--- title: "Adjusted angles and bearings, angles to the left" subject: "Surveying I" chapter: 4 key: "semester_3/surveying_i:adjusted-angles-bearings-left-angles-aw-dx" source: https://www.bctengineer.com/q/semester_3/surveying_i/adjusted-angles-bearings-left-angles-aw-dx site: BCT Engineering --- # Adjusted angles and bearings, angles to the left Surveying I · Chapter 4: The Compass - Asked in: 2061 Baisakh, 10 marks - Video: [Azimuths and Bearings Traverse Surveying: A Complete Solved Land Problem in Civil & Construction](https://www.youtube.com/watch?v=mXMEPFJLpKo) — Kelvin Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/adjusted-angles-bearings-left-angles-aw-dx - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4 ## Question For the traverse of measured angles (angle to the left) given in the table below, compute the adjusted angles and bearings to the nearest 0.1′. | Station | Measured angles | |---|---| | A | 208° 7.6′ | | B | 101° 35.1′ | | C | 89° 05.3′ | | D | 17° 11.9′ | Known bearings: Line AW = 234° 17.6′; Line DX = 358° 18.5′. ## Short answer Angular misclosure +0.8', distributed as -0.2' per angle. Adjusted angles: A 208°07.4', B 101°34.9', C 89°05.1', D 17°11.7'. Adjusted bearings: AB 26°10.2', BC 104°35.3', CD 195°30.2', DX 358°18.5'. ## Answer The traverse starts on the known line AW and ends on the known line DX, so the observed angles can be checked and adjusted. **Rule for angles measured to the left** (counter-clockwise from the back line to the forward line): $$ \text{Bearing of forward line} = \text{Bearing of back line} - \text{angle to the left} $$ where the back line is the line from the station towards the previous station. The back line at the next station is the forward line $\pm 180^\circ$. #### Step 1: Compute the bearings from the observed angles | Station | Back line bearing | Angle (left) | Bearing of forward line | |---|---|---|---| | A | AW = 234° 17.6′ | 208° 07.6′ | AB = 26° 10.0′ | | B | BA = 206° 10.0′ | 101° 35.1′ | BC = 104° 34.9′ | | C | CB = 284° 34.9′ | 89° 05.3′ | CD = 195° 29.6′ | | D | DC = 15° 29.6′ | 17° 11.9′ | DX = 358° 17.7′ (computed) | #### Step 2: Angular misclosure Known bearing of DX = 358° 18.5′; computed = 358° 17.7′. $$ \text{Misclosure} = 358^\circ 18.5' - 358^\circ 17.7' = +0.8' $$ To raise the computed bearing by 0.8′, the angles to the left must be reduced. There are 4 angles, so the correction is $-0.8'/4 = -0.2'$ per angle. #### Step 3: Adjusted angles and bearings | Station | Observed angle | Correction | Adjusted angle | Adjusted bearing of forward line | |---|---|---|---|---| | A | 208° 07.6′ | −0.2′ | 208° 07.4′ | AB = 26° 10.2′ | | B | 101° 35.1′ | −0.2′ | 101° 34.9′ | BC = 104° 35.3′ | | C | 89° 05.3′ | −0.2′ | 89° 05.1′ | CD = 195° 30.2′ | | D | 17° 11.9′ | −0.2′ | 17° 11.7′ | DX = 358° 18.5′ ✓ | Check: the bearing of DX obtained from the adjusted angles equals the known 358° 18.5′. **Answer:** adjusted angles A = 208° 07.4′, B = 101° 34.9′, C = 89° 05.1′, D = 17° 11.7′; adjusted bearings AB = 26° 10.2′, BC = 104° 35.3′, CD = 195° 30.2′, DX = 358° 18.5′. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For angles measured to the left, how is the forward bearing found? A. Back line bearing + angle B. Back line bearing − angle C. Back line bearing + angle − 180° always D. Forward line = previous forward bearing Answer: B. Angles to the left are counter-clockwise, so forward bearing = back bearing − angle (with 360° adjustments). (Key point: Angle-left rule) 2. AB is computed as 26°10.0′. What is the back line used at B? A. 26°10.0′ B. 206°10.0′ C. 116°10.0′ D. 333°50.0′ Answer: B. The back line at B is BA = AB + 180° = 206°10.0′. (Key point: Back bearing at next station) 3. The known bearing of DX is 358°18.5′ and the computed one is 358°17.7′. What is the misclosure (known − computed)? A. −0.8′ B. +0.8′ C. +0.2′ D. +1.6′ Answer: B. 358°18.5′ − 358°17.7′ = +0.8′. (Key point: Misclosure) 4. Why is −0.2′ applied to each left angle and not +0.2′? A. Reducing left angles raises each forward bearing, which increases the computed DX B. Left angles always get negative corrections C. The correction is subtracted because there are 4 angles D. Bearings decrease when the angle decreases Answer: A. Since forward bearing = back − angle, reducing the angle raises the computed bearing, which must increase by 0.8′. So −0.8′/4 = −0.2′ per angle. 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