A magnetic needle can freely rotate in a vertical plane that makes...

Let the true dip be $\delta and the apparent dip be$\delta'$.<br><br>The relation between true dip and apparent dip in a plane making an angle$\alpha$with the magnetic meridian is given by<br><br>$\tan \delta' = \dfrac{\tan \delta}{\cos \alpha}$<br><br>Given$\delta' = \tan^{-1} \left(\dfrac{2}{\sqrt{3}}\right), we have $\tan \delta' =$\dfrac{2}{\sqrt{3}}.

The angle with the magnetic meridian is $\alpha = 60^\circ$, so $\cos \alpha =$\dfrac{1}{2}.

Substituting these values into the formula, we get

\dfrac{2}{\sqrt{3}} = \dfrac{\tan \delta}{1/2}

$\tan \delta =$\dfrac{2}{\sqrt{3}} \times \dfrac{1}{2} = \dfrac{1}{\sqrt{3}}

$\delta = 30^\circ$