Question

The plane of dip circle is set in the geographic meridian and the apparent dip is $\theta_{1}$. It is then set in a vertical plane perpendicular to the geographic meridian. Now, the apparent dip is $\theta_{2}$. The angle of declination $\alpha$ at that place is

• A. $\tan \alpha=\sqrt{\tan \theta_{1} \tan \theta_{2}}$
• B. $\tan \alpha=\sqrt{\left(\tan \theta_{1}\right)^{2}+\left(\tan \theta_{2}\right)^{2}}$
• C. $\tan \alpha=\frac{\tan \theta_{1}}{\tan \theta_{2}}$
• D. $\tan \alpha=\frac{\tan \theta_{2}}{\tan \theta_{1}}$

Answer 1

Answer: (C)

$\tan \theta_{1}=\frac{\tan \delta}{\cos \alpha} $
or $ \cos \alpha=\frac{\tan \delta}{\tan \theta_{1}}\,\,\,(i)$
Again, $\tan \theta_{2}=\frac{\tan \delta}{\tan \theta_{2}}$
or $\sin \theta=\frac{\tan \delta}{\tan \theta_{2}}\,\,\,(ii)$
$\frac{(i)}{\text { (ii) }}$ gives $\tan \alpha=\frac{\tan \theta_{1}}{\tan \theta_{2}}$