A magnet hung at 45 ° with magnetic meridian makes an angle of 60 °...

Given here, $\theta ' = 60 °$ and apparent value of dip, $\alpha = 45 °$.

If $H$ is horizontal component of earth's magnetic field in magnetic meridian, then $tan{\theta}^{'} = \dfrac{V}{H}$.

Let ${H}_{1}$ be component of $H$ at $60 °$ to magnetic meridian, then $tan\theta = \dfrac{V}{{H}_{1}} = \dfrac{V}{Hcos\alpha} = \dfrac{tan{\theta}^{'}}{cos\alpha}$

$tan\theta = \tan 60 ° \times \cos 45 °$

$⇒ tan\theta = \sqrt{3} \times \dfrac{1}{\sqrt{2}}$

Thus, the actual value of the angle of dip is $\theta = \tan^{- 1}\sqrt{\dfrac{3}{2}}$