If `DeltaABC,angleA=52^@`, Its sides AB and AC are produced to the points D and E respectively. If the bisectors of the `angleCBD and angleBCE` meet at point O, then `angleBOC` is equal to:

Options

A. `106^@`
B. `64^@`
C. `16^@`
D. `32^@`

Answer

To solve the problem, we need to find the angle \( \angle BOC \) given that \( \angle A = 52^\circ \) in triangle \( ABC \) and the sides \( AB \) and \( AC \) are extended to points \( D \) and \( E \) respectively.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- We know that \( \angle A = 52^\circ \).
- Let \( \angle B = \angle 1 \) and \( \angle C = \angle 2 \).

2. **Using the Exterior Angle Theorem**:
- The exterior angle \( \angle CBD \) (which is \( \angle 1 + \angle 2 \)) is equal to the sum of the opposite interior angles:
\[ \angle CBD = \angle A + \angle 2 = 52^\circ + \angle 2 \]
- Similarly, for \( \angle BCE \):
\[ \angle BCE = \angle A + \angle 1 = 52^\circ + \angle 1 \]

3. **Setting Up the Angles at Point O**:
- Let \( \angle BOC = x \) and \( \angle OBC = y \).
- Since \( O \) is the intersection of the angle bisectors, we have:
\[ \angle OBC = \frac{1}{2} \angle CBD = \frac{1}{2} (52^\circ + \angle 2) = \frac{52^\circ + \angle 2}{2} \]
\[ \angle BCO = \frac{1}{2} \angle BCE = \frac{1}{2} (52^\circ + \angle 1) = \frac{52^\circ + \angle 1}{2} \]

4. **Using the Triangle Sum Property**:
- In triangle \( ABC \), the sum of angles is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
\[ 52^\circ + \angle 1 + \angle 2 = 180^\circ \]
\[ \angle 1 + \angle 2 = 128^\circ \]

5. **Finding \( \angle BOC \)**:
- Now we can express \( \angle BOC \) in terms of \( x \) and \( y \):
\[ \angle OBC + \angle BCO + \angle BOC = 180^\circ \]
\[ \frac{52^\circ + \angle 2}{2} + \frac{52^\circ + \angle 1}{2} + x = 180^\circ \]
- Substituting \( \angle 1 + \angle 2 = 128^\circ \):
\[ \frac{52^\circ + \angle 2 + 52^\circ + \angle 1}{2} + x = 180^\circ \]
\[ \frac{104^\circ + 128^\circ}{2} + x = 180^\circ \]
\[ \frac{232^\circ}{2} + x = 180^\circ \]
\[ 116^\circ + x = 180^\circ \]
\[ x = 180^\circ - 116^\circ = 64^\circ \]

Thus, \( \angle BOC = 64^\circ \).