A dip needle lies initially in the magnetic meridian when it shows an angle of dip θ at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip θ. Then (tan θ' / tan θ) is:(a) (1/cos x)(b) (1/sin x)(c) (1/tan x)(d) cos x

In first case

\(tan \theta =\frac{B_{V}}{B_{H}}\)

In second case

\(tan \theta {}’=\frac{B_{V}}{B_{H}Cosx}\\\frac{tan\theta {}’}{tan\theta }=\frac{B_{V}}{B_{H}Cosx}\times \frac{B_{H}}{B_{V}}\)

We get,

\(\frac{tan\theta {}’}{tan\theta }=\frac{1}{Cosx}\)

Hence, the correct option is (a)

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