1. **Problem statement:** Calculate the algebraic moment of the force $20\sqrt{2}$ newton with respect to point A. 2. **Given:** - Force magnitude: $F = 20\sqrt{2}$ N - Horizontal distance from A: $d_x = 0.2$ m - Vertical distance from A: $d_y = 1.2$ m - Force direction: $45^\circ$ from vertical, pointing outward and up to the right. 3. **Formula for moment:** $$ M = F \times d_\perp $$ where $d_\perp$ is the perpendicular distance from point A to the line of action of the force. 4. **Calculate components of the force:** Since the force is at $45^\circ$ from vertical, its components are: $$ F_x = F \sin 45^\circ = 20\sqrt{2} \times \frac{\sqrt{2}}{2} = 20 $$ $$ F_y = F \cos 45^\circ = 20\sqrt{2} \times \frac{\sqrt{2}}{2} = 20 $$ 5. **Calculate moment about A:** Moment is positive if it causes counterclockwise rotation. $$ M = F_x \times d_y - F_y \times d_x $$ Substitute values: $$ M = 20 \times 1.2 - 20 \times 0.2 = 24 - 4 = 20 $$ 6. **Answer:** The algebraic moment of the force with respect to point A is $20$ newton.meter. Therefore, the correct choice is **a) 20**.