---
title: "Closed traverse ABCDA missing figures (C=2500)"
subject: "Surveying II"
chapter: 1
key: "semester_4/surveying_ii:closed-traverse-abcda-missing-2072m"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/closed-traverse-abcda-missing-2072m
site: BCT Engineering
---

# Closed traverse ABCDA missing figures (C=2500)

Surveying II · Chapter 1: Traversing

- Asked in: 2072 Magh, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/closed-traverse-abcda-missing-2072m
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1

## Question

In a four sided closed traverse ABCDA the following informations are given.

| Side | Length (m) | Deflection angle | Bearing | Coordinates |
|---|---|---|---|---|
| AB | 280.00 | ? | S50°W | ? |
| BC | 360.00 | 85°00' Left | ? | ? |
| CD | 320.00 | 135°00' Left | ? | C = 2500 mE, 2500 mN |
| DA | ? | ? | ? | ? |

i) Compute all missing figures.
ii) Compute the coordinates of other points with respect to C.

## Short answer

Length DA = 166.667 m, bearing DA = 343°25' (N 16°35' W). Left deflections: at A 113°25', at D 26°35'. Coordinates relative to C (2500 N, 2500 E): A (2974.875, 2508.005), B (2794.895, 2293.512), D (2815.138, 2555.567).

## Answer

All the given deflections are to the left, so the traverse runs anticlockwise and **bearing of a line = bearing of previous line − left deflection**. For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$, which gives the length and bearing of the last line.

#### Step 1: Bearings of the known lines
- AB: S50°W = 230°00'
- BC: 230°00' − 85°00' = 145°00'
- CD: 145°00' − 135°00' = 10°00'

#### Step 2: Latitudes and departures of the known lines
$$
\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta
$$

| Line | L (m) | Bearing | Latitude | Departure |
|---|---|---|---|---|
| AB | 280.00 | 230°00' | -179.981 | -214.492 |
| BC | 360.00 | 145°00' | -294.895 | +206.488 |
| CD | 320.00 | 10°00' | +315.138 | +55.567 |
| **Sum** |  |  | -159.737 | +47.562 |

#### Step 3: Length and bearing of DA
For closure the missing line must cancel the sums:

$$
\begin{aligned}
\text{Lat}_{DA}&=-(-159.737)=+159.737\ \text{m}\\
\text{Dep}_{DA}&=-(+47.562)=-47.562\ \text{m}\\
L_{DA}&=\sqrt{(+159.737)^2+(-47.562)^2}=166.667\ \text{m}\\
\tan\theta&=\frac{-47.562}{+159.737}\ \Rightarrow\ \theta_{DA}=343^\circ 25'
\end{aligned}
$$

(Positive latitude and negative departure put DA in the north-west quadrant, N 16°35' W.)

#### Step 4: Missing deflection angles
Left deflection at a station = bearing of previous line − bearing of next line (add 360° if negative).

- At A (DA to AB): 343°25' − 230°00' = 113°25'
- At D (CD to DA): 10°00' − 343°25' + 360° = 26°35'

Check: sum of left deflections = 85°00' + 135°00' + 113°25' + 26°35' = 360°00' (anticlockwise traverse).

#### Step 5: Coordinates (C = 2500 mN, 2500 mE)
Independent coordinates = previous coordinates + latitude/departure of the line.

| Station | Northing (m) | Easting (m) |
|---|---|---|
| A | 2974.875 | 2508.005 |
| B | 2794.895 | 2293.512 |
| C | 2500.000 | 2500.000 |
| D | 2815.138 | 2555.567 |

Check: B plus line BC gives (2500.000 N, 2500.000 E), which is the given point, so the traverse closes.

#### Summary of the table

| Line | Length (m) | Deflection at start (left) | Bearing (WCB) | Latitude | Departure |
|---|---|---|---|---|---|
| AB | 280.000 | 113°25' (computed) | 230°00' | -179.981 | -214.492 |
| BC | 360.000 | 85°00' (given) | 145°00' | -294.895 | +206.488 |
| CD | 320.000 | 135°00' (given) | 10°00' | +315.138 | +55.567 |
| DA | 166.667 (computed) | 26°35' (computed) | 343°25' (computed) | +159.737 | -47.562 |

## Check yourself

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

1. For this anticlockwise traverse, how is the bearing of a line found from the previous line?
   A. Previous bearing + left deflection
   B. Previous bearing - left deflection
   C. Previous bearing + 180° - deflection
   D. Previous bearing - 180°
   Answer: B. A left deflection turns the line anticlockwise, which reduces the bearing: bearing = previous - left deflection. (Key point: Bearing rule for left deflections)

2. AB is S50°W (230°). What are the bearings of BC and CD?
   A. BC = 145°, CD = 10°
   B. BC = 315°, CD = 180°
   C. BC = 145°, CD = 280°
   D. BC = 35°, CD = 170°
   Answer: A. BC = 230° - 85° = 145° and CD = 145° - 135° = 10°. (Key point: Bearing of BC and CD)

3. How are the length and bearing of DA found?
   A. Its latitude and departure are equal to the sums of the other lines
   B. Its latitude and departure cancel the sums of the other three lines (ΣLat = ΣDep = 0)
   C. From the deflection angle at D only
   D. By averaging the other lengths
   Answer: B. For closure ΣLat = 0 and ΣDep = 0, so DA = -(sum of the known latitudes and departures); length from Pythagoras, bearing from tan^-1. (Key point: Closure condition for missing line)

4. Lat(DA) = +159.737 and Dep(DA) = -47.562. Where does DA lie and what is its bearing?
   A. North-east quadrant, 16°35'
   B. North-west quadrant, WCB 343°25'
   C. South-west quadrant, 196°35'
   D. South-east quadrant, 163°25'
   Answer: B. Positive latitude and negative departure put the line in the NW quadrant, with WCB 360° - 16°35' = 343°25'. (Key point: Quadrant of DA)

5. What check applies to the left deflections of this anticlockwise closed traverse?
   A. They sum to 180°
   B. They sum to 360°
   C. They sum to (2n-4) x 90° = 360° as interior angles
   D. They are all equal
   Answer: B. For a closed polygon the algebraic sum of deflections is 360° (all left here); 85 + 135 + 113°25' + 26°35' = 360°. (Key point: Deflection angle check)

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