Mark against the correct answer:
In an isosceles `Delta` ABC, the bisectors of `/_B` and `/_C` meet at a point O. If `/_A = 40 ^(@)`, then `/_BOC=?`

Options

A. `110^(@)`
B. `70^(@)`
C. `130^(@)`
D. `150^(@)`

Answer

To solve the problem step by step, we will follow the properties of isosceles triangles and the concept of angle bisectors.

### Step-by-Step Solution:

1. **Identify the Triangle Properties**:
Since triangle ABC is isosceles, we know that two of its angles are equal. Let's denote the angles as follows:
- Angle A = 40°
- Angle B = Angle C (since it's isosceles)

2. **Use the Triangle Sum Property**:
The sum of the angles in any triangle is 180°. Therefore, we can write the equation:
\[ \text{Angle A} + \text{Angle B} + \text{Angle C} = 180° \]
Substituting the known values:
\[ 40° + \text{Angle B} + \text{Angle B} = 180° \]

3. **Simplify the Equation**:
Since Angle B = Angle C, we can replace Angle C with Angle B:
\[ 40° + 2 \times \text{Angle B} = 180° \]

4. **Isolate Angle B**:
Rearranging the equation gives:
\[ 2 \times \text{Angle B} = 180° - 40° \]
\[ 2 \times \text{Angle B} = 140° \]

5. **Solve for Angle B**:
Dividing both sides by 2:
\[ \text{Angle B} = \frac{140°}{2} = 70° \]
Thus, Angle B = 70° and Angle C = 70°.

6. **Determine the Angles at Point O**:
The angle bisectors of Angle B and Angle C meet at point O. Therefore, each bisector divides the respective angles into two equal parts:
- Angle B bisector = \(\frac{70°}{2} = 35°\)
- Angle C bisector = \(\frac{70°}{2} = 35°\)

7. **Calculate Angle BOC**:
Now, we can find Angle BOC using the triangle sum property again. In triangle BOC:
\[ \text{Angle B} + \text{Angle C} + \text{Angle BOC} = 180° \]
Substituting the known values:
\[ 35° + 35° + \text{Angle BOC} = 180° \]
\[ 70° + \text{Angle BOC} = 180° \]

8. **Isolate Angle BOC**:
Rearranging gives:
\[ \text{Angle BOC} = 180° - 70° = 110° \]

### Final Answer:
Thus, the value of Angle BOC is \(110°\).

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