1. **State the problem:** We have two forces, 6N and 4N, acting at an angle of 60 degrees to each other. We need to find the resultant force and the angle $\alpha$ it makes with the 6N force using $\tan \alpha$. 2. **Formula used:** The magnitude of the resultant force $R$ when two forces $F_1$ and $F_2$ act at an angle $\theta$ is given by the law of cosines: $$R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta}$$ The angle $\alpha$ between $F_1$ and the resultant $R$ is found using the law of sines or by: $$\tan \alpha = \frac{F_2 \sin \theta}{F_1 + F_2 \cos \theta}$$ 3. **Substitute values:** - $F_1 = 6$N - $F_2 = 4$N - $\theta = 60^\circ$ Calculate $R$: $$R = \sqrt{6^2 + 4^2 + 2 \times 6 \times 4 \times \cos 60^\circ}$$ $$= \sqrt{36 + 16 + 48 \times 0.5}$$ $$= \sqrt{36 + 16 + 24} = \sqrt{76}$$ $$\approx 8.7178 \text{ N}$$ 4. **Calculate $\tan \alpha$:** $$\tan \alpha = \frac{4 \times \sin 60^\circ}{6 + 4 \times \cos 60^\circ}$$ $$= \frac{4 \times 0.8660}{6 + 4 \times 0.5} = \frac{3.464}{6 + 2} = \frac{3.464}{8} = 0.433$$ 5. **Find $\alpha$:** $$\alpha = \tan^{-1}(0.433) \approx 23.41^\circ$$ **Final answer:** - Resultant force $R \approx 8.72$ N - Angle $\alpha \approx 23.41^\circ$ with the 6N force