---
title: "Mean angles BAC, CAD, DAE at A"
subject: "Surveying I"
chapter: 7
key: "semester_3/surveying_i:mean-horizontal-angles-bac-cad-dae"
source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-bac-cad-dae
site: BCT Engineering
---

# Mean angles BAC, CAD, DAE at A

Surveying I · Chapter 7: Transit and Theodolite

- Asked in: 2059 Chaitra, 10 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-bac-cad-dae
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7

## Question

Find the mean horizontal angles BAC, CAD and DAE in the following cases.

| Instrument Stn | Target Stn | F.L. | F.R. |
|---|---|---|---|
| A | B | 0°10′10″ | 180°10′20″ |
| | C | 62°10′10″ | 242°10′30″ |
| | D | 135°20′15″ | 315°20′35″ |
| | E | 250°30′20″ | 70°30′40″ |

## Short answer

Compute mean direction = (FL + FR – 180°)/2. Mean directions: B=0°10′15″, C=62°10′20″, D=135°20′25″, E=250°30′30″. Subtract successive values: BAC=62°00′05″, CAD=73°10′05″, DAE=115°10′05″.

## Answer

**Method.** The instrument is at A and four targets B, C, D and E are observed on both faces. Mean direction of each target = [F.L. + (F.R. - 180°)]/2. The angle between two targets = difference of their mean directions.

| Inst. stn | Target | F.L. | F.R. | F.R. ∓ 180° | Mean direction |
|---|---|---|---|---|---|
| A | B | 0°10′10″ | 180°10′20″ | 0°10′20″ | 0°10′15″ |
| A | C | 62°10′10″ | 242°10′30″ | 62°10′30″ | 62°10′20″ |
| A | D | 135°20′15″ | 315°20′35″ | 135°20′35″ | 135°20′25″ |
| A | E | 250°30′20″ | 70°30′40″ | 250°30′40″ | 250°30′30″ |

$$
\begin{aligned}
\angle BAC &= 62°10′20″ - 0°10′15″ = 62°00′05″\\
\angle CAD &= 135°20′25″ - 62°10′20″ = 73°10′05″\\
\angle DAE &= 250°30′30″ - 135°20′25″ = 115°10′05″
\end{aligned}
$$

(Check: BAE = 250°20′15″ = 62°00′05″ + 73°10′05″ + 115°10′05″.)

**Answer:** BAC = **62°00′05″**, CAD = **73°10′05″**, DAE = **115°10′05″**.

## 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. With mean directions B = 0°10′15″ and C = 62°10′20″, angle BAC is:
   A. 62°10′20″
   B. 62°00′05″
   C. 62°10′05″
   D. 61°59′55″
   Answer: B. 62°10′20″ - 0°10′15″ = 62°00′05″. (Key point: Angle BAC)

3. With D = 135°20′25″ and E = 250°30′30″, angle DAE is:
   A. 115°10′15″
   B. 115°09′55″
   C. 115°10′05″
   D. 250°30′30″
   Answer: C. 250°30′30″ - 135°20′25″ = 115°10′05″. (Key point: Angle DAE)

4. What is the check on the angles BAC, CAD and DAE?
   A. Their sum equals 180°
   B. Their sum equals 360°
   C. Each equals 60°
   D. Their sum equals the single angle BAE (250°20′15″)
   Answer: D. Adjacent angles at one station add up to the overall angle. (Key point: Check)

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