The plane of a dip circle is set in the geographic meridian and the apparent dip is $δ_{1}$ . It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $δ_{2}$ . The declination q at the place is

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${q = tan}^{- 1} (tan δ_{1} tan δ_{2})$

${q = tan}^{- 1} (tan δ_{1} - tan δ_{2})$

${q = tan}^{- 1} (\frac{tan δ_{1}}{tan δ_{2}})$

${q = tan}^{- 1} (tan δ_{1} +tan δ_{2})$

answer is C.

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Detailed Solution

Let, q be the declination at the place.
$\begin{array}{l} \tan δ_{1} = \frac{V}{H \cos q} \\ a n d \tan δ_{2} = \frac{V}{H \cos (90^{0} - q)} = \frac{V}{H \sin q} \\ \Rightarrow \frac{\tan δ_{1}}{\tan δ_{2}} = \frac{\sin q}{\cos q} = \tan q \\ \Rightarrow q = \tan^{- 1} (\frac{\tan δ_{1}}{\tan δ_{2}}) \end{array}$