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 delta_(1) = (B_(V))/(B_(H) cos phi)`
`tan delta_(1) = (tan delta)/( cos phi)`
`cos phi = (tan delta)/(tan phi_(1)) " " … (i)`
Angle of dip on the meridian 2 is
`tan delta_(2) = (B_(V))/(B_(H) sin phi) " " cos (90^(@) - phi) = sin phi `
`tan delta_(2) = (tan delta)/(sin phi)`
`sin phi = (tan delta)/(tan delta_(2)) " " ... (ii)`
Squaring and adding equation (i) and (ii)
`sin^(2) phi + cos^(2) phi = (tan^(2) delta)/(tan^(2) delta_(2)) + (tan^(2) delta)/(tan^(2) delta_(1))`
`(1)/(tan^(2) delta) = (1)/(tan^(2) delta_(1)) + (1)/(tan^(2) delta_(2))`
`cos^(2) delta = cot^(2) delta_(1) + cot^(2) delta_(2)`