---
title: "Local attraction ABCDEA by included angles"
subject: "Surveying I"
chapter: 4
key: "semester_3/surveying_i:local-attraction-abcdea-included-angle"
source: https://www.bctengineer.com/q/semester_3/surveying_i/local-attraction-abcdea-included-angle
site: BCT Engineering
---

# Local attraction ABCDEA by included angles

Surveying I · Chapter 4: The Compass

- Asked in: 2081 Bhadra, 8 marks; 2068 Baisakh, 8 marks (asked 2 times)
- Video: [Finding Local Attraction and Correct Magnetic Bearings Using Included Angle Method|Solved Problem](https://www.youtube.com/watch?v=3pWrw1Kr8vc) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/local-attraction-abcdea-included-angle
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4

## Question

The following fore and back bearings were observed in a closed traverse ABCDEA where local attraction was suspected. Find which stations are affected by local attraction and work out the correct bearings of the lines by the included angle method. Also calculate the included angles and correct them if necessary.

| Line | F.B. | B.B. |
|---|---|---|
| AB | 191°30′ | 13°0′ |
| BC | 69°30′ | 246°30′ |
| CD | 32°15′ | 210°30′ |
| DE | 262°45′ | 80°45′ |
| EA | 230°15′ | 53°00′ |

## Short answer

All stations affected. Included angles: 138°30′, 56°30′, 145°45′, 52°15′, 149°30′; sum 542°30′, error +2°30′ distributed equally to 138°00′, 56°00′, 145°15′, 51°45′, 149°00′. Adopted AB reference bearing 192°15′; corrected FB: AB 192°15′, BC 68°15′, CD 33°30′, DE 265°15′, EA 234°15′.

## Answer

Because the FB and BB of every line differ from 180° by 1°30′ to 3°00′, **all the stations (A, B, C, D, E) are affected by local attraction** (no line is free of it). The included angle method is therefore used, since an angle at a station is not affected by local attraction at that station.

#### Step 1: Check each line

| Line | FB | BB | Difference between FB and BB | Remark |
|---|---|---|---|---|
| AB | 191°30′ | 13°00′ | 178°30′ | Differs from 180° by 1°30′ |
| BC | 69°30′ | 246°30′ | 177°00′ | Differs from 180° by 3°00′ |
| CD | 32°15′ | 210°30′ | 178°15′ | Differs from 180° by 1°45′ |
| DE | 262°45′ | 80°45′ | 182°00′ | Differs from 180° by 2°00′ |
| EA | 230°15′ | 53°00′ | 177°15′ | Differs from 180° by 2°45′ |

#### Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

- A: FB of AB − BB of EA = 191°30′ − 53°00′ = 138°30′
- B: FB of BC − BB of AB = 69°30′ − 13°00′ = 56°30′
- C: 360° − (BB of BC − FB of CD) = 360° − (246°30′ − 32°15′) = 145°45′
- D: FB of DE − BB of CD = 262°45′ − 210°30′ = 52°15′
- E: FB of EA − BB of DE = 230°15′ − 80°45′ = 149°30′

The sum of the interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

| Station | Angle between | Observed | Corrected |
|---|---|---|---|
| A | AE and AB | 138°30′ | 138°00′ |
| B | BA and BC | 56°30′ | 56°00′ |
| C | CB and CD | 145°45′ | 145°15′ |
| D | DC and DE | 52°15′ | 51°45′ |
| E | ED and EA | 149°30′ | 149°00′ |

Sum observed = 542°30′; error = +2°30′; correction per angle = −0°30′ (distributed equally, since each angle is measured with the same care).

No line has a difference of exactly 180°. Line AB has the smallest discrepancy (1°30′), so it is taken as the most reliable; its adopted bearing is the mean of the FB and (BB + 180°): (191°30′ + 193°00′)/2 = 192°15′.

#### Step 3: Corrected bearings from the corrected angles

Rule used (traverse run anticlockwise): FB of next line = BB of previous line + interior angle (subtract 360° if the result exceeds 360°).

- BC: 12°15′ + 56°00′ = 68°15′
- CD: 248°15′ + 145°15′ − 360° = 33°30′
- DE: 213°30′ + 51°45′ = 265°15′
- EA: 85°15′ + 149°00′ = 234°15′
- Check, AB: 54°15′ + 138°00′ = 192°15′ (agrees with the adopted value)

#### Step 4: Corrected bearings

| Line | Observed FB | Correction | Corrected FB | Observed BB | Corrected BB |
|---|---|---|---|---|---|
| AB | 191°30′ | +0°45′ | 192°15′ | 13°00′ | 12°15′ |
| BC | 69°30′ | −1°15′ | 68°15′ | 246°30′ | 248°15′ |
| CD | 32°15′ | +1°15′ | 33°30′ | 210°30′ | 213°30′ |
| DE | 262°45′ | +2°30′ | 265°15′ | 80°45′ | 85°15′ |
| EA | 230°15′ | +4°00′ | 234°15′ | 53°00′ | 54°15′ |

Check: corrected FB and corrected BB differ by exactly 180° for every line.

**Answer:** all stations A to E are affected. Corrected included angles: A = 138°00′, B = 56°00′, C = 145°15′, D = 51°45′, E = 149°00′ (sum 540°). The corrected bearings are given in the last table (FB: AB 192°15′, BC 68°15′, CD 33°30′, DE 265°15′, EA 234°15′).

*Assumption:* no line in the traverse has FB and BB differing by exactly 180°, so the line with the smallest discrepancy (AB) is taken as the reference and its mean bearing is adopted.

## Check yourself

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

1. Why is the included angle method used here?
   A. Because bearings are never wrong at the first station
   B. An angle at a station is not affected by local attraction at that station
   C. Because it needs no angle sum check
   D. Because local attraction changes angles by equal amounts at all stations
   Answer: B. Both bearings at a station have the same error, which cancels in the angle between them. (Key point: Why included angle method)

2. What should the sum of the interior angles of the 5-sided traverse be?
   A. 360°
   B. 450°
   C. 540°
   D. 720°
   Answer: C. (2n - 4) x 90° = (10 - 4) x 90° = 540°. (Key point: Angle sum check)

3. Observed angles sum to 542°30'. What correction is applied to each angle?
   A. +0°30'
   B. -2°30'
   C. -1°15'
   D. -0°30'
   Answer: D. Error +2°30' is spread equally over 5 angles, giving -0°30' each, because all were measured with equal care. (Key point: Angle correction)

## Related in Surveying I

- [Local attraction in closed traverse ABCDEFA](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-abcdefa-216-30) — Surveying I, chapter 4
- [Interior angles and bearings, closed traverse ABCD](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Ainterior-angles-closed-traverse-104-30) — Surveying I, chapter 4
- [Local attraction in closed traverse PQRSTP](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-pqrstp-188-45) — Surveying I, chapter 4
- [Local attraction, S 40°30′ W traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-quadrantal-s40-30w) — Surveying I, chapter 4
- [Local attraction, bearings AB 305°00′ (EA 45°)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-305-00) — Surveying I, chapter 4, asked 2×
- [Local attraction, bearings AB 305°30′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-bb-44-00) — Surveying I, chapter 4, asked 2×

## Related in other subjects

- [Closed traverse missing EA bearing, BC length](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aclosed-traverse-abcdea-missing-ea-bc) — Surveying II, chapter 1