In a circle, chord AB subtends an angle of 60° at the center. What is the angle subtended by the same chord AB at any point C on the major arc (opposite side of the center)?

Correct — A. With C on the major arc, ∠ACB stands on the minor arc AB, the same arc as the angle at the centre.

Given at the centre: ∠AOB = 60°
Rule: angle at the circumference = ½ × angle at the centre on the same arc
Halve: ∠ACB = ½ × 60° = 30° → option (a)

Why the others are wrong

• (b)
  60° — 60° is the angle at the centre, which the stem gives. A point on the circle sees the same arc at half that angle, 30°.
• (c)
  90° — 90° is the angle in a semicircle. It needs AB to be a diameter, subtending 180° at the centre. Here AB subtends 60°.
• (d)
  120° — 120° is 2 × 60°, the rule run backwards. Doubling takes an angle from the circle to the centre. From the centre to C you halve, giving 30°.

An angle at the circumference is half the angle at the centre standing on the same arc.

With C on the major arc, ∠AOB and ∠ACB both stand on minor arc AB, so ∠ACB = ½ × 60° = 30°.

A point D on the minor arc stands on the major arc instead. It sees AB at 180° − 30° = 150°, because ACBD is a cyclic quadrilateral.

Key facts

• Angle at the centre = 2 × angle at the circumference on the same arc.
• A chord that subtends 60° at the centre subtends 30° at every point of the major arc.
• The same chord subtends 150° at every point of the minor arc.
• A chord that subtends 60° at the centre equals the radius, because triangle AOB is equilateral.

Common traps

• Answering 60°, the central angle the stem gives, without halving it.
• Using the half-angle for a point on the minor arc. There the angle is 180° − 30° = 150°.

No directly related past PYQ was found.