Question

The needle of a dip circle shows an apparent dip of ${45}^0$ in a particular position and ${53}^0$ when the circle is rotated through ${90}^0$. Find the true dip.

Answer 1

If ${\delta }_1$ and ${\delta }_2$ be the apparent dips shown by the dip circle in the 2 perpendicular positions, the true dip $\delta$ is given by

$Co{t}^2\delta =Co{t}^2\delta 1+Co{t}^2\delta 2$

$\Rightarrow Co{t}^2\delta =Co{t}^2{45}^0+Co{t}^2{53}^0$

$Co{t}^2\delta$ = 1.56

$Co{t}^2\delta$ = 1.56

$\delta ={38.6}^0\approx {39}^0$.