Triangle PQR with incentre O and point S on side QR such that OS is perpendicular to QR. Angle QOS = 15 degrees.

In a triangle where O is the incentre, the angle $\angle QOS = 90°- \frac{\angle P}{2}$ is not directly applicable because O is the incentre, but rather $\angle OQS = \frac{\angle PQR}{2}$. Given $\triangle OQS$ is right-angled at S, $\angle OQS = 90°- 15°= 75°$. Since $OQ$ bisects $\angle PQR$, $\angle PQR = 2 \times 75°= 150°$.

Students often confuse the angle subtended by the incentre with the angle subtended by the circumcentre or orthocentre, leading to incorrect angle relationships.

Similar questions often appear in exams involving circumcentres or orthocentres where property-based angle relationships (like $90°+ A/2$ or $180°- A$) are required.