From a point P outside a circle of radius 5 cm two tangents are drawn, touching the circle at A and at B, and ∠APB = 60°. What is the length of the chord AB?

Tangents from an external point are equal, so PA = PB and triangle PAB is isosceles; with the apex angle 60° the base angles are 60° too, so PAB is equilateral and AB = PA. A radius meets a tangent at 90°, so in the right triangle OAP we have ∠OPA = 30° and OA = 5, giving OP = 10 and PA = √$(100-25)=5$√3 cm. Checking from the centre: ∠AOB = 360° − 90° − 90° − 60° = 120°, so AB = $2\times 5$ × sin 60° = 5√3 cm. The trap 10 cm is OP (the same size as the diameter), not the chord.