---
title: "Sandstone attitude: A N45W 900 m, C S20E 700 m"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:bh-sandstone-a-n45w-c-s20e"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/bh-sandstone-a-n45w-c-s20e
site: BCT Engineering
---

# Sandstone attitude: A N45W 900 m, C S20E 700 m

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

- Asked in: 2075 Bhadra, 3+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/bh-sandstone-a-n45w-c-s20e
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Three boreholes A, B, C were drilled in a flat terrain to investigate depth of bedrock. Borehole A lies N45°W from borehole B at a distance of 900 m and borehole C lies S20°E from borehole B at a distance of 700 m. A sandstone bedrock is encountered at the following depth of each borehole.
Borehole A: Top (-350 m), Bottom (-410 m)
Borehole B: Top (-310 m), Bottom (-370 m)
Borehole C: Top (-390 m), Bottom (-450 m)
Find out the attitude of the sandstone bedrock with true thickness.

## Short answer

Solving the plane through top levels gives gradients a=0.2900 and b=0.2272. Sandstone attitude is N38.1°W/20.2° SW (dip direction 232°); true thickness is 56.3 m.

## Answer

A bed is a plane, so its level changes uniformly in plan: $z = z_B + a x + b y$, where $a$ and $b$ are the rates of change of level per metre east and north. This is the analytical form of the three-point (strike-line) method.

#### Coordinates (B at origin)
$$
\begin{aligned}
A &: x = 900\sin 315^\circ = -636.40\ \text{m},\ y = 900\cos 315^\circ = 636.40\ \text{m} \\
C &: x = 700\sin 160^\circ = 239.41\ \text{m},\ y = 700\cos 160^\circ = -657.78\ \text{m}
\end{aligned}
$$
Levels of the top: A = -350 m, B = -310 m, C = -390 m. The bed is 60 m thick in each hole, $t_v = 60$ m.

#### Plane of the top surface
$z = -310 + a x + b y$:
$$
\begin{aligned}
-350 &= -310 - 636.40a + 636.40b \\
-390 &= -310 + 239.41a - 657.78b
\end{aligned}
$$
Solving: $a = 0.2900$, $b = 0.2272$.

#### Dip, dip direction and strike
$$
\tan\delta = \sqrt{a^2+b^2} = 0.3684 \Rightarrow \delta = 20.23^\circ
$$
The bed descends toward the direction $\arctan(-a/-b)$ = **231.9°** (SW), so the strike is $\ 231.9^\circ - 90^\circ = 141.9^\circ$, i.e. **N38.1°W**.

#### True thickness
All three boreholes show the same vertical thickness, 60 m, so
$$
t = t_v\cos\delta = 60\cos 20.23^\circ = 56.30\ \text{m}
$$

**Answer: attitude of the sandstone = strike N38.1°W, dip 20.2° towards SW (dip direction 232°); true thickness = 56.3 m.**

## Check yourself

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

1. How is A located relative to B?
   A. x = 900, y = 0
   B. x = 0, y = 900
   C. x = 636, y = 636
   D. x = 900 sin315° = −636.4, y = 900 cos315° = 636.4
   Answer: D. Bearings are converted into east and north components. (Key point: Coordinates)

2. What is solved to get a and b?
   A. Two equations from levels at A and C with B as origin
   B. One equation only
   C. The thickness
   D. Dip directly
   Answer: A. Two unknowns need two equations. (Key point: Plane fit)

3. What is the dip?
   A. 29°
   B. 20.2°, from tan = sqrt(0.29² + 0.2272²)
   C. 0.368°
   D. 45°
   Answer: B. tan(dip) = 0.3684. (Key point: Dip)

4. What is the true thickness?
   A. 60 m
   B. 63.9 m
   C. 60 cos20.23° = 56.3 m
   D. 20 m
   Answer: C. True thickness equals vertical thickness times cos(dip). (Key point: True thickness)

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