1. **Problem Statement:** We have a trapezium ABCD with sides AD = DC = CB = 1 m and AB = 2 m. Forces act along sides AD (1 N), DC (3 N), CB (6 N), and AC (2\sqrt{3} N). We need to find the magnitude of the resultant force and the angle it makes with side AB. 2. **Approach:** Resolve each force into components along the x-axis (along AB) and y-axis (perpendicular to AB). Then sum all components to find the resultant vector. 3. **Coordinate Setup:** Let AB lie along the positive x-axis. Angles are given relative to AB or known angles in the trapezium. 4. **Force Components:** - Force along AD (1 N): AD makes 30° with AB (given \angle DAB = 30°). Components: $$F_{AD,x} = 1 \cos 30^\circ = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ $$F_{AD,y} = 1 \sin 30^\circ = 1 \times \frac{1}{2} = \frac{1}{2}$$ - Force along DC (3 N): DC makes 150° with AB (since \angle ADC = 120°, and AB is horizontal, DC is 180° - 30° = 150° from AB). Components: $$F_{DC,x} = 3 \cos 150^\circ = 3 \times (-\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2}$$ $$F_{DC,y} = 3 \sin 150^\circ = 3 \times \frac{1}{2} = \frac{3}{2}$$ - Force along CB (6 N): CB makes 60° with AB (given \angle ABC = 60°). Since force acts along CB from C to B, direction is 180° - 60° = 120° from AB. $$F_{CB,x} = 6 \cos 120^\circ = 6 \times (-\frac{1}{2}) = -3$$ $$F_{CB,y} = 6 \sin 120^\circ = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$$ - Force along AC (2\sqrt{3} N): AC makes 30° with AB (given \angle CAB = 30°). Components: $$F_{AC,x} = 2\sqrt{3} \cos 30^\circ = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 2 \times \frac{3}{2} = 3$$ $$F_{AC,y} = 2\sqrt{3} \sin 30^\circ = 2\sqrt{3} \times \frac{1}{2} = \sqrt{3}$$ 5. **Sum Components:** $$F_x = \frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} - 3 + 3 = \left(\frac{\sqrt{3}}{2} - \frac{3\sqrt{3}}{2}\right) + ( -3 + 3 ) = -\sqrt{3} + 0 = -\sqrt{3}$$ $$F_y = \frac{1}{2} + \frac{3}{2} + 3\sqrt{3} + \sqrt{3} = 2 + 4\sqrt{3}$$ 6. **Resultant Magnitude:** $$F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-\sqrt{3})^2 + (2 + 4\sqrt{3})^2} = \sqrt{3 + (2 + 4\sqrt{3})^2}$$ Calculate: $$(2 + 4\sqrt{3})^2 = 2^2 + 2 \times 2 \times 4\sqrt{3} + (4\sqrt{3})^2 = 4 + 16\sqrt{3} + 16 \times 3 = 4 + 16\sqrt{3} + 48 = 52 + 16\sqrt{3}$$ So, $$F = \sqrt{3 + 52 + 16\sqrt{3}} = \sqrt{55 + 16\sqrt{3}}$$ 7. **Angle with AB:** $$\theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) = \tan^{-1} \left( \frac{2 + 4\sqrt{3}}{-\sqrt{3}} \right)$$ Since $F_x$ is negative and $F_y$ positive, the vector lies in the second quadrant. Calculate inside tangent: $$\frac{2 + 4\sqrt{3}}{-\sqrt{3}} = -\frac{2 + 4\sqrt{3}}{\sqrt{3}}$$ Numerically, $$2 + 4\sqrt{3} \approx 2 + 4 \times 1.732 = 2 + 6.928 = 8.928$$ $$\frac{8.928}{1.732} \approx 5.154$$ So, $$\theta = 180^\circ - \tan^{-1}(5.154) \approx 180^\circ - 79^\circ = 101^\circ$$ **Final answers:** - Magnitude of resultant force: $$\boxed{\sqrt{55 + 16\sqrt{3}} \text{ N} \approx 9.7 \text{ N}}$$ - Angle with AB: $$\boxed{101^\circ}$$