Three forces are acting on a body. Two of the forces are 10 N due East and 5 N at 60 degrees North-East. If the body is in equilibrium, what is the third force?

Let $\vec{F_1}$, $\vec{F_2}$, and $\vec{F_3}$ be the three forces.
Given:

• $\vec{F_1} = 10\,\text{N}$ due East.
• $\vec{F_2} = 5\,\text{N}$ at $60^\circ$ North-East. Since the body is in equilibrium, the vector sum of the forces is zero:

Therefore, the third force $\vec{F_3}$ is the negative of the resultant of the first two forces:

• For $\vec{F_1}$ ($10\,\text{N}$ due East, angle $0^\circ$): $$F_{1x} = 10 \cos(0^\circ) = 10 \times 1 = 10\,\text{N}$$ $$F_{1y} = 10 \sin(0^\circ) = 10 \times 0 = 0\,\text{N}$$
• For $\vec{F_2}$ ($5\,\text{N}$ at $60^\circ$ North-East, angle $60^\circ$): $$F_{2x} = 5 \cos(60^\circ) = 5 \times 0.5 = 2.5\,\text{N}$$ $$F_{2y} = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 5 \times 0.8660 = 4.330\,\text{N}$$

Let $\vec{R} = \vec{F_1} + \vec{F_2}$.

Since $\vec{F_3} = -\vec{R}$:

The magnitude $F_3$ is given by:

The direction $\theta_3$ is found using the tangent function. First, find the reference angle $\alpha$:

Since both $F_{3x}$ and $F_{3y}$ are negative, the third force lies in the third quadrant. The angle measured counter-clockwise from the positive x-axis is:

Alternatively, this direction can be described as $19.1^\circ$ South of West.

The magnitude of the third force is approximately $\boxed{\text{13.2 N}}$ and its direction is approximately $\boxed{\text{199.1}^\circ \text{ from the positive x-axis (or 19.1}^\circ \text{ South of West)}}$.