Apparent dip angle if the plane of a dip circle is set in the geographic meridian, $\tan {\delta }_1=\frac{\tan \varphi }{\cos \alpha }\Rightarrow \cos \alpha =\frac{\tan \varphi }{\tan {\delta }_1} \cdots \left(1\right)$

Apparent dip angle if the vertical plane perpendicular to the geographic meridian,

$\tan {\delta }_2=\frac{\tan \varphi }{\sin \alpha }\Rightarrow \sin \alpha =\frac{\tan \varphi }{\tan {\delta }_2} \cdots \left(2\right)$

From equations (1) and (2) we have

$$\begin{gathered} \Rightarrow \frac{\sin \alpha }{\cos \alpha }=\tan \alpha =\frac{\tan {\delta }_1}{\tan {\delta }_2} \\ \Rightarrow \alpha ={\tan }^{-1}\left(\frac{\tan {\delta }_1}{\tan {\delta }_2}\right) \end{gathered}$$