The sides AB and AC of ABC have been produced to D and E reapectively. The bisectors of `/_CBD` and `/_BCE` meet at O. If `/_A=50^(@)`, find `/_BOC`.

Options

A. `50^(@)`
B. `55^(@)`
C. `60^(@)`
D. `65^(@)`

Answer

To find the angle \( \angle BOC \) in triangle \( ABC \) where \( \angle A = 50^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Triangle**: Start by sketching triangle \( ABC \) with \( A \) at the top, \( B \) on the left, and \( C \) on the right. Extend side \( AB \) to point \( D \) and side \( AC \) to point \( E \).

2. **Label the Angles**: Mark \( \angle A = 50^\circ \). Let \( \angle B = P \) and \( \angle C = Q \).

3. **Use the Angle Sum Property**: According to the angle sum property of triangles, we have:
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the known value:
\[ 50^\circ + P + Q = 180^\circ \]
This simplifies to:
\[ P + Q = 130^\circ \quad \text{(Equation 1)} \]

4. **Identify Exterior Angles**: The angle \( \angle CBD \) is an exterior angle for triangle \( ABC \):
\[ \angle CBD = \angle A + \angle C = 50^\circ + Q \]
Since \( \angle CBD \) is bisected by line \( BO \), we have:
\[ \angle CBD = 2x \quad \text{where } x = \text{the angle on either side of the bisector} \]
Thus:
\[ 2x = 50^\circ + Q \quad \Rightarrow \quad x = \frac{1}{2}(50^\circ + Q) \]

5. **Similarly for \( \angle BCE \)**: The angle \( \angle BCE \) is also an exterior angle:
\[ \angle BCE = \angle A + \angle B = 50^\circ + P \]
This angle is bisected by line \( CO \):
\[ 2y = 50^\circ + P \quad \Rightarrow \quad y = \frac{1}{2}(50^\circ + P) \]

6. **Find \( \angle BOC \)**: In triangle \( BOC \), we have:
\[ \angle BOC + \angle OBC + \angle BCO = 180^\circ \]
Substituting \( OBC = x \) and \( BCO = y \):
\[ \angle BOC + x + y = 180^\circ \]
Therefore:
\[ \angle BOC = 180^\circ - (x + y) \]

7. **Substitute Values of \( x \) and \( y \)**:
\[ \angle BOC = 180^\circ - \left(\frac{1}{2}(50^\circ + Q) + \frac{1}{2}(50^\circ + P)\right) \]
Combining the terms inside the parentheses:
\[ \angle BOC = 180^\circ - \frac{1}{2}(100^\circ + P + Q) \]
From Equation 1, we know \( P + Q = 130^\circ \):
\[ \angle BOC = 180^\circ - \frac{1}{2}(100^\circ + 130^\circ) \]
Simplifying gives:
\[ \angle BOC = 180^\circ - \frac{1}{2}(230^\circ) = 180^\circ - 115^\circ = 65^\circ \]

### Final Answer:
\[ \angle BOC = 65^\circ \]