True dip from two apparent dips

Principle: the apparent dip in a direction making an angle $\theta$ with the true dip direction satisfies $\tan\delta_a = \tan\delta\cos\theta$. Using the two apparent dips as components of the gradient of the bed gives the true dip.

Given

Apparent slope 1: $\tan\delta_1 = 1/12 = 0.0833$ along N30°W (330°). Apparent slope 2: $\tan\delta_2 = 1/16 = 0.0625$ along N10°W (350°).

Solution

Let the true dip gradient components be $g_E$ (east) and $g_N$ (north), so that for a direction of bearing $\beta$:

$$g_E\sin\beta + g_N\cos\beta = \tan\delta_a$$ $$\begin{aligned} g_E\sin 330^\circ + g_N\cos 330^\circ &= 0.08333 \\ g_E\sin 350^\circ + g_N\cos 350^\circ &= 0.06250 \end{aligned}$$

Solving: $g_E = -0.08169$, $g_N = 0.04906$. Note that these are components in the sense of the direction of downward dip (the bed dips towards the quadrant of both bearings, NW).

$$\tan\delta = \sqrt{g_E^2 + g_N^2} = 0.0953 \Rightarrow \delta = 5.44^\circ$$ $$\text{dip amount} = \frac{1}{0.0953} = 1:10.5$$ $$\text{direction} = \arctan\left(\frac{g_E}{g_N}\right) = \arctan\left(\frac{-0.08169}{0.04906}\right) = -59.0^\circ \Rightarrow 301.0^\circ = \text{N}59^\circ\text{W}$$

Answer: true dip = 1:10.5 (5.44°), directed N59°W (dip direction 301°); the strike is N31°E.

Check: apparent dip along 330° = $\tan^{-1}(0.0953\cos 29^\circ) = 4.76^\circ$ = 1:12. Along 350° = $\tan^{-1}(0.0953\cos 49^\circ) = 3.58^\circ$ = 1:16. Both agree with the given data.