Given: Circle O(r), AB and AC are tangents; B and C are points of tangency, \angle OAB = 30°, AB = 5 cm. Find: BC.

Solution:

1. Properties of tangents: The tangent is perpendicular to the radius at the point of tangency. Therefore, \angle OBA = 90°.
2. Right-angled triangle OAB: In the right-angled triangle OAB, we are given \angle OAB = 30° and AB = 5 cm.
3. Using trigonometry: We can find OB and OA.
   - \tan(30°) = OB / AB => OB = AB * \tan(30°) = 5 * (1/\sqrt{3}) = 5/\sqrt{3} cm.
   - \cos(30°) = AB / OA => OA = AB / \cos(30°) = 5 / (\sqrt{3}/2) = 10/\sqrt{3} cm.
4. Quadrilateral OBAC: OB = OC (radii). OA bisects \angle BAC and \angle BOC. Also, OA is the perpendicular bisector of BC.
5. Triangle OBC: OA is the angle bisector of \angle BOC. Since \triangle OBC is isosceles (OB = OC), OA is also the altitude to BC. Therefore, \angle BOC = 2 * \angle BOA.
6. Finding \angle BOA: In \triangle OAB, \angle BOA = 90° - \angle OAB = 90° - 30° = 60°.
7. Finding \angle BOC: \angle BOC = 2 * \angle BOA = 2 * 60° = 120°.
8. Finding BC: In \triangle OBC, we can use the Law of Cosines:
   BC² = OB² + OC² - 2 * OB * OC * \cos(120°)
   BC² = (5/\sqrt{3})² + (5/\sqrt{3})² - 2 * (5/\sqrt{3}) * (5/\sqrt{3}) * (-1/2)
   BC² = 25/3 + 25/3 + 25/3 = 75/3 = 25.
   BC = \sqrt{25} = 5 cm.
9. Alternative method for BC: In \triangle OBС, draw a perpendicular from O to BC, let's call the intersection point M. \triangle OBM is a right-angled triangle with \angle BOM = 120°/2 = 60°.
   BM = OB * \sin(60°) = (5/\sqrt{3}) * (\sqrt{3}/2) = 5/2 cm.
   BC = 2 * BM = 2 * (5/2) = 5 cm.

Ответ: 5 см