---
title: "Mean horizontal angles, station O (P, Q, R)"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:mean-horizontal-angles-o-pqrp"
source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-o-pqrp
site: BCT Engineering
---

# Mean horizontal angles, station O (P, Q, R)

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2078 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-o-pqrp
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

Calculate the mean horizontal angles. If necessary, adjust them also.

| Instrument | Sighted to | HCR Face Left | HCR Face right |
|---|---|---|---|
| O | P | 00°00′00″ | 179°59′40″ |
| | Q | 294°29′50″ | 114°30′20″ |
| | R | 137°54′20″ | 317°54′40″ |
| | P | 00°00′40″ | 180°00′20″ |

## Short answer

Adjusted angles: POQ=65°29′58.33″, QOR=156°35′48.33″, ROP=137°54′13.33″. Mean directions from face left and right give the angles; error of -40″ distributed equally.

## Answer

**Method.** The mean direction of each target is found from the two faces: mean direction = [Face L + (Face R - 180°)]/2.

The readings decrease from P to Q to R and back to P, so the telescope was turned anticlockwise (the circle is graduated clockwise). Each angle is therefore (previous mean direction) - (next mean direction). The three angles at O must add up to 360°, and the second observation of P closes the horizon.

| Inst. | Sighted to | Face L | Face R | Mean direction |
|---|---|---|---|---|
| O | P | 00°00′00″ | 179°59′40″ | 359°59′50″ |
| | Q | 294°29′50″ | 114°30′20″ | 294°30′05″ |
| | R | 137°54′20″ | 317°54′40″ | 137°54′30″ |
| | P (closing) | 00°00′40″ | 180°00′20″ | 0°00′30″ |

#### Mean angles
$$
\begin{aligned}
\angle POQ &= 359°59′50″ - 294°30′05″ = 65°29′45″\\
\angle QOR &= 294°30′05″ - 137°54′30″ = 156°35′35″\\
\angle ROP &= 137°54′30″ - 0°00′30″ = 137°54′00″
\end{aligned}
$$

Sum = 359°59′20″; required = 360°00′00″; error = -0°00′40″ (the sum is 0°00′40″ less than 360°). The error is small, so it is distributed equally: correction = +0°00′13.33″ per angle.

| Angle | Mean angle | Correction | Adjusted angle |
|---|---|---|---|
| POQ | 65°29′45″ | +0°00′13.33″ | **65°29′58.33″** |
| QOR | 156°35′35″ | +0°00′13.33″ | **156°35′48.33″** |
| ROP | 137°54′00″ | +0°00′13.33″ | **137°54′13.33″** |
| Sum | 359°59′20″ | +0°00′40″ | 360°00′00″ |

**Answer:** adjusted angles POQ = 65°29′58.33″, QOR = 156°35′48.33″, ROP = 137°54′13.33″.

## Check yourself

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

1. The readings decrease from P to Q to R. What does this show?
   A. Telescope was turned anticlockwise, so angle = previous direction - next direction
   B. Telescope was turned clockwise
   C. The circle is faulty
   D. The face was not changed
   Answer: A. Decreasing readings show anticlockwise turning, so each angle is the previous mean direction minus the next. (Key point: Direction of turning)

2. Mean directions of P and Q are 359°59'50'' and 294°30'05''. What is angle POQ?
   A. 294°30'05''
   B. 654°29'55''
   C. 65°29'45''
   D. 114°30'05''
   Answer: C. 359°59'50'' - 294°30'05'' = 65°29'45''. (Key point: Angle POQ)

3. The angles sum to 359°59'20''. What is the correction per angle?
   A. -0°00'13.33''
   B. +0°00'40''
   C. +0°00'20''
   D. +0°00'13.33''
   Answer: D. The sum is 40'' short of 360°, so add 40''/3 = 13.33'' to each angle. (Key point: Adjustment)

## Related in Surveying I

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