--- title: "Suspension bridge triangle ABC; span AB" subject: "Surveying I" chapter: 7 key: "semester_3/surveying_i:suspension-bridge-triangle-abc-span-ab" source: https://www.bctengineer.com/q/semester_3/surveying_i/suspension-bridge-triangle-abc-span-ab site: BCT Engineering --- # Suspension bridge triangle ABC; span AB Surveying I · Chapter 7: Transit and Theodolite - Asked in: 2073 Shrawan, 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/suspension-bridge-triangle-abc-span-ab - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7 ## Question During the survey of a suspension bridge the following observations were made in triangle ABC. AB is the bridge axis. The least count of the instrument is 01′00″. | Inst. Station | Sighted to | HCR Face Left | HCR Face Right | |---|---|---|---| | A | B | 0°00′00″ | 180°00′20″ | | | C | 54°38′20″ | 234°38′00″ | | B | C | 0°00′00″ | 179°59′50″ | | | A | 89°20′40″ | 269°21′00″ | | C | A | 0°00′00″ | 180°00′00″ | | | B | 36°01′00″ | 215°58′20″ | Compute the angles by mean direction method and correct them if necessary. If the length of line BC is 58.232 m, find the span of bridge axis AB. ## Short answer Triangle closure error −1′25″ distributed equally; corrected angles A=54°38′28.33″, B=89°21′23.33″, C=36°00′08.33″. Span AB by sine rule: AB = 58.232 × sin C / sin A = 41.972 m. ## Answer **Method.** Mean directions are obtained from the two faces, angles from differences of directions, the triangle sum is checked against 180°, the error is distributed equally, and the side AB is then found by the sine rule. | Inst. stn | Sighted to | Face L | Face R | Mean direction | Angle | |---|---|---|---|---|---| | A | B | 0°00′00″ | 180°00′20″ | 0°00′10″ | | | | C | 54°38′20″ | 234°38′00″ | 54°38′10″ | A = **54°38′00″** | | B | C | 0°00′00″ | 179°59′50″ | 359°59′55″ | | | | A | 89°20′40″ | 269°21′00″ | 89°20′50″ | B = **89°20′55″** | | C | A | 0°00′00″ | 180°00′00″ | 0°00′00″ | | | | B | 36°01′00″ | 215°58′20″ | 35°59′40″ | C = **35°59′40″** | #### Check and correction Sum of angles = 179°58′35″; required = 180°00′00″; error = -0°01′25″ (sum is less than 180°). The permissible error is about the least count times $\sqrt{3}$ = 1′ × 1.73 ≈ 1′44″, so the error is acceptable and is distributed equally: correction = +0°00′28.33″ per angle. | Angle | Observed | Correction | Corrected | |---|---|---|---| | A | 54°38′00″ | +0°00′28.33″ | **54°38′28.33″** | | B | 89°20′55″ | +0°00′28.33″ | **89°21′23.33″** | | C | 35°59′40″ | +0°00′28.33″ | **36°00′08.33″** | | Sum | 179°58′35″ | +0°01′25″ | 180°00′00″ | #### Span AB In triangle ABC, the side AB is opposite to the angle C and BC is opposite to angle A. $$ \begin{aligned} \frac{AB}{\sin C} &= \frac{BC}{\sin A}\\ AB &= BC\,\frac{\sin C}{\sin A} = 58.232 \times \frac{\sin 36°00′08.33″}{\sin 54°38′28.33″} = 41.972\ \text{m} \end{aligned} $$ **Answer:** corrected angles A = 54°38′28.33″, B = 89°21′23.33″, C = 36°00′08.33″; span of the bridge AB = **41.972 m** (about 41.97 m, since the instrument reads only to 1′). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The three computed angles total 179°58′35″. What is the error and its sign? A. -1′25″ (sum is less than 180°) B. +1′25″ C. -1′35″ D. 0, as it is within 1 minute Answer: A. 180°00′00″ - 179°58′35″ = 85″ short. (Key point: Triangle angle sum) 2. How is this error corrected? A. -28.33″ taken from each angle B. +28.33″ added to each of the three angles C. +85″ added to angle B D. +42.5″ added to two angles Answer: B. 85″/3 = 28.33″ added to each angle so that they total 180°. (Key point: Correction) 3. Why is the error of 1′25″ accepted for a least count of 1′? A. It is smaller than the least count of 1′ B. It is greater than 1′ 44″ C. It is less than about least count x sqrt(3) = 1′44″ D. Any triangle error is acceptable Answer: C. The permissible triangle error is about LC x sqrt(n) with n = 3, i.e. about 1′44″. (Key point: Permissible error) 4. Which expression gives the span AB from BC? A. AB = BC x sin A / sin C B. AB = BC x sin B / sin A C. AB = BC x cos C / cos A D. AB = BC x sin C / sin A Answer: D. AB is opposite angle C and BC is opposite angle A, so AB/sin C = BC/sin A. This gives 41.972 m. 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