---
title: "Booking format for two-set horizontal angle"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:booking-format-two-set-horizontal-angle"
source: https://www.bctengineer.com/q/semester_3/surveying_i/booking-format-two-set-horizontal-angle
site: BCT Engineering
---

# Booking format for two-set horizontal angle

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2069 Chaitra, 4 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/booking-format-two-set-horizontal-angle
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

Develop a booking format for recording 2 set horizontal angle with appropriate numerical example. Calculate the mean horizontal angle also.

## Short answer

In the direction method, two sets with different zero settings (e.g., 0° and 90°) are recorded with FL and FR readings for each target. Mean direction = (FL + FR ± 180°)/2, and the mean of the two set angles gives the final result; example mean horizontal angle AOB = 62°24′45.0″.

## Answer

In the direction method with two sets, the zero setting of the circle is changed between the sets (e.g. 0° for set 1 and 90° for set 2) so that different parts of the circle are used. Each set has face left and face right observations.

#### Booking format (angle AOB, 2 sets)
| Inst. at | Set | Sighted to | Face L (D M S) | Face R (D M S) | Mean direction | Angle for the set |
|---|---|---|---|---|---|---|
| O | I | A | 0°00′00″ | 180°00′20″ | 0°00′10.0″ | |
| | | B | 62°24′40″ | 242°25′00″ | 62°24′50.0″ | 62°24′40.0″ |
| | II | A | 90°00′20″ | 270°00′00″ | 90°00′10.0″ | |
| | | B | 152°24′50″ | 332°25′10″ | 152°25′00.0″ | 62°24′50.0″ |

**Mean direction** = [Face L + (Face R - 180°)] / 2.

Set I: A = (0°00′00″ + 0°00′20″)/2 = 0°00′10.0″, B = (62°24′40″ + 62°25′00″)/2 = 62°24′50.0″, angle = 62°24′40.0″.

Set II: A = (90°00′20″ + 90°00′00″)/2 = 90°00′10.0″, B = (152°24′50″ + 152°25′10″)/2 = 152°25′00.0″, angle = 62°24′50.0″.

$$
\text{Mean angle AOB} = \frac{62°24′40.0″ + 62°24′50.0″}{2} = 62°24′45.0″
$$

**Answer:** mean horizontal angle AOB = **62°24′45.0″**. The two sets agree within 10 seconds, so the result is accepted.

## Check yourself

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

1. Why is the zero setting changed between sets (e.g. 0° and 90°)?
   A. To avoid reading both faces
   B. So different parts of the circle are used, reducing graduation error
   C. To make the angle larger
   D. To remove centring error
   Answer: B. Using different parts of the circle averages out graduation errors. (Key point: Why change zero between sets)

2. In Set I, A reads 0°00'00'' on face left and 180°00'20'' on face right. What is the mean direction of A?
   A. 0°00'20''
   B. 0°00'10''
   C. 90°00'10''
   D. 0°00'05''
   Answer: B. (0°00'00'' + (180°00'20'' - 180°))/2 = 10''. (Key point: Mean direction)

3. Set I gives AOB = 62°24'40'' and Set II gives 62°24'50''. What is the mean angle?
   A. 62°24'50''
   B. 62°24'40''
   C. 62°24'45''
   D. 62°24'30''
   Answer: C. (62°24'40'' + 62°24'50'')/2 = 62°24'45''. (Key point: Mean of sets)

## 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
- [Horizontal angle by repetition and direction methods](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Arepetition-and-direction-methods-field-notes) — Surveying I, chapter 7, asked 2×
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- [Mean direction, two sets, stations A and B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-two-sets-o-a-b) — Surveying I, chapter 7

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