1. **Problem statement:** Given two weights each of mass $M$ hanging from a horizontal string (string 2) with tension 30 N, and two angled strings (strings 1 and 3) each making an angle $\alpha = 40^\circ$ with the horizontal, find the mass $M$. 2. **Setup and formula:** The system is in equilibrium, so the forces balance. The tension in the horizontal string (string 2) is $T_2 = 30$ N. 3. The angled strings support the weights. Each angled string has tension $T_1 = T_3$ and makes angle $\alpha$ with the horizontal. 4. The vertical components of the tensions in strings 1 and 3 balance the weight of the masses: $$2 T_1 \sin \alpha = 2 M g$$ where $g = 9.8$ m/s$^2$ is acceleration due to gravity. 5. The horizontal components of the tensions in strings 1 and 3 balance the tension in string 2: $$2 T_1 \cos \alpha = T_2$$ 6. From the horizontal balance, solve for $T_1$: $$T_1 = \frac{T_2}{2 \cos \alpha} = \frac{30}{2 \cos 40^\circ}$$ Calculate $\cos 40^\circ \approx 0.7660$: $$T_1 = \frac{30}{2 \times 0.7660} = \frac{30}{1.532} \approx 19.59 \text{ N}$$ 7. Substitute $T_1$ into the vertical balance to find $M$: $$2 \times 19.59 \times \sin 40^\circ = 2 M \times 9.8$$ Calculate $\sin 40^\circ \approx 0.6428$: $$2 \times 19.59 \times 0.6428 = 2 M \times 9.8$$ $$25.19 = 19.6 M$$ 8. Solve for $M$: $$M = \frac{25.19}{19.6} \approx 1.29 \text{ kg}$$ **Final answer:** $$\boxed{M \approx 1.29 \text{ kg}}$$