Given:

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

We need to find $\log_e \left( \frac{s}{r} \right)$.

Solution:

1. Set Up Triangle $\triangle OCP$: Since $C$ is a point on the circle with center $O$, $OC$ is a radius of this circle. Therefore, $r = OC = 2$.

2. Applying the Cosine Rule in $\triangle OCP$: We know:

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

   Let $CP = s$. Using the cosine rule in $\triangle OCP$: $s^2 = r^2 + OP^2 - 2 \cdot r \cdot OP \cdot \cos(60^\circ)$

3. Plugging in Values: Substitute $r = 2$, $OP = 6$, and $\cos(60^\circ) = \frac{1}{2}$: $s^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \frac{1}{2}$ Simplify each term: $s^2 = 4 + 36 - 2 \cdot 2 \cdot 6 \cdot \frac{1}{2}$