A magnet hung at 45º with magnetic meridian makes an angle of 60º with the horizontal. The actual value of the angle of dip is

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

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

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

$\tan^{- 1} (\sqrt{6})$

answer is A.

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Detailed Solution

$\begin{array}{l} \tan θ^{'} = \frac{\tan θ}{\cos α} \\ θ^{'} = 60^{\circ} \\ α = 45^{\circ} \\ \sqrt{3} = \frac{\tan θ}{\frac{1}{\sqrt{2}}} \\ \tan θ = \sqrt{\frac{3}{2}} \\ θ = \tan^{- 1} \sqrt{\frac{3}{2}} \end{array}$

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