Question

Horizontal component of earth's magnetic field at a place is $\sqrt{3}$ times its vertical component. What is the value of angle of dip at that place?

Answer 1

Given $H=\sqrt{3}V\Rightarrow \tan \theta =\frac{V}{H}=\frac{1}{\sqrt{3}}$
$\Rightarrow$ Angle of dip, $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^o$