1. **Problem statement:** Calculate the components of the 100 N force in each scenario based on the given angles. 2. **Formula and rules:** The force components along the directions can be found using trigonometry: $$F_x = F \cos(\theta)$$ $$F_y = F \sin(\theta)$$ where $F$ is the force magnitude and $\theta$ is the angle with respect to the horizontal or vertical axis. 3. **Scenario (a):** Given: $F = 100$ N, angle $\theta = 30^\circ$ from horizontal. Calculate horizontal and vertical components: $$F_x = 100 \cos(30^\circ) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \approx 86.6$$ $$F_y = 100 \sin(30^\circ) = 100 \times \frac{1}{2} = 50$$ 4. **Scenario (b):** Given: $F = 100$ N, angles $45^\circ$ with horizontal and $60^\circ$ with vertical. Calculate components relative to horizontal: $$F_x = 100 \cos(45^\circ) = 100 \times \frac{\sqrt{2}}{2} = 50\sqrt{2} \approx 70.7$$ $$F_y = 100 \sin(45^\circ) = 100 \times \frac{\sqrt{2}}{2} = 50\sqrt{2} \approx 70.7$$ 5. **Scenario (c):** Given: $F = 100$ N, angles $90^\circ$ (horizontal left) and $60^\circ$ upwards. Calculate components: Horizontal component (left): $$F_x = 100 \cos(90^\circ) = 0$$ Vertical component: $$F_y = 100 \sin(90^\circ) = 100$$ For the $60^\circ$ line: $$F_x = 100 \cos(60^\circ) = 100 \times \frac{1}{2} = 50$$ $$F_y = 100 \sin(60^\circ) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \approx 86.6$$ **Final answers:** - (a) $F_x \approx 86.6$ N, $F_y = 50$ N - (b) $F_x \approx 70.7$ N, $F_y \approx 70.7$ N - (c) Horizontal line: $F_x = 0$, $F_y = 100$ N; $60^\circ$ line: $F_x = 50$ N, $F_y \approx 86.6$ N