The plane of dip circle is set in the geographical meridian and the apparent dip is `theta_(1)` it is then set in a vertical plane prependicular to the geopgraphical meridian the apparent dip becomes `theta_(2)` the angle of declination `alpha` is given by

Options

A. `tan alpha=sqrt(tan theta_(1)xxtantheta_(2))`
B. `tan alpha=sqrt(tan theta_(1)xxtan theta_(2))^(2)`
C. `tan alpha=tan theta_(1)//tan theta_(2)`
D. `tan alpha=tan thetasqrt(tan)theta_(1)`

Answer

C

Explanation

Here in geographcal meridian
`tan theta_(1)=(V)/(H cos alpha)`
`tan theta_(2)=(V)/(H cos (90^(@)-alpha)) =(v)/(Hsin alpha)`
`therefore (tan theta_(1))/(tan theta_(2))=(sin alpha)/(cos alpha)=tan alpha`