--- title: "Bearings of BC and CD in traverse ABCDE" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:traverse-abcde-bearings-bc-cd" source: https://www.bctengineer.com/q/semester_4/surveying_ii/traverse-abcde-bearings-bc-cd site: BCT Engineering --- # Bearings of BC and CD in traverse ABCDE Surveying II · Chapter 1: Traversing - Asked in: 2065 Chaitra (old course), 9 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-bc-cd - 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 BC and CD. | Line | Length in (m) | Bearing | |---|---|---| | AB | 306.00 | 164°00' | | BC | 432.00 | ? | | CD | 324.00 | ? | | DE | 302.40 | 328°00' | | EA | 629.43 | 269°06' | ## Short answer Compute closing vector from lines AB, DE, EA (magnitude 706.858 m, bearing 86°08'). Form a triangle with BC (432 m) and CD (324 m) using the cosine rule. Two solutions exist; the pair yielding whole degrees is BC bearing 104°00' (S 76°00' E) and CD bearing 62°00' (N 62°00' E). ## Answer For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Here the bearings of BC and CD are unknown, but their lengths are known. The two unknown lines together must cancel the resultant of the other lines, so BC, CD 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 | 306.00 | 164°00' | -294.146 | +84.345 | | DE | 302.40 | 328°00' | +256.450 | -160.248 | | EA | 629.43 | 269°06' | -9.887 | -629.352 | | **Sum** | | | -47.583 | -705.255 | $$ \begin{aligned} \Sigma\text{Lat}&=-47.583\ \text{m},\qquad \Sigma\text{Dep}=-705.255\ \text{m}\\ \text{Closing vector}&:\ \text{Lat}=+47.583,\ \text{Dep}=+705.255\\ r&=\sqrt{47.583^2+705.255^2}=706.858\ \text{m}\\ \phi&=\tan^{-1}\frac{+705.255}{+47.583}=86^\circ 08' \end{aligned} $$ #### Step 2: Triangle of BC, CD and the closing line Sides: $r=706.858$, $l_i=432.00$ (BC), $l_j=324.00$ (CD). $$ \cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{706.858^2+432.00^2-324.00^2}{2\times706.858\times432.00}=0.95181\ \Rightarrow\ \alpha=17^\circ 52' $$ $\alpha$ is the angle between the closing line and BC, so the bearing of BC is $\phi\pm\alpha$: - $\phi+\alpha=104^\circ 00'$ and $\phi-\alpha=68^\circ 17'$, giving two geometrically possible triangles (the figure can lie on either side of the closing line). - The bearing of CD follows from the remaining vector: for BC = 104°00', the vector left for CD is (+152.088, +286.086), so CD = 62°00'; for BC = 68°17', CD = 110°17'. Both pairs satisfy closure; they are mirror images about the closing line, and the field sketch of the traverse decides which one applies. The first pair is adopted here, since it gives whole-degree bearings, which is clearly the intended answer. #### Step 3: Check with the chosen bearings $$ \Sigma\text{Lat}=-47.583+(-104.505)+(+152.088)=0.000,\qquad \Sigma\text{Dep}=-705.255+(+419.169)+(+286.086)=0.000 $$ Both are zero, so the traverse closes. **Answer:** bearing of BC = 104°00' (S 76°00' E); bearing of CD = 62°00' (N 62°00' E). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. BC and CD have known lengths but unknown bearings. How are they found? A. By Bowditch's rule B. From the interior angle sum C. By subtracting 180° from AB D. Using the closing vector of the known lines and a triangle of sides r, BC, CD solved by the cosine rule Answer: D. Closure requires BC + CD to cancel the other lines' resultant; the three known sides define a triangle. (Key point: Method) 2. Known lines give sum Lat = -47.583 and sum Dep = -705.255. The closing vector has bearing phi in which quadrant? A. Northeast (both positive), phi = 86°08' B. Southeast C. Southwest D. Northwest Answer: A. The closing vector is (+47.583, +705.255), both positive, so phi = tan^-1(705.255/47.583) = 86°08' and r = 706.858 m. (Key point: Closing vector) 3. With r = 706.858, BC = 432.00, CD = 324.00, alpha is 17°52'. The bearing of BC equals: A. alpha only B. 180° - alpha C. phi ± alpha D. phi x alpha Answer: C. alpha is the angle between the closing line and BC, so BC = phi ± alpha: 104°00' or 68°17'. (Key point: Angle alpha) 4. How is the correct pair chosen between the two solutions? A. The larger bearing is always right B. The field sketch decides; here BC = 104°00', CD = 62°00' is adopted C. Average the two D. Either, as they are identical Answer: B. Both pairs close the traverse as mirror images, so the actual shape from the field sketch decides. (Key point: Choosing a solution) ## Close variants set in other exams - [Bearings of DE and EA in traverse ABCDE](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atraverse-abcde-bearings-de-ea) — 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, 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, 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°30′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-bb-44-00) — Surveying I, chapter 4, asked 2× - [Corrected and true bearings, 8°40′E](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atrue-bearings-declination-8-40e) — Surveying I, chapter 4