Let C be a circle with centre O and PQ be a minor arc on C. Let R and S be two points on the major arc of PQ on C. If PQ subtends an angle 80° at O, find ∠ PRQ and ∠ PSQ, respectively.

Correct — D. PQ is a chord of the circle and it subtends 80° at the centre O.

Rule: the angle a chord subtends at the centre is twice the angle it subtends at any point on the major arc.

R lies on the major arc, so ∠PRQ = 80° ⁄ 2 = 40°
S also lies on the major arc, so the same halving applies
∠PSQ = 80° ⁄ 2 = 40°

Angles in the same segment are equal, so ∠PRQ = ∠PSQ = 40° and 40° → option (d)

Why the others are wrong

• (a)
  40° and 50° — The first angle is right, but R and S sit on the same arc, so nothing in the figure can tell them apart. Angles in the same segment are equal, and 50° would need S moved to a different arc or a different chord.
• (b)
  50° and 80° — 80° is the angle at the centre, not at the circumference. Copying the central angle onto a point of the circle skips the halving the theorem demands, and 50° is not half of anything the stem gives.
• (c)
  40° and 60° — Both points lie on the major arc of PQ, so both angles must come out at 40°. 60° is neither half of 80° nor equal to the 40° the option itself concedes for R.

Two theorems settle every question of this shape.

Angle at the centre = 2 × angle at the circumference, when both stand on the same arc. PQ subtends 80° at O, so any point on the major arc sees PQ at 40°.

Angles in the same segment are equal. R and S both lie on the major arc, so ∠PRQ and ∠PSQ are one angle measured twice.

The minor-arc side behaves differently. A point T on the minor arc would see PQ at 140°, because PRQT is then a cyclic quadrilateral and its opposite angles sum to 180°.

Key facts

• The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circle.
• Angles in the same segment of a circle are equal.
• PQ subtends 80° at O, so ∠PRQ = ∠PSQ = 40° for R and S on the major arc.
• A point on the minor arc would see PQ at 140°, since opposite angles of a cyclic quadrilateral sum to 180°.

Common traps

• Halving one angle and then hunting for a different value for the second.
• Using 80° itself as the inscribed angle.
• Forgetting that the halving holds for points on the major arc, the arc away from the chord.

No directly related past PYQ was found.