In triangle PQR, the perpendicular bisectors OX, OY, and OZ meet at O (the circumcenter).\nGiven ∠QPR = 65° and ∠PQR = 60°, find the value (in degrees) of ∠QOR + ∠POR.

Introduction / Context:
This problem uses the circumcenter O of triangle PQR. The perpendicular bisectors of the three sides meet at O. For any arc in a circle, the central angle at O is twice the inscribed angle that subtends the same chord. We will convert given vertex angles into corresponding central angles and then add them appropriately.

Given Data / Assumptions:

• ∠QPR = 65° (angle at P)
• ∠PQR = 60° (angle at Q)
• O is the circumcenter; ∠QOR subtends arc QR; ∠POR subtends arc PR.

Concept / Approach:
In a circumcircle, a central angle equals twice the inscribed angle that subtends the same arc. Therefore, ∠QOR = 2 * ∠QPR and ∠POR = 2 * ∠PQR. Sum them to obtain the target value.

Step-by-Step Solution:

Verification / Alternative check:
Using triangle angle sum, ∠PRQ = 180° − (65° + 60°) = 55°. The three corresponding central angles would be 130°, 120°, and 110°, which sum to 360°, consistent with a full circle. Our selected pair (130° and 120°) adds to 250° as computed.

Why Other Options Are Wrong:

• 180: Would require ∠QPR + ∠PQR = 90°, which is not the case.
• 210 or 125: Do not match 2*(65°) + 2*(60°).

Common Pitfalls:

• Confusing the inscribed angle with the central angle (forgetting the factor 2).
• Using the wrong vertex angle for a given arc.

Final Answer:
250