Bedrock attitude: A 700 m N23E, C S71W

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.

Reading the data

The table headings are interchanged: the top is the shallower level (A = -240, B = -280, C = -300 m) and the bottom is deeper (-280, -320, -340 m). The bed is 40 m thick in every hole, so $t_v = 40$ m. The distance BC is not printed; BC = 500 m is assumed (the method is the same for any value).

Coordinates (B at origin)

$$\begin{aligned} A &: x = 700\sin 23^\circ = 273.51\ \text{m},\ y = 700\cos 23^\circ = 644.35\ \text{m} \\ C &: x = 500\sin 251^\circ = -472.76\ \text{m},\ y = 500\cos 251^\circ = -162.78\ \text{m} \end{aligned}$$

Plane of the top of bedrock

$z = -280 + a x + b y$:

$$\begin{aligned} -240 &= -280 + 273.51a + 644.35b \\ -300 &= -280 - 472.76a - 162.78b \end{aligned}$$

Solving: $a = 0.0245$, $b = 0.0517$.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.0572 \Rightarrow \delta = 3.27^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 205.4° (SSW), so the strike is $\ 205.4^\circ - 90^\circ = 115.4^\circ$, i.e. N64.6°W.

True thickness

All three boreholes show the same vertical thickness, 40 m, so

$$t = t_v\cos\delta = 40\cos 3.27^\circ = 39.93\ \text{m}$$

Answer (with BC = 500 m): bedrock attitude = strike N64.6°W, dip 3.3° towards SSW (dip direction 205°); true thickness = 39.9 m. The bed is nearly flat, so the true thickness is almost equal to 40 m.