1. **Problem Statement:** A body is in equilibrium under three forces. Two forces are given: one is 10 N acting due east, and the second is 5 N acting at 60° north of east. We need to find the magnitude and direction of the third force. 2. **Concept:** For a body to be in equilibrium, the vector sum of all forces must be zero. This means the third force must exactly balance the resultant of the first two forces. 3. **Step 1: Represent forces as vectors.** - Force 1: $\vec{F_1} = 10\hat{i}$ (east direction) - Force 2: $\vec{F_2} = 5(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = 5(0.5\hat{i} + 0.866\hat{j}) = 2.5\hat{i} + 4.33\hat{j}$ 4. **Step 2: Calculate resultant of first two forces.** $$\vec{R} = \vec{F_1} + \vec{F_2} = (10 + 2.5)\hat{i} + 4.33\hat{j} = 12.5\hat{i} + 4.33\hat{j}$$ 5. **Step 3: Find the third force.** Since the body is in equilibrium, the third force $\vec{F_3}$ must satisfy: $$\vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \implies \vec{F_3} = -\vec{R} = -12.5\hat{i} - 4.33\hat{j}$$ 6. **Step 4: Calculate magnitude of $\vec{F_3}$.** $$|\vec{F_3}| = \sqrt{(-12.5)^2 + (-4.33)^2} = \sqrt{156.25 + 18.75} = \sqrt{175} \approx 13.23$$ 7. **Step 5: Calculate direction of $\vec{F_3}$.** Direction angle $\theta$ is measured from east (positive x-axis) towards north (positive y-axis): $$\theta = \tan^{-1} \left( \frac{-4.33}{-12.5} \right) = \tan^{-1} (0.3464) = 19.1^\circ$$ Since both components are negative, $\vec{F_3}$ points to the southwest quadrant, so the angle is $180^\circ - 19.1^\circ = 160.9^\circ$ from east, or equivalently $19.1^\circ$ south of west. **Final answer:** - Magnitude of third force: $13.23$ N - Direction: $19.1^\circ$ south of west (or $160.9^\circ$ from east)