---
title: "Theodolite traverse ABCDA, booking format with VCR"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:traverse-abcda-booking-format-vcr-120-14-20"
source: https://www.bctengineer.com/q/semester_3/surveying_i/traverse-abcda-booking-format-vcr-120-14-20
site: BCT Engineering
---

# Theodolite traverse ABCDA, booking format with VCR

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2074 Chaitra, 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/traverse-abcda-booking-format-vcr-120-14-20
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

The following observations were recorded in a theodolite traverse ABCDA. Compute the mean horizontal angles and the missing readings by entering the following readings in a standard booking format.

| Instrument Stations | Target Stations | HCR Direct | HCR Reversed | VCR FL | VCR FR |
|---|---|---|---|---|---|
| A | D | 89°59′50″ | 270°00′10″ | | |
| | B | 160°12′40″ | 340°12′30″ | | |
| B | A | 90°00′00″ | 269°59′40″ | 120°14′20″ | ? |
| | C | 179°08′40″ | 359°08′30″ | | |
| C | B | 90°00′00″ | 269°59′50″ | | |
| | D | 200°25′40″ | 20°25′20″ | | |
| D | C | 90°00′10″ | 270°00′00″ | ? | 308°51′20″ |
| | A | 180°16′10″ | 00°16′00″ | | |

## Short answer

Mean directions computed; observed sum 360°02′55″, error +2′55″ distributed equally. Corrected angles: A=70°11′51.25″, B=89°08′01.25″, C=110°24′51.25″, D=90°15′16.25″. Missing VCR: FR (B-A)=239°45′40″, FL (D-C)=51°08′40″; vertical angles: −30°14′20″ and +38°51′20″.

## Answer

**Assumption:** closed traverse ABCDA (four stations), angle sum 360°. The readings are entered in the standard booking format below.

**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 | 89°59′50″ | 270°00′10″ | 90°00′00″ | |
| | B | 160°12′40″ | 340°12′30″ | 160°12′35″ | **70°12′35″** |
| B | A | 90°00′00″ | 269°59′40″ | 89°59′50″ | |
| | C | 179°08′40″ | 359°08′30″ | 179°08′35″ | **89°08′45″** |
| C | B | 90°00′00″ | 269°59′50″ | 89°59′55″ | |
| | D | 200°25′40″ | 20°25′20″ | 200°25′30″ | **110°25′35″** |
| D | C | 90°00′10″ | 270°00′00″ | 90°00′05″ | |
| | A | 180°16′10″ | 00°16′00″ | 180°16′05″ | **90°16′00″** |

#### 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 | 70°12′35″ | -0°00′43.75″ | **70°11′51.25″** |
| B | 89°08′45″ | -0°00′43.75″ | **89°08′01.25″** |
| C | 110°25′35″ | -0°00′43.75″ | **110°24′51.25″** |
| D | 90°16′00″ | -0°00′43.75″ | **90°15′16.25″** |
| Sum | 360°02′55″ | -0°02′55″ | 360°00′00″ |

**Answer:** corrected horizontal angles: A = 70°11′51.25″, B = 89°08′01.25″, C = 110°24′51.25″, D = 90°15′16.25″.

#### Missing VCR readings
**Missing VCR.** For a vertical circle graduated as zenith angle, the sum of the face-left and face-right readings of the same target is 360° when the instrument has no index error. So the missing reading = 360° - the given reading. The vertical angle is $90^\circ - Z$ (+ elevation, - depression).

| Observation | VCR Face left | VCR Face right | Vertical angle (90° - Z) |
|---|---|---|---|
| B to A | 120°14′20″ | **239°45′40″** | -30°14′20″ |
| D to C | **51°08′40″** | 308°51′20″ | +38°51′20″ |

## Check yourself

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

1. How is the mean direction to a target found from the two face readings?
   A. [Face L + (Face R -/+ 180°)] / 2
   B. Face L + Face R
   C. Face L - Face R
   D. (Face L + Face R) / 360°
   Answer: A. Face R differs from Face L by 180°, so it is brought back by 180° and averaged with Face L. (Key point: Mean direction formula)

2. What is the required sum of the four interior angles, and what is the observed error here?
   A. 180°; observed error is +2′55″
   B. 360°; the observed sum is 360°02′55″, error +2′55″
   C. 360°; observed sum is 359°57′05″
   D. 720°; observed error is -2′55″
   Answer: B. (4 - 2) x 180° = 360°. The observed sum is 2′55″ too large. (Key point: Angle sum of ABCDA)

3. What correction is applied to each angle?
   A. +43.75″
   B. -35″
   C. -43.75″
   D. -2′55″
   Answer: C. 175″ / 4 = 43.75″, subtracted because the sum is too large. (Key point: Correction per angle)

4. VCR face left B to A is 120°14′20″. The face right reading and vertical angle are:
   A. FR = 59°45′40″, VA = +30°14′20″
   B. FR = 239°45′40″, VA = +30°14′20″
   C. FR = 240°14′20″, VA = -30°14′20″
   D. FR = 239°45′40″, VA = -30°14′20″
   Answer: D. FR = 360° - 120°14′20″ = 239°45′40″; VA = 90° - 120°14′20″ = -30°14′20″, a depression. (Key point: Missing VCR)

## Related in Surveying I

- [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 (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 (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
- [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×
- [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

## Related in other subjects

- [Closed traverse ABCDA missing data, C(1200, 2000)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aclosed-traverse-abcda-c-1200-2000) — Surveying II, chapter 1
- [Closed traverse ABCDA missing figures (C=2500)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aclosed-traverse-abcda-missing-2072m) — Surveying II, chapter 1
- [Closed traverse ABCDA, missing data](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aclosed-traverse-abcda-missing-2069b) — Surveying II, chapter 1
- [Coordinates of O from S and D (FL/FR)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aintersection-o-south-of-s-and) — Surveying II, chapter 5