1. Find KL.

Solution:

1. In the right triangle formed by the radius to the point of tangency (let's call it R), the segment from the center to the external point (O to L), and the tangent segment (KL), we have a right angle at K.
2. The radius is given as 6. The angle at the center is 60 degrees.
3. In triangle OKL, angle OKL = 90 degrees. Angle KOL = 60 degrees.
4. We can use trigonometry. We know the adjacent side (OK = 6) and want to find the opposite side (KL). However, we don't have enough information to directly use trigonometry for KL without knowing OL or another angle.
5. Let's reconsider the diagram. The angle 60 degrees is between the radius OK and the segment OL. Therefore, angle KOL = 60 degrees.
6. In right triangle OKL, we have OK = 6 (radius).
7. We can find OL using cosine: \( \cos(60^{\circ}) = \frac{OK}{OL} \) => \( \frac{1}{2} = \frac{6}{OL} \) => \( OL = 12 \).
8. Now we can find KL using the Pythagorean theorem: \( KL^2 + OK^2 = OL^2 \) => \( KL^2 + 6^2 = 12^2 \) => \( KL^2 + 36 = 144 \) => \( KL^2 = 108 \) => \( KL = \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \).

Answer: KL = $$6\sqrt{3}$$.