1. **Stating the problem:** We have a force $P = 100$ KN acting downward at point A in a triangle ABC where angle $\angle B = 90^\circ$ and angle $\angle C = 35^\circ$. We need to find the components of force $P$ along AB and AC, and then find the magnitude and direction of the resultant force. 2. **Understanding the geometry:** Since $\angle B = 90^\circ$ and $\angle C = 35^\circ$, the remaining angle at A is $\angle A = 180^\circ - 90^\circ - 35^\circ = 55^\circ$. 3. **Force components:** The force $P$ acts vertically downward. We want to resolve $P$ into components along AB and AC. 4. **Using the angles to find components:** - The component along AB is $P_{AB} = P \times \cos(35^\circ)$ because AB is adjacent to angle C. - The component along AC is $P_{AC} = P \times \cos(90^\circ - 35^\circ) = P \times \sin(35^\circ)$ because AC is adjacent to angle B. 5. **Calculate components:** $$P_{AB} = 100 \times \cos(35^\circ) = 100 \times 0.8192 = 81.92 \text{ KN}$$ $$P_{AC} = 100 \times \sin(35^\circ) = 100 \times 0.574 = 57.4 \text{ KN}$$ 6. **Resultant force magnitude:** The components $P_{AB}$ and $P_{AC}$ are perpendicular (since $\angle B = 90^\circ$), so the resultant magnitude is: $$R = \sqrt{P_{AB}^2 + P_{AC}^2} = \sqrt{81.92^2 + 57.4^2} = \sqrt{6707.5 + 3294.8} = \sqrt{10002.3} = 100.01 \text{ KN}$$ 7. **Resultant force direction:** The angle $\theta$ of the resultant relative to AB is: $$\theta = \tan^{-1}\left(\frac{P_{AC}}{P_{AB}}\right) = \tan^{-1}\left(\frac{57.4}{81.92}\right) = \tan^{-1}(0.7) = 35^\circ$$ **Final answers:** - Component along AB: $81.92$ KN - Component along AC: $57.4$ KN - Resultant magnitude: $100.01$ KN (approximately $100$ KN) - Resultant direction: $35^\circ$ from AB