1. **State the problem:** We need to find the magnitude and direction of the resultant force from two forces: $f_1 = 50$N at northeast by 37° and $f_2 = 100$N at northwest by 60°. 2. **Understand directions:** Northeast means 45° from east towards north. "Due northeast by 37°" means we rotate 37° from northeast towards north, so the angle from east is $45° + 37° = 82°$. Northwest means 135° from east (90° + 45°). "Due northwest by 60°" means rotate 60° from northwest towards north, so angle from east is $135° - 60° = 75°$. 3. **Resolve forces into components:** - For $f_1$: $$F_{1x} = 50 \cos 82°$$ $$F_{1y} = 50 \sin 82°$$ - For $f_2$: $$F_{2x} = 100 \cos 75°$$ $$F_{2y} = 100 \sin 75°$$ 4. **Calculate components:** - $F_{1x} = 50 \times 0.1392 = 6.96$ - $F_{1y} = 50 \times 0.9903 = 49.52$ - $F_{2x} = 100 \times 0.2588 = 25.88$ - $F_{2y} = 100 \times 0.9659 = 96.59$ 5. **Sum components:** - $R_x = F_{1x} + F_{2x} = 6.96 + 25.88 = 32.84$ - $R_y = F_{1y} + F_{2y} = 49.52 + 96.59 = 146.11$ 6. **Find magnitude of resultant:** $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{32.84^2 + 146.11^2} = \sqrt{1078.6 + 21348.1} = \sqrt{22426.7} = 149.75$$ 7. **Find direction of resultant:** $$\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right) = \tan^{-1} \left( \frac{146.11}{32.84} \right) = \tan^{-1} (4.45) = 77.3°$$ This angle is measured from the east axis towards north. **Final answer:** The resultant force has magnitude approximately $149.75$N and direction $77.3°$ north of east.