---
title: "Horizontal angles from field notes, triangle ABC"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:theodolite-field-notes-abc-64-jestha"
source: https://www.bctengineer.com/q/semester_3/surveying_i/theodolite-field-notes-abc-64-jestha
site: BCT Engineering
---

# Horizontal angles from field notes, triangle ABC

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2064 Jestha, 10 marks
- Video: [Mean angle method |Surveying-1 |Theodolite #surveying #theodolite](https://www.youtube.com/watch?v=spp_2OSoWdQ) —  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/theodolite-field-notes-abc-64-jestha
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

The following are field notes of theodolite survey. Compute horizontal angles and apply check.

| Station | Sighted to | Direct (D M S) | Reverse (D M S) |
|---|---|---|---|
| A | C | 0 00 10 | 180 00 00 |
| | B | 30 00 20 | 210 00 00 |
| B | A | 0 00 00 | 180 00 20 |
| | C | 60 00 00 | 240 00 10 |
| C | B | 0 00 10 | 180 00 00 |
| | A | 90 00 05 | 270 00 10 |

## Short answer

Mean angles from two-face readings: A = 30°00′05″, B = 59°59′55″, C = 90°00′02.5″. Sum 180°00′02.5″, excess +2.5″; equal correction -0.83″ gives A = 30°00′04.17″, B = 59°59′54.17″, C = 90°00′01.67″.

## Answer

**Check.** The three stations A, B, C form a triangle, so the sum of the three angles must be 180°.

**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 | C | 0 00 10 | 180 00 00 | 0°00′05″ | |
| | B | 30 00 20 | 210 00 00 | 30°00′10″ | **30°00′05″** |
| B | A | 0 00 00 | 180 00 20 | 0°00′10″ | |
| | C | 60 00 00 | 240 00 10 | 60°00′05″ | **59°59′55″** |
| C | B | 0 00 10 | 180 00 00 | 0°00′05″ | |
| | A | 90 00 05 | 270 00 10 | 90°00′07.5″ | **90°00′02.5″** |

#### Check and adjustment
Required sum = $(3-2) \times 180^\circ$ = 180°00′00″.
Observed sum = 180°00′02.5″.
Angular error = 180°00′02.5″ - 180°00′00″ = +0°00′02.5″.
Correction per angle = -0°00′00.83″ (equal for all 3 angles).

| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 30°00′05″ | -0°00′00.83″ | **30°00′04.17″** |
| B | 59°59′55″ | -0°00′00.83″ | **59°59′54.17″** |
| C | 90°00′02.5″ | -0°00′00.83″ | **90°00′01.67″** |
| Sum | 180°00′02.5″ | -0°00′02.5″ | 180°00′00″ |

**Answer:** corrected horizontal angles: A = 30°00′04.17″, B = 59°59′54.17″, C = 90°00′01.67″.

## Check yourself

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

1. What check is applied to the three angles A, B and C?
   A. Their sum must be 180°
   B. Their sum must be 360°
   C. Their sum must be 540°
   D. Each must equal 60°
   Answer: A. The three stations form a triangle: (3 - 2) x 180° = 180°. (Key point: Triangle check)

2. At B, mean directions are A = 0°00′10″ and C = 60°00′05″. The angle at B is:
   A. 60°00′15″
   B. 59°59′55″
   C. 60°00′05″
   D. 59°59′45″
   Answer: B. 60°00′05″ - 0°00′10″ = 59°59′55″. (Key point: Angle at B)

3. The observed sum is 180°00′02.5″. What is the error?
   A. -2.5″
   B. +5″
   C. +2.5″
   D. 0″
   Answer: C. Sum - 180° = +2.5″. (Key point: Error)

4. What is the correction to each angle?
   A. +0.83″
   B. -2.5″
   C. -1.25″
   D. -0.83″
   Answer: D. 2.5″/3 = 0.83″, subtracted from each. (Key point: Correction)

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