---
title: "Mean angles; VCR and VA from A"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:traverse-abcdea-vcr-va-from-a-to-e-and-b"
source: https://www.bctengineer.com/q/semester_3/surveying_i/traverse-abcdea-vcr-va-from-a-to-e-and-b
site: BCT Engineering
---

# Mean angles; VCR and VA from A

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2076 Chaitra, 6 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/traverse-abcdea-vcr-va-from-a-to-e-and-b
- 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. Also calculate the VCR and VA when sighted from station A to target stations E and B.

| Inst. Station | Target Stn | HCR Face Left | HCR Face Right | VCR Face Left | VCR Face Right |
|---|---|---|---|---|---|
| A | E | 0°0′0″ | 180°0′40″ | 65°10′30″ | ?? |
| | B | 128°47′20″ | 308°47′40″ | ?? | 297°25′40″ |
| B | A | 0°0′0″ | 180°0′40″ | | |
| | C | 102°6′40″ | 282°6′40″ | | |
| C | B | 0°0′0″ | 180°0′20″ | | |
| | D | 108°52′20″ | 288°53′0″ | | |
| D | C | 0°0′0″ | 180°0′0″ | | |
| | E | 91°0′0″ | 271°0′0″ | | |
| E | D | 0°0′0″ | 180°0′0″ | | |
| | A | 109°11′20″ | 289°12′0″ | | |

## Short answer

Mean directions from two faces, interior angles summed to 539°57′40″, error −2′20″ distributed equally. Corrected horizontal angles: A=128°47′38″, B=102°06′48″, C=108°52′58″, D=91°00′28″, E=109°12′08″. Missing VCR: FR to E=294°49′30″, FL to B=62°34′20″; vertical angles: +24°49′30″ (E) and +27°25′40″ (B).

## Answer

**Assumption:** closed pentagon ABCDEA, so the sum of the interior angles is $(5-2) \times 180^\circ = 540^\circ$.

**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 | E | 0°0′0″ | 180°0′40″ | 0°00′20″ | |
| | B | 128°47′20″ | 308°47′40″ | 128°47′30″ | **128°47′10″** |
| B | A | 0°0′0″ | 180°0′40″ | 0°00′20″ | |
| | C | 102°6′40″ | 282°6′40″ | 102°06′40″ | **102°06′20″** |
| C | B | 0°0′0″ | 180°0′20″ | 0°00′10″ | |
| | D | 108°52′20″ | 288°53′0″ | 108°52′40″ | **108°52′30″** |
| D | C | 0°0′0″ | 180°0′0″ | 0°00′00″ | |
| | E | 91°0′0″ | 271°0′0″ | 91°00′00″ | **91°00′00″** |
| E | D | 0°0′0″ | 180°0′0″ | 0°00′00″ | |
| | A | 109°11′20″ | 289°12′0″ | 109°11′40″ | **109°11′40″** |

#### Check and adjustment
Required sum = $(5-2) \times 180^\circ$ = 540°00′00″.
Observed sum = 539°57′40″.
Angular error = 539°57′40″ - 540°00′00″ = -0°02′20″.
Correction per angle = +0°00′28″ (equal for all 5 angles).

| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 128°47′10″ | +0°00′28″ | **128°47′38″** |
| B | 102°06′20″ | +0°00′28″ | **102°06′48″** |
| C | 108°52′30″ | +0°00′28″ | **108°52′58″** |
| D | 91°00′00″ | +0°00′28″ | **91°00′28″** |
| E | 109°11′40″ | +0°00′28″ | **109°12′08″** |
| Sum | 539°57′40″ | +0°02′20″ | 540°00′00″ |

**Answer:** corrected horizontal angles: A = 128°47′38″, B = 102°06′48″, C = 108°52′58″, D = 91°00′28″, E = 109°12′08″.

#### VCR and vertical angles from A
For a zenith-angle vertical circle, FL + FR = 360°.

| Target | VCR Face left | VCR Face right | Zenith distance Z | Vertical angle VA = 90° - Z |
|---|---|---|---|---|
| E | 65°10′30″ | **294°49′30″** | 65°10′30″ | **+24°49′30″** |
| B | **62°34′20″** | 297°25′40″ | 62°34′20″ | **+27°25′40″** |

**Answer:** VCR (FR) to E = 294°49′30″; VCR (FL) to B = 62°34′20″; VA to E = 24°49′30″ (elevation); VA to B = 27°25′40″ (elevation).

## Check yourself

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

1. For the closed pentagon ABCDEA, what is the required sum of the interior angles?
   A. 540°
   B. 360°
   C. 720°
   D. 900°
   Answer: A. (n - 2) x 180° = 3 x 180° = 540°. (Key point: Interior angle sum)

2. The observed sum is 539°57′40″. What is applied to each angle?
   A. -28″, since the error is positive
   B. +28″, since the error is -2′20″ shared by 5 angles
   C. +2′20″ to each angle
   D. +35″, as 2′20″ is divided by 4
   Answer: B. The sum is short by 140″; 140″/5 = 28″ is added to each angle. (Key point: Angular error and correction)

3. How is the angle at station A computed from the readings?
   A. Face L reading to B + Face R reading to E
   B. Mean direction of E + mean direction of B
   C. Mean direction of the forward station - mean direction of the back station
   D. Face R reading of B - 180°
   Answer: C. The angle between two lines is the difference of their mean directions; here B (128°47′30″) minus E (0°00′20″) = 128°47′10″. (Key point: Angle at a station)

4. VCR face left to E is 65°10′30″. What are VCR face right and the vertical angle?
   A. FR = 114°49′30″, VA = +65°10′30″
   B. FR = 294°49′30″, VA = -24°49′30″
   C. FR = 245°10′30″, VA = +24°49′30″
   D. FR = 294°49′30″, VA = +24°49′30″
   Answer: D. FR = 360° - 65°10′30″ = 294°49′30″; VA = 90° - Z = 24°49′30″, positive so it is an elevation. (Key point: Missing VCR and VA)

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