---
title: "Mean angles, traverse ABCDA (2076)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:mean-horizontal-angles-abcda-hcr-90-00-20-c-b"
source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-abcda-hcr-90-00-20-c-b
site: BCT Engineering
---

# Mean angles, traverse ABCDA (2076)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2076 Asoj, 6 marks
- Video: [FIND THE MEAN HORIZONTAL ANGLE - SURVEY- II DIPLOMA IN ENGINEERING 4TH SEMESTER @Er.dipesh186](https://www.youtube.com/watch?v=suMVMgn8UYs) — Er. Dipesh Bhatt
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-abcda-hcr-90-00-20-c-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

The following observations were recorded in a theodolite traverse ABCDA. Rule out the proper field book and compute horizontal angles and adjust them if necessary.

| Instrument station | Target station | HCR Direct | HCR Reversed |
|---|---|---|---|
| A | D | 90°00′10″ | 269°59′40″ |
| | B | 200°25′40″ | 20°25′30″ |
| B | A | 89°59′30″ | 270°00′10″ |
| | C | 180°16′10″ | 00°16′00″ |
| C | B | 90°00′10″ | 270°00′10″ |
| | D | 179°08′40″ | 359°08′20″ |
| D | C | 89°59′50″ | 270°00′00″ |
| | A | 160°12′40″ | 340°12′30″ |

## Short answer

Mean directions computed; sum 360°02′55″, error +2′55″ equally distributed. Corrected angles: A=110°24′56.25″, B=90°15′31.25″, C=89°07′36.25″, D=70°11′56.25″.

## Answer

**Assumption:** closed traverse ABCDA with four stations; angle sum must be 360°. Readings are booked in the standard field book form (instrument station, target station, face L, face R, mean direction, angle).

**Method.** For each target the mean direction is found from the two faces:

$$
\text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2}
$$

The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care).

#### Field book and mean angles
| Inst. stn | Sighted to | Face L | Face R | Mean direction | Mean angle |
|---|---|---|---|---|---|
| A | D | 90°00′10″ | 269°59′40″ | 89°59′55″ | |
| | B | 200°25′40″ | 20°25′30″ | 200°25′35″ | **110°25′40″** |
| B | A | 89°59′30″ | 270°00′10″ | 89°59′50″ | |
| | C | 180°16′10″ | 00°16′00″ | 180°16′05″ | **90°16′15″** |
| C | B | 90°00′10″ | 270°00′10″ | 90°00′10″ | |
| | D | 179°08′40″ | 359°08′20″ | 179°08′30″ | **89°08′20″** |
| D | C | 89°59′50″ | 270°00′00″ | 89°59′55″ | |
| | A | 160°12′40″ | 340°12′30″ | 160°12′35″ | **70°12′40″** |

#### Check and adjustment
Required sum = $(4-2) \times 180^\circ$ = 360°00′00″.
Observed sum = 360°02′55″.
Angular error = 360°02′55″ - 360°00′00″ = +0°02′55″.
Correction per angle = -0°00′43.75″ (equal for all 4 angles).

| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 110°25′40″ | -0°00′43.75″ | **110°24′56.25″** |
| B | 90°16′15″ | -0°00′43.75″ | **90°15′31.25″** |
| C | 89°08′20″ | -0°00′43.75″ | **89°07′36.25″** |
| D | 70°12′40″ | -0°00′43.75″ | **70°11′56.25″** |
| Sum | 360°02′55″ | -0°02′55″ | 360°00′00″ |

**Answer:** corrected horizontal angles: A = 110°24′56.25″, B = 90°15′31.25″, C = 89°07′36.25″, D = 70°11′56.25″.

## Check yourself

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

1. What columns does the field book for this theodolite traverse contain?
   A. Instrument station, target station, face L, face R, mean direction, angle
   B. Only Face L readings and the angle
   C. Bearing, latitude and departure
   D. Staff readings and reduced levels
   Answer: A. Both faces are booked, the mean direction is found and the angle computed from the differences. (Key point: Standard field book)

2. For C to B, Face L is 90°00′10″ and Face R is 270°00′10″. What is the mean direction?
   A. 180°00′20″
   B. 90°00′10″
   C. 90°00′00″
   D. 89°59′50″
   Answer: B. (90°00′10″ + 90°00′10″)/2 = 90°00′10″. (Key point: Mean direction)

3. Mean angles total 360°02′55″. What is the angular error?
   A. -2′55″
   B. +3′10″
   C. +2′55″
   D. +2′50″
   Answer: C. Sum minus 360° = 175″ too large. (Key point: Angular error)

4. Each angle is corrected by:
   A. +43.75″
   B. -175″
   C. -58.3″
   D. -43.75″
   Answer: D. 175″/4 = 43.75″, subtracted from each. (Key point: Correction)

## Close variants set in other exams

- [Mean direction method, traverse ABCDEA (2080)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-abcda-hcr-90-00-20) — Surveying I, chapter 7, asked 2×

## Related in Surveying I

- [Theodolite traverse ABCDA, booking format with VCR](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atraverse-abcda-booking-format-vcr-120-14-20) — Surveying I, chapter 7
- [Mean angles, traverse ABCDA (2072)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-face-left-right-b-a) — Surveying I, chapter 7
- [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
- [Mean angles, traverse ABCDA (2074 Asoj)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-hcr-90-00-00-a-d) — Surveying I, chapter 7, asked 2×
- [Mean angles, traverse ABCDA (2068 Baisakh)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-hcr-209-25-40) — Surveying I, chapter 7, asked 2×
- [Horizontal angles from field notes ABCDA (2065)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atheodolite-field-notes-abcda-65-shrawan) — Surveying I, chapter 7

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