In a `Delta ABC, angleBOC=130^(@)`, if O is orthocentre . Find `angle A`.

Answer

To solve the problem, we need to find the value of angle A in triangle ABC, given that angle BOC = 130° and O is the orthocenter of the triangle.

### Step-by-Step Solution:

1. **Understanding the Orthocenter**:
The orthocenter (O) of a triangle is the point where all three altitudes intersect. In triangle ABC, the altitudes from vertices A, B, and C meet at point O.

2. **Identifying Angles**:
Since O is the orthocenter, angles BOC and AOB are related to angles A and B in triangle ABC. The angles at O are right angles because they are formed by the altitudes from A and C to line BC (angle AOB and angle AOC are both 90°).

3. **Using Vertically Opposite Angles**:
Angle BOC = 130°. By the property of vertically opposite angles, angle AOP (where P is the foot of the altitude from A) is also equal to 130°.

4. **Setting Up the Quadrilateral**:
Consider quadrilateral AOPQ (where Q is the foot of the altitude from B). The sum of the interior angles of a quadrilateral is 360°. Therefore, we can write:
\[ \text{Angle A} + \text{Angle O} + \text{Angle P} + \text{Angle Q} = 360° \]
Here, angle O (angle AOB) = 90° and angle Q (angle BOC) = 90°.

5. **Substituting Known Values**:
We know:
- Angle O = 90°
- Angle P (angle BOC) = 130°
- Angle Q = 90°

Thus, we can substitute these values into the equation:
\[ \text{Angle A} + 90° + 90° + 130° = 360° \]

6. **Simplifying the Equation**:
Combine the known angles:
\[ \text{Angle A} + 310° = 360° \]

7. **Solving for Angle A**:
Now, isolate angle A:
\[ \text{Angle A} = 360° - 310° = 50° \]

### Final Answer:
Angle A = 50°.