If `theta_1` and `theta_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, `theta` is given by `cot^2theta=cot^2theta+cot^2theta`.

Options

A. `cot^(2)theta=cot^(2)theta_(1)+cot^(2)theta_(2)`
B. `tan^(2)theta=tan^(2)theta_(1)+tan^(2)theta_(2)`
C. `cot^(2)theta=cot^(2)theta_(1)-cot^(2)theta_(2)`
D. `tan^(2)theta=tan^(2)theta_(1)-tan^(2)theta_(2)`

Answer

A

Explanation

Let the `B_(H) andB_(V)` be the horizontal and vertical component of earth 's magnetic field B.
`tantheta =(B_(V))/(B_(H))rArrcottheta=(B_(H))/(B_(V))`

Let plane 1 and 2 are mutually perpendicular planes making angle `theta and (90^(@)-theta)` with magnetic meridian.The vertical component of earth of earth's magnetic field remain same in two plane but effective horizontal component in the two planes is given by
`B_(1)=B_(H) cos theta`
and " " `B_(2)=B_(H) sin theta`
Then , `tan theta_(1)= (B_(V))/(B_(1))=(B_(V))/(B_(H)cos theta') rArrcot theta_(1)=(B_(H)cos theta')/(B_(V))`
Similarly,
`rArr tan theta_(2)= (B_(V))/(B_(1))=(B_(V))/(B_(H)sin theta' rArrcot theta_(2)=(B_(H)sin theta)'/(B_(V))`
From Eq (iv) and Eq (v)
`rArr cot^(2) theta_(1)+cot^(2)theta_(2)=(B_(H)^(2)cos^(2)theta)/(B_(V)^(2))+(B_(H)^(2)sin^(2)theta)/(B_(V)^(2))`
`cot^(2) theta_(1)+cot^(2)theta_(2)=(B_(H)^(2))/(B_(V)^(2))cos^(2)theta+sin^(2 theta)`
`rArr cot^(2) theta_(1)+cot^(2)theta_(2)=cot^(2)theta`