A triangle ABC is inscribed in a circle with center O. Lines from vertex B and C are extended to meet O. Find `angleBOC`. Given `angleABC = 60^@ and angleBCA = 65^@` .
(a)`200^@`
(b)`110^@`
(c)`120^@`
(d)`55^@`

Options

A. `200^@`
B. `110^@`
C. `120^@`
D. `55^@`

Answer

To find the angle \( \angle BOC \) in the triangle \( ABC \) inscribed in a circle with center \( O \), we can follow these steps:

### Step 1: Identify the angles given in the triangle
We are given:
- \( \angle ABC = 60^\circ \)
- \( \angle BCA = 65^\circ \)

### Step 2: Calculate the third angle \( \angle A \)
Using the property that the sum of angles in a triangle is \( 180^\circ \):
\[ \angle A = 180^\circ - \angle ABC - \angle BCA \]
Substituting the known values:
\[ \angle A = 180^\circ - 60^\circ - 65^\circ = 180^\circ - 125^\circ = 55^\circ \]

### Step 3: Use the inscribed angle theorem
According to the inscribed angle theorem, the angle at the center of the circle \( \angle BOC \) is twice the angle at the circumference \( \angle A \):
\[ \angle BOC = 2 \times \angle A \]
Substituting the value of \( \angle A \):
\[ \angle BOC = 2 \times 55^\circ = 110^\circ \]

### Conclusion
Thus, the measure of \( \angle BOC \) is \( 110^\circ \).

### Answer
(b) \( 110^\circ \)

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