Assume the dipole model for Earth's magnetic field $B$,which is given by:
$B_v = \text{vertical component of magnetic field} = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \text{horizontal component of magnetic field} = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which the dip angle is $\pm 45^{\circ}$.

(A) The dip angle $\delta$ is defined by the relation $\tan \delta = \frac{B_v}{B_H}$.
Given the expressions for the vertical component $B_v$ and the horizontal component $B_H$:
$B_v = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
Substituting these into the expression for $\tan \delta$:
$\tan \delta = \frac{\frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}}{\frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}} = \frac{2 \cos \theta}{\sin \theta} = 2 \cot \theta$
We are looking for the loci of points where the dip angle $\delta = \pm 45^{\circ}$.
Since $\tan(\pm 45^{\circ}) = \pm 1$,we have:
$\pm 1 = 2 \cot \theta$
$\cot \theta = \pm 0.5$
$\tan \theta = \pm 2$
Since $\theta = 90^{\circ} - \lambda$ (where $\lambda$ is the latitude),the locus of points is defined by the condition $\tan(90^{\circ} - \lambda) = \pm 2$,which simplifies to $\cot \lambda = \pm 2$ or $\tan \lambda = \pm 0.5$.