Question

In a vertical plane $p_1$ making angle ${30}^o$ with magnetic meridian, apparent angle of dip in vertical plane $p_2$ is ${60}^o$ , which is perpendicular to plane $p_1$ is:

• A. ${30}^o$
• B. ${45}^o$
• C. ${\tan }^{-1}\left(3\right)$
• D. ${\tan }^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is D ${\tan }^{-1}\left(\frac{2}{3}\right)$

The true angle of the dip is

$\varphi ={30}^{\circ }$

The angle made by vertical plane $p_2$ with magnetic meridian is ${60}^{\circ }$

Then the apparent angle of dip is

$\tan {\varphi }^{\prime }=\frac{\tan \varphi }{\cos \theta }$

Then the apparent dip in the plane is

$\tan {\varphi }^{\prime }=\frac{\tan {30}^{\circ }}{\cos {60}^{\circ }}$

$\tan {\varphi }^{\prime }=\frac{\tan {30}^{\circ }}{\cos {60}^{\circ }}$

${\varphi }^{\prime }={\tan }^{-1}\left(\frac{2}{3}\right)$