If `theta_1` of `theta_2` be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip `theta` given by

Options

A. `tan^2 theta = tan^2theta_1 + tan ^2theta_2`
B. `cot^2 theta = cot^2 theta_1 - cot^2 theta_2`
C. `tan^2theta = tan^2 theta_1 - tan^2 theta_2`
D. `cot theta= cot^2 theta_1 + cot^2 theta_2`

Answer

D

Explanation

Let `B_H` and `B_v` be the horizontal and vetical components of earth.s magnetic field `vec B`. Since `theta` is the angle of dip
`therefore tan theta = (B_V)/(B_H) or cot theta = (B_H)/(B_V) ....(i)`
Suppose planes 1 and 2 are two mutually perpendicualr planes and respectively make angle `theta` and `90^@ - theta` withe the magnetic meridian. The vetical compunets of earth.s magnetic field remains same in the the two planes but the effective horizontal components in the planes will be
`B_1 = B_(H) sin theta`
and `B_2 = B_H sin theta` The angles of dip `theta_1` and `theta_2` in the two planes are given by
`tan theta_1 = (B_V)/(B_1)`
`tan theta_1 = (B_V)/(B_H cos theta)`
or `cot theta_1 = (B_H cos theta)/(B_V)`
Similary,
`cot theta_2 = (B_H sin theta)/(B_V)`
From eqns . (ii) and (iii)
`cot^2 theta_1 + cot^2 theta_2 = (B_H^2)/(B_V^2)(cos ^2 theta + sin ^2 theta) = (B_H^2)/(B_V^2)`
`therefore cot^2 theta_1 + cot^2 theta_2 = cot^2 theta`