Q.A magnet makes an angle of $45^\circ$ with the horizontal in a plane making an angle of $30^\circ$ with the magnetic meridian. Find the true value of the dip angle at the place.

[!TLDR] Using the apparent-vs-true dip relation $\tan\phi_{apparent}=\tan\phi_{true}/\cos\theta$ with $\phi_{apparent}=45^\circ$ and $\theta=30^\circ$ gives $\tan\phi_{true}=\cos30^\circ=0.866$. [!ANSWER] $\phi_{true}=\tan^{-1}(0.866)\approx40.9^\circ$

When a dip circle (or magnet) is set up in a vertical plane that makes some angle $\theta$ with the true magnetic meridian, the dip angle $\phi_{apparent}$ it reads is related to the TRUE dip $\phi_{true}$ (the dip measured exactly IN the magnetic meridian) by $\tan\phi_{apparent}=\dfrac{\tan\phi_{true}}{\cos\theta}$. This is because only the component of the horizontal field lying within the observation plane, $B_H\cos\theta$, is what the instrument actually measures, while the full vertical component $B_V$ still acts undiminished; the ratio $\tan\phi_{apparent}=B_V/(B_H\cos\theta)$ is therefore larger than the true $\tan\phi_{true}=B_V/B_H$ by the factor $1/\cos\theta$. Given $\phi_{apparent}=45^\circ$ and $\theta=30^\circ$: $\tan45^\circ=\dfrac{\tan\phi_{true}}{\cos30^\circ}\Rightarrow\tan\phi_{true}=\cos30^\circ\times\tan45^\circ=0.866\times1=0.866$. So the true dip angle is $\phi_{true}=\tan^{-1}(0.866)\approx40.9^\circ$. [!ANSWER] $\phi_{true}=\tan^{-1}(0.866)\approx40.9^\circ$

Apply the standard relation between the dip observed in a vertical plane at angle $\theta$ to the magnetic meridian and the true dip, $\tan\phi_{apparent}=\tan\phi_{true}/\cos\theta$, and solve for $\phi_{true}$.

Treating the $45^\circ$ reading directly as the true dip, without correcting for the fact that it was measured in a plane $30^\circ$ away from the magnetic meridian.