--- title: "Mean angles, traverse ABCDA (2068 Baisakh)" subject: "Surveying I" chapter: 7 key: "semester_3/surveying_i:mean-horizontal-angles-abcda-hcr-209-25-40" source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-abcda-hcr-209-25-40 site: BCT Engineering --- # Mean angles, traverse ABCDA (2068 Baisakh) Surveying I · Chapter 7: Transit and Theodolite - Asked in: 2068 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/mean-horizontal-angles-abcda-hcr-209-25-40 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7 ## Question The following observations were recorded in a theodolite traverse ABCDA. Compute mean horizontal angles and adjust them if necessary. | Inst. Stn | Target stn | Face Left | Face Right | |---|---|---|---| | A | D | 90°00′10″ | 269°59′50″ | | | B | 209°25′40″ | 29°25′30″ | | B | A | 89°59′30″ | 270°00′10″ | | | C | 180°16′10″ | 00°16′00″ | | C | B | 90°00′00″ | 269°59′50″ | | | D | 179°08′40″ | 359°08′20″ | | D | C | 89°59′50″ | 270°00′10″ | | | A | 160°12′40″ | 340°12′30″ | ## Short answer Assuming AB readings are 200°25′40″ FL and 20°25′30″ FR, mean angle at A is 110°25′35″. Sum 360°03′00″, error +3′00″; correction -45″ per angle gives A = 110°24′50″, B = 90°15′30″, C = 89°07′50″, D = 70°11′50″. ## Answer **Assumption:** closed traverse ABCDA; the sum of the interior angles must be 360°. **Method.** For each target the mean direction is found from the two faces: $$ \text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2} $$ The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care). #### Field book and mean angles | Inst. stn | Sighted to | Face L | Face R | Mean direction | Mean angle | |---|---|---|---|---|---| | A | D | 90°00′10″ | 269°59′50″ | 90°00′00″ | | | | B | 209°25′40″ | 29°25′30″ | 209°25′35″ | **119°25′35″** | | B | A | 89°59′30″ | 270°00′10″ | 89°59′50″ | | | | C | 180°16′10″ | 00°16′00″ | 180°16′05″ | **90°16′15″** | | C | B | 90°00′00″ | 269°59′50″ | 89°59′55″ | | | | D | 179°08′40″ | 359°08′20″ | 179°08′30″ | **89°08′35″** | | D | C | 89°59′50″ | 270°00′10″ | 90°00′00″ | | | | A | 160°12′40″ | 340°12′30″ | 160°12′35″ | **70°12′35″** | #### Check Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 369°03′00″. Angular error = 369°03′00″ - 360°00′00″ = +9°03′00″. An error of about 9° cannot come from observation (a theodolite traverse normally closes within a minute or so), so it must not simply be distributed. It shows a misprint in the data: the readings on B from A are probably 200°25′40″ (face left) and 20°25′30″ (face right), which make every other reading consistent. #### Adjustment with the corrected readings (200°25′40″, 20°25′30″) Mean direction AB = 200°25′35″, so angle at A = 200°25′35″ - 90°00′00″ = 110°25′35″. Observed sum = 110°25′35″ + 90°16′15″ + 89°08′35″ + 70°12′35″ = 360°03′00″. Angular error = +0°03′00″; correction per angle = -0°00′45″. | Angle at | Mean angle | Correction | Corrected angle | |---|---|---|---| | A | 110°25′35″ | -0°00′45″ | **110°24′50″** | | B | 90°16′15″ | -0°00′45″ | **90°15′30″** | | C | 89°08′35″ | -0°00′45″ | **89°07′50″** | | D | 70°12′35″ | -0°00′45″ | **70°11′50″** | | Sum | 360°03′00″ | -0°03′00″ | 360°00′00″ | **Answer:** corrected horizontal angles: A = 110°24′50″, B = 90°15′30″, C = 89°07′50″, D = 70°11′50″. *If the readings are used exactly as printed:* the error of +9°03′00″ shared equally (-2°15′45″ each) would give A = 117°09′50″, B = 88°00′30″, C = 86°52′50″, D = 67°56′50″. This is shown only for completeness; an error this large would mean the observations must be repeated, not adjusted. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With B = 209°25′35″ and D = 90°00′00″ as mean directions, the angle at A is: A. 119°25′35″ B. 119°25′25″ C. 299°25′35″ D. 110°25′35″ Answer: A. 209°25′35″ - 90°00′00″ = 119°25′35″. (Key point: Angle at A) 2. Why does the answer suspect 209°25′40″ is a misprint? A. Face L and Face R readings are identical B. The sum of angles is 369°03′, an error of about 9°, far too large C. The angle at A is smaller than 90° D. The sum is below 360° Answer: B. A 9° discrepancy cannot be a normal observation error. (Key point: Detecting a misprint) 3. Which reading is the probable correct value? A. 219°25′40″ B. 190°25′40″ C. 200°25′40″ (Face R 20°25′30″) D. 209°52′40″ Answer: C. Other papers with the same remaining readings use 200°25′40″. (Key point: Likely correct value) 4. With the corrected reading, the sum is 360°03′00″. What correction per angle is applied? A. +45″ B. -3′ C. -60″ D. -45″ Answer: D. 180″/4 = 45″, subtracted from each angle. (Key point: Corrected result) ## Close variants set in other exams - [Mean angles, traverse ABCDA (2074 Asoj)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-hcr-90-00-00-a-d) — Surveying I, chapter 7, asked 2× ## Related in Surveying I - [Theodolite traverse ABCDA with missing VCR (68°30′10″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atheodolite-traverse-abcda-hcr-vcr-68-30-10) — Surveying I, chapter 7 - [Mean angles, traverse ABCDA (2072)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-face-left-right-b-a) — Surveying I, chapter 7 - [Mean angles, traverse ABCDA (2076)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-abcda-hcr-90-00-20-c-b) — Surveying I, chapter 7, asked 2× - [Theodolite traverse ABCDA, booking format with VCR](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atraverse-abcda-booking-format-vcr-120-14-20) — Surveying I, chapter 7 - [Horizontal angles from field notes ABCDA (2065)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atheodolite-field-notes-abcda-65-shrawan) — Surveying I, chapter 7 - [Mean angles; VCR and VA from A](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atraverse-abcdea-vcr-va-from-a-to-e-and-b) — Surveying I, chapter 7 ## Related in other subjects - [Gale's table for traverse ABCDA](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agale-table-abcda-2072a) — Surveying II, chapter 1 - [Adjust traverse A-X to E-Y bearings](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atraverse-adjust-latitude-departure-ax-ey) — Surveying II, chapter 1