Question

Dip circle is not set into the magnetic meridian, and the angle at which it is inclined to the magnetic meridian is unknown. $\delta '{\text{ and }}\delta ''$ are apparent dips at a place in which the dip circle is kept in transverse positions. The dip is:
$
A.\cot \delta = \cot \delta ' + \cot \delta '' \\
B.{\tan ^2}\delta ' = \tan \delta ' + {\tan ^2}\delta '' \\
C.{\cos ^2}\delta = {\cos ^2}\delta ' + {\cos ^2}\delta '' \\
D.{\cot ^2}\delta = {\cot ^2}\delta ' + {\cot ^2}\delta '' \\
 $

Answer 1

Hint:The angle of dip or magnetic dip is defined as the angle, which is made by the earth’s magnetic field lines with the horizontal. A dip circle or dip needle is an instrument for measuring the inclination or dip of the earth’s magnetic field. It consists essentially of a magnetic needle pivoted at the center of the graduated circle.

Complete step by step answer:
 Actually, the dip angle is measured by using the dip circle. Now, if the dip circle has set in the magnetic meridian, then the angle that is read by the dip circle will be equal to the angle of dip.
Now, if the dip circle is inclined at an angle $\alpha $with the magnetic meridian and dip circle reads, $\delta '$ then it is called the apparent angle of dip. Now, the major thing is that what would be the value of the true dip angle; it would be $\delta $.
The relation among \[\alpha ,\delta {\text{ and }}\delta '\] is given as follows,
$\tan \delta = \tan \delta '\cos \alpha $
$\cos \alpha = \dfrac{{\tan \delta }}{{\tan \delta '}} - - - - (i)$
If the dip circle is making an angle $\alpha $ with the magnetic meridian is an unknown quantity, now let’s rotate the dip circle by $\dfrac{\pi }{2}$ radian or ${90^0}$after noticing the value of the apparent dip $\delta '$.
Now, let’s read is $\delta ''$ at the corresponding rotated position. In this case, the true value of dip is given by,
$
\Rightarrow \tan \delta = \tan \delta ''\sin \alpha \\
\Rightarrow \sin \alpha = \dfrac{{\tan \delta }}{{\tan \delta ''}} - - - - (ii) \\
$
We know from trigonometry,
${\sin ^2}\alpha + {\cos ^{}}\alpha = 1 - - - - (iii)$
Putting the values of $\cos \alpha $ and $\sin \alpha $ respectively in the equation (iii) from (i) and (ii)
So,
$
\Rightarrow {\left( {\dfrac{{\tan \delta }}{{\tan \delta ''}}} \right)^2} + {\left( {\dfrac{{\tan \delta }}{{\tan \delta '}}} \right)^2} = 1 \\
\Rightarrow \dfrac{{{{\tan }^2}\delta }}{{{{\tan }^2}\delta ''}} + \dfrac{{{{\tan }^2}\delta }}{{{{\tan }^2}\delta '}} = 1 \\
\Rightarrow {\tan ^2}\delta \left( {\dfrac{1}{{{{\tan }^2}\delta ''}} + \dfrac{1}{{{{\tan }^2}\delta '}}} \right) = 1 \\
\Rightarrow \dfrac{1}{{{{\tan }^2}\delta ''}} + \dfrac{1}{{{{\tan }^2}\delta '}} = \dfrac{1}{{{{\tan }^2}\delta }} \\
\Rightarrow {\cot ^2}\delta '' + {\cot ^2}\delta ' = {\cot ^2}\delta \\
 $
Hence, the dip is ${\cot ^2}\delta = {\cot ^2}\delta ' + {\cot ^2}\delta ''$ and option (D) is correct.

Note:The vertical circular scale of the dip circle is divided into four quadrants, each graduated from ${0^0}$ to ${90^0}$ with \[{0^0} - {0^0}\] in horizontal and \[{90^0} - {90^0}\] in vertical positions. The horizontal circular scale at the bottom is graduated from ${0^0}$ to ${360^0}.$