--- title: "RL of tower top, base inaccessible" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:tower-inaccessible-base-rl-two-stations" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tower-inaccessible-base-rl-two-stations site: BCT Engineering --- # RL of tower top, base inaccessible Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2073 Bhadra, 6 marks - Video: [Problem on determination of RL of top Of Object (Trigonometric levelling)](https://www.youtube.com/watch?v=DEIcbWQaAwc) — Tech Reinforce - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/tower-inaccessible-base-rl-two-stations - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question Calculate the RL of top of tower P, base of which was not accessible from the two instrumentation stations X and Y. The top of the tower and the instrument stations were not in the same vertical plane. The observed zenithal angles from X to P and Y to P were 30°30' and 29°20' respectively. H.I. of instruments at X and Y were 1.57 m and 1.50 m respectively and distance between them was 200 m. The horizontal angles observed were: PXY = 45°45' XYP = 60°30' respectively. The back sights taken to BM with RL of 1000.00 m were 1.8 m and 0.8 m from X and Y respectively. ## Short answer Horizontal distances XP = 181.315 m, YP = 149.222 m from solving triangle XYP. RL from X = 1309.611 m, from Y = 1266.348 m; the two values disagree, so the mean RL of tower top = 1287.98 m. ## Answer Assumption: the back sights are taken on the BM (RL 1000.00 m), so the axis RL of an instrument = 1000.00 + back sight. The zenith angles are used as given (elevation $= 90^\circ - z$). #### Plan triangle XYP $\angle XPY = 180^\circ - (45^\circ45' + 60^\circ30') = 73°45'$ ```text P / \ XP / \ YP / \ X ----- Y 200 m ``` $$ XP = \frac{200\sin 60^\circ30'}{\sin 73°45'} = 181.315\ \text{m}, \qquad YP = \frac{200\sin 45^\circ45'}{\sin 73°45'} = 149.222\ \text{m} $$ #### Axis levels - Axis of X = 1000.000 + 1.8 = 1001.800 m - Axis of Y = 1000.000 + 0.8 = 1000.800 m #### RL of P - Elevation from X $= 90^\circ - 30^\circ30' = 59^\circ30'$: $h_X = 181.315\tan 59^\circ30' = 307.811$ m, so $RL_P = 1001.800 + 307.811 = 1309.611$ m - Elevation from Y $= 90^\circ - 29^\circ20' = 60^\circ40'$: $h_Y = 149.222\tan 60^\circ40' = 265.548$ m, so $RL_P = 1000.800 + 265.548 = 1266.348$ m The two values differ by 43.263 m, which shows that the given data are not consistent, so the mean of the two is taken. **Answer: RL of the top of the tower P = 1287.98 m (mean of 1309.61 m and 1266.35 m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In triangle XYP, angle PXY = 45°45' and angle XYP = 60°30'. What is angle XPY? A. 106°15' B. 73°45' C. 14°15' D. 60°30' Answer: B. The angles of a triangle sum to 180°, so XPY = 180° - (45°45' + 60°30') = 73°45'. (Key point: Angle at P in the plan) 2. Which expression gives XP, the horizontal distance X to P? A. 200 sin 45°45' / sin 73°45' B. 200 sin 73°45' / sin 60°30' C. 200 sin 60°30' / sin 73°45' D. 200 sin 60°30' / sin 45°45' Answer: C. XP is opposite the angle at Y (60°30'), and the base XY = 200 m is opposite angle P (73°45'). (Key point: Sine rule) 3. Back sights of 1.8 m (at X) and 0.8 m (at Y) are taken on the BM (RL 1000.00 m). What are the axis RLs? A. X 998.200 m, Y 999.200 m B. X 1001.570 m, Y 1001.500 m C. X 1000.800 m, Y 1001.800 m D. X 1001.800 m, Y 1000.800 m Answer: D. The axis is above the BM by the back sight, so axis = 1000.00 + BS. (Key point: Axis levels from back sights) 4. The RL of P is 1309.611 m from X but 1266.348 m from Y. What does the answer do? A. Takes the mean, 1287.98 m, since the data are not perfectly consistent B. Takes the higher value, 1309.61 m C. Takes the lower value, 1266.35 m D. Takes the difference, 43.26 m Answer: A. The two values differ by 43.263 m, so the mean is reported as the RL of P. (Key point: Inconsistent data) ## Related in Surveying II - [RL of tower top Q from stations P and R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atower-top-q-from-p-and) — Surveying II, chapter 3 - [RL of hill top from two stations (A-B 118 m)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-two-stations-2079j) — Surveying II, chapter 3, asked 4× - [Inaccessible object, stations in different vertical plane](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ainaccessible-object-different-vertical-plane) — Surveying II, chapter 3, asked 3× - [RL of hill top Q from stations P and R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-q-from-p-and) — Surveying II, chapter 3, asked 4× - [RL of temple top from stations A and B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atemple-top-rl-two-stations) — Surveying II, chapter 3 - [RL of building top from stations A and B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Abuilding-top-rl-two-stations-rod) — Surveying II, chapter 3 ## Related in other subjects - [Accessible and inaccessible exploration methods](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Aaccessible-inaccessible-exploration) — Foundation Engineering, chapter 2 - [Level book, B and C, RL 50.250 base](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-50-800-51-000) — Surveying I, chapter 5, asked 3× - [Moment of guy wire forces about tower base](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aguy-wires-tower-moment-about-base) — Applied Mechanics, chapter 3 - [Reciprocal levelling; RL of closed traverse stations](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-closed-traverse-bcdefgb) — Surveying I, chapter 5