A dip circle shows an apparent dip of 45° at a place where the true dip is 30°. If the dip circle is rotated through 90°, what apparent dip will it show x degrees. Then the value of 10 x.$\left[{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)=35.2^0\right]$

Let ${\mathrm{\theta }}_1$ and ${\mathrm{\theta }}_2$ are the angles of dip in two arbitrary planes which are perpendicular to each other.
Here, ${\mathrm{\theta }}_1={45}^{\circ }$ and $\mathrm{\theta }={30}^{\circ }\mathrm{AS},{\cot }^2{30}^{\circ }={\cot }^2{\mathrm{\theta }}_1+{\cot }^2{\mathrm{\theta }}_2$ Where, $\mathrm{\theta }$ is true dip.
$\therefore {\cot }^2{30}^0={\cot }^2{45}^0+{\cot }^2{\mathrm{\theta }}_2{\cot }^2{\mathrm{\theta }}_2=3-1=2\Rightarrow \cot {\mathrm{\theta }}_2=1.414\therefore {\mathrm{\theta }}_2=35{.2}^0$