---
title: "Theodolite traverse ABCDA with missing VCR (68°30′10″)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:theodolite-traverse-abcda-hcr-vcr-68-30-10"
source: https://www.bctengineer.com/q/semester_3/surveying_i/theodolite-traverse-abcda-hcr-vcr-68-30-10
site: BCT Engineering
---

# Theodolite traverse ABCDA with missing VCR (68°30′10″)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2081 Baisakh, 5+1 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/theodolite-traverse-abcda-hcr-vcr-68-30-10
- 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 correct horizontal angles and find the missing readings from the following readings in the given table.

| Instrument station | Target station | HCR Face left | HCR Face right | VCR Face left | VCR Face right |
|---|---|---|---|---|---|
| A | D | 90°00′00″ | 269°59′50″ | 68°30′10″ | ? |
| | B | 160°12′40″ | 340°12′30″ | | |
| B | A | 90°00′10″ | 270°00′20″ | ? | 315°45′40″ |
| | C | 200°25′50″ | 20°26′10″ | | |
| C | B | 90°00′00″ | 269°59′40″ | 94°05′40″ | ? |
| | D | 179°08′40″ | 359°08′30″ | | |
| D | C | 89°59′50″ | 270°00′10″ | ? | 270°14′20″ |
| | A | 180°16′10″ | 00°16′10″ | | |

## Short answer

Corrected horizontal angles: A=70°11′50″, B=110°24′55″, C=89°07′55″, D=90°15′20″. Missing VCR: A to D face right=291°29′50″, B to A face left=44°14′20″, C to B face right=265°54′20″, D to C face left=89°45′40″.

## Answer

**Assumption:** the traverse ABCDA is a closed four-sided figure, so the sum of the interior angles must be 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′00″ | 269°59′50″ | 89°59′55″ | |
| | B | 160°12′40″ | 340°12′30″ | 160°12′35″ | **70°12′40″** |
| B | A | 90°00′10″ | 270°00′20″ | 90°00′15″ | |
| | C | 200°25′50″ | 20°26′10″ | 200°26′00″ | **110°25′45″** |
| C | B | 90°00′00″ | 269°59′40″ | 89°59′50″ | |
| | D | 179°08′40″ | 359°08′30″ | 179°08′35″ | **89°08′45″** |
| D | C | 89°59′50″ | 270°00′10″ | 90°00′00″ | |
| | A | 180°16′10″ | 00°16′10″ | 180°16′10″ | **90°16′10″** |

#### Check and adjustment
Required sum = $(4-2) \times 180^\circ$ = 360°00′00″.
Observed sum = 360°03′20″.
Angular error = 360°03′20″ - 360°00′00″ = +0°03′20″.
Correction per angle = -0°00′50″ (equal for all 4 angles).

| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 70°12′40″ | -0°00′50″ | **70°11′50″** |
| B | 110°25′45″ | -0°00′50″ | **110°24′55″** |
| C | 89°08′45″ | -0°00′50″ | **89°07′55″** |
| D | 90°16′10″ | -0°00′50″ | **90°15′20″** |
| Sum | 360°03′20″ | -0°03′20″ | 360°00′00″ |

**Answer:** corrected horizontal angles: A = 70°11′50″, B = 110°24′55″, C = 89°07′55″, D = 90°15′20″.

#### 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) |
|---|---|---|---|
| A to D | 68°30′10″ | **291°29′50″** | +21°29′50″ |
| B to A | **44°14′20″** | 315°45′40″ | +45°45′40″ |
| C to B | 94°05′40″ | **265°54′20″** | -4°05′40″ |
| D to C | **89°45′40″** | 270°14′20″ | +0°14′20″ |

## Check yourself

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

1. What should the sum of interior angles of the closed four-sided traverse be?
   A. 180°
   B. 540°
   C. 360°
   D. 720°
   Answer: C. Sum = (n - 2) x 180° = 2 x 180° = 360°. (Key point: Angle sum condition)

2. How is the angle at a station found?
   A. Mean direction of the forward station - mean direction of the back station
   B. Sum of Face L and Face R
   C. Face L - 180°
   D. Mean of the VCR readings
   Answer: A. Angle = forward mean direction - back mean direction. (Key point: Angle at a station)

3. The observed sum is 360°03'20''. What is the correction per angle?
   A. -0°00'25''
   B. -0°00'50''
   C. -0°03'20''
   D. -0°01'40''
   Answer: B. Error is 200'', divided equally among 4 angles gives 50'' each, subtracted. (Key point: Distribution of error)

4. D to C has FR VCR 270°14'20''. What is the missing FL?
   A. 89°45'40''
   B. 90°14'20''
   C. 270°14'20''
   D. 0°14'20''
   Answer: A. FL = 360° - 270°14'20'' = 89°45'40''. (Key point: Missing VCR)

## 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 (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 (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×
- [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
- [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

## 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
- [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