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

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

Step 1: Strike line

Along AB the altitude rises uniformly from 400 m (B) to 500 m (A), i.e. 100 m in 600 m (0.1667 m per m). The point D on AB with altitude 400 m (equal to C) is at a distance from B of

$$BD = \frac{400-400}{500-400} \times 600 = 0\ \text{m}$$

So D coincides with B, and line BC, which joins two points at 400 m, is itself the strike line.

Step 2: Strike direction

BC runs due east, so the strike is E-W (N90°E).

Cross-check: taking east as +x and north as +y, the eastward rise is $a = (400-400)/300 = 0$ and the northward rise is $b = (500-400)/600 = 0.1667$. The strike line is perpendicular to the direction of rise (due north), giving an azimuth of 90°.

Step 3: Dip direction and dip amount

The bed rises towards A (north), so it dips the opposite way (down the slope), at 90° to the strike. Dip direction azimuth = 180°, which is due south.

$$\tan\delta = \sqrt{a^2 + b^2} = \sqrt{0^2 + 0.1667^2} = 0.1667 \Rightarrow \delta = 9.46°$$

Answer: Strike E-W (N90°E); dip direction due south; dip amount 9.5°.