1. **Problem statement:** Find the components of the 200 N force acting at 45° downward from horizontal, the resultant magnitude, and the angle when combined with a 300 N horizontal force acting right. 2. **Formula and rules:** - Components of a force $F$ at angle $\theta$ are: - Horizontal component: $F_x = F \cos \theta$ - Vertical component: $F_y = F \sin \theta$ - Resultant vector magnitude: $R = \sqrt{(F_{x1} + F_{x2})^2 + (F_{y1} + F_{y2})^2}$ - Resultant angle $\alpha = \tan^{-1}\left(\frac{\text{sum of vertical components}}{\text{sum of horizontal components}}\right)$ 3. **Calculate components of each force:** - Force 1: 300 N horizontal right - $F_{x1} = 300$ N - $F_{y1} = 0$ N - Force 2: 200 N at 45° downward - $F_{x2} = 200 \cos 45^\circ = 200 \times \frac{\sqrt{2}}{2} = 141.42$ N - $F_{y2} = -200 \sin 45^\circ = -200 \times \frac{\sqrt{2}}{2} = -141.42$ N (negative because downward) 4. **Sum components:** - Horizontal: $F_x = 300 + 141.42 = 441.42$ N - Vertical: $F_y = 0 - 141.42 = -141.42$ N 5. **Resultant magnitude:** $$ R = \sqrt{441.42^2 + (-141.42)^2} = \sqrt{194851 + 20000} = \sqrt{214851} \approx 463.56 \text{ N} $$ 6. **Resultant angle:** $$ \alpha = \tan^{-1}\left(\frac{-141.42}{441.42}\right) = \tan^{-1}(-0.32) \approx -17.7^\circ $$ This means the resultant force is $17.7^\circ$ below the horizontal to the right. **Final answers:** - Components of 200 N force: $F_x = 141.42$ N, $F_y = -141.42$ N - Resultant magnitude: $463.56$ N - Resultant angle: $17.7^\circ$ below horizontal right