For apparent dip, \(\tan {\phi }' = \frac{{{B_V}}}{{{B_H}\cos \beta }} = \frac{{{B_V}}}{{{B_H}\cos {{30}^o}}} = \frac{{2{B_V}}}{{\sqrt 3 {B_H}}}\) or
\(\tan {45^o} = \frac{2}{{\sqrt 3 }}.\tan \phi \) or
\(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\)