--- title: "Mean angles and VCR, station O (reading 179°59′40″)" subject: "Surveying I" chapter: 7 key: "semester_3/surveying_i:mean-direction-o-abcd-first-reversed-179-59-40" source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-o-abcd-first-reversed-179-59-40 site: BCT Engineering --- # Mean angles and VCR, station O (reading 179°59′40″) Surveying I · Chapter 7: Transit and Theodolite - Asked in: 2079 Bhadra, 5+1 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-direction-o-abcd-first-reversed-179-59-40 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7 ## Question A theodolite is set over station O to measure direction to stations A, B, C and D. The observed circle readings are as follows. Compute the mean horizontal angle by mean direction method and adjust them if necessary. Also calculate the missing data of vertical circle reading. | Instrument | Target Station | Telescope | Horizontal Circle Readings | Vertical Circle Readings | |---|---|---|---|---| | O | A | Direct | 00°00′00″ | 120°15′10″ | | | | Reversed | 179°59′40″ | ? | | | B | Direct | 60°55′10″ | ? | | | | Reversed | 240°55′20″ | 308°51′40″ | | | C | Direct | 140°50′50″ | ? | | | | Reversed | 320°51′20″ | 269°15′10″ | | | D | Direct | 270°20′10″ | 177°20′10″ | | | | Reversed | 90°20′20″ | ? | | | A | Direct | 00°00′20″ | - | | | | Reversed | 180°00′30″ | - | ## Short answer Mean horizontal angles are found by averaging FL and FR readings, reducing directions to zero at A, and distributing the 35″ closing error equally; adjusted angles are AOB=60°55′16.25″, BOC=79°55′41.25″, COD=129°29′01.25″, DOA=89°40′01.25″. Missing vertical readings are A(rev)=239°44′50″, B(dir)=51°08′20″, C(dir)=90°44′50″, D(rev)=182°39′50″. ## Answer **Method.** The mean direction of each target is found from the two faces: mean = (Direct + (Reversed - 180°)) / 2. The directions are referred to the first reading on A (taken as 0°00′00″). The second reading on A closes the horizon, so the difference from 360° is the closing error, which is distributed equally. #### Mean horizontal directions | Station | Face | Reading L (D) | Reading R (Rev) | R - 180 | Mean direction | Direction from A | Correction | Corrected direction | |---|---|---|---|---|---|---|---|---| | A | | 00°00′00″ | 179°59′40″ | 359°59′40″ | 359°59′50.0″ | 0°00′00.0″ | - | 0°00′00.00″ | | B | | 60°55′10″ | 240°55′20″ | 60°55′20″ | 60°55′15.0″ | 60°55′25.0″ | -0°00′08.75″ | 60°55′16.25″ | | C | | 140°50′50″ | 320°51′20″ | 140°51′20″ | 140°51′05.0″ | 140°51′15.0″ | -0°00′17.50″ | 140°50′57.50″ | | D | | 270°20′10″ | 90°20′20″ | 270°20′20″ | 270°20′15.0″ | 270°20′25.0″ | -0°00′26.25″ | 270°19′58.75″ | | A (closing) | | 00°00′20″ | 180°00′30″ | 0°00′30″ | 0°00′25.0″ | 0°00′35.0″ | -0°00′35.00″ | 360°00′00″ | #### Mean angles and adjustment | Angle | Mean angle (before adjustment) | Correction | Adjusted angle | |---|---|---|---| | AOB | 60°55′25.0″ | -0°00′08.75″ | **60°55′16.25″** | | BOC | 79°55′50.0″ | -0°00′08.75″ | **79°55′41.25″** | | COD | 129°29′10.0″ | -0°00′08.75″ | **129°29′01.25″** | | DOA | 89°40′10.0″ | -0°00′08.75″ | **89°40′01.25″** | | Sum | 360°00′35.0″ | -0°00′35.0″ | 360°00′00″ | Closing error = 0°00′35.0″ (second reading on A larger than the first). The sum of the four angles at O must be 360°, and it is 360°00′35.0″; the error of 0°00′35.0″ is small and is therefore distributed equally: correction = -0°00′08.75″ per angle. #### Missing vertical circle readings For a theodolite with a zenith-angle vertical circle, FL + FR = 360° for a correct instrument. So the missing reading = 360° - the other face. The vertical angle is $90° - Z$ (positive = elevation, negative = depression). | Target | VCR Direct (FL) | VCR Reversed (FR) | Mean zenith distance | Vertical angle (90° - Z) | |---|---|---|---|---| | A | 120°15′10″ | **239°44′50″** | 120°15′10″ | -30°15′10″ | | B | **51°08′20″** | 308°51′40″ | 51°08′20″ | +38°51′40″ | | C | **90°44′50″** | 269°15′10″ | 90°44′50″ | -0°44′50″ | | D | 177°20′10″ | **182°39′50″** | 177°20′10″ | -87°20′10″ | **Answer:** adjusted angles AOB = 60°55′16.25″, BOC = 79°55′41.25″, COD = 129°29′01.25″, DOA = 89°40′01.25″; missing VCR: A (reversed) 239°44′50″, B (direct) 51°08′20″, C (direct) 90°44′50″, D (reversed) 182°39′50″. *Note:* the last pair of readings on A has no vertical circle readings (shown as - in the table), so only the four targets are completed. In this question the first reversed reading on A is 179°59′40″, which gives a closing error of 35″. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A reads 0°00'00'' direct and 179°59'40'' reversed. What is the mean direction of A? A. 0°00'10'' B. 359°59'50'' C. 89°59'50'' D. 0°00'20'' Answer: B. (0 + (179°59'40'' - 180°))/2 = -20''/2 = -10'', i.e. 359°59'50''. (Key point: Mean direction of A) 2. The four angles at O sum to 360°00'35''. What is the correction per angle? A. -0°00'35'' B. -0°00'17.5'' C. -0°00'8.75'' D. +0°00'8.75'' Answer: C. Error is +35'' so subtract 35''/4 = 8.75'' from each angle. (Key point: Closing error) 3. Target D has direct VCR 177°20'10''. What is the missing reversed reading? A. 182°39'50'' B. 177°20'10'' C. 87°20'10'' D. 2°39'50'' Answer: A. FR = 360° - 177°20'10'' = 182°39'50''. (Key point: Missing VCR) 4. AOB before adjustment is 60°55'25.0'' with a correction of -8.75''. What is the adjusted angle? A. 60°55'33.75'' B. 60°55'16.25'' C. 60°55'25'' D. 60°55'08.75'' Answer: B. 60°55'25.0'' - 8.75'' = 60°55'16.25''. (Key point: Adjusted angle) ## Close variants set in other exams - [Mean angles and VCR, station O (A-D)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-vcr-120-15-10-ls10) — Surveying I, chapter 7, asked 3× ## Related in Surveying I - [Direction theodolite, least count 6″, station D](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Adirection-theodolite-ls-6-sec-station-d) — Surveying I, chapter 7 - [Mean direction angles with VCR (125°25′20″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-traverse-abcda-vcr-125-25-20) — Surveying I, chapter 7 - [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 - [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; 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 - [Mean horizontal angles, station O (P, Q, R)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-o-pqrp) — Surveying I, chapter 7 ## Related in other subjects - [Taper turning: set-over and compound angle](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Ataper-turning-setover-numerical) — Workshop Technology, chapter 6 - [RL of hill station P from stations R and Q](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-signal-r-q-p) — Surveying II, chapter 3