Let $\mathbf{v}_0$ be a vector.  The vector $\mathbf{v}_0$ is rotated about the origin by an angle of $42^\circ$ counter-clockwise, taking it to vector $\mathbf{v}_1.$  The vector $\mathbf{v}_1$ is then reflected over the line with direction vector $\begin{pmatrix} \cos 108^\circ \\ \sin 108^\circ \end{pmatrix},$ taking it to vector $\mathbf{v}_2.$

The vector $\mathbf{v}_2$ can also be produced by reflecting the vector $\mathbf{v}_0$ over the line with direction vector $\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix},$ where $\theta$ is an acute angle.  Find $\theta.$
The matrix for the rotation is given by
\[\begin{pmatrix} \cos 42^\circ & -\sin 42^\circ \\ \sin 42^\circ & \cos 42^\circ \end{pmatrix}.\]In general, the matrix for reflecting over the line with direction vector $\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}$ is given by $\begin{pmatrix} \cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & -\cos 2 \theta \end{pmatrix}.$  So here, the matrix for the reflection is
\[\begin{pmatrix} \cos 216^\circ & \sin 216^\circ \\ \sin 216^\circ & -\cos 216^\circ \end{pmatrix}.\]Hence, the matrix that takes $\mathbf{v}_0$ to $\mathbf{v}_2$ is
\begin{align*}
\begin{pmatrix} \cos 216^\circ & \sin 216^\circ \\ \sin 216^\circ & -\cos 216^\circ \end{pmatrix} \begin{pmatrix} \cos 42^\circ & -\sin 42^\circ \\ \sin 42^\circ & \cos 42^\circ \end{pmatrix} &= \begin{pmatrix} \cos 216^\circ \cos 42^\circ + \sin 216^\circ \sin 42^\circ & -\cos 216^\circ \sin 42^\circ + \sin 216^\circ \cos 42^\circ \\ \sin 216^\circ \cos 42^\circ - \cos 216^\circ \sin 42^\circ & -\sin 216^\circ \sin 42^\circ - \cos 216^\circ \cos 42^\circ \end{pmatrix} \\
&= \begin{pmatrix} \cos (216^\circ - 42^\circ) & \sin (216^\circ - 42^\circ) \\ \sin (216^\circ - 42^\circ) & -\cos (216^\circ - 42^\circ) \end{pmatrix} \\
&= \begin{pmatrix} \cos 174^\circ & \sin 174^\circ \\ \sin 174^\circ & -\cos 174^\circ \end{pmatrix}.
\end{align*}Thus, $\theta = 174^\circ/2 = \boxed{87^\circ}.$