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 ?

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Solution

$t a n δ_{1} = \frac{tan δ}{cos θ}$

$∴ t a n δ = t a n δ_{1} c o s θ$

= $t a n (t a n^{- 1} \frac{2}{\sqrt{3}}) c o s 60^{0}$

= $\frac{2}{\sqrt{3}} \times \frac{1}{2} = \frac{1}{3}$

$∴ δ = 30^{0}$

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A magnetic needle is free to rotate in a vertical plane which makes an angle of 600 with the magnetic meridian. If the needle stays in a direction making an angle of tan −1(2√3) with the horizontal, what would be the dip at that place ?

tan δ1=tan δcos θ ∴tan δ=tan δ1 cos θ = tan(tan−12√3) cos 600 = 2√3×12=13 ∴δ=300

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 ?

$t a n δ_{1} = \frac{tan δ}{cos θ}$

$∴ t a n δ = t a n δ_{1} c o s θ$

= $t a n (t a n^{- 1} \frac{2}{\sqrt{3}}) c o s 60^{0}$

= $\frac{2}{\sqrt{3}} \times \frac{1}{2} = \frac{1}{3}$

$∴ δ = 30^{0}$