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(2sqrt(3))` with the horizontal, what would be the dip at that place?

Answer 1

Given :
Angle made by the magnetic meridian with the plane of rotation of the needle , `θ = 60^circ`

Angle made by the needle with the horizontal, `δ_1 = tan^-1(2/sqrt(3))`

If δ is the angle of dip , then 

`tan  δ_1 = tanδ/cosθ`

⇒ `tan  δ = tan  δ_1  cos θ`

⇒ `tan  δ = tan (tan^-1  2/sqrt3) cos60^circ`

⇒ `tan  δ = 2/sqrt(3) xx 1/2 = 1/sqrt(3)`

⇒ `δ = 30^circ`