Question

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

• A. $\delta ={\tan }^{-1}\left(\tan {\theta }_1/\tan {\theta }_2\right)$
• B. $\delta ={\tan }^{-1}(\tan {\theta }_1.\tan {\theta }_2)$
• C. $\delta ={\tan }^{-1}\left(\tan {\theta }_2/\tan {\theta }_1\right)$
• D. $\delta ={\tan }^{-1}\left(\sqrt{\tan {\theta }_1\tan {\theta }_2}\right)$

Answer 1

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

First case:
$\cot {\theta }_1=\cos a\times cot\theta$
$a$ is the angle made by geometric meridian with the magnetic meridian
${\theta }_1,\theta$ are apparent and true dip respectively

Second case:
$\cot {\theta }_2=\cos (90-a)\times \cot \theta$
$\cot {\theta }_2=\sin a\times \cot \theta$

Divide both the equations and we get
$\tan a=\frac{\cot {\theta }_2}{\cot {\theta }_1}$
$\tan a=\frac{\tan {\theta }_1}{\tan {\theta }_2}$