---
title: "Mean direction method, traverse ABCDEA (2080)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:mean-direction-abcda-hcr-90-00-20"
source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-abcda-hcr-90-00-20
site: BCT Engineering
---

# Mean direction method, traverse ABCDEA (2080)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2080 Baisakh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-abcda-hcr-90-00-20
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

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

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

## Short answer

Corrected horizontal angles: A=110°24′52.5″, B=90°15′37.5″, C=89°07′37.5″, D=70°11′52.5″.

## Answer

**Assumption:** the title says ABCDEA, but only four instrument stations (A, B, C, D) are given, so the traverse is taken as the closed quadrilateral ABCDA with angle sum 360°.

**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′20″ | 269°59′40″ | 90°00′00″ | |
| | B | 200°25′40″ | 20°25′30″ | 200°25′35″ | **110°25′35″** |
| B | A | 89°59′30″ | 270°00′10″ | 89°59′50″ | |
| | C | 180°16′10″ | 00°16′10″ | 180°16′10″ | **90°16′20″** |
| 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′10″ | 90°00′00″ | |
| | A | 160°12′40″ | 340°12′30″ | 160°12′35″ | **70°12′35″** |

#### Check and adjustment
Required sum = $(4-2) \times 180^\circ$ = 360°00′00″.
Observed sum = 360°02′50″.
Angular error = 360°02′50″ - 360°00′00″ = +0°02′50″.
Correction per angle = -0°00′42.5″ (equal for all 4 angles).

| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 110°25′35″ | -0°00′42.5″ | **110°24′52.5″** |
| B | 90°16′20″ | -0°00′42.5″ | **90°15′37.5″** |
| C | 89°08′20″ | -0°00′42.5″ | **89°07′37.5″** |
| D | 70°12′35″ | -0°00′42.5″ | **70°11′52.5″** |
| Sum | 360°02′50″ | -0°02′50″ | 360°00′00″ |

**Answer:** corrected horizontal angles: A = 110°24′52.5″, B = 90°15′37.5″, C = 89°07′37.5″, D = 70°11′52.5″.

## Check yourself

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

1. Only stations A, B, C, D are given although the title says ABCDEA. What is assumed?
   A. A five-sided traverse with sum 540°
   B. A closed quadrilateral ABCDA with angle sum 360°
   C. An open traverse
   D. A triangle with sum 180°
   Answer: B. With four instrument stations the traverse is taken as a closed quadrilateral, so (4 - 2) x 180° = 360°. (Key point: Assumption of closed figure)

2. The mean angles sum to 360°02'50''. What is the correction per angle?
   A. -0°01'25''
   B. -0°02'50''
   C. -0°00'42.5''
   D. +0°00'42.5''
   Answer: C. 170'' / 4 = 42.5'' is subtracted from each angle. (Key point: Angular error)

3. At A, mean direction to D is 90°00'00'' and to B is 200°25'35''. What is the angle?
   A. 110°25'35''
   B. 290°25'35''
   C. 90°25'35''
   D. 200°25'35''
   Answer: A. Angle = forward - back direction = 200°25'35'' - 90°00'00'' = 110°25'35''. (Key point: Mean angle at A)

## Close variants set in other exams

- [Mean angles, traverse ABCDA (2076)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-hcr-90-00-20-c-b) — Surveying I, chapter 7, asked 2×

## Related in Surveying I

- [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, 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
- [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; VCR and VA from A](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atraverse-abcdea-vcr-va-from-a-to-e-and-b) — Surveying I, chapter 7
- [Direction theodolite, least count 6″, station D](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Adirection-theodolite-ls-6-sec-station-d) — Surveying I, chapter 7

## Related in other subjects

- [Coordinates of instrument station O and station A](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-pwd-instrument-station-a) — Surveying II, chapter 5
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- [Gale's table for traverse ABCDEA](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agale-table-abcdea-2078b) — Surveying II, chapter 1