---
title: "Open traverse ABCDEFG, deflection angles"
subject: "Surveying I"
chapter: 4
key: "semester_3/surveying_i:open-traverse-deflection-angles-40-45"
source: https://www.bctengineer.com/q/semester_3/surveying_i/open-traverse-deflection-angles-40-45
site: BCT Engineering
---

# Open traverse ABCDEFG, deflection angles

Surveying I · Chapter 4: The Compass

- Asked in: 2079 Bhadra, 8 marks
- Video: [Open traverse numerical (Calculation & correction of fore bearing when deflection angle is given )](https://www.youtube.com/watch?v=PHl6saFXC7o) —  The civil classroom
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/open-traverse-deflection-angles-40-45
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4

## Question

The FB of line AB of an open traverse ABCDEFG is 40°45′. The deflection angles between the lines were measured with a theodolite and were as follows: 26°37′ (R) at B, 66°45′ (L) at C, 20°56′ (R) at D, 33°54′ (R) at E and 26°54′ (L) at F. If the BB of the last line FG observed was 209°33′, check whether the observations for deflection angles were correct or not. If not, compute the correct bearings of all the lines and the correct deflection angles.

## Short answer

Misclosure +1°00′, observations incorrect. Corrected bearings: AB 40°45′, BC 67°34′, CD 1°01′, DE 22°09′, EF 56°15′, FG 29°33′. Corrected deflection angles: B 26°49′ R, C 66°33′ L, D 21°08′ R, E 34°06′ R, F 26°42′ L.

## Answer

For an open traverse whose last line has been observed independently, the bearing of the last line computed through the deflection angles must agree with the observed bearing. Right deflection is taken as $+$ and left as $-$.

**Rule (deflection angles):** $\text{FB of next line} = \text{FB of previous line} \pm \text{deflection}$, with $+$ for a right deflection and $-$ for a left deflection.

#### Step 1: Bearings computed from the observed angles

| Line | Working | Computed FB |
|---|---|---|
| AB | given | 40°45′ |
| BC | 40°45′ + 26°37′ | 67°22′ |
| CD | 67°22′ − 66°45′ | 0°37′ |
| DE | 0°37′ + 20°56′ | 21°33′ |
| EF | 21°33′ + 33°54′ | 55°27′ |
| FG | 55°27′ − 26°54′ | 28°33′ |

Computed FB of FG = 28°33′. The observed back bearing of FG is 209°33′, so its observed FB is 209°33′ − 180° = 29°33′.

#### Step 2: Angular misclosure

$$
\text{Misclosure} = \text{observed} - \text{computed} = 29^\circ 33' - 28^\circ 33' = +1^\circ 00'
$$

The computed and observed bearings do not agree, so the observations of the deflection angles are **not correct**. The misclosure of +1°00′ is distributed equally among the 5 angles (+12′ each).

#### Step 3: Adjustment

Correction per angle = +1°00′ / 5 = +0°12′. The correction to a line is +0°12′ multiplied by the number of angles used to reach it.

| Line | Computed FB | Correction | Corrected FB | Corrected BB |
|---|---|---|---|---|
| AB | 40°45′ | 0 | 40°45′ | 220°45′ |
| BC | 67°22′ | +0°12′ | 67°34′ | 247°34′ |
| CD | 0°37′ | +0°24′ | 1°01′ | 181°01′ |
| DE | 21°33′ | +0°36′ | 22°09′ | 202°09′ |
| EF | 55°27′ | +0°48′ | 56°15′ | 236°15′ |
| FG | 28°33′ | +1°00′ | 29°33′ | 209°33′ |

Check: the corrected BB of the last line equals the observed value 209°33′.

#### Corrected deflection angles

The correction of +12′ is applied to each signed deflection angle (right $+$, left $-$), so a right deflection becomes 12′ larger and a left deflection 12′ smaller.

| Station | Observed deflection | Correction (to signed angle) | Corrected deflection |
|---|---|---|---|
| B | 26°37′ (R) | +0°12′ | 26°49′ (R) |
| C | 66°45′ (L) | +0°12′ | 66°33′ (L) |
| D | 20°56′ (R) | +0°12′ | 21°08′ (R) |
| E | 33°54′ (R) | +0°12′ | 34°06′ (R) |
| F | 26°54′ (L) | +0°12′ | 26°42′ (L) |

**Answer:** misclosure = +1°00′ (observations not correct). Corrected bearings: AB 40°45′, BC 67°34′, CD 1°01′, DE 22°09′, EF 56°15′, FG 29°33′. Corrected deflection angles: B 26°49′ R, C 66°33′ L, D 21°08′ R, E 34°06′ R, F 26°42′ L.

## Check yourself

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

1. The FB of the next line from deflection angles is:
   A. FB previous + 180°
   B. FB previous + left deflection
   C. BB previous - deflection
   D. FB previous + right deflection or - left deflection
   Answer: D. Right is +, left is -. (Key point: Deflection rule)

2. Observed BB of FG = 209°33'. The observed FB is:
   A. 29°33'
   B. 209°33'
   C. 119°33'
   D. 389°33'
   Answer: A. BB > 180° so FB = BB - 180° = 29°33'. (Key point: Observed FB of FG)

3. Computed FB of FG = 28°33', observed 29°33'. Misclosure is:
   A. -1°00'
   B. +1°00'
   C. +12'
   D. +5°
   Answer: B. Given - computed = +1°00'. (Key point: Misclosure)

4. Distributing +1°00' over 5 angles gives:
   A. +1°00' per angle
   B. +5' per angle
   C. +12' per angle
   D. +20' per angle
   Answer: C. 60'/5 = 12'; added to signed angles, so a left angle becomes smaller. (Key point: Correction)

## Close variants set in other exams

- [Open traverse ABCDEFGH, deflection angles](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aopen-traverse-deflection-angles-81-45) — Surveying I, chapter 4, asked 2×

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- [Angular misclosure, link traverse ABCDEFG](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alink-traverse-initial-154-50-last-335-30) — Surveying I, chapter 4
- [Link traverse ABCDEF, last line 108°05′40″](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alink-traverse-abcdef-108-05-40) — Surveying I, chapter 4, asked 3×
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- [Theodolite traverse ABCDA with missing VCR (68°30′10″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atheodolite-traverse-abcda-hcr-vcr-68-30-10) — Surveying I, chapter 7
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