1. **Problem:** Determine the magnitude and direction of the resultant of the vectors: 8.0 N at 40°, 9.0 N at 120°, and 12.0 N at 200°, angles measured from the positive x-axis. 2. **Formula and rules:** - Use component method: resolve each vector into x and y components. - For a vector $V$ with magnitude $V$ and angle $\theta$, components are: $$V_x = V \cos\theta$$ $$V_y = V \sin\theta$$ - Sum all $x$ components and all $y$ components separately. - Resultant magnitude: $$R = \sqrt{(\sum V_x)^2 + (\sum V_y)^2}$$ - Resultant direction (angle from positive x-axis): $$\alpha = \tan^{-1}\left(\frac{\sum V_y}{\sum V_x}\right)$$ 3. **Calculate components:** - Vector 1: 8.0 N at 40° $$V_{1x} = 8.0 \cos 40^\circ = 8.0 \times 0.7660 = 6.128$$ $$V_{1y} = 8.0 \sin 40^\circ = 8.0 \times 0.6428 = 5.142$$ - Vector 2: 9.0 N at 120° $$V_{2x} = 9.0 \cos 120^\circ = 9.0 \times (-0.5) = -4.5$$ $$V_{2y} = 9.0 \sin 120^\circ = 9.0 \times 0.8660 = 7.794$$ - Vector 3: 12.0 N at 200° $$V_{3x} = 12.0 \cos 200^\circ = 12.0 \times (-0.9397) = -11.276$$ $$V_{3y} = 12.0 \sin 200^\circ = 12.0 \times (-0.3420) = -4.104$$ 4. **Sum components:** $$\sum V_x = 6.128 - 4.5 - 11.276 = 6.128 - 15.776 = -9.648$$ $$\sum V_y = 5.142 + 7.794 - 4.104 = 8.832$$ 5. **Calculate resultant magnitude:** $$R = \sqrt{(-9.648)^2 + (8.832)^2} = \sqrt{93.06 + 77.99} = \sqrt{171.05} = 13.08$$ 6. **Calculate resultant direction:** $$\alpha = \tan^{-1}\left(\frac{8.832}{-9.648}\right) = \tan^{-1}(-0.915)$$ Since $\sum V_x$ is negative and $\sum V_y$ positive, angle is in second quadrant: $$\alpha = 180^\circ - 42.4^\circ = 137.6^\circ$$ **Final answer:** - Magnitude of resultant vector: $13.1$ N (rounded) - Direction: $137.6^\circ$ from positive x-axis