If a magnet is suspended at an angle of ${30}^0$ to the magnetic meridian, the dip needle makes an angle of ${60}^0$ with the horizontal. The real dip is 

$\tan \theta =\left(B_V/B_H\right)$  [real dip is $\theta$ ]…………………(1)
In the place situated at ${30}^0$ with the magnetic meridian, the horizontal component of the 
earth’s magnetic field will be $B_H\cos {30}^0$ while the vertical component will be  $B_v$. The angle 
of dip in this place is
$\tan {60}^o=\frac{B_v}{B_H\cos {30}^o}$ 
 $\Rightarrow t\mathrm{an}{60}^o.\cos {30}^o=\frac{B_v}{B_H}$

From equation (1), 

 $\Rightarrow \tan \theta =\sqrt{3}.\frac{\sqrt{3}}{2}$
 $\Rightarrow \tan \theta =\frac{3}{2}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{2}\right)$ .