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