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

Answer

D

Explanation

As , `tan delta'=( tan delta)/(cos theta )=(tan 45^(@))/(cos 30^(@))`
`tan delta'=(1)/(sqrt(3)//2)=(2)/(sqrt(3))`
`implies delta'=tan^(-1)""((2)/(sqrt(3)))`