The plane of dip circle is set in the geographic meridian and the apparent dip is `theta_(1)`. It is then set in a vertical plane perpendicular to the geographic meridian. Now, the apparent dip is `theta_(2)`. The angle of declination `theta` at that place is

Options

A. `tan alpha=sqrt(tantheta_(1)xxtantheta_(2))`
B. `tan alpha=sqrt((tantheta_(1))^(2)xx(tantheta_(2))^(2))`
C. `tan alpha=(tantheta_(1))/(tantheta_(2))`
D. `tan alpha=(tantheta_(2))/(tantheta_(1))`

Answer

C

Explanation

Here, in geographical meridian,
`tantheta_(1)=V/(H cos alpha)`
In a plane , perpendicular to the magnetic meridian, tan
`theta_(2)=V/(H cos(90^(@)-alpha))=V/(H sin alpha)`
`:. (tantheta_(1))/(tan theta_(2))=(sin alpha)/(cos alpha)=tan alpha`