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

If the vertical plane in which angle of dip is `phi_(1)`, subtends an angle a with magnetic meridian, the other vertical plane in which angle of dip is `phi_(2)` and which is perpendicular to first vertical plane will subtend an angle `(90-alpha)` with magnetic meridian, so,
`tan phi_(1)=B’_(v)/B’_(H)cos alpha=B_(v)/B_(H)cos alpha=tan phi/cos alpha rArr cos alpha =tan phi/tan phi_(1)`
and`tan phi_(2)=B’’_(v)/B’’_(H)cos alpha=B_(v)/B_(H)cos (90 –alpha) =tan phi/sin alpha=tan phi/tan phi_(2) `
Now,`sin^(2)a+cos^(2)a=1 rArr tan^(2)phi/tan^(2)phi_(1)+ tan^(2)phi/tan^(2)phi_(2)=1`