Q. 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 $\delta_{2}$. The declination $\theta$ at the place is
A) $\theta = \tan^{- 1}{(tan}\delta_{1}\tan{\delta_{2})}$
B) $\theta = \tan^{- 1}{(tan}\delta_{1}{+ \ tan}{\delta_{2})}$
C) $\theta = \tan^{- 1}\left( \ \frac{\tan\delta_{1}}{\tan\delta_{2}} \right)$
D) $\theta = \tan^{- 1}{(tan}\delta_{1}{- \ tan}{\delta_{2})}$
Explanation
(c) Let $\theta$ be the declination at the place. As is clear from figure.
[image:5_image44.emf]
$\tan\delta_{1} = \frac{V}{H\ cos\theta}$
$\tan\delta_{2} = \frac{V}{H\ cos(90{^\circ} - \theta)} = \frac{V}{H\ sin\theta}$
$\ \frac{\tan\delta_{1}}{\tan\delta_{2}} = \frac{sin\theta}{cos\theta} = tan\theta$
$\therefore\theta = \tan^{- 1}\left( \ \frac{\tan\delta_{1}}{\tan\delta_{2}} \right)$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.