1. **Problem statement:** Find the magnitude and direction of the resultant of three coplanar forces: i) 40 N at 45° east of south ii) 10 N west iii) 70 N at 35° east of north 2. **Formula and approach:** To find the resultant, resolve each force into its components along the east-west (x-axis) and north-south (y-axis) directions. - East and north are positive directions. - South and west are negative directions. Use trigonometry: For a force $F$ at angle $\theta$ from a reference axis, $$F_x = F \cos \theta, \quad F_y = F \sin \theta$$ 3. **Resolve each force:** i) 40 N, 45° east of south: - South is negative y, east is positive x. - Angle from south towards east is 45°, so components: $$F_{1x} = 40 \sin 45^\circ = 40 \times 0.7071 = 28.28$$ $$F_{1y} = -40 \cos 45^\circ = -40 \times 0.7071 = -28.28$$ ii) 10 N west: - West is negative x direction. $$F_{2x} = -10, \quad F_{2y} = 0$$ iii) 70 N, 35° east of north: - North is positive y, east is positive x. $$F_{3x} = 70 \sin 35^\circ = 70 \times 0.574 = 40.18$$ $$F_{3y} = 70 \cos 35^\circ = 70 \times 0.819 = 57.33$$ 4. **Sum components:** $$R_x = F_{1x} + F_{2x} + F_{3x} = 28.28 - 10 + 40.18 = 58.46$$ $$R_y = F_{1y} + F_{2y} + F_{3y} = -28.28 + 0 + 57.33 = 29.05$$ 5. **Calculate magnitude of resultant:** $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{58.46^2 + 29.05^2} = \sqrt{3417.5 + 843.9} = \sqrt{4261.4} = 65.28$$ 6. **Calculate direction:** Direction angle $\theta$ from north towards east: $$\theta = \tan^{-1} \left( \frac{R_x}{R_y} \right) = \tan^{-1} \left( \frac{58.46}{29.05} \right) = \tan^{-1}(2.01) = 63.7^\circ$$ So the resultant is approximately 65 N at 64° east of north. 7. **Answer choice:** Closest to option e: 65 N 65 degrees east of north. **Final answer:** 65 N 65 degrees east of north.