1. **State the problem:** We need to find the tension forces in cables AC and BC supporting a 300 kg weight at point C in equilibrium. 2. **Identify the forces:** - Weight force $W = mg = 300 \times 9.8 = 2940$ N acting vertically downward at C. - Tension in cable AC, $T_{AC}$, at an angle of 40° above the horizontal. - Tension in cable BC, $T_{BC}$, at an angle of 20° above the horizontal. 3. **Set equilibrium equations:** Since the beam is in equilibrium, both horizontal and vertical force components must balance: - Horizontal: $T_{AC} \cos 40° = T_{BC} \cos 20°$ - Vertical: $T_{AC} \sin 40° + T_{BC} \sin 20° = 2940$ 4. **Express $T_{AC}$ in terms of $T_{BC}$:** $$T_{AC} = T_{BC} \frac{\cos 20°}{\cos 40°}$$ 5. **Substitute into vertical equation:** $$T_{BC} \frac{\cos 20°}{\cos 40°} \sin 40° + T_{BC} \sin 20° = 2940$$ 6. **Simplify:** $$T_{BC} \left( \frac{\cos 20° \sin 40°}{\cos 40°} + \sin 20° \right) = 2940$$ Calculate numeric values: - $\cos 20° \approx 0.9397$ - $\sin 40° \approx 0.6428$ - $\cos 40° \approx 0.7660$ - $\sin 20° \approx 0.3420$ So, $$T_{BC} \left( \frac{0.9397 \times 0.6428}{0.7660} + 0.3420 \right) = 2940$$ Calculate inside parentheses: $$\frac{0.6041}{0.7660} + 0.3420 = 0.789\ + 0.3420 = 1.131$$ 7. **Solve for $T_{BC}$:** $$T_{BC} = \frac{2940}{1.131} \approx 2599.3 \text{ N}$$ 8. **Calculate $T_{AC}$:** $$T_{AC} = 2599.3 \times \frac{0.9397}{0.7660} = 2599.3 \times 1.2275 \approx 3193.1 \text{ N}$$ **Final answer:** - Tension in cable AC: $\boxed{3193 \text{ N}}$ - Tension in cable BC: $\boxed{2599 \text{ N}}$