If `theta_1` and `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` is given by

Options

A. `cos^(2) varphi=cos^(2) varphi_(1)+cos^(2) varphi_(2)`
B. `sec^(2) varphi=sec^(2) varphi_(1)+sec^(2) varphi_(2)`
C. `tan^(2) varphi=tan^(2) varphi_(1)+tan^(2) varphi_(2)`
D. `cot^(2) varphi=cot^(2) varphi_(1)+cot^(2) varphi_(2)`

Answer

D

Explanation

Let `alpha` be the angle which one of the planes make with the magnetic meridian the other plane makes an angle `(90^(@)- alpha)` with it. The components of H in these planes will be `H cos alpha` and `H sin alpha` respectively. If `varphi_(1)` and `varphi_(2)` are the apparent dips in these two planes, then

`tan varphi_(1)=V/(H cos alpha)` so `cos alpha=V/(H tan varphi_(1))`
`tan varphi_(2)=V/(H sin alpha)` so `sin alpha=V/(H tan varphi_(2))`
squaring and adding (i) and (ii), we get
`cos^(2)alpha+sin^(2)alpha=(V/H)^(2)(1/(tan^(2) varpha_(1))+1/(tan^(2) varphi_(2)))`
or `1=V^(2)/H^(2)(cot^(2) varphi_(1)+cot^(2) varphi_(2))`
or `H^(2)/V^(2)=cot^(2) varphi_(1)+cot^(2) varphi_(2)`
or `cot^(2) varphi=cot^(2) varphi_(1)+cot^(2) varphi_(2)`
This is the required result.