1. **State the problem:** A bead Q on a horizontal wire AB experiences three forces: a force of magnitude $F$ N at $70^\circ$ above the horizontal to the left, a 10 N force at $30^\circ$ below the horizontal to the left, and a 20 N force horizontally to the right. The resultant force $R$ N acts at $15^\circ$ below the horizontal to the right. Find $F$ and $R$. 2. **Set up coordinate system and resolve forces:** Take right as positive x-direction and up as positive y-direction. - Force $F$ at $70^\circ$ above left: components are $$F_x = -F \cos 70^\circ, \quad F_y = F \sin 70^\circ$$ - Force 10 N at $30^\circ$ below left: $$10_x = -10 \cos 30^\circ, \quad 10_y = -10 \sin 30^\circ$$ - Force 20 N horizontally right: $$20_x = 20, \quad 20_y = 0$$ - Resultant $R$ at $15^\circ$ below right: $$R_x = R \cos 15^\circ, \quad R_y = -R \sin 15^\circ$$ 3. **Write equations for resultant components:** Sum of x-components: $$-F \cos 70^\circ - 10 \cos 30^\circ + 20 = R \cos 15^\circ$$ Sum of y-components: $$F \sin 70^\circ - 10 \sin 30^\circ + 0 = -R \sin 15^\circ$$ 4. **Substitute known values of cos and sin:** $$\cos 70^\circ \approx 0.3420, \quad \sin 70^\circ \approx 0.9397$$ $$\cos 30^\circ \approx 0.8660, \quad \sin 30^\circ = 0.5$$ $$\cos 15^\circ \approx 0.9659, \quad \sin 15^\circ \approx 0.2588$$ 5. **Rewrite equations:** $$-0.3420F - 8.660 + 20 = 0.9659 R$$ $$0.9397F - 5 = -0.2588 R$$ Simplify first: $$-0.3420F + 11.34 = 0.9659 R \quad (1)$$ $$0.9397F - 5 = -0.2588 R \quad (2)$$ 6. **Express $R$ from (1):** $$R = \frac{-0.3420F + 11.34}{0.9659}$$ 7. **Substitute into (2):** $$0.9397F - 5 = -0.2588 \times \frac{-0.3420F + 11.34}{0.9659}$$ Multiply both sides by 0.9659: $$0.9659 \times (0.9397F - 5) = -0.2588 (-0.3420F + 11.34)$$ Calculate left: $$0.9075F - 4.8295 = 0.0885F - 2.935$$ 8. **Bring terms to one side:** $$0.9075F - 0.0885F = -2.935 + 4.8295$$ $$0.819F = 1.8945$$ 9. **Solve for $F$:** $$F = \frac{1.8945}{0.819} \approx 2.31$$ 10. **Find $R$ using (1):** $$R = \frac{-0.3420 \times 2.31 + 11.34}{0.9659} = \frac{-0.790 + 11.34}{0.9659} = \frac{10.55}{0.9659} \approx 10.92$$ **Final answers:** $$F \approx 2.31 \text{ N}, \quad R \approx 10.92 \text{ N}$$