A pair of straight lines from an external point F intersects a circle at A and B (FA < FB), and touches the circle at C. O is the centre of the circle. Given that ∠ ACF = 50° and ∠ AFC = 30°, find ∠ AOB.

Correct — C. Two circle theorems settle this, and no length is needed.

Tangent–chord (alternate segment): the angle between tangent FC and chord CA equals the angle CA subtends from the far arc, so ∠ABC = ∠ACF = 50°.

∠CAB is the exterior angle of △AFC at A, so ∠CAB = ∠AFC + ∠ACF = 30° + 50° = 80°.

In △ABC: ∠ACB = 180° − 80° − 50° = 50°.
Angle at the centre is twice the angle at the circumference on arc AB, so ∠AOB = 2 × 50° = 100° → option (c).

Why the others are wrong

• (a)
  80° — 80° is ∠CAB, the angle the working passes through on the way to the answer. It is an angle at the circumference of the wrong arc, and the central angle is twice ∠ACB, not ∠CAB itself.
• (b)
  90° — 90° would need ∠ACB = 45°. The two given angles force ∠ACB to 50°, so AB never subtends a right angle at O here.
• (d)
  40° — 40° is what ∠AOB would be if ∠ACB were 20°. Twenty degrees is only the difference 50° − 30°, and that difference plays no part in either theorem used.

Two standard circle results carry this item. The tangent–chord angle (alternate segment theorem) says the angle between a tangent and a chord drawn from the point of contact equals the inscribed angle that the chord subtends from the other arc.

The angle-at-the-centre theorem says a central angle is double the inscribed angle standing on the same arc.

Between them you convert the two angles given at F and C into an angle at B, then into an angle at C inside triangle ABC, then double it. The triangle angle sum and the exterior-angle shortcut do the joining.

Key facts

• Alternate segment theorem: the angle between a tangent and a chord at the point of contact equals the inscribed angle subtended by that chord from the alternate segment.
• The angle subtended by an arc at the centre is twice the angle it subtends anywhere on the remaining circumference.
• An exterior angle of a triangle equals the sum of the two remote interior angles, which turns 30° and 50° into 80° in one line.
• Here ∠ABC = 50°, ∠CAB = 80° and ∠ACB = 50°, so triangle ABC is isosceles with AB = AC, the two sides opposite the equal 50° angles.

Common traps

• Reading ∠ACF as an ordinary inscribed angle inside the circle rather than a tangent–chord angle.
• Placing B between F and A, which reverses the supplementary step and loses 80°.
• Stopping at ∠CAB = 80° and reporting it, since it is one of the printed options.

No directly related past PYQ was found.