At a place, the dip angle is 30° when dip circle is in geographical meridian and the dip angle is found to be 45° when the dip circle is in a plane perpendicular to the geographical meridian. The dip angle in magnetic meridian is

Solution:

Let $\delta$ be the true dip angle in the magnetic meridian.

Let $\delta_1$ be the apparent dip angle in the geographical meridian, so $\delta_1 = 30^\circ$.

Let $\delta_2$ be the apparent dip angle in the plane perpendicular to the geographical meridian, so $\delta_2 = 45^\circ$.

Let $\phi$ be the angle between the geographical meridian and the magnetic meridian.

The formula for apparent dip is given by $\tan \delta' = \frac{\tan \delta}{\cos \alpha}$, where $\alpha$ is the angle between the plane of the dip circle and the magnetic meridian.

1. When the dip circle is in the geographical meridian, the angle between the geographical meridian and the magnetic meridian is $\phi$.

   So, $\tan \delta_1 = \frac{\tan \delta}{\cos \phi}$

   $\tan 30^\circ = \frac{\tan \delta}{\cos \phi}$ (Equation 1)

2. When the dip circle is in a plane perpendicular to the geographical meridian, the angle between this plane and the magnetic meridian is $(90^\circ - \phi)$.

   So, $\tan \delta_2 = \frac{\tan \delta}{\cos (90^\circ - \phi)}$

   $\tan 45^\circ = \frac{\tan \delta}{\sin \phi}$ (since $\cos(90^\circ - \phi) = \sin \phi$) (Equation 2)

From Equation 1: $\cos \phi = \frac{\tan \delta}{\tan 30^\circ}$

From Equation 2: $\sin \phi = \frac{\tan \delta}{\tan 45^\circ}$

Using the identity $\sin^2 \phi + \cos^2 \phi = 1$:

$(\frac{\tan \delta}{\tan 45^\circ})^2 + (\frac{\tan \delta}{\tan 30^\circ})^2 = 1$

$\tan^2 \delta \left( \frac{1}{\tan^2 45^\circ} + \frac{1}{\tan^2 30^\circ} \right) = 1$

We know $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

$\tan^2 \delta \left( \frac{1}{1^2} + \frac{1}{(1/\sqrt{3})^2} \right) = 1$

$\tan^2 \delta \left( 1 + \frac{1}{1/3} \right) = 1$

$\tan^2 \delta (1 + 3) = 1$

$4 \tan^2 \delta = 1$

$\tan^2 \delta = \frac{1}{4}$

$\tan \delta = \frac{1}{2}$ (Since dip angle is positive)

Therefore, the dip angle in the magnetic meridian is $\delta = \tan^{-1}(\frac{1}{2})$.