---
title: "True dip from two apparent dips"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:true-dip-from-apparent-dips"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/true-dip-from-apparent-dips
site: BCT Engineering
---

# True dip from two apparent dips

Engineering Geology II · Chapter 6: Measurement, analysis and interpretation of structural geological data

- Asked in: 2070 Magh, 4 marks
- Video: [STRUCTURAL GEOLOGY - PRACTICAL - Calculation of True dip from two apparent dip.](https://www.youtube.com/watch?v=lzlm6QPvXYM) — Success Guru
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/true-dip-from-apparent-dips
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

The apparent dip amount of an inclined bed is 1:12 and 1:16 along N30°W and N10°W respectively. Calculate the true dip amount and direction.

## Short answer

By solving the gradient components from the two apparent dips (1:12 at N30°W, 1:16 at N10°W), the true dip is 1:10.5 (5.44°) directed N59°W.

## Answer

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the solution find the true dip from two apparent dips?
   A. It averages the two apparent dips
   B. It adds the two dip angles
   C. It takes the larger apparent dip as the true dip
   D. It treats them as components of the bed's gradient and solves two simultaneous equations for gE and gN
   Answer: D. Each apparent dip gives gE sin(b) + gN cos(b) = tan(apparent); solving gives the gradient vector. (Key point: Method)

2. What true dip and direction were obtained?
   A. 1:12 (4.76°) towards N30°W
   B. 1:10.5 (5.44°) towards N59°W
   C. 1:16 (3.58°) towards N10°W
   D. 1:10.5 towards N31°E
   Answer: B. Gradient 0.0953 gives 5.44° or 1:10.5, directed 301° (N59°W); strike N31°E is perpendicular. (Key point: Result)

3. How can the answer be checked?
   A. Check that the true dip is smaller than both apparent dips
   B. Check that the directions add to 90°
   C. Check that the strike equals the average bearing
   D. Compute tan^-1(0.0953 cos θ) along each given bearing and compare with 1:12 and 1:16
   Answer: D. cos 29° and cos 49° reproduce 4.76° (1:12) and 3.58° (1:16). (Key point: Check)

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