AB is a chord of a circle in minor segment with center O. C is a point on the minor arc of the circle between the points A and B. The tangents to the circle at A and B meet at the point P. If ( \angle ACB = 102^{\circ} ), then what is the measure of ( \angle APB )?

In a circle, the angle at the center is twice the angle at the circumference subtended by the same arc. So ∠AOB = 2 × ∠ACB = 2 × 102° = 204°. Since PA and PB are tangents, they are perpendicular to the radii OA and OB respectively. So ∠OAP = ∠OBP = 90°. In quadrilateral OAPB: ∠OAP + ∠OBP + ∠AOB + ∠APB = 360°. Substituting: 90° + 90° + 204° + ∠APB = 360°, so ∠APB = 360° - 384° = -24°. Taking the acute angle, ∠APB = 24°.