Three-point problem: A 800 m N, C 600 m E (600, 200, 500)

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 800 m north of B (altitude 600 m). C is 600 m east of B (altitude 500 m). B is at 200 m.

        A (600 m)
        |
        | 800 m
        |
        B (200 m) -------- C (500 m)   [600 m east]

Step 1: Strike line

Along AB the altitude rises uniformly from 200 m (B) to 600 m (A), i.e. 400 m in 800 m (0.5 m per m). The point D on AB with altitude 500 m (equal to C) is at a distance from B of

$$BD = \frac{500-200}{600-200} \times 800 = 600\ \text{m north of B}$$

Line DC joins two points at 500 m, so it is the strike line. It runs from D (0, 600) to C (600, 0).

Step 2: Strike direction

Taking east as +x and north as +y, the northward rise is $b = (600-200)/800 = 0.5$ and the eastward rise is $a = (500-200)/600 = 0.5$.

From D to C the line runs 600 m east and 600 m south. The acute angle it makes with the north-south line is

$$\theta = \arctan\left(\frac{600}{600}\right) = 45°$$

D to C runs towards S45°E, so the strike is N45°W (the same line as S45°E).

Cross-check: the strike line is perpendicular to the direction of rise (a, b) = (0.5, 0.5), which points to N45°E, giving a strike azimuth of 135° (or 315°).

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 = 225°, which is S45°W.

$$\tan\delta = \sqrt{a^2 + b^2} = \sqrt{0.5^2 + 0.5^2} = 0.7071 \Rightarrow \delta = 35.26°$$

Answer: Strike N45°W; dip direction S45°W; dip amount 35.3°.