4. In the figure, point O is the center of the circle. The angle ACB = 60 degrees. Find AB.

This problem involves a circle with center O and an inscribed angle ACB which measures 60 degrees. The angle ACB subtends the arc AB.

The central angle subtended by the same arc AB is angle AOB. The relationship between an inscribed angle and a central angle that subtend the same arc is that the central angle is twice the inscribed angle. Therefore, angle AOB = 2 * angle ACB.

Angle AOB = 2 * 60 degrees = 120 degrees.

Now consider triangle AOB. OA and OB are radii of the circle. From the figure, it seems that OA = 4. If OA = 4, then OB = 4 as well.

We can find the length of AB using the Law of Cosines in triangle AOB:

AB² = OA² + OB² - 2 * OA * OB * cos(AOB)

AB² = 4² + 4² - 2 * 4 * 4 * cos(120°)

AB² = 16 + 16 - 32 * (-0.5)

AB² = 32 + 16

AB² = 48

AB = sqrt(48) = sqrt(16 * 3) = 4 * sqrt(3)

Ответ: 4√3