When the plane of a dip needle makes an angle of  ${45}^{\mathrm{o}}$ with magnetic meridian, the dip is found to be  ${30}^{\mathrm{o}}$ what will be the dip when the plane of dip needle makes an angle of ${60}^{\mathrm{o}}$  with magnetic meridian ?

If ${\mathrm{\delta }}_{\mathrm{T}}$  be the true dip,  $\mathrm{\delta }$ be the apparent dip and $\mathrm{\theta }$ be the angle between plane of dip needle and magnetic meridian, 
Then
$\tan \mathrm{\delta }=\frac{\tan {\mathrm{\delta }}_{\mathrm{T}}}{\cos \mathrm{\theta }}\Rightarrow \mathrm{tan\delta }\propto \frac{1}{\cos \mathrm{\theta }}$

$\therefore \frac{\tan {\mathrm{\delta }}_2}{\tan {\mathrm{\delta }}_1}=\frac{\cos {\mathrm{\theta }}_1}{\cos {\mathrm{\theta }}_2}\Rightarrow \frac{{\mathrm{tan\delta }}_2}{\tan {30}^0}=\frac{\cos {45}^0}{\cos {60}^0}$

$\Rightarrow \tan {\mathrm{\delta }}_2=\sqrt{\frac{2}{3}}\Rightarrow {\mathrm{\delta }}_2={\tan }^{-1}\sqrt{\frac{2}{3}}$