Let $P$ be a point outside a circle $\Gamma$ centered at point $O$, and let $PA$ and $PB$ be tangent lines to circle $\Gamma$. Let segment $PO$ intersect circle $\Gamma$ at $C$. A tangent to circle $\Gamma$ through $C$ intersects $PA$ and $PB$ at points $E$ and $F$, respectively. Given that $EF=8$ and $\angle{APB}=60^\circ$, compute the area of $\triangle{AOC}$.

[i]2020 CCA Math Bonanza Individual Round #6[/i]
1. **Identify the given information and the goal:**
   - Point \( P \) is outside circle \( \Gamma \) centered at \( O \).
   - \( PA \) and \( PB \) are tangent lines to \( \Gamma \).
   - Segment \( PO \) intersects \( \Gamma \) at \( C \).
   - A tangent to \( \Gamma \) through \( C \) intersects \( PA \) and \( PB \) at points \( E \) and \( F \), respectively.
   - Given \( EF = 8 \) and \( \angle APB = 60^\circ \).
   - We need to compute the area of \( \triangle AOC \).

2. **Analyze the geometry of the problem:**
   - Since \( PA \) and \( PB \) are tangents from \( P \) to \( \Gamma \), \( PA = PB \).
   - \( \angle APB = 60^\circ \) implies that \( \triangle APB \) is isosceles with \( \angle PAB = \angle PBA = 60^\circ \).
   - Therefore, \( \triangle APB \) is equilateral.

3. **Determine the relationship between the triangles:**
   - Since \( \triangle APB \) is equilateral, \( PA = PB = AB \).
   - Let \( PA = PB = AB = x \).

4. **Use the given length \( EF = 8 \):**
   - The tangent at \( C \) intersects \( PA \) and \( PB \) at \( E \) and \( F \), respectively.
   - Since \( \angle APB = 60^\circ \), \( \angle AOB = 120^\circ \) because \( \angle AOB = 180^\circ - \angle APB \).

5. **Calculate the length \( PC \):**
   - Since \( \angle AOB = 120^\circ \), \( \triangle AOB \) is isosceles with \( AO = BO = R \) (radius of \( \Gamma \)).
   - Using the Law of Cosines in \( \triangle AOB \):
     \[
     AB^2 = AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos(120^\circ)
     \]
     \[
     x^2 = R^2 + R^2 - 2 \cdot R \cdot R \cdot (-\frac{1}{2})
     \]
     \[
     x^2 = 2R^2 + R^2 = 3R^2
     \]
     \[
     x = R\sqrt{3}
     \]

6. **Determine the area of \( \triangle AOC \):**
   - Since \( \triangle AOC \) is equilateral with side length \( R \sqrt{3} \):
     \[
     \text{Area of } \triangle AOC = \frac{\sqrt{3}}{4} \cdot (R \sqrt{3})^2
     \]
     \[
     = \frac{\sqrt{3}}{4} \cdot 3R^2
     \]
     \[
     = \frac{3\sqrt{3}}{4} R^2
     \]

7. **Use the given \( EF = 8 \) to find \( R \):**
   - Since \( EF = 8 \) and \( \triangle PEC \) and \( \triangle POA \) are similar \( 30-60-90 \) triangles:
     \[
     \frac{x}{4} = \frac{x + 4 \sqrt{3}}{8}
     \]
     Solving for \( x \):
     \[
     8x = 4(x + 4 \sqrt{3})
     \]
     \[
     8x = 4x + 16 \sqrt{3}
     \]
     \[
     4x = 16 \sqrt{3}
     \]
     \[
     x = 4 \sqrt{3}
     \]
     Therefore, \( R = 4 \).

8. **Calculate the final area:**
   \[
   \text{Area of } \triangle AOC = \frac{3\sqrt{3}}{4} \cdot (4)^2
   \]
   \[
   = \frac{3\sqrt{3}}{4} \cdot 16
   \]
   \[
   = 12 \sqrt{3}
   \]

The final answer is \(\boxed{12 \sqrt{3}}\)