(b) $\tan\phi = \frac{\tan\phi}{\cos\beta};$ where $\phi =$ Apparent angle of dip, $\phi =$ True angle of dip, $\beta =$ Angle made by vertical plane with magnetic meridian
$\Rightarrow \tan\phi = \frac{\tan{60{^\circ}}}{\cos{30{^\circ}}} = 2 \Rightarrow \phi = \tan^{- 1}{(2)}$