1. **Problem Statement:** We have a force diagram with two tension forces $T_A$ and $T_B$ each making a 60° angle with the vertical weight force $W$. We want to find the tensions $T_A$ and $T_B$ given the weight $W=294$ N. 2. **Horizontal Force Equilibrium:** The sum of horizontal forces must be zero: $$-T_A \cos 60^\circ + T_B \cos 60^\circ = 0$$ This implies: $$T_A \cos 60^\circ = T_B \cos 60^\circ$$ Since $\cos 60^\circ$ is the same for both, it follows that: $$T_A = T_B$$ 3. **Vertical Force Equilibrium:** The sum of vertical forces must balance the weight: $$T_A \sin 60^\circ + T_B \sin 60^\circ = W$$ Using $T_A = T_B$, this becomes: $$2 T_A \sin 60^\circ = 294$$ 4. **Solving for $T_A$:** $$T_A = \frac{294}{2 \sin 60^\circ}$$ Recall $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$: $$T_A = \frac{294}{2 \times 0.866} = \frac{294}{1.732} \approx 169.74$$ 5. **Conclusion:** Both tensions are equal: $$T_A = T_B \approx 169.74 \text{ N}$$ This means each tension force supports half the vertical component needed to balance the weight, considering the 60° angle.