1. **Stating the problem:** We have a net of fish at point A suspended by ropes AB and AC. The rope AB makes an angle $\theta = 14^\circ$ with the horizontal, and the rope AC makes an angle $\phi = 19^\circ$ with the vertical. We want to find the forces $F_{AB}$ and $F_{AC}$ in the ropes. 2. **Understanding the setup:** - The net at A is in equilibrium, so the sum of forces must be zero. - The forces in the ropes act along the ropes. - Let $F_{AB}$ be the tension in rope AB and $F_{AC}$ be the tension in rope AC. 3. **Setting up the force components:** - For rope AB, angle $\theta=14^\circ$ with horizontal means: - Horizontal component: $F_{AB} \cos \theta$ - Vertical component: $F_{AB} \sin \theta$ - For rope AC, angle $\phi=19^\circ$ with vertical means: - Vertical component: $F_{AC} \cos \phi$ - Horizontal component: $F_{AC} \sin \phi$ 4. **Equilibrium conditions:** - Horizontal forces sum to zero: $$F_{AB} \cos \theta = F_{AC} \sin \phi$$ - Vertical forces sum to zero (assuming weight $W$ acts downward at A): $$F_{AB} \sin \theta + F_{AC} \cos \phi = W$$ 5. **Solving for $F_{AB}$ and $F_{AC}$:** From horizontal equilibrium: $$F_{AB} = \frac{F_{AC} \sin \phi}{\cos \theta}$$ Substitute into vertical equilibrium: $$\frac{F_{AC} \sin \phi}{\cos \theta} \sin \theta + F_{AC} \cos \phi = W$$ Simplify: $$F_{AC} \left( \frac{\sin \phi \sin \theta}{\cos \theta} + \cos \phi \right) = W$$ 6. **Final expressions:** $$F_{AC} = \frac{W}{\frac{\sin \phi \sin \theta}{\cos \theta} + \cos \phi}$$ $$F_{AB} = \frac{F_{AC} \sin \phi}{\cos \theta}$$ 7. **Explanation:** - We used the equilibrium of forces in horizontal and vertical directions. - Angles were used to resolve forces into components. - The weight $W$ must be known to find numeric values. **Note:** Without the weight $W$, we can only express $F_{AB}$ and $F_{AC}$ in terms of $W$.