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O is the incentre of ∆ PQR and ∠QPR = 50°, then the measure of ∠QOR is :
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- 125°
- 100°
- 130°
- 115°
- 125°
Correct Option: D
As per the given in question , we draw a figure of triangle ABC 
∠QPR = 50°
∴ ∠PQR + ∠PRQ = 180° – 50° = 130°
| ∴ | ∠PQR + | ∠PRQ = 65° | ||
| 2 | 2 |
The point of intersection of internal bisectors of angles is in-centre.
| ∴ ∠OQR = | ∠PQR ...... ( 1 ) | |
| 2 |
| ∠ORQ = | ∠PRQ ...... ( 2 ) | |
| 2 |
In ∆ OQR,
∠OQR + ∠QOR + ∠ORQ = 180°
⇒ ∠QOR = 180° – 65° = 115° { From ( 1 ) + ( 2 ) }
