At an angle of 30° to the magnetic meridian, the apparent dip is 45°. Find the true dip.

Given, $\mathrm{\theta }$ = 30°

Apparent dip, $\mathrm{\delta }\text{'}={45}^{\circ }$

Since,   $\tan \mathrm{\delta }\text{'}=\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}\cos \mathrm{\theta }}$

$\begin{array}{l}\Rightarrow \tan {45}^{\circ }=\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}\cos {30}^{\circ }}\\ \Rightarrow 1=\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}\left(\frac{\sqrt{3}}{2}\right)} \left(\because \cos {30}^{\circ }=\frac{\sqrt{3}}{2}\right)\\ \Rightarrow 1=\frac{2{\mathrm{B}}_{\mathrm{V}}}{\sqrt{3}{\mathrm{B}}_{\mathrm{H}}}\Rightarrow \sqrt{3}{\mathrm{B}}_{\mathrm{H}}=2{\mathrm{B}}_{\mathrm{V}}\\ \Rightarrow {\mathrm{B}}_{\mathrm{V}}=\frac{\sqrt{3}}{2}{\mathrm{B}}_{\mathrm{H}} ...\left(\mathrm{i}\right)\end{array}$

If d be the true angle of dip, then

$\begin{array}{l}\tan \mathrm{\delta }=\frac{{\mathrm{B}}_{\mathrm{V}}}{{\mathrm{B}}_{\mathrm{H}}}\\ \tan \mathrm{\delta }=\frac{\frac{\sqrt{3}}{2}{\mathrm{B}}_{\mathrm{H}}}{{\mathrm{B}}_{\mathrm{H}}} [\mathrm{From} \mathrm{Eq}. (\mathrm{i}\left)\right]\\ \Rightarrow \tan \mathrm{\delta }=\frac{\sqrt{3}}{2}\Rightarrow \mathrm{\delta }={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)\end{array}$