Question

A magnetic needle is free to rotate in a vertical plane which makes an angle of 60° 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?

Answer 1

If the needle stays in a direction making an angle of  $\tan ^{-1} \frac{2}{\sqrt{3}}$ with the horizontal,then the dip at that place is $\delta=30^{\circ}$

Explanation:

Given data in the question  

Angle made from the magnetic meridian with the needle's rotation axis,

θ = 60°

Angle made with horizontal needle, $\delta_{1}=\tan ^{-1} \frac{2}{\sqrt{3}}$

When index is the dip angle,

Dip angle, or Magnetic affiliations, is the orientation of the Earth's magnetic area panel made horizontal. This angle increase at various points on the surface of the Earth.

$\tan \delta_{1}=\frac{\tan \delta}{\cos \theta}$

$\tan \delta=\tan \delta_{1} \times \cos \theta$

$\tan \delta=\tan \times\left(\tan ^{-1} \frac{2}{\sqrt{3}} \times \cos 60^{\circ}\right)$

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

$\delta=30^{\circ}$

Answer 2

Answer:

30 degree

Explanation:

tanФ=tanФ1\cosQ

tanФ1=tanФ*cosQ=2\√3

=1\√3