From a point T, a tangent TP at point P, is drawn to a circle with centre O. A secant TQR (point Q is near to point T) is drawn from the point T. ∆PQR is inscribed into the circle by joining the points P, Q and R. Draw lines OQ and OR. If ∠ PTQ is 27° and ∠ TPQ = 55°, what is the degree measure of ∠ ROQ?

Correct — B. Two circle facts do it: the tangent-chord angle, then the doubling at the centre.

Tangent-chord (alternate segment): ∠TPQ = ∠PRQ, so ∠PRQ = 55°.

Exterior angle of ∆TPQ at Q: ∠PQR = ∠PTQ + ∠TPQ = 27° + 55° = 82°.

In ∆PQR: ∠QPR = 180° − 82° − 55° = 43°.

∠QPR stands on arc QR at the circumference and ∠ROQ stands on the same arc at the centre, so ∠ROQ = 2 × 43° = 86° → option (b).

Why the others are wrong

• (a)
  94 — 94° is 180° − 86°. An inscribed angle and the central angle on the same arc are not supplementary — the central angle is double the inscribed one, and 2 × 43° = 86°.
• (c)
  98 — 98° is the third angle of ∆TPQ, ∠TQP = 180° − 27° − 55°. It lies on the T side of the secant; the angle inside the circle at Q is its supplement, 82°.
• (d)
  82 — 82° is ∠PQR itself, an inscribed angle standing on arc PR. ∠ROQ is a central angle on arc QR, so it doubles ∠QPR, not ∠PQR.

Two theorems carry almost every tangent-and-secant item.

Tangent-chord: the angle between a tangent and a chord at the point of contact equals the inscribed angle in the alternate segment. That is what turns the given 55° at P into 55° at R.

Central angle: an arc subtends at the centre twice what it subtends anywhere on the remaining circumference, so ∠ROQ = 2∠QPR.

The bridge between them is the exterior-angle rule on ∆TPQ, which converts the 27° at the outside point into the 82° at Q.

Key facts

• The tangent-chord angle equals the inscribed angle in the alternate segment.
• An arc subtends at the centre twice the angle it subtends at the circumference.
• Here ∠PQR = ∠PTQ + ∠TPQ = 82°, the exterior angle of triangle TPQ at Q.
• ∠ROQ = 2 × ∠QPR = 86°, and triangle OQR is isosceles with base angles of 47° each.

Common traps

• Using ∠TQP = 98° as the inscribed angle instead of its supplement, ∠PQR = 82°.
• Doubling ∠PQR — the central angle ∠ROQ doubles ∠QPR, the angle at P.
• Reading the tangent-chord angle as the angle in the same segment rather than the alternate one.

No directly related past PYQ was found.