If a magnet is suspended at an angle of 30^o to the magnetic meridian, the dip needle makes an angle of 45^o with the horizontal. The real dip is

Let the actual angle of dip (in the magnetic meridian) is $\mathrm{\theta }$. lf B_H and, B_V be the horizontal and vertical components of earth's magnetic field respectively, then
$\tan \mathrm{\theta }=\left({\mathrm{B}}_{\mathrm{V}}/{\mathrm{B}}_{\mathrm{H}}\right)$
In the plane situated at 30. with the magnetic meridian, see fig.(9) , the horizontal component of earth's magnetic
field will be B_H cos 30^o while the vertical component will be B_V. The angle of dip in this plane is
$\begin{array}{l}\tan {45}^{\circ }=\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}\cos {30}^{\circ }}\\ \therefore \frac{\tan \mathrm{\theta }}{\tan {45}^{\circ }}=\cos {30}^{\circ }\text{ or }\frac{\tan \mathrm{\theta }}{1}=\frac{\sqrt{3}}{2}\\ \therefore \mathrm{\theta }={\tan }^{-1}(\sqrt{3}/2)\end{array}$