P and Q are the middle points of two chords (not diameters) AB and AC respectively of a circle with centre at a point O. The lines OP and OQ are produced to meet the circle respectively at the points R and S. T is any point on the major arc between the points R and S of the circle. If `angle BAC = 32^(@), angle RTS` = ?

Options

A. `32^(@)`
B. `74^(@)`
C. `106^(@)`
D. `64^(@)`

Answer

To solve the problem, we need to find the angle \( \angle RTS \) given that \( \angle BAC = 32^\circ \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a circle with center \( O \).
- Chords \( AB \) and \( AC \) intersect at point \( A \).
- Points \( P \) and \( Q \) are the midpoints of chords \( AB \) and \( AC \), respectively.
- Lines \( OP \) and \( OQ \) are extended to meet the circle at points \( R \) and \( S \).

2. **Identifying Angles**:
- Since \( P \) and \( Q \) are midpoints of the chords, \( OP \) and \( OQ \) are perpendicular to the chords \( AB \) and \( AC \). Therefore, \( \angle OPA = 90^\circ \) and \( \angle OQA = 90^\circ \).
- We know that \( \angle BAC = 32^\circ \).

3. **Using the Inscribed Angle Theorem**:
- The angle \( \angle BAC \) is an inscribed angle that subtends the arc \( BC \).
- The angle at the center \( O \) that subtends the same arc \( BC \) is \( \angle BOC \).
- By the inscribed angle theorem, \( \angle BOC = 2 \times \angle BAC = 2 \times 32^\circ = 64^\circ \).

4. **Finding \( \angle RTS \)**:
- The angles \( \angle RTS \) and \( \angle BOC \) are related because \( T \) is on the major arc \( RS \) which subtends the same arc \( BC \).
- Therefore, \( \angle RTS \) is equal to half of \( \angle BOC \) because \( \angle RTS \) is also an inscribed angle subtending the same arc \( BC \).
- Thus, \( \angle RTS = \frac{1}{2} \times \angle BOC = \frac{1}{2} \times 64^\circ = 32^\circ \).

5. **Conclusion**:
- The value of \( \angle RTS \) is \( 32^\circ \).

### Final Answer:
\[ \angle RTS = 32^\circ \]