Question

The plane of a dip circle is set in the geographic meridian and the apparent dip is ${\delta }_1$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is ${\delta }_2$. Then declination $\theta$ at the place is

• A. $\theta ={\tan }^{-1}\left(\tan {\delta }_1\tan {\delta }_2\right)$
• B. $\theta ={\tan }^{-1}(\tan {\delta }_1+\tan {\delta }_2)$
• C. $\theta =ta{n}^{-1}\left(\frac{\tan {\delta }_1}{\tan {\delta }_2}\right)$
• D. $\theta ={\tan }^{-1}(\tan {\delta }_1-\tan {\delta }_2)$

Answer 1

The correct option is C $\theta =ta{n}^{-1}\left(\frac{\tan {\delta }_1}{\tan {\delta }_2}\right)$

Geographic meridian is at $\theta$ angle from magnetic meridian therefore, angle of dip is $\theta$.
Now angle of dip in a plane at any angle $\alpha$ from the magnetic meridian is
$\tan \varphi =\frac{B_H\cos \alpha }{B_V}$
therefore, Now angle of dip in geographic meridian is
$\tan {\delta }_1=\frac{B_H\cos \theta }{B_V}.............................\left(1\right)$
and angle of dip in a plane perpendicular to geographic meridian is
$\tan {\delta }_2=\frac{B_H\sin \theta }{B_V}................\left(2\right)$
Dividing eq $\left(1\right)$ from $\left(2\right),$ we get,
$\tan \theta =\frac{\tan {\delta }_1}{\tan {\delta }_2}$
Therefore, option(C)