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A magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{0}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{- 1} \frac{2}{\sqrt{3}}$ with the horizontal, what would be the dip at that place ? ___________in degree.

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answer is 30.

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Detailed Solution

$t a n δ^{1} = \frac{\tan δ}{\cos θ} \Rightarrow \frac{2}{\sqrt{3}} = \frac{\tan δ}{\cos 60^{0}}, ∴ t a n δ = \frac{1}{\sqrt{3}} \Rightarrow δ = 30^{0} .$

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