Question

A magnet hung at $45^{\circ }$ with magnetic meridian makes an angle of $60^{\circ }$ with the horizontal. The actual value of the angle of dip is

• A. ${\tan }^{-1}\left(\sqrt{\frac{3}{2}}\right)$
• B. ${\tan }^{-1}(\sqrt{6})$
• C. ${\tan }^{-1}\left(\sqrt{\frac{2}{3}}\right)$
• D. ${\tan }^{-1}\left(\sqrt{\frac{1}{2}}\right)$

Answer 1

Answer: (A)

$\tan {\theta }^{\prime }=\frac{\tan \theta }{\cos \alpha }$
${\theta }^{\prime }=60^{\circ }$
$\alpha =45^{\circ }$
$\sqrt{3}=\frac{\tan \theta }{\frac{1}{\sqrt{2}}}$
$\tan \theta =\sqrt{\frac{3}{2}}$
$\theta ={\tan }^{-1}\sqrt{\frac{3}{2}}$