A dip circle shows an apparent dip of 60^o at a place where the true dip is 45^o . If the dip circle is rotated through 90^o , apparent dip will be ${\cot }^{-1}\sqrt{\frac{2}{\mathrm{b}}}$. The value of 'b' is

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

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\cot }^2\mathrm{\delta }={\cot }^2{\mathrm{\delta }}_1+{\cot }^2{\mathrm{\delta }}_2\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\cot }^2{45}^0={\cot }^2{60}^0+{\cot }^2{\mathrm{\delta }}_2\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\cot }^2{\mathrm{\delta }}_2=2/3\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ or, }\cot {\mathrm{\delta }}_2=0.816\text{ giving }{\mathrm{\delta }}_2={51}^0\end{array}$