At `45^(@)` to the magnetic meridian the apparent dip is `60^(@)`. The true dip is

Options

A. `tan^(-1)sqrt(3)`
B. `tan^(-1)"1/sqrt(3)`
C. `tan^(-1)"sqrt(3)/2`
D. `tan^(-1)"sqrt(1/6)`

Answer

C

Explanation

Apparent dip `delta^(')` is given by
`tandelta^(')=V/H^(')=V/(H cos 45^(@))=V/Hxx1/(cos 45^(@))`
`=tandelta.sqrt(2)` But `tan 60^(@)=tandelta.sqrt(2)`
`sqrt(3)=tandelta.sqrt(2)`
`tandelta=sqrt(3)//2 :. delta=tan^(-1) sqrt(3)//2`