---
title: "Open traverse ABCDEFGH, deflection angles"
subject: "Surveying I"
chapter: 4
key: "semester_3/surveying_i:open-traverse-deflection-angles-81-45"
source: https://www.bctengineer.com/q/semester_3/surveying_i/open-traverse-deflection-angles-81-45
site: BCT Engineering
---

# Open traverse ABCDEFGH, deflection angles

Surveying I · Chapter 4: The Compass

- Asked in: 2074 Chaitra, 8 marks
- Video: [PROBLEMS ON DEFLECTION ANGLE SURVEY UNIT-1 SESSION 10](https://www.youtube.com/watch?v=13ItxDlOu0A) — CHANDRASHEKAR HANUMANTHAIAH
- 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-81-45
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4

## Question

The fore bearing of line AB of an open traverse ABCDEFGH is 81°45′. The deflection angles between the lines were measured with a theodolite and were as follows: 25°30′ (R) at B, 37°45′ (L) at C, 45°15′ (R) at D, 55°30′ (L) at E, 75°15′ (L) at F and 80°00′ (R) at G. If the FB of the last line observed was 63°00′, check whether the observations for deflection angles are correct or not. If not, compute the correct bearings of all the lines.

## Short answer

Misclosure −1°00′, observations are incorrect. Corrected bearings: AB 81°45′, BC 107°05′, CD 69°10′, DE 114°15′, EF 58°35′, FG 343°10′, GH 63°00′.

## Answer

The bearing of the last line, computed from the first bearing and the observed deflection angles, is compared with its 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 | 81°45′ |
| BC | 81°45′ + 25°30′ | 107°15′ |
| CD | 107°15′ − 37°45′ | 69°30′ |
| DE | 69°30′ + 45°15′ | 114°45′ |
| EF | 114°45′ − 55°30′ | 59°15′ |
| FG | 59°15′ − 75°15′ + 360° | 344°00′ |
| GH | 344°00′ + 80°00′ − 360° | 64°00′ |

Computed FB of GH = 64°00′; observed FB of GH = 63°00′ (given directly).

#### Step 2: Angular misclosure

$$
\text{Misclosure} = \text{observed} - \text{computed} = 63^\circ 00' - 64^\circ 00' = -1^\circ 00'
$$

The computed and observed bearings do not agree, so the deflection angles are **not correct**. The misclosure of −1°00′ is distributed equally among the 6 angles (−10′ each).

#### Step 3: Adjustment

Correction per angle = −1°00′ / 6 = −0°10′. The correction to a line is −0°10′ multiplied by the number of angles used to reach it.

| Line | Computed FB | Correction | Corrected FB | Corrected BB |
|---|---|---|---|---|
| AB | 81°45′ | 0 | 81°45′ | 261°45′ |
| BC | 107°15′ | −0°10′ | 107°05′ | 287°05′ |
| CD | 69°30′ | −0°20′ | 69°10′ | 249°10′ |
| DE | 114°45′ | −0°30′ | 114°15′ | 294°15′ |
| EF | 59°15′ | −0°40′ | 58°35′ | 238°35′ |
| FG | 344°00′ | −0°50′ | 343°10′ | 163°10′ |
| GH | 64°00′ | −1°00′ | 63°00′ | 243°00′ |

Check: the corrected FB of the last line equals the observed value 63°00′.

#### Corrected deflection angles

The correction of −10′ is applied to each signed deflection angle (right $+$, left $-$), so a right deflection becomes 10′ smaller and a left deflection 10′ larger.

| Station | Observed deflection | Correction (to signed angle) | Corrected deflection |
|---|---|---|---|
| B | 25°30′ (R) | −0°10′ | 25°20′ (R) |
| C | 37°45′ (L) | −0°10′ | 37°55′ (L) |
| D | 45°15′ (R) | −0°10′ | 45°05′ (R) |
| E | 55°30′ (L) | −0°10′ | 55°40′ (L) |
| F | 75°15′ (L) | −0°10′ | 75°25′ (L) |
| G | 80°00′ (R) | −0°10′ | 79°50′ (R) |

**Answer:** misclosure = −1°00′ (observations not correct). Corrected bearings: AB 81°45′, BC 107°05′, CD 69°10′, DE 114°15′, EF 58°35′, FG 343°10′, GH 63°00′.

## Check yourself

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

1. FB of FG = 59°15' - 75°15' (left). Result:
   A. -16°00'
   B. 134°30'
   C. 16°00'
   D. 344°00' (after adding 360°)
   Answer: D. Negative result is brought into range by adding 360°. (Key point: Bearing computation)

2. Computed GH = 64°00', observed 63°00'. Misclosure is:
   A. -1°00'
   B. +1°00'
   C. -10'
   D. +2°
   Answer: A. Given - computed = -1°00'. (Key point: Misclosure)

3. With 6 angles the correction per angle is:
   A. -12'
   B. -10'
   C. -1°
   D. -15'
   Answer: B. -60'/6 = -10'. (Key point: Correction per angle)

4. Left angle at C observed 37°45' becomes:
   A. 37°35' (L)
   B. 37°45' (L)
   C. 37°55' (L)
   D. 37°55' (R)
   Answer: C. Correction is on signed angle (L negative): -37°45' - 10' = -37°55', i.e. 37°55' L. (Key point: Corrected deflection)

## Close variants set in other exams

- [Open traverse ABCDEFG, deflection angles](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aopen-traverse-deflection-angles-40-45) — Surveying I, chapter 4, asked 2×

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- [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×
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- [Local attraction: AB 69°30′ compass traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-69-30-reordered) — 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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