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At a place on the surface of earth, true dip is $60^{o}$ . when the dip circle is rotated through an angle $θ$ , the apparent dip is found to be $\tan^{- 1} (2)$ . Then $θ$ is equal to

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45o

tan⁡−123

30o

tan⁡−1(32)

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detailed solution

Correct option is C

BVBH= tanδo after rotation BvBHcosθ= tanδ or, tanδ0cosθ=tanδ here δ0=600 =3 and tanδ=2 hence cosθ=32 θ=300

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At a place on the surface of earth, true dip is 60o. when the dip circle is rotated through an angle θ, the apparent dip is found to be tan⁡−1(2). Then θ is equal to

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At a place on the surface of earth, true dip is $60^{o}$ . when the dip circle is rotated through an angle $θ$ , the apparent dip is found to be $\tan^{- 1} (2)$ . Then $θ$ is equal to