Limestone attitude: square 900 m, depth at D

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

Layout (plan, north up)

   B(50) ------- A(200)      B is west of A
     |             |         C is south of B
     |             |         D is south of A
   C(100) ------ D(?)

Origin at A; x east, y north: A = (0, 0), B = (-900, 0), C = (-900, -900), D = (0, -900). Depths: A = 200 m, B = 50 m, C = 100 m.

Gradients of depth

$$\begin{aligned} a &= \frac{d_A - d_B}{900} = \frac{200 - 50}{900} = 0.1667\ \text{(deeper eastward)} \\ b &= \frac{d_B - d_C}{900} = \frac{50 - 100}{900} = -0.0556\ \text{(deeper southward)} \end{aligned}$$

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.1757 \Rightarrow \delta = 9.96^\circ$$

The bed deepens towards the direction with east component $a = 0.1667$ and north component $b = -0.0556$, i.e. $180^\circ - \arctan(0.1667/0.0556) =$ 108.4° (ESE). So the strike is $108.4^\circ - 90^\circ = 18.4^\circ$, i.e. N18.4°E.

Depth at D

$$d_D = d_A + b(-900) = 200 + (-0.0556)(-900) = 250\ \text{m}$$

Check with the parallelogram rule: $d_D = d_A + d_C - d_B = 200 + 100 - 50 = 250$ m.

Answer: limestone attitude = N18.4°E / 10.0° towards ESE (dip direction 108°); depth at D = 250 m.