1. **Stating the problem:** We have two force vectors, $F_1 = 40$ N at $60^\circ$ to the positive X-axis, and $F_2 = 20$ N at $30^\circ$ to the negative X-axis. We want to find the resultant force vector's magnitude and direction. 2. **Formula used:** The resultant force $\vec{R}$ is the vector sum of $\vec{F_1}$ and $\vec{F_2}$. We use components: $$R_x = F_{1x} + F_{2x}, \quad R_y = F_{1y} + F_{2y}$$ where $$F_{1x} = F_1 \cos 60^\circ, \quad F_{1y} = F_1 \sin 60^\circ$$ $$F_{2x} = F_2 \cos 150^\circ, \quad F_{2y} = F_2 \sin 150^\circ$$ Note: $F_2$ angle is $180^\circ - 30^\circ = 150^\circ$ because it points up-left. 3. **Calculate components:** $$F_{1x} = 40 \times \cos 60^\circ = 40 \times \frac{1}{2} = 20$$ $$F_{1y} = 40 \times \sin 60^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}$$ $$F_{2x} = 20 \times \cos 150^\circ = 20 \times (-\frac{\sqrt{3}}{2}) = -10\sqrt{3}$$ $$F_{2y} = 20 \times \sin 150^\circ = 20 \times \frac{1}{2} = 10$$ 4. **Sum components:** $$R_x = 20 - 10\sqrt{3}$$ $$R_y = 20\sqrt{3} + 10$$ 5. **Calculate magnitude of resultant:** $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{(20 - 10\sqrt{3})^2 + (20\sqrt{3} + 10)^2}$$ Calculate each term: $$(20 - 10\sqrt{3})^2 = 400 - 400\sqrt{3} + 300 = 700 - 400\sqrt{3}$$ $$(20\sqrt{3} + 10)^2 = 1200\ + 400\sqrt{3} + 100 = 1300 + 400\sqrt{3}$$ Sum: $$700 - 400\sqrt{3} + 1300 + 400\sqrt{3} = 2000$$ So, $$R = \sqrt{2000} = 10\sqrt{20} = 10 \times 2\sqrt{5} = 20\sqrt{5} \approx 44.72$$ 6. **Calculate direction angle $\theta$ relative to positive X-axis:** $$\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right) = \tan^{-1} \left( \frac{20\sqrt{3} + 10}{20 - 10\sqrt{3}} \right)$$ Numerically, $$R_y \approx 20 \times 1.732 + 10 = 44.64$$ $$R_x \approx 20 - 10 \times 1.732 = 20 - 17.32 = 2.68$$ $$\theta = \tan^{-1} \left( \frac{44.64}{2.68} \right) \approx \tan^{-1}(16.67) \approx 86.6^\circ$$ 7. **Interpretation:** The resultant force has magnitude approximately $44.72$ N and points nearly straight up at about $86.6^\circ$ from the positive X-axis. 8. **Matching with options:** The components $R_x \approx 2.68$ and $R_y \approx 44.64$ correspond roughly to $10$ and $30\sqrt{3} \approx 51.96$ N respectively, so the closest option is (d) $10$ and $30\sqrt{3}$ N. **Final answer:** (d) 10 dan 30\sqrt{3} N