The circum-centre of a triangle ABC is O. If `angle BAC = 85 ^(@)`, `angle BCA =75^(@)`, then `angle OAC` is of

Options

A. `70^(@)`
B. `72^(@)`
C. `75^(@)`
D. `74^(@)`

Answer

To find the angle \( OAC \) in triangle \( ABC \) with given angles \( \angle BAC = 85^\circ \) and \( \angle BCA = 75^\circ \), we can follow these steps:

### Step 1: Calculate \( \angle ABC \)
We know that the sum of angles in a triangle is \( 180^\circ \). Therefore, we can find \( \angle ABC \) as follows:

\[ \angle ABC = 180^\circ - \angle BAC - \angle BCA \]
Substituting the given values:

\[ \angle ABC = 180^\circ - 85^\circ - 75^\circ = 20^\circ \]

### Step 2: Understanding the Angles at the Circumcenter
The circumcenter \( O \) of triangle \( ABC \) is the point where the perpendicular bisectors of the sides intersect. The angle \( OAC \) is related to the angles at the vertices of the triangle.

### Step 3: Relate \( \angle OAC \) to \( \angle ABC \)
The angle \( OAC \) is half of the angle \( ABC \) because the angle subtended at the center (which is \( O \)) is twice that at the circumference (which is \( A \)). Therefore, we can express this relationship as:

\[ \angle OAC = \frac{1}{2} \times \angle ABC \]

### Step 4: Substitute the Value of \( \angle ABC \)
Now, substituting the value we found for \( \angle ABC \):

\[ \angle OAC = \frac{1}{2} \times 20^\circ = 10^\circ \]

### Step 5: Conclusion
Thus, the angle \( OAC \) is \( 10^\circ \).

### Final Answer
\[ \angle OAC = 10^\circ \]

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