If `phi_(1)` and `phi_(2)` be the apparent angles of dip observed in two vertical planes at right angles to each other , then show that the true angle of dip `phi` is given by `cot^(2) phi = cot^(2) phi_(1) + cot^(2) phi_(2)`.

Explanation

Angle of dip on meridian one is
`tan del _(1) = (B_(v))/(B _(H) cos phi)`
` tan del_(1) = (tan del )/(cos phi)`
` cos phi = (tan del )/( tan del _(1))" "…(i)`
Angle of dip on the meridian 2 is
`tan del _(2) = (B_(v))/(B _(H) sin phi ) cos (90^(@) phi ) = sin phi `
` tan del _(2) = (tan del )/( sin phi)`
`sin phi = (tan del )/(tan del _(2))" "...(ii)`
Squaring and adding equation (i) and (ii).
`sin^(2) phi + cos ^(2) phi = (tan ^(2)del )/(tan ^(2) del _(2)) + (tan ^(2) del)/(tan ^(2) del _(1))`
`(1)/(tan ^(2)del)=(1)/(tan ^(2) del _(1))+ (1)/( tan ^(2) del _(2))`
` cos ^(2) del = cot ^(2) del _(1) + cot ^(2) del _(2)`