a b c d

d

Angle of dip $\delta = 45^0$ $\Rightarrow \delta' = \frac{tan\delta}{cos\theta}= \frac{tan 45^0}{cos 30^0} = \frac{1}{\sqrt 3 /2} = \frac{2}{\sqrt 3}$ $\delta'= tan^{-1}\frac{2}{\sqrt 3}$