Vectors $\mathbf{a}$ and $\mathbf{b}$ satisfy $\|\mathbf{a}\| = 5$ and $\|\mathbf{b}\| = 4.$  Also, the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is $60^\circ.$  Find $\|\mathbf{a} - \mathbf{b}\|.$
We have that
\[\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos 60^\circ = 5 \cdot 4 \cdot \frac{1}{2} = 10.\]Then
\begin{align*}
\|\mathbf{a} - \mathbf{b}\|^2 &= (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \\
&= \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \\
&= \|\mathbf{a}\|^2 - 2 \mathbf{a} \cdot \mathbf{b} + \|\mathbf{b}\|^2 \\
&= 5^2 - 2 \cdot 10 + 4^2 \\
&= 21.
\end{align*}Therefore, $\|\mathbf{a} - \mathbf{b}\| = \boxed{\sqrt{21}}.$