---
title: "Missing bearings of BC and DE in traverse"
subject: "Surveying II"
chapter: 1
key: "semester_4/surveying_ii:traverse-missing-bearings-bc-de"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-missing-bearings-bc-de
site: BCT Engineering
---

# Missing bearings of BC and DE in traverse

Surveying II · Chapter 1: Traversing

- Asked in: 2076 Bhadra, 8 marks
- Video: [Calculation of Bearing from Included Angles | Close Traverse | Theodolite Traversing](https://www.youtube.com/watch?v=nEI2bq4qzS4) — Civil Engineering Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-missing-bearings-bc-de
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1

## Question

The following traverse notes were taken during a field survey in which the bearing of lines BC and DE could not be taken due to some reasons. Compute the missing data.

| Line | AB | BC | CD | DE | EF | FA |
|---|---|---|---|---|---|---|
| Length (m) | 872 | 322 | 770 | 406 | 1079 | 1480 |
| Q.B | N41°35'W | ?? | N63°12'W | ?? | S26°39'E | N53°30'E |

## Short answer

The resultant of the known lines is 1002.05 m, exceeding the sum of BC and DE lengths (728 m), so no triangle can be formed. Therefore, with the given data, the bearings of BC and DE cannot be computed; field notes must be rechecked.

## Answer

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. The lengths of BC and DE are known but their bearings are not, so the two lines must together cancel the resultant of the four lines AB, CD, EF and FA. The lines BC, DE and that resultant form a triangle, which is solved by the cosine rule.

#### Step 1: Resultant of the known lines
| Line | Length (m) | Bearing | Latitude | Departure |
|---|---|---|---|---|
| AB | 872 | 318°25' | +652.248 | -578.754 |
| CD | 770 | 296°48' | +347.176 | -687.291 |
| EF | 1079 | 153°21' | -964.370 | +483.974 |
| FA | 1480 | 53°30' | +880.338 | +1189.708 |
| **Sum** |  |  | +915.391 | +407.637 |

$$
\begin{aligned}
r&=\sqrt{915.391^2+407.637^2}=1002.05\ \text{m}\\
\phi&=\tan^{-1}\frac{-407.637}{-915.391}=204^\circ 00'\ \ (\text{direction of the vector BC + DE})
\end{aligned}
$$

#### Step 2: Triangle test
BC and DE must together make up a vector of length $r$, which is possible only if

$$
|DE-BC|=84\ \text{m}\ \le\ r\ \le\ BC+DE=728\ \text{m}
$$

Here $r=1002.05$ m is greater than $BC+DE=728$ m. A triangle with sides 322 m, 406 m and 1002.05 m does not exist, so **no real bearings of BC and DE satisfy closure with the lengths and bearings as recorded**. The cosine rule gives

$$
\cos\alpha=\frac{r^2+l_{BC}^2-l_{DE}^2}{2\,r\,l_{BC}}=\frac{1002.05^2+322^2-406^2}{2\times1002.05\times322}=1.461
$$

A cosine cannot exceed 1, so the triangle is impossible. With the printed figures the closure check fails by a large margin, so one or more of the recorded lengths or bearings (AB, CD, EF, FA, BC, DE) must be wrong and need to be re-observed.

#### Method to apply once the data are correct
1. Compute $r$ and $\phi$ as in Step 1.
2. Compute $\alpha$ from the cosine rule above; then bearing of BC $=\phi\pm\alpha$.
3. Find the bearing of DE from the remaining vector: $\tan\theta_{DE}=\dfrac{-(\Sigma\text{Dep}+\text{Dep}_{BC})}{-(\Sigma\text{Lat}+\text{Lat}_{BC})}$, placing it in the correct quadrant.
4. Check that $\Sigma\text{Lat}=\Sigma\text{Dep}=0$.

**Answer:** with the data as given, the bearings of BC and DE cannot be computed, because the resultant of the known lines (1002.05 m) exceeds the sum of the two unknown lengths (728 m); the field notes need to be rechecked.

## Check yourself

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

1. When the lengths of BC and DE are known but not their bearings, how is the problem solved?
   A. BC and DE together with the resultant of the other four lines form a triangle solved by the cosine rule
   B. Both bearings are taken equal to the mean of adjacent bearings
   C. Interior angles are subtracted from 540°
   D. The Bowditch rule is applied to BC and DE
   Answer: A. BC and DE must cancel the resultant of the known lines, so BC, DE and the resultant form a triangle. (Key point: Method for two unknown bearings)

2. The sums of the known lines are Lat = +915.391 and Dep = +407.637. What is the resultant length r?
   A. about 500 m
   B. about 728 m
   C. about 1002 m
   D. about 1323 m
   Answer: C. r = sqrt(915.391^2 + 407.637^2) = 1002.05 m. (Key point: Resultant of known lines)

3. What condition must r satisfy for real bearings of BC and DE to exist?
   A. r must equal BC + DE
   B. |DE - BC| <= r <= BC + DE
   C. r must exceed BC + DE
   D. r must be less than |DE - BC|
   Answer: B. A triangle with sides BC, DE and r exists only when r lies between the difference and the sum of the other two sides. (Key point: Triangle existence test)

4. Here BC + DE = 728 m and r = 1002.05 m. What is the conclusion?
   A. BC and DE are computed with two possible bearings
   B. Bearings can be found after Bowditch adjustment
   C. No triangle exists (cos alpha = 1.461 > 1), so the field notes must be rechecked
   D. BC and DE are parallel to the resultant
   Answer: C. Since r > BC + DE the cosine comes out greater than 1, which is impossible; the recorded data are inconsistent. (Key point: Conclusion)

## Related in Surveying II

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