---
title: "Mean angles, traverse ABCDA (2072)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:mean-horizontal-angles-abcda-face-left-right-b-a"
source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-abcda-face-left-right-b-a
site: BCT Engineering
---

# Mean angles, traverse ABCDA (2072)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2072 Chaitra, 6 marks
- Video: [Calculate length and bearing by theodolite| how to calculate bearing|how to calculate length](https://www.youtube.com/watch?v=rdtOjdbTrnU) — Hamro Engineering
- 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-face-left-right-b-a
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

The following observations were recorded in a theodolite traverse ABCDEA. Compute the mean horizontal angles and adjust them if necessary.

| Inst. Stn | Target stn | 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′10″ |
| C | B | 90°00′0″ | 269°59′50″ |
| | D | 179°08′40″ | 359°08′20″ |
| D | C | 89°59′50″ | 270°00′10″ |
| | A | 160°12′40″ | 340°12′30″ |

## Short answer

Mean directions from two faces; sum 360°03′10″, error +3′10″ equally distributed. Corrected angles: A=110°24′52.5″, B=90°15′32.5″, C=89°07′47.5″, D=70°11′47.5″.

## 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′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′10″ | 180°16′10″ | **90°16′20″** |
| C | B | 90°00′0″ | 269°59′50″ | 89°59′55″ | |
| | D | 179°08′40″ | 359°08′20″ | 179°08′30″ | **89°08′35″** |
| 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°03′10″.
Angular error = 360°03′10″ - 360°00′00″ = +0°03′10″.
Correction per angle = -0°00′47.5″ (equal for all 4 angles).

| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 110°25′40″ | -0°00′47.5″ | **110°24′52.5″** |
| B | 90°16′20″ | -0°00′47.5″ | **90°15′32.5″** |
| C | 89°08′35″ | -0°00′47.5″ | **89°07′47.5″** |
| D | 70°12′35″ | -0°00′47.5″ | **70°11′47.5″** |
| Sum | 360°03′10″ | -0°03′10″ | 360°00′00″ |

**Answer:** corrected horizontal angles: A = 110°24′52.5″, B = 90°15′32.5″, C = 89°07′47.5″, D = 70°11′47.5″.

## Check yourself

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

1. The question says ABCDEA but gives only four instrument stations. How is the traverse treated?
   A. As a closed quadrilateral ABCDA with angle sum 360°
   B. As a pentagon with angle sum 540°
   C. As an open traverse with no check
   D. As a triangle with angle sum 180°
   Answer: A. Only A, B, C and D are occupied, so the figure is a quadrilateral. (Key point: Assumption on stations)

2. At B, mean directions are A = 89°59′50″ and C = 180°16′10″. What is the angle at B?
   A. 90°16′10″
   B. 90°16′20″
   C. 89°43′40″
   D. 270°16′20″
   Answer: B. 180°16′10″ - 89°59′50″ = 90°16′20″. (Key point: Mean angle at B)

3. The mean angles total 360°03′10″. What is the error?
   A. -3′10″
   B. +3′50″
   C. +3′10″, since the sum is too large
   D. Zero
   Answer: C. 360°03′10″ - 360° = +3′10″. (Key point: Error found)

4. What correction is applied to each angle?
   A. +47.5″
   B. -190″
   C. -38″
   D. -47.5″
   Answer: D. 190″ / 4 = 47.5″, subtracted because the sum exceeds 360°. (Key point: Correction per angle)

## Related in Surveying I

- [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×
- [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×
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## Related in other subjects

- [Gale's table for traverse ABCDA](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agale-table-abcda-2072a) — Surveying II, chapter 1