O is the circumcentre of `DeltaABC`, given `/_BAC = 85^@ and /_BCA = 55^@`, find `/_OAC`.

Options

A. `40^@`
B. `50^@`
C. `60^@`
D. `80^@`

Answer

To find the angle \( \angle OAC \) in triangle \( \Delta ABC \) where \( O \) is the circumcenter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Angles**:
- We are given \( \angle BAC = 85^\circ \) and \( \angle BCA = 55^\circ \).

2. **Calculate Angle ABC**:
- We know that the sum of angles in a triangle is \( 180^\circ \). Therefore, we can find \( \angle ABC \) using the formula:
\[ \angle ABC = 180^\circ - \angle BAC - \angle BCA \]
- Substituting the known values:
\[ \angle ABC = 180^\circ - 85^\circ - 55^\circ = 40^\circ \]

3. **Understanding the Circumcenter**:
- The circumcenter \( O \) of triangle \( ABC \) is the point where the perpendicular bisectors of the sides intersect. It is also the center of the circumcircle of the triangle.

4. **Identify Angles in Triangle PAC**:
- In triangle \( PAC \), we need to find \( \angle OAC \). We know that the angle \( \angle APC \) is \( 90^\circ \) because \( O \) lies on the perpendicular bisector.

5. **Apply the Triangle Angle Sum Property**:
- Using the triangle angle sum property in triangle \( PAC \):
\[ \angle PAC + \angle APC + \angle ACB = 180^\circ \]
- Substituting the known angles:
\[ \angle PAC + 90^\circ + 55^\circ = 180^\circ \]

6. **Solve for Angle PAC**:
- Rearranging the equation gives:
\[ \angle PAC = 180^\circ - 90^\circ - 55^\circ = 35^\circ \]

7. **Conclusion**:
- Since \( \angle OAC \) is equal to \( \angle PAC \), we find:
\[ \angle OAC = 35^\circ \]

### Final Answer:
\[ \angle OAC = 35^\circ \]