Sandstone attitude: A N45W 900 m, C S20E 700 m

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.

Coordinates (B at origin)

$$\begin{aligned} A &: x = 900\sin 315^\circ = -636.40\ \text{m},\ y = 900\cos 315^\circ = 636.40\ \text{m} \\ C &: x = 700\sin 160^\circ = 239.41\ \text{m},\ y = 700\cos 160^\circ = -657.78\ \text{m} \end{aligned}$$

Levels of the top: A = -350 m, B = -310 m, C = -390 m. The bed is 60 m thick in each hole, $t_v = 60$ m.

Plane of the top surface

$z = -310 + a x + b y$:

$$\begin{aligned} -350 &= -310 - 636.40a + 636.40b \\ -390 &= -310 + 239.41a - 657.78b \end{aligned}$$

Solving: $a = 0.2900$, $b = 0.2272$.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.3684 \Rightarrow \delta = 20.23^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 231.9° (SW), so the strike is $\ 231.9^\circ - 90^\circ = 141.9^\circ$, i.e. N38.1°W.

True thickness

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

$$t = t_v\cos\delta = 60\cos 20.23^\circ = 56.30\ \text{m}$$

Answer: attitude of the sandstone = strike N38.1°W, dip 20.2° towards SW (dip direction 232°); true thickness = 56.3 m.