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To solve this problem, we need to calculate the moment (torque) of the \( 80 \sqrt{3} \, \text{N} \) force about point \( A \). 

### Step-by-Step Solution:

1. **Identify the Force and Distance**:
   - The force is \( 80 \sqrt{3} \, \text{N} \) acting downward at point \( C \).
   - Point \( A \) is the pivot point.
   - The distance from \( A \) to \( C \) can be found using points \( A \), \( B \), and \( C \).

2. **Calculate the Perpendicular Distance (Moment Arm)**:
   - From the diagram:
     - \( AB = 8 \, \text{m} \) (horizontal distance from \( A \) to \( B \))
     - \( BC = 4 \, \text{m} \) at an angle of \( 120^\circ \) with the horizontal.
   - The perpendicular distance from point \( A \) to the line of action of the force is the projection of \( AC \) in a direction perpendicular to the force.

3. **Using Trigonometry**:
   - The angle between \( AB \) and \( BC \) is \( 120^\circ \).
   - The perpendicular distance from \( A \) to the line of force (let’s call it \( d \)) is given by:
     \[
     d = AB + BC \cos(120^\circ)
     \]
   - Since \( \cos(120^\circ) = -\frac{1}{2} \):
     \[
     d = 8 + 4 \left(-\frac{1}{2}\right) = 8 - 2 = 6 \, \text{m}
     \]

4. **Calculate the Moment**:
   - Moment about \( A \) is calculated by:
     \[
     M = F \cdot d
     \]
   - Substituting \( F = 80 \sqrt{3} \, \text{N} \) and \( d = 6 \, \text{m} \):
     \[
     M = 80 \sqrt{3} \times 6 = 480 \sqrt{3} \, \text{N} \cdot \text{m}
     \]
   - Since the force would cause a clockwise rotation about \( A \), the moment is negative:
     \[
     M = -480 \sqrt{3} \, \text{N} \cdot \text{m}
     \]

### Answer:
The correct option is:
**(b) \(-800\sqrt{3}\)**.

The algebraic measure of the moment of the 80√3 N force about A = ... N·m
Options:

a) -80√3
b) -800√3
c) -480
d) -640

Learn how to calculate the moment of an 80√3 N force about point A using trigonometry and vector projection. This physics problem covers essential concepts for students in Grades 11-12.

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