To solve this problem, we can analyze the given geometry and use trigonometric relationships.

Step-by-Step Solution

1. Given Information:

   • $OP = 6$
   • $OC = 2$
   • $\angle AOP = 60^\circ$

2. Relation Between $OC$ and $OP$: Since $OC = 2$, $C$ is a point on the circle with center $O$ and radius $r$. Therefore, $OC = r = 2$.

3. Using the Cosine Rule in Triangle $OCP$: In $\triangle OCP$, we know:

   • $OP = 6$
   • $OC = r = 2$
   • $\angle OCP = 60^\circ$

   To find $s = CP$, we use the Cosine Rule: $CP^2 = OP^2 + OC^2 - 2 \cdot OP \cdot OC \cdot \cos(60^\circ)$ Substituting the values: $s^2 = 6^2 + 2^2 - 2 \cdot 6 \cdot 2 \cdot \cos(60^\circ)$ $= 36 + 4 - 2 \cdot 6 \cdot 2 \cdot \frac{1}{2}$ $= 40 - 12 = 28$ $s = \sqrt{28} = 2\sqrt{7}$

4. Calculate $\ln\left(\frac{s}{r}\right)$: We need to find $\ln\left(\frac{s}{r}\right)$.

   Since $r = 2$ and $s = 2\sqrt{7}$, we have: $\frac{s}{r} = \frac{2\sqrt{7}}{2} = \sqrt{7}$ Thus, $\ln\left(\frac{s}{r}\right) = \ln\left(\sqrt{7}\right)$ Using the property $\ln(\sqrt{a}) = \frac{1}{2} \ln(a)$, we get: $\ln\left(\sqrt{7}\right) = \frac{1}{2} \ln(7)$

Final Answer:

$\ln\left(\frac{s}{r}\right) = \frac{1}{2} \ln(7)$

Would you like further clarification on any of the steps?

---

Follow-up Questions:

1. How does the cosine rule apply in non-right triangles?
2. What other trigonometric methods can help solve geometric problems involving circles?
3. Can the solution be extended if the angle $\angle AOP$ were different?
4. What if the radius $s$ were given? How would that simplify the problem?
5. How is logarithmic simplification applied in different mathematical contexts?

Tip: Always check if given angles allow using symmetry or special triangle properties to simplify complex problems.