In a triangle ABC if `angleA=theta` and perpendiculars drawn from vertices B and C to respective opposite sides meet in P then find the value of `angleBPC` in terms of `theta`.

Answer

To solve the problem, we need to find the angle \( \angle BPC \) in triangle \( ABC \) where \( \angle A = \theta \) and \( P \) is the point where the perpendiculars from vertices \( B \) and \( C \) to their respective opposite sides meet.

### Step-by-Step Solution:

1. **Understanding the Triangle and Angles**:
We have triangle \( ABC \) with \( \angle A = \theta \). The angles at vertices \( B \) and \( C \) can be denoted as \( \angle B \) and \( \angle C \) respectively. According to the triangle angle sum property, we know that:
\[ \angle A + \angle B + \angle C = 180^\circ \]
Therefore, we can express \( \angle B + \angle C \) as:
\[ \angle B + \angle C = 180^\circ - \theta \]

2. **Identifying the Orthocenter**:
The point \( P \) is the orthocenter of triangle \( ABC \), which is the point where the altitudes of the triangle intersect. The altitudes from vertices \( B \) and \( C \) meet at point \( P \).

3. **Finding \( \angle BPC \)**:
In triangle \( ABC \), the angle \( \angle BPC \) can be determined using the property of the orthocenter. The angle \( \angle BPC \) is related to the angles at vertices \( B \) and \( C \):
\[ \angle BPC = 180^\circ - \angle A \]
Since \( \angle A = \theta \), we can substitute this into the equation:
\[ \angle BPC = 180^\circ - \theta \]

4. **Final Result**:
Thus, the value of \( \angle BPC \) in terms of \( \theta \) is:
\[ \angle BPC = 180^\circ - \theta \]