Points P, Q and Rare on a circle such that `/_PQR=40^@ and /_QRP = 60^@`. Then the subtended angle by arc QR at the centre is :

Options

A. `80^@`
B. `120^@`
C. `140^@`
D. `160^@`

Answer

To solve the problem, we need to find the angle subtended by arc QR at the center of the circle, given that angles PQR and QRP are 40° and 60°, respectively.

### Step-by-Step Solution:

1. **Identify the Angles in Triangle PQR:**
- We know that in triangle PQR, the angles are given as:
- Angle PQR = 40°
- Angle QRP = 60°
- We need to find angle P (∠P).

2. **Use the Angle Sum Property of a Triangle:**
- The sum of angles in a triangle is always 180°.
- Therefore, we can write:
\[ \text{Angle P} + \text{Angle Q} + \text{Angle R} = 180° \]
- Substituting the known angles:
\[ \text{Angle P} + 40° + 60° = 180° \]
- Simplifying this gives:
\[ \text{Angle P} + 100° = 180° \]
- Thus, we find:
\[ \text{Angle P} = 180° - 100° = 80° \]

3. **Relate the Angle at the Center to the Angle at the Circumference:**
- According to the property of angles in a circle, the angle subtended at the center (∠QOR) is twice the angle subtended at the circumference (∠P).
- Therefore, we can express this relationship as:
\[ \text{Angle QOR} = 2 \times \text{Angle P} \]

4. **Calculate the Angle QOR:**
- Now substituting the value of angle P:
\[ \text{Angle QOR} = 2 \times 80° = 160° \]

5. **Final Answer:**
- The angle subtended by arc QR at the center is **160°**.