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At a point on the surface of earth true dip measured by a dip needle is $45^{o}$ . Now the plane of the dip needle is rotated through an angle of $θ$ and the new dip measured by the dip needle is $60^{o}$ . Then $θ$ is

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$\cos^{- 1} (\frac{1}{\sqrt{3}})$

$\cos^{- 1} (\frac{\sqrt{3}}{2})$

$\cos^{- 1} (\frac{1}{2})$

$\tan^{- 1} (\frac{\sqrt{3}}{2})$

answer is A.

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Detailed Solution

$\tan δ^{1} = \frac{\tan δ}{\cos θ} \Rightarrow \cos θ = \frac{\tan δ}{\tan δ^{1}} = \frac{\tan 45^{o}}{\tan 60^{o}}$

$\Rightarrow \cos θ = \frac{1}{\sqrt{3}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{3}})$

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