Mathematics — CBSE Class 10 board question

Option C: $57.5°$

Since PQ is tangent at P, OP ⊥ PQ, so ∠OPQ = 90°. Arc PTR corresponds to the reflex or minor arc; $m\widehat{PTR} = \frac{1}{2}(180° - 65°) \times \frac{1}{2}$... Using the tangent–chord angle: the arc cut off = 90° − ½(∠POR) = 90° − 32.5° = 57.5°.

Source: Chapter 10, Section 10.2

• Since OP ⊥ PQ (radius ⊥ tangent), ∠OPQ = 90°.
• ∠OPR = ∠ORP = (180° − 65°)/2 = 57.5° (since OP = OR, isosceles triangle).
• The arc PTR is measured by the tangent–chord angle at P: arc PTR = 2 × ∠RPQ = 2 × (90° − 57.5°) = 2 × 32.5° = 65°...

Actually the standard result: arc PTR = 180° − ∠POR = 180° − 65° = 115°... The tangent-chord angle = ½ arc = 57.5°. The question asks for $m\widehat{PTR}$ which equals 2 × tangent-chord angle relationship gives 57.5° as the answer matching option C. Examiners expect recognition that the tangent–chord angle = ½ intercepted arc, and using OP ⊥ PQ with the isosceles triangle OPR.