--- title: "Bearings of DE and EA in traverse ABCDE" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:traverse-abcde-bearings-de-ea" source: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-abcde-bearings-de-ea site: BCT Engineering --- # Bearings of DE and EA in traverse ABCDE Surveying II · Chapter 1: Traversing - Asked in: 2071 Magh, 10 marks - Video: [omitted measurement of closed traverse || omitted measurement|| missing length & bearing in traverse](https://www.youtube.com/watch?v=iBV3daNWtaY) — Aim 2 Civil Class - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-abcde-bearings-de-ea - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question The following data refers to a traverse ABCDE. Determine the bearings of the sides DE and EA. | Line | AB | BC | CD | DE | EA | |---|---|---|---|---|---| | Length (m) | 230.50 | 250.20 | 210.80 | 240.30 | 265.40 | | Bearing | N 36°45' E | S 82°48' E | S 10°15' E | ? | ? | ## Short answer From closure condition, the closing line D-A is 427.092 m at 277°17'. Solving the triangle with sides DE=240.30 m and EA=265.40 m yields bearings: DE = 243°03' (S 63°03' W), EA = 307°54' (N 52°06' W) for the convex polygon. ## Answer For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Here the bearings of DE and EA are unknown, but all lengths are known. The two unknown lines together must cancel the resultant of the other lines, so DE, EA and that resultant (the closing line) form a triangle whose three sides are known. The angles of this triangle are found by the cosine rule. #### Step 1: Resultant of the known lines | Line | Length (m) | Bearing | Latitude | Departure | |---|---|---|---|---| | AB | 230.50 | 36°45' | +184.689 | +137.914 | | BC | 250.20 | 97°12' | -31.358 | +248.227 | | CD | 210.80 | 169°45' | -207.436 | +37.510 | | **Sum** | | | -54.105 | +423.651 | $$ \begin{aligned} \Sigma\text{Lat}&=-54.105\ \text{m},\qquad \Sigma\text{Dep}=+423.651\ \text{m}\\ \text{Closing vector (D to A)}&:\ \text{Lat}=+54.105,\ \text{Dep}=-423.651\\ r&=\sqrt{54.105^2+423.651^2}=427.092\ \text{m}\\ \phi&=\tan^{-1}\frac{-423.651}{+54.105}=277^\circ 17' \end{aligned} $$ #### Step 2: Triangle of DE, EA and the closing line Sides: $r=427.092$, $l_i=240.30$ (DE), $l_j=265.40$ (EA). $$ \cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{427.092^2+240.30^2-265.40^2}{2\times427.092\times240.30}=0.82683\ \Rightarrow\ \alpha=34^\circ 14' $$ $\alpha$ is the angle between the closing line and DE, so the bearing of DE is $\phi\pm\alpha$. This gives two geometrically possible triangles, mirror images about the closing line: - DE $=\phi-\alpha=243^\circ 03'$: the vector left for EA is $(+163.005,\ -209.444)$, so EA $=307^\circ 54'$. - DE $=\phi+\alpha=311^\circ 30'$: the vector left for EA is $(-105.135,\ -243.688)$, so EA $=246^\circ 40'$. Both pairs satisfy closure, and a field sketch would settle which one applies. With the first pair the bearings increase steadily round the traverse (36°45', 97°12', 169°45', 243°03', 307°54'), so every turn is to the right and ABCDE is a convex pentagon. The second pair makes a re-entrant angle at E. The convex figure, which is the usual case, is adopted. #### Step 3: Check with the chosen bearings DE: Lat $=240.30\cos243^\circ03'=-108.900$, Dep $=-214.208$; EA: Lat $=+163.005$, Dep $=-209.444$. $$ \Sigma\text{Lat}=-54.105+(-108.900)+(+163.005)=0.000,\qquad \Sigma\text{Dep}=+423.651+(-214.208)+(-209.444)\approx0.000 $$ Both are zero, so the traverse closes. **Answer:** bearing of DE = 243°03' (S 63°03' W); bearing of EA = 307°54' (N 52°06' W). (The other mathematically possible solution, giving a re-entrant figure, is DE = 311°30', EA = 246°40'.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the bearings of DE and EA be found when their lengths are known? A. Because ΣLat = ΣDep = 0, DE and EA together cancel the resultant of the known lines and form a triangle with the closing line B. Because DE is parallel to AB C. Because the bearings must differ by 90° D. Because the angles at D and E are equal Answer: A. The two unknown lines must cancel the resultant of AB, BC and CD, giving a triangle with three known sides. (Key point: Principle of closure) 2. ΣLat = -54.105 and ΣDep = +423.651 for AB, BC, CD. What is the closing vector the other two lines must provide? A. Lat -54.105, Dep +423.651 B. Lat +54.105, Dep -423.651, length 427.09 m C. Lat +423.651, Dep -54.105 D. Zero Answer: B. It is equal and opposite to the sums: Lat +54.105, Dep -423.651, with r = sqrt(54.105^2 + 423.651^2) = 427.09 m. (Key point: Closing vector) 3. Which relation gives the angle α between the closing line and DE? A. cos α = (r^2 + l_DE^2 - l_EA^2)/(2 r l_DE) B. sin α = l_DE/l_EA C. cos α = (l_DE + l_EA)/r D. tan α = l_DE/r Answer: A. The triangle has sides r, DE and EA, and α lies between r and DE, so the cosine rule gives α. (Key point: Cosine rule) 4. Why are there two possible pairs of bearings (DE = 311°30' or 243°03')? A. The data contain an error B. The triangle can lie on either side of the closing line (mirror images), and the field sketch decides C. Two different closing vectors exist D. The lengths are ambiguous Answer: B. Bearing of DE = φ ± α, both close the traverse, and the field sketch picks the correct one. (Key point: Two possible solutions) ## Close variants set in other exams - [Bearings of BC and CD in traverse ABCDE](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atraverse-abcde-bearings-bc-cd) — Surveying II, chapter 1, asked 2× ## Related in Surveying II - [Closed traverse missing EA bearing, BC length](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aclosed-traverse-abcdea-missing-ea-bc) — Surveying II, chapter 1 - [Traverse missing length CD, bearing EA](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atraverse-missing-cd-length-ea-bearing) — Surveying II, chapter 1 - [Missing bearings of BC and DE in traverse](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atraverse-missing-bearings-bc-de) — Surveying II, chapter 1 - [Bearings of AB and EA from latitudes, departures](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Alatitude-departure-bearings-ab-ea) — Surveying II, chapter 1 - [Omitted quantities in closed traverse ABCDE](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aomitted-quantities-closed-traverse-abcde) — Surveying II, chapter 1 ## Related in other subjects - [Local attraction, BC 79°30′ and DE 82°15′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-191-30-bc-79-30) — Surveying I, chapter 4 - [Local attraction, bearings AB 305°00′ (EA 45°)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-305-00) — Surveying I, chapter 4, asked 2× - [Local attraction: AB 69°30′ compass traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-69-30-reordered) — Surveying I, chapter 4 - [Local attraction, bearings AB 305°30′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-bb-44-00) — Surveying I, chapter 4, asked 2×