1. **Problem statement:** We have two forces: one is 700 N at 15° to the negative x-axis (upward-left), and another force $F$ at angle $\theta$ above the positive x-axis (upward-right). The resultant force has magnitude 500 N directed along the positive y-axis. We need to find the magnitude $F$ and angle $\theta$. 2. **Set up coordinate system and components:** - Force 1 (700 N) makes an angle of 15° with the negative x-axis, so its components are: $$F_{1x} = -700 \cos 15^\circ$$ $$F_{1y} = 700 \sin 15^\circ$$ - Force 2 ($F$) makes an angle $\theta$ with the positive x-axis, so its components are: $$F_{2x} = F \cos \theta$$ $$F_{2y} = F \sin \theta$$ 3. **Resultant force components:** The resultant force $\vec{R}$ has magnitude 500 N along positive y-axis, so: $$R_x = 0$$ $$R_y = 500$$ 4. **Write equations for components:** - For x-components: $$F_{1x} + F_{2x} = 0 \implies -700 \cos 15^\circ + F \cos \theta = 0$$ - For y-components: $$F_{1y} + F_{2y} = 500 \implies 700 \sin 15^\circ + F \sin \theta = 500$$ 5. **Calculate known trigonometric values:** $$\cos 15^\circ \approx 0.9659$$ $$\sin 15^\circ \approx 0.2588$$ 6. **Rewrite equations:** $$-700 \times 0.9659 + F \cos \theta = 0 \implies F \cos \theta = 675.13$$ $$700 \times 0.2588 + F \sin \theta = 500 \implies 181.16 + F \sin \theta = 500$$ $$F \sin \theta = 500 - 181.16 = 318.84$$ 7. **Square and add both equations to solve for $F$:** $$ (F \cos \theta)^2 + (F \sin \theta)^2 = 675.13^2 + 318.84^2$$ $$F^2 (\cos^2 \theta + \sin^2 \theta) = 675.13^2 + 318.84^2$$ $$F^2 = 675.13^2 + 318.84^2$$ $$F = \sqrt{675.13^2 + 318.84^2}$$ $$F \approx \sqrt{455796 + 101670} = \sqrt{557466} \approx 746.63$$ 8. **Find angle $\theta$:** $$\tan \theta = \frac{F \sin \theta}{F \cos \theta} = \frac{318.84}{675.13} \approx 0.472$$ $$\theta = \arctan(0.472) \approx 25.3^\circ$$ **Final answer:** - Magnitude of force $F$ is approximately **746.63 N**. - Direction $\theta$ is approximately **25.3°** above the positive x-axis.