---
title: "Borehole attitude: square 360 m, M N O P"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:bh-attitude-square-360-mnop"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/bh-attitude-square-360-mnop
site: BCT Engineering
---

# Borehole attitude: square 360 m, M N O P

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

- Asked in: 2079 Chaitra, 4.5 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-attitude-square-360-mnop
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Four boreholes M, N, O and P are proposed at the corners of the square land. The sides of the square land are 360 m. Borehole M is west of N and P is north of N. A quartzite bed is encountered at 160 m depth in M, 60 m depth in N and 240 m depth in P. Determine the attitude of the quartzite bed. Another borehole O is north of M corner of the square land. Calculate at what depth the borehole O encounters the same quartzite bed.

## Short answer

Using three-point plane method, depth gradients are -0.2778 m/m east and 0.5 m/m north. Quartzite attitude is N61°E/29.8° NNW (dip direction 331°); borehole O at (-360,360) encounters it at 340 m depth.

## Answer

**Method:** the bed is a plane; its depth varies linearly with position. Take N as origin, x east, y north.

#### Data
- M is west of N: M = (-360, 0), depth 160 m
- N = (0, 0), depth 60 m
- P is north of N: P = (0, 360), depth 240 m
- O is north of M: O = (-360, 360)

#### Depth gradients
$$
\begin{aligned}
\text{east-west:}\quad & a = \frac{60 - 160}{360} = -0.2778\ \text{m/m (depth decreases eastward, M to N)} \\
\text{north-south:}\quad & b = \frac{240 - 60}{360} = 0.5\ \text{m/m (depth increases northward)}
\end{aligned}
$$
Depth $d = 60 - 0.2778\,x + 0.5\,y$ (m) - here $x,y$ in metres. Check M: $60 + 0.2778 \times 360 = 160$ m. Check P: $60 + 0.5 \times 360 = 240$ m.

#### Attitude
$$
\tan\delta = \sqrt{0.2778^2 + 0.5^2} = 0.572 \Rightarrow \delta = 29.8^\circ
$$
The bed deepens toward the direction $(-0.2778\ \text{east},\ +0.5\ \text{north})$, i.e. $\arctan(0.2778/0.5) = 29.05^\circ$ west of north: dip direction $= 330.95^\circ$ (N29°W). The strike is perpendicular: $240.95^\circ$, i.e. N61°E (to S61°W).

**Attitude of the quartzite: strike N61°E, dip 29.8° towards N29°W (NNW).**

#### Depth at O
$$
d_O = 60 - 0.2778(-360) + 0.5(360) = 60 + 100 + 180 = 340\ \text{m}
$$
Check using the parallelogram rule: $d_O = d_M + d_P - d_N = 160 + 240 - 60 = 340$ m.

**Answer: attitude N61°E/29.8° NNW (dip direction 331°); borehole O meets the quartzite at 340 m depth.**

## Check yourself

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

1. How is the north-south depth gradient found?
   A. (240 − 60)/360 = 0.5 using N and P
   B. (160 − 60)/360 using M and P
   C. 360/240
   D. 60/360
   Answer: A. N and P lie on a north-south line; depth rises from 60 to 240 over 360 m. (Key point: Depth gradient method)

2. How is the dip calculated?
   A. tan(dip) = 0.5 + 0.2778
   B. tan(dip) = sqrt(0.2778² + 0.5²) = 0.572, dip 29.8°
   C. dip = 0.572°
   D. tan(dip) = 0.2778
   Answer: B. The resultant gradient of the plane is the square root of the sum of squares. (Key point: Dip)

3. What is the depth at O and by what quick rule?
   A. 220 m
   B. 400 m
   C. 340 m by dM + dP − dN
   D. 100 m
   Answer: C. 160 + 240 − 60 = 340 m. (Key point: Depth at O)

4. Why is the dip direction about N29°W?
   A. Depth increases eastward
   B. Depth increases only southward
   C. Dip direction equals strike
   D. Depth increases westward and northward, so the bed deepens to NNW
   Answer: D. The deepening direction is arctan(0.2778/0.5) west of north. (Key point: Dip direction)

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