Question

Dip circle is not set into magnetic meridian and the angle at which it is inclined to the magnetic meridian is unknown. ${\delta }^{\prime }$ and ${\delta }^{\prime \prime }$ are apparent dips at a place in which dip circle is kept in transverse positions. The dip is:

• A. $\cot \delta =\cot {\delta }^{\prime }+\cos {\delta }^{\prime \prime }$
• B. ${\tan }^2{\delta }^{\prime }=\tan {\delta }^{\prime }+{\tan }^2{\delta }^{\prime \prime }$
• C. ${\cos }^2\delta ={\cos }^2{\delta }^{\prime }+{\cos }^2{\delta }^{\prime \prime }$
• D. ${\cot }^2\delta ={\cot }^2{\delta }^{\prime }+{\cot }^2{\delta }^{\prime \prime }$

Answer 1

The correct option is D ${\cot }^2\delta ={\cot }^2{\delta }^{\prime }+{\cot }^2{\delta }^{\prime \prime }$

Angle of dip is measured using dip circle. If dip circle is set in magnetic meridian then the angle read by dip circle is angle of dip.
If the dip circle is inclined at an angle $\theta$ with magnetic meridian and dip circle reads ${\delta }^{\prime }$ called apparent angle of dip. True dip $\delta$ is then given by $\tan \delta =\tan {\delta }^{\prime }\cos \theta .$
If the angle $\theta$ the dip circle makes with magnetic meridian is unknown, then rotate the dip circle by ${90}^{\circ }$ after noting apparent dip ${\delta }^{\prime }$.
At rotated position it reads dip $\delta "$. The true dip $\delta$ is given by $\tan \delta =\tan {\delta }^{\prime \prime }\sin \theta .$
$\cos \theta =\frac{\tan \delta }{\tan {\delta }^{\prime }}=\frac{\cot {\delta }^{\prime }}{\cot \delta }$
Similarly $\sin \theta =\frac{\cot {\delta }^{\prime \prime }}{\cot \delta }$
${\cos }^2\theta +{\sin }^2\theta =1$
$\therefore (\frac{\cot {\delta }^{\prime }}{\cot \delta }{)}^2+(\frac{\cot {\delta }^{\prime \prime }}{\cot \delta }{)}^2=1$
$\therefore {\cot }^2\delta ={\cot }^2{\delta }^{\prime }+{\cot }^2{\delta }^{\prime \prime }$