1. **State the problem:** A 44 kg object is suspended by two cables making angles of 32° and 40° with the horizontal. We need to find the tensions $T_1$ and $T_2$ in the cables. 2. **Known values:** Mass $m = 44$ kg, gravity $g = 9.8$ m/s$^2$, angles $\theta_1 = 32^\circ$, $\theta_2 = 40^\circ$. 3. **Calculate the weight:** $$W = mg = 44 \times 9.8 = 431.2 \text{ N}$$ 4. **Set up equilibrium equations:** Since the object is at rest, the sum of forces in both horizontal and vertical directions is zero. - Horizontal components: $$T_1 \cos 32^\circ = T_2 \cos 40^\circ$$ - Vertical components: $$T_1 \sin 32^\circ + T_2 \sin 40^\circ = W$$ 5. **Express $T_1$ in terms of $T_2$ from horizontal equation:** $$T_1 = T_2 \frac{\cos 40^\circ}{\cos 32^\circ}$$ 6. **Substitute into vertical equation:** $$\left(T_2 \frac{\cos 40^\circ}{\cos 32^\circ}\right) \sin 32^\circ + T_2 \sin 40^\circ = 431.2$$ 7. **Simplify:** $$T_2 \left( \frac{\cos 40^\circ}{\cos 32^\circ} \sin 32^\circ + \sin 40^\circ \right) = 431.2$$ Calculate the trigonometric values: $$\cos 40^\circ \approx 0.7660, \cos 32^\circ \approx 0.8480, \sin 32^\circ \approx 0.5299, \sin 40^\circ \approx 0.6428$$ Calculate the coefficient: $$\frac{0.7660}{0.8480} \times 0.5299 + 0.6428 = 0.4785 + 0.6428 = 1.1213$$ 8. **Solve for $T_2$:** $$T_2 = \frac{431.2}{1.1213} \approx 384.5 \text{ N}$$ 9. **Find $T_1$:** $$T_1 = 384.5 \times \frac{0.7660}{0.8480} = 384.5 \times 0.9035 \approx 347.3 \text{ N}$$ **Final answer:** $$T_1 \approx 347.3 \text{ N}, \quad T_2 \approx 384.5 \text{ N}$$