Rock layer attitude: A 900 m N, C 800 m W

A bed is a plane, so its level changes uniformly in plan: $z = z_B + a x + b y$, where $a$ and $b$ are the rates of change of level per metre east and north. This is the analytical form of the three-point (strike-line) method.

Data and coordinates (B at origin, x east, y north)

A = (0, 900), C = (-800, 0), B = (0, 0). Levels of the top: A = -350 m, B = -310 m, C = -390 m. The layer is 60 m thick in each hole (-350 to -410, -310 to -370, -390 to -450), so $t_v = 60$ m.

Gradients of the top surface

$$\begin{aligned} b &= \frac{z_A - z_B}{900} = \frac{-350 - (-310)}{900} = -0.0444 \\ a &= \frac{z_B - z_C}{800} = \frac{-310 - (-390)}{800} = 0.1000 \end{aligned}$$

The level falls northward and rises eastward, so the bed descends towards the NW/WNW.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.1094 \Rightarrow \delta = 6.25^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 294.0° (WNW), so the strike is $\ 294.0^\circ - 90^\circ = 204.0^\circ$, i.e. N24.0°E.

True thickness

All three boreholes show the same vertical thickness, 60 m, so

$$t = t_v\cos\delta = 60\cos 6.25^\circ = 59.64\ \text{m}$$

Answer: attitude of the layer = strike N24.0°E, dip 6.2° towards WNW (dip direction 294°); true thickness = 59.6 m.