Question

At a certain place, horizontal component is $\sqrt{3}$ times the vertical component. The angle of dip at this place is:

• A. $0$
• B. $\pi /3$
• C. $\pi /6$
• D. none of these

Answer 1

The correct option is C $\pi /6$

$\tan \delta =\frac{B_V}{B_H}$
$=\frac{B_V}{\sqrt{3}{B}_V}=\frac{1}{\sqrt{3}}=\tan {30}^{\circ }$

$\Rightarrow \delta ={30}^{\circ }=\frac{\pi }{6}$