If O is the circum-centre in `DeltaPQR and /_QOR = 40^@`, then what is the value (in degrees) of `/_QPR`?

Options

A. 40
B. 60
C. 80
D. 20

Answer

To find the value of angle \( \angle QPR \) in triangle \( \Delta PQR \) given that \( O \) is the circumcenter and \( \angle QOR = 40^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circumcenter**:
- The circumcenter \( O \) of a triangle is the point where the perpendicular bisectors of the sides intersect. It is also the center of the circumcircle, which means all vertices of the triangle lie on this circle.

2. **Identify the Given Angle**:
- We are given that \( \angle QOR = 40^\circ \).

3. **Using the Inscribed Angle Theorem**:
- According to the inscribed angle theorem, the angle subtended at the center of the circle (in this case, \( \angle QOR \)) is twice the angle subtended at any point on the circumference (in this case, \( \angle QPR \)).
- Therefore, we can express this relationship mathematically as:
\[ \angle QPR = \frac{1}{2} \angle QOR \]

4. **Substituting the Known Value**:
- Now, substituting the value of \( \angle QOR \):
\[ \angle QPR = \frac{1}{2} \times 40^\circ = 20^\circ \]

5. **Final Answer**:
- Thus, the value of \( \angle QPR \) is \( 20^\circ \).

### Summary:
The value of \( \angle QPR \) is \( 20^\circ \).