The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:

Options

A. `tan^(-1)((sqrt3)/(2))`
B. `tan^(-1)(sqrt3)`
C. `tan^(-1)(sqrt((3)/(2)))`
D. `tan^(-1)((2)/(sqrt3))`

Answer

D

Explanation

`tan delta'=(tan delta)/(cos alpha)`
`delta=`true dip, `delta'=`dip with the field `alpha=`angle made by the meridian