At a place true magnetic dip is 60°. The apparent dip when plane of dip circle is placed at angle of 60° with the magnetic meridian is

Solution:

Let the true magnetic dip be $\delta$ and the apparent dip be $\delta'$.

The angle between the plane of the dip circle and the magnetic meridian is $\theta$.

The relationship between true dip ($\delta$), apparent dip ($\delta'$), and the angle ($\theta$) is given by the formula:

$\tan \delta' = \frac{\tan \delta}{\cos \theta}$

Given:

True magnetic dip, $\delta = 60^\circ$

Angle of the plane of dip circle with the magnetic meridian, $\theta = 60^\circ$

Substitute the values into the formula:

$\tan \delta' = \frac{\tan 60^\circ}{\cos 60^\circ}$

We know that $\tan 60^\circ = \sqrt{3}$ and $\cos 60^\circ = \frac{1}{2}$.

$\tan \delta' = \frac{\sqrt{3}}{1/2}$

$\tan \delta' = 2\sqrt{3}$

Therefore, the apparent dip is:

$\delta' = \tan^{-1}(2\sqrt{3})$

Thus, option (4) is the correct answer.