---
title: "Length, bearing of EF between midpoints"
subject: "Surveying II"
chapter: 1
key: "semester_4/surveying_ii:traverse-abcda-length-bearing-ef-midpoints"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-abcda-length-bearing-ef-midpoints
site: BCT Engineering
---

# Length, bearing of EF between midpoints

Surveying II · Chapter 1: Traversing

- Asked in: 2071 Bhadra, 10 marks
- Video: [How to Find distance and Bearing by coordinates | Most easy  method | RAAK Engineer](https://www.youtube.com/watch?v=mJe1PxCUXVk) — RAAK Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-abcda-length-bearing-ef-midpoints
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1

## Question

A traverse ABCDA was conducted and the following data were obtained. It was required to connect the midpoint E of CD to the midpoint F of AB. Find the length and bearing of EF.

| Line | Length | Bearing |
|---|---|---|
| AB | 610.00 | N 80°10' E |
| BC | 510.00 | N 13°00' E |
| CD | 1130.00 | S 80°10' W |
| DA | 450.00 | S 15°30' E |

## Short answer

After adjusting the closed traverse by Bowditch's rule, coordinates of midpoints E (of CD) and F (of AB) are (1520.535, 1332.168) and (1054.971, 1331.853). Length EF = √(ΔN²+ΔE²) = 465.56 m, bearing = 180°02' (S 0°02' W).

## Answer

The positions of E (midpoint of CD) and F (midpoint of AB) are found from the coordinates of A, B, C and D. The traverse is first computed and balanced, and EF is then found from the coordinates of its two end points. A is taken as the origin (1000 N, 1000 E) to keep the coordinates positive.

#### Step 1: Latitudes and departures
| Line | Length (m) | Bearing | Latitude | Departure | Corr. lat | Corr. dep |
|---|---|---|---|---|---|---|
| AB | 610.000 | 80°10' | +104.177 | +601.038 | +5.7638 | +62.6669 |
| BC | 510.000 | 13°00' | +496.929 | +114.725 | +4.8189 | +52.3937 |
| CD | 1130.000 | 260°10' | -192.985 | -1113.399 | +10.6772 | +116.0879 |
| DA | 450.000 | 164°30' | -433.634 | +120.257 | +4.2520 | +46.2297 |
| **Sum** | 2700.000 |  | -25.512 | -277.378 | +25.5120 | +277.3782 |

The sums are not zero, so the traverse has a closing error of $e=\sqrt{(-25.512)^2+(-277.378)^2}=278.55$ m on a perimeter of 2700 m. The error is very large (about 1 in 10), so the field data should be rechecked; for the computation it is distributed by Bowditch's rule so that the traverse closes.

#### Step 2: Adjusted lines and coordinates
| Line | Adj. lat | Adj. dep | Adj. length (m) | Adj. bearing |
|---|---|---|---|---|
| AB | +109.941 | +663.705 | 672.749 | 80°36' |
| BC | +501.748 | +167.119 | 528.847 | 18°25' |
| CD | -182.307 | -997.311 | 1013.837 | 259°38' |
| DA | -429.382 | +166.487 | 460.529 | 158°48' |

| Station | Northing (m) | Easting (m) |
|---|---|---|
| A | 1000.000 | 1000.000 |
| B | 1109.941 | 1663.705 |
| C | 1611.689 | 1830.824 |
| D | 1429.382 | 833.513 |

#### Step 3: Midpoints
$$
\begin{aligned}
E&=\left(\frac{N_C+N_D}{2},\frac{E_C+E_D}{2}\right)=(1520.535,\ 1332.168)\\
F&=\left(\frac{N_A+N_B}{2},\frac{E_A+E_B}{2}\right)=(1054.971,\ 1331.853)
\end{aligned}
$$

#### Step 4: Length and bearing of EF
$$
\begin{aligned}
\Delta N&=N_F-N_E=-465.565,\qquad \Delta E=E_F-E_E=-0.316\\
EF&=\sqrt{\Delta N^2+\Delta E^2}=465.56\ \text{m}\\
\theta_{EF}&=\tan^{-1}\frac{-0.316}{-465.565}=180^\circ 02'
\end{aligned}
$$

Both differences are negative, so EF points almost due south (S 0°02' W), whole circle bearing 180°02'.

**Answer:** EF = 465.56 m, bearing 180°02' (S 0°02' W). (If the unbalanced values of AB, BC and CD are used directly, i.e. EF taken as the line closing F-B-C-E, then EF = 474.12 m at 162°38'.)

## Check yourself

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

1. What is the basic approach to find length and bearing of EF?
   A. Find coordinates of A, B, C, D, then coordinates of midpoints E and F, then join them
   B. Average the bearings of AB and CD
   C. Measure EF from a plotted sketch
   D. Use the cosine rule on triangle ABC
   Answer: A. The midpoints are obtained from station coordinates, and the length and bearing follow from the coordinate differences. (Key point: Method to find EF)

2. The traverse sums are ΣLat = -25.512 and ΣDep = -277.378. What does this indicate and what is done?
   A. It closes well, no action needed
   B. A large error (about 1 in 10): field data should be rechecked, and for computation it is balanced by Bowditch's rule
   C. It must be balanced by the transit rule only
   D. It means the bearings are reversed
   Answer: B. The closing error is about 278.6 m in 2700 m, far too big; it is distributed by Bowditch's rule so the polygon closes. (Key point: Closing error and balancing)

3. How are the coordinates of E, midpoint of CD, found?
   A. Mean of the northings and mean of the eastings of C and D
   B. Difference of the coordinates of C and D
   C. Coordinates of C plus the length CD
   D. Coordinates of D plus half of DA
   Answer: A. E = ((N_C + N_D)/2, (E_C + E_D)/2). (Key point: Midpoint coordinates)

4. ΔN = N_F - N_E = -465.565 and ΔE = -0.316. What is the bearing of EF?
   A. About 0°02' (north)
   B. About 180°02' (almost due south)
   C. About 90°
   D. About 359°58'
   Answer: B. Both differences are negative, so the line is in the south-west quadrant, nearly due south: WCB 180°02'. (Key point: Bearing of EF)

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