Question

The plane of a dip circle is set in the geographic meridian. Then the apparent dip is (${\delta }_1$ is the actual dip and ${\delta }_2$ is the declination):

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

Answer 1

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

$\tan {\delta }_1=\frac{V}{H\cos \theta }........\left(1\right)$

$\tan {\delta }_2=\frac{V}{H\cos (90-\theta )}=\frac{V}{H\sin \theta }........\left(1\right)$

Dividing (1) by (2), we get

$\frac{\tan {\delta }_1}{\tan {\delta }_2}=\tan \theta$

$\theta ={\tan }^{-1}\left(\frac{\tan {\delta }_1}{\tan {\delta }_2}\right)$