1. **State the problem:** We have a smooth ring R of mass 0.8 kg threaded on a string fixed at points A and B, with A vertically above B. The string sections AR and BR make angles 50° and 20° with the horizontal, respectively. A horizontal force of magnitude $X$ newtons acts on R. The ring is in equilibrium. We need to find (i) the tension in the string and (ii) the value of $X$ using Lami's theorem. 2. **Identify forces acting on the ring:** - Tension $T$ in the string acts along AR and BR (same magnitude since string is continuous and light). - Weight $W = mg = 0.8 \times 9.8 = 7.84$ N acts vertically downward. - Horizontal force $X$ acts horizontally. 3. **Resolve angles for Lami's theorem:** Lami's theorem states for three coplanar, concurrent forces in equilibrium: $$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$$ where $\alpha, \beta, \gamma$ are the angles opposite to forces $F_1, F_2, F_3$ respectively. 4. **Determine angles between forces:** - The tension forces along AR and BR make angles 50° and 20° with horizontal. - The angle between the two tension forces is $50° + 20° = 70°$. - The weight acts vertically downward. - The horizontal force $X$ acts horizontally. 5. **Find angles opposite to each force:** - Angle opposite to $T$ along AR is angle between $X$ and $W$. - Angle opposite to $T$ along BR is angle between $X$ and $W$. - The three forces are $T$ (along AR), $T$ (along BR), and resultant of $X$ and $W$. 6. **Calculate angles for Lami's theorem:** - Angle between $T$ along AR and $W$ is $90° - 50° = 40°$ (since AR is 50° above horizontal, weight is vertical). - Angle between $T$ along BR and $W$ is $90° - 20° = 70°$. - Angle between $T$ along AR and $T$ along BR is $70°$. 7. **Apply Lami's theorem:** Let tensions along AR and BR be $T$. The three forces are $T$, $T$, and the resultant of $X$ and $W$. The angles opposite to these forces are: - Opposite to $T$ along AR: $70°$ - Opposite to $T$ along BR: $40°$ - Opposite to resultant force $R$: $70°$ Using Lami's theorem: $$\frac{T}{\sin 70°} = \frac{T}{\sin 40°} = \frac{R}{\sin 70°}$$ Since tensions are equal, this implies the resultant force $R$ is equal to $T$. 8. **Calculate resultant force $R$ from $X$ and $W$:** Since $X$ is horizontal and $W$ is vertical downward, $$R = \sqrt{X^2 + W^2}$$ 9. **Equate tensions and resultant:** $$T = R = \sqrt{X^2 + 7.84^2}$$ 10. **Resolve horizontal and vertical components of tension:** Horizontal components: $$T \cos 50° + T \cos 20° = X$$ Vertical components: $$T \sin 50° = T \sin 20° + 7.84$$ 11. **Calculate $T$ from vertical components:** $$T \sin 50° - T \sin 20° = 7.84$$ $$T (\sin 50° - \sin 20°) = 7.84$$ Calculate sines: $$\sin 50° \approx 0.7660, \sin 20° \approx 0.3420$$ $$T (0.7660 - 0.3420) = 7.84$$ $$T \times 0.424 = 7.84$$ $$T = \frac{7.84}{0.424} \approx 18.49 \text{ N}$$ 12. **Calculate $X$ from horizontal components:** $$X = T \cos 50° + T \cos 20° = 18.49 (0.6428 + 0.9397) = 18.49 \times 1.5825 \approx 29.26 \text{ N}$$ **Final answers:** (i) Tension in the string $T \approx 18.49$ N (ii) Horizontal force $X \approx 29.26$ N