The plane of the dip circle is set in the geographical meridian and apparent dip is $θ_{1}$ . It is then set in a vertical plane perpendicular to the geographical meridian, the apparent dip becomes $θ_{2}$ . The angle of declination $α$ at that place is given by

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$\tan α = \sqrt{\tan θ_{1} \tan θ_{2}}$

$\tan α = \sqrt{\tan^{2} θ_{1} \tan^{2} θ_{2}}$

$\tan α = \frac{\tan θ_{1}}{\tan θ_{2}}$

$\tan α = \frac{\tan θ_{2}}{\tan θ_{1}}$

answer is C.

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Detailed Solution

$\begin{array}{l} \tan θ_{1} = \frac{\tan θ}{\cos α} \\ a n d \tan θ_{2} = \frac{\tan θ}{\cos (90^{0} - α)} = \frac{\tan θ}{\sin α} \\ \Rightarrow \cos α = \frac{\tan θ}{\tan θ_{1}} ................. (1) \\ a n d \sin α = \frac{\tan θ}{\tan θ_{2}} ................. (2) \\ D i v i d i n g (2) b y (1), w e h a v e \\ \tan α = \frac{\tan θ_{1}}{\tan θ_{2}} \end{array}$