---
title: "Closed traverse ABCDA missing data, C(1200, 2000)"
subject: "Surveying II"
chapter: 1
key: "semester_4/surveying_ii:closed-traverse-abcda-c-1200-2000"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/closed-traverse-abcda-c-1200-2000
site: BCT Engineering
---

# Closed traverse ABCDA missing data, C(1200, 2000)

Surveying II · Chapter 1: Traversing

- Asked in: 2076 Baisakh, 8 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-c-1200-2000
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1

## Question

In a four legged closed traverse ABCDA, the following data were taken.

| Leg | Length (m) | Deflection angle | Bearing | Co-ordinate |
|---|---|---|---|---|
| AB | 280.00 | ? | S 50°00'00" W | ? |
| BC | 360.00 | 85°00'00" left | ? | ? |
| CD | 320.00 | 135°00'00" left | ? | C (1200 mN, 2000 mE) |
| DA | ? | ? | ? | ? |

Find out the missing data and independent coordinates of all remaining station.

## Short answer

Using ΣLat=0 and ΣDep=0, missing line DA length is 166.667 m, bearing 343°25'. Missing left deflections: A=113°25', D=26°35'. Coordinates from C: A (1674.875 N, 2008.005 E), B (1494.895 N, 1793.512 E), D (1515.138 N, 2055.567 E).

## Answer

The deflection angles 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°00'00"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:

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

(The signs of the latitude and departure put the line in the north-west quadrant.)

#### Step 4: Missing deflection angles
Left deflection at a station = bearing of previous line − bearing of next line (add 360° if the result is 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 = 1200 mN, 2000 mE)
Independent coordinates = previous coordinates + latitude/departure of the line.

| Station | Northing (m) | Easting (m) |
|---|---|---|
| A | 1674.875 | 2008.005 |
| B | 1494.895 | 1793.512 |
| C | 1200.000 | 2000.000 |
| D | 1515.138 | 2055.567 |

Check: B plus line BC gives (1200.000 N, 2000.000 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. In this anticlockwise traverse with left deflections, 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° + deflection
   Answer: B. For an anticlockwise traverse with left deflections, bearing = bearing of previous line minus the left deflection (e.g. 230° - 85° = 145°). (Key point: Bearing rule for left deflection)

2. Which condition lets us find the length and bearing of the missing line DA?
   A. Sum of interior angles = 360°
   B. Sum of latitudes = 0 and sum of departures = 0
   C. Latitude of DA equals departure of DA
   D. Sum of the bearings = 360°
   Answer: B. For a closed traverse the missing line must cancel the sums of latitudes and departures of the other lines. (Key point: Closure condition)

3. The known lines give sum Lat = -159.737 and sum Dep = +47.562. What are the latitude and departure of DA?
   A. -159.737 and +47.562
   B. +159.737 and +47.562
   C. +159.737 and -47.562
   D. -159.737 and -47.562
   Answer: C. DA must cancel the sums, so its latitude and departure are the negatives of the sums: +159.737 and -47.562. (Key point: Latitude and departure of DA)

4. With Lat = +159.737 and Dep = -47.562, in which direction does DA lie and what is the length?
   A. North-west quadrant, bearing 343°25', length 166.667 m
   B. North-east quadrant, bearing 16°35', length 166.667 m
   C. South-west quadrant, bearing 196°35', length 207.3 m
   D. North-west quadrant, bearing 343°25', length 207.3 m
   Answer: A. Positive latitude and negative departure put DA in the NW quadrant (WCB 343°25'), and L = sqrt(159.737^2 + 47.562^2) = 166.667 m. (Key point: Bearing and length of DA)

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