is:
\[ \angle OAC = 50^\circ \]