1. **Stating the problem:** We have a right triangle with a vertical side of 1.2 m, a horizontal base of 2.3 m, and a hypotenuse with a distributed load of 5 kN/m acting downward. There is a 6 kN horizontal force acting left at the top of the vertical side. The bottom-left corner has a pinned support with a vertical reaction force, and the bottom-right corner has a fixed support. Additionally, a rod extends from the bottom-left support at 25° and a wire from the top of the rod at 45°, with a 1 kN vertical load at the wire's joint. We need to find the forces in the rod and wire. 2. **Calculate the length of the hypotenuse:** Using Pythagoras theorem: $$\text{hypotenuse} = \sqrt{1.2^2 + 2.3^2} = \sqrt{1.44 + 5.29} = \sqrt{6.73} \approx 2.594 \, m$$ 3. **Calculate total distributed load on hypotenuse:** Distributed load = 5 kN/m over 2.594 m: $$F_{dist} = 5 \times 2.594 = 12.97 \, kN$$ This acts downward at the midpoint of the hypotenuse. 4. **Resolve the 6 kN horizontal force:** It acts left at the top of the vertical side. 5. **Analyze the rod and wire geometry:** Rod angle = 25° from horizontal at bottom-left. Wire angle = 45° from horizontal at top of rod. 6. **Set up equilibrium equations:** Let $F_r$ be the force in the rod, $F_w$ the force in the wire. - Horizontal equilibrium: $$F_r \cos 25^\circ = F_w \cos 45^\circ + 6$$ - Vertical equilibrium: $$F_r \sin 25^\circ + F_w \sin 45^\circ = 1 + R_v$$ where $R_v$ is the vertical reaction force at the pinned support. 7. **Moment equilibrium about bottom-left support:** Sum moments to zero to solve for unknowns. 8. **Calculate components:** $$\cos 25^\circ \approx 0.9063, \sin 25^\circ \approx 0.4226$$ $$\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071$$ 9. **From horizontal equilibrium:** $$F_r \times 0.9063 = F_w \times 0.7071 + 6$$ 10. **From vertical equilibrium (assuming $R_v$ balances remaining vertical forces):** $$F_r \times 0.4226 + F_w \times 0.7071 = 1 + R_v$$ 11. **Assuming $R_v$ is unknown, focus on rod and wire forces:** Express $F_w$ from horizontal equation: $$F_w = \frac{F_r \times 0.9063 - 6}{0.7071}$$ 12. **Substitute into vertical equation:** $$F_r \times 0.4226 + \left(\frac{F_r \times 0.9063 - 6}{0.7071}\right) \times 0.7071 = 1 + R_v$$ Simplify: $$F_r \times 0.4226 + F_r \times 0.9063 - 6 = 1 + R_v$$ $$F_r (0.4226 + 0.9063) = 7 + R_v$$ $$F_r \times 1.3289 = 7 + R_v$$ 13. **If $R_v$ is reaction force, to find $F_r$ and $F_w$ we need $R_v$ or additional info.** 14. **Summary:** - Length hypotenuse = 2.594 m - Distributed load total = 12.97 kN - Rod force $F_r$ and wire force $F_w$ related by: $$F_w = \frac{0.9063 F_r - 6}{0.7071}$$ - Vertical equilibrium relates $F_r$, $R_v$, and $F_w$. Without $R_v$, exact $F_r$ and $F_w$ cannot be numerically determined. **Final answer:** The forces in the rod and wire satisfy: $$F_w = \frac{0.9063 F_r - 6}{0.7071}$$ and $$F_r \times 1.3289 = 7 + R_v$$ where $R_v$ is the vertical reaction force at the pinned support. Further information about $R_v$ or additional constraints are needed for numeric values.