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//sqrt3)` with the horizontal, what would be the true dip at that place?

Answer

`30^@`

Explanation

Here, `theta=60^@`, `tan delta^'2/sqrt3`, `delta=?`
`tan delta=tan delta^'(cos theta)`
`=2/sqrt3cos60^@=2/sqrt3xx1/2=1/sqrt3`
`:. delta=30^@`