Given that angle AOB = 120 degrees and AB = 10, find AC.

Given:

• Angle AOB = 120°
• AB = 10

To find: AC

Solution:

• In triangle AOB, OA = OB (radii of the same circle).
• Since angle AOB = 120°, triangle AOB is an isosceles triangle.
• We can use the Law of Cosines in triangle AOB to find the radius (OA or OB). However, the problem states that AB = 10, which is a chord.
• Let's consider the properties of tangents from an external point to a circle. AC and BC are tangents to the circle at points A and B respectively.
• Therefore, OA is perpendicular to AC and OB is perpendicular to BC. This means angle OAC = 90° and angle OBC = 90°.
• In quadrilateral OACB, the sum of angles is 360°.
• So, angle ACB + angle OAC + angle AOC + angle OBC = 360°
• angle ACB + 90° + angle AOB + 90° = 360°
• angle ACB + angle AOB + 180° = 360°
• angle ACB + angle AOB = 180°
• Given angle AOB = 120°, then angle ACB = 180° - 120° = 60°.
• In triangle ACB, OA = OB (radii), and AC = BC (tangents from an external point are equal). Thus, triangle ACB is an isosceles triangle.
• Since angle ACB = 60°, and triangle ACB is isosceles, it must be an equilateral triangle. Therefore, AC = BC = AB.
• Given AB = 10, then AC = 10.

Final Answer: The final answer is $$oxed{10}$$