---
title: "Mean direction, two sets, stations A and B"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:mean-direction-two-sets-o-a-b"
source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-two-sets-o-a-b
site: BCT Engineering
---

# Mean direction, two sets, stations A and B

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2075 Chaitra, 6 marks
- Video: [Calculating Mean Horizontal Angles In A Total Station Survey](https://www.youtube.com/watch?v=jKgbIFTVlZ0) — Redefined Horizons Learning
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-two-sets-o-a-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

Using mean direction method, calculate the mean horizontal angle.

| Instrument at | Sighted to | Set | HCR Left Face | HCR Right Face |
|---|---|---|---|---|
| O | A | I | 00°00′00″ | 179°59′30″ |
| | B | I | 121°00′00″ | 301°00′20″ |
| | A | II | 90°00′10″ | 269°59′40″ |
| | B | II | 211°00′40″ | 31°00′20″ |

## Short answer

Using mean direction method, set I angle=121°00′25″, set II=121°00′35″; mean horizontal angle AOB = 121°00′30″.

## Answer

**Method.** In each set the mean direction of each target is found from the two faces, mean direction = [Face L + (Face R - 180°)]/2. The angle AOB in the set is the difference of the mean directions of B and A. The mean of the two sets is the final angle.

| Set | Sighted to | Face L | Face R | Face R - 180° | Mean direction | Angle AOB |
|---|---|---|---|---|---|---|
| I | A | 00°00′00″ | 179°59′30″ | 359°59′30″ | 359°59′45″ | |
| | B | 121°00′00″ | 301°00′20″ | 121°00′20″ | 121°00′10″ | **121°00′25″** |
| II | A | 90°00′10″ | 269°59′40″ | 89°59′40″ | 89°59′55″ | |
| | B | 211°00′40″ | 31°00′20″ | 211°00′20″ | 211°00′30″ | **121°00′35″** |

(Face R - 180° for A in set I is 359°59′30″, i.e. -0°00′30″, and the mean of 0°00′00″ and -0°00′30″ is -0°00′15″ = 359°59′45″. For B in set II, 31°00′20″ + 180° = 211°00′20″.)

$$
\text{Mean angle AOB} = \frac{121°00′25″ + 121°00′35″}{2} = 121°00′30″
$$

The two sets differ by 0°00′10″, which is small, so the result is accepted.

**Answer:** mean horizontal angle AOB = **121°00′30″**.

## 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. In set I, mean directions are A = 359°59′45″ and B = 121°00′10″. What is angle AOB in this set?
   A. 120°59′35″
   B. 121°00′25″
   C. 121°00′10″
   D. 121°00′55″
   Answer: B. Angle = direction of B - direction of A = 121°00′10″ - (-0°00′15″) = 121°00′25″. (Key point: Angle in one set)

3. How is the final value of angle AOB obtained from the two sets (121°00′25″ and 121°00′35″)?
   A. Take the sum of the two
   B. Take the difference, 10″
   C. Take the mean of the two set values, 121°00′30″
   D. Use only set II
   Answer: C. Repeating in two sets reduces errors, and the average of the sets is the result. (Key point: Final angle)

4. In set I the face R reading for A is 179°59′30″. What is Face R - 180° for A?
   A. +0°00′30″
   B. 179°59′30″
   C. 0°00′00″
   D. -0°00′30″ (359°59′30″)
   Answer: D. 179°59′30″ - 180° is 30″ short of zero, i.e. -0°00′30″, which equals 359°59′30″. (Key point: Handling readings near 0°)

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