1. **State the problem:** Find the angle between the vectors $\mathbf{u} = 3\mathbf{i} + 2\mathbf{j}$ and $\mathbf{v} = 5\mathbf{i} - \mathbf{j}$.\n\n2. **Formula used:** The angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by the formula:\n$$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}$$\nwhere $\mathbf{u} \cdot \mathbf{v}$ is the dot product and $\|\mathbf{u}\|$, $\|\mathbf{v}\|$ are the magnitudes of the vectors.\n\n3. **Calculate the dot product:**\n$$\mathbf{u} \cdot \mathbf{v} = (3)(5) + (2)(-1) = 15 - 2 = 13$$\n\n4. **Calculate the magnitudes:**\n$$\|\mathbf{u}\| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$$\n$$\|\mathbf{v}\| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$$\n\n5. **Substitute into the formula:**\n$$\cos \theta = \frac{13}{\sqrt{13} \times \sqrt{26}} = \frac{13}{\sqrt{338}}$$\n\n6. **Simplify the fraction:**\n$$\cos \theta = \frac{\cancel{13}}{\sqrt{\cancel{13} \times 26}} = \frac{1}{\sqrt{26}}$$\n\n7. **Calculate the angle:**\n$$\theta = \cos^{-1} \left( \frac{1}{\sqrt{26}} \right)$$\nUsing a calculator,\n$$\theta \approx 78.69^\circ$$\n\n**Final answer:** The angle between $\mathbf{u}$ and $\mathbf{v}$ is approximately $78.69^\circ$.