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

Answer

D

Explanation

Angle of dip `delta = 45^(@)`
`therefore tan delta =(tan delta)/(cos theta)= (tan 45)/(cos 30^(@)) =(1)/(sqrt(3)//2) = (2)/(sqrt(3))`
`therefore ` Real dip `delta = tan^(-1) (2//sqrt(3))`