
Given:
True thickness of vertical dyke = \(t=2\ \text{m}\).
Intercepted core length along the borehole = \(L=4\ \text{m}\).
Borehole is inclined from the vertical by angle \(\theta\). We assume the borehole intersects the dyke so that its horizontal projection is normal to the dyke (i.e. it cuts across the thickness directly).
Geometry & formula:
Let \(\mathbf{n}\) be the unit normal to the (vertical) dyke (horizontal direction). If the borehole direction makes angle \(\theta\) from the vertical, its horizontal component (towards the dyke normal) has magnitude \(\sin\theta\). The length of core inside the dyke equals the true thickness divided by the component of the borehole direction along the normal: \[ L=\frac{t}{\sin\theta}. \] Compute: \[ 4=\frac{2}{\sin\theta}\quad\Rightarrow\quad \sin\theta=\frac{2}{4}=0.5. \] Thus \[ \theta=\arcsin(0.5)=30^\circ. \] Answer (integer): \(\boxed{30^\circ}\)
