First we will find the eigenvector corresponding to the eigenvalue 1. Remember that the only fixed points (i.e. eigenvectors corresponding to the eigenvalue 1) of a reflection are those which are on the line used for the reflection. Therefore is on the line whose angle we are looking for. Let us denote our line in the generic form . Since the line must pass through the origin, thus and since is on the line, satisfies . Therefore, the line we are looking for is given by the equation .

We know that the slope of the line is the same as the tangent of the angle: therefore .

The other eigenvector will be orthogonal to the previous one: this vectors on this line will be inverted under the reflection and thus will correspond to the eigenvalue . The angle orthogonal to which lies in the given interval is .

**Answer:** .