A magnetic needle is free to rotate in a vertical plance which makes an angle of `60^@` with the magnetic meridian. If the needle stays in a direction making an angle of `tan^(-1) (2/sqrt(3))` with the horizontal, what would be the dip at that place?

Explanation

`tan delta_(1) = (tan delta)/(cos theta)`
`:. Tan delta = tan delta _(1) cos theta `
`= tan (tan ^(-1)(2)/(sqrt(3)))cos 60^(@)`
`= (2)/(sqrt(3)) xx (1)/(2) = (1)/(3)`
`:. Delta = 30^(@)`