1. **Problem Statement:** Find the resultant of three concurrent coplanar forces: i) 60 N acting 20 degrees east of north ii) 30 N east iii) 150 N acting 40 degrees east of south 2. **Formula and Approach:** To find the resultant force, we resolve each force into its components along the east (x-axis) and north (y-axis) directions, then sum these components. - For a force $F$ at an angle $\theta$ from north towards east, components are: $$F_x = F \sin(\theta)$$ $$F_y = F \cos(\theta)$$ - For a force at an angle from south towards east, the north component is negative: $$F_x = F \sin(\theta)$$ $$F_y = -F \cos(\theta)$$ 3. **Resolve each force:** i) 60 N, 20° east of north: $$F_{1x} = 60 \sin(20^\circ) = 60 \times 0.3420 = 20.52$$ $$F_{1y} = 60 \cos(20^\circ) = 60 \times 0.9397 = 56.38$$ ii) 30 N east: $$F_{2x} = 30$$ $$F_{2y} = 0$$ iii) 150 N, 40° east of south: $$F_{3x} = 150 \sin(40^\circ) = 150 \times 0.6428 = 96.42$$ $$F_{3y} = -150 \cos(40^\circ) = -150 \times 0.7660 = -114.90$$ 4. **Sum components:** $$R_x = 20.52 + 30 + 96.42 = 146.94$$ $$R_y = 56.38 + 0 - 114.90 = -58.52$$ 5. **Calculate magnitude of resultant:** $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{146.94^2 + (-58.52)^2} = \sqrt{21588.56 + 3423.06} = \sqrt{25011.62} = 158.13$$ 6. **Calculate direction:** Angle $\theta$ east of south (since $R_y$ is negative and $R_x$ positive, vector points southeast): $$\theta = \tan^{-1}\left(\frac{|R_x|}{|R_y|}\right) = \tan^{-1}\left(\frac{146.94}{58.52}\right) = \tan^{-1}(2.51) = 68^\circ$$ 7. **Final answer:** The resultant force is approximately 160 N, 68 degrees east of south. **Answer:** c. 160 N 68 degrees east of south