At ${45}^0$ to the magnetic meridian, the apparent dip is ${30}^0.$ The true dip is

At $45°$ to the magnetic meridian, the effective horizontal component of the earth’s magnetic field is 

$B_H^{\mathrm{\prime }}=B_H\cos {45}^{\circ }=\frac{1}{\sqrt{2}}{B}_H$

The apparent dip ${\delta }^1$ is given by

$\tan {\delta }^1=\frac{B_v}{B_H^{\mathrm{\prime }}}=\frac{\sqrt{2}{B}_v}{B_H}=\sqrt{2}\tan \delta$ where $\delta$ is 

the true dip Thus $\tan {30}^{\circ }=\sqrt{2}\tan \delta$ or

$\delta ={\tan }^{-1}\sqrt{\frac{1}{6}}$