In a `Delta PQR`, the sides PQ and PR and are produced to S and T respectively. Bisectors of `angle SQR and angle QRT` meet at the point O. If `angle P = 66^@`, then what is the value of `angle QOR`?

Options

A. `47^@`
B. `50^@`
C. `57^@`
D. `67^@`

Answer

To find the value of angle QOR in triangle PQR, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Information**:
- We have triangle PQR with angle P = 66°.
- The sides PQ and PR are extended to points S and T respectively.
- The angle bisectors of angles SQR and QRT meet at point O.

2. **Use the Exterior Angle Theorem**:
- According to the exterior angle theorem, the exterior angle (angle SQR) is equal to the sum of the two opposite interior angles (angle P and angle Q).
- Therefore, angle SQR = angle P + angle Q = 66° + β (where β is angle Q).

3. **Similarly for Angle QRT**:
- By the same theorem, angle QRT = angle P + angle R = 66° + α (where α is angle R).

4. **Apply the Angle Sum Property in Triangle PQR**:
- The sum of angles in triangle PQR is 180°.
- Thus, we have:
\[ 66° + α + β = 180° \]
- Rearranging gives us:
\[ α + β = 180° - 66° = 114° \]

5. **Determine Angles at Point O**:
- Since O is the intersection of the angle bisectors, we can express the angles at O:
- Angle OQR = Angle OQS = (66° + β) / 2
- Angle ORQ = Angle ORT = (66° + α) / 2

6. **Sum of Angles in Triangle RQO**:
- In triangle RQO, we apply the angle sum property again:
\[ Angle OQR + Angle ORQ + Angle QOR = 180° \]
- Substituting the expressions we found:
\[ \left(33° + \frac{β}{2}\right) + \left(33° + \frac{α}{2}\right) + Angle QOR = 180° \]

7. **Combine and Simplify**:
- This simplifies to:
\[ 66° + \frac{α + β}{2} + Angle QOR = 180° \]
- We know from step 4 that α + β = 114°, so:
\[ 66° + \frac{114°}{2} + Angle QOR = 180° \]
- This further simplifies to:
\[ 66° + 57° + Angle QOR = 180° \]
- Therefore:
\[ Angle QOR = 180° - 123° = 57° \]

### Final Answer:
The value of angle QOR is **57°**.

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