Three-point problem: A 600 m N, C 300 m E (500, 100, 400)

Method: the three-point problem. Three points at known altitudes define a plane. Find a point on the line AB with the same altitude as C; the line joining it to C is a strike line (a horizontal line on the bed).

Given: B is the origin. A is 600 m north of B (altitude 500 m). C is 300 m east of B (altitude 400 m). B is at 100 m.

        A (500 m)
        |
        | 600 m
        |
        B (100 m) -------- C (400 m)   [300 m east]

Step 1: Strike line

Along AB the altitude rises uniformly from 100 m (B) to 500 m (A), i.e. 400 m in 600 m (0.6667 m per m). The point D on AB with altitude 400 m (equal to C) is at a distance from B of

$$BD = \frac{400-100}{500-100} \times 600 = 450\ \text{m north of B}$$

Line DC joins two points at 400 m, so it is the strike line. It runs from D (0, 450) to C (300, 0).

Step 2: Strike direction

Taking east as +x and north as +y, the northward rise is $b = (500-100)/600 = 0.6667$ and the eastward rise is $a = (400-100)/300 = 1.0$.

From D to C the line runs 300 m east and 450 m south. The acute angle it makes with the north-south line is

$$\theta = \arctan\left(\frac{300}{450}\right) = 33.69°$$

D to C runs towards S33.7°E, so the strike is N33.7°W (the same line as S33.7°E).

Cross-check: the strike line is perpendicular to the direction of rise (a, b) = (1.0, 0.6667), which points to N56.3°E, giving a strike azimuth of 146.31° (or 326.31°).

Step 3: Dip direction and dip amount

The bed rises towards the north-east (A and C are higher than B), so it dips the opposite way (down the slope), at 90° to the strike. Dip direction azimuth = 236.31°, which is S56.3°W.

$$\tan\delta = \sqrt{a^2 + b^2} = \sqrt{1.0^2 + 0.6667^2} = 1.2019 \Rightarrow \delta = 50.24°$$

Answer: Strike N33.7°W; dip direction S56.3°W; dip amount 50.2°.