--- title: "RL of hill station P (A-B 125 m, RL B 1280.50)" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:hill-p-rod-2m-at-b" source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-p-rod-2m-at-b site: BCT Engineering --- # RL of hill station P (A-B 125 m, RL B 1280.50) Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2075 Bhadra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-p-rod-2m-at-b - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question Compute RL of a hill station 'P' from two instrument stations A and B at very different level with same line of sight to that of target from the following information. | Inst st'n | Inst ht (m) | Target | Zenith angle FL | Zenith angle FR | Distance, Rod ht and RL | |---|---|---|---|---|---| | A | 1.47 | P | 62°18' | 297°36' | Distance between A and B = 125 m | | B | 1.42 | P | 64°52' | 239°00' | Rod ht at B = 2.0 m, RL of B = 1280.50 m | | A | 1.47 | B | 96°10' | 263°40' | | ## Short answer From RL of B and instrument heights, compute RL of axes: B 1281.920 m, A 1296.190 m. Using elevation angles 27°39' and 25°04', solve horizontal distance x = 911.767 m, then RL of hill station P = 1708.376 m. ## Answer The face right reading of B to P is printed as 239°00', which does not match the face left reading (64°52'); it is taken as 295°00' (a misprint). #### Mean vertical angles | Sight | Face left | 360° − FR | Mean zenith | Elevation | |---|---|---|---|---| | A to P | 62°18' | 62°24' | 62°21'00" | 27°39'00" | | B to P | 64°52' | 65°00' | 64°56'00" | 25°04'00" | | A to B | 96°10' | 96°20' | 96°15'00" | -6°15'00" | A to B is a depression, so A is higher than B. #### RL of instrument axes - Axis of B = RL of B + HI = 1280.500 + 1.42 = 1281.920 m - Sight A to B: angle -6°15' (depression). The vane at B (2 m above the foot) is at 1280.500 + 2 = 1282.500 m, so the axis of A = 1282.500 − 125 tan(-6°15') = 1296.190 m. Elevation angles to P: from B = 25°04'00", from A = 27°39'00". Take D = 125 m as the horizontal distance BA. #### Distance and RL of P Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 125$ m. ```text P _.--'| _.--' | A _.--' aA | B _.--' aB - - - - - - - - | |<--- D --->|<--- (x - D) --->| |<------------ x ------------>| ``` Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations: $$ \begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned} $$ With $\alpha_B = 25°04'00"$ ($\tan = 0.46772$), $\alpha_A = 27°39'00"$ ($\tan = 0.52390$), axis RL of B = 1281.920 m and of A = 1296.190 m: $$ x = \frac{(1296.190 - 1281.920) - 125\times 0.52390}{0.46772 - 0.52390} = 911.767\ \text{m} $$ $$ RL_{P} = 1281.920 + 911.767\times 0.46772 = 1708.376\ \text{m} $$ Check from A: $1296.190 + (911.767 - 125)\times 0.52390 = 1708.376$ m (same). **Answer: RL of the hill station P = 1708.376 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The FR of B to P is printed 239°00' but FL is 64°52'. How does the answer handle it? A. It uses 239°00' as given B. It takes it as 295°00' since 360° - FR should be near FL C. It ignores face right and drops the observation D. It uses 180° - 239°00' Answer: B. 360° - 295° = 65°00' agrees with FL 64°52', so 239°00' is treated as a misprint of 295°00'. (Key point: Misprinted reading) 2. A to B has a mean zenith of 96°15'. What does it show? A. Depression of 6°15', so A is higher than B B. Elevation of 6°15', so B is higher C. Depression of 96°15' D. Both stations are at the same level Answer: A. A zenith angle above 90° is a depression of 6°15' from A to B, so A is higher. (Key point: Which station is higher) 3. The vane at B is 2 m above the foot, i.e. at 1282.500 m. How is the axis of A found? A. 1282.500 + 125 tan 6°15' = 1296.190 m B. 1282.500 - 125 tan 6°15' C. 1281.920 + 125 tan 6°15' D. 1280.500 + 1.47 Answer: A. The vane is 125 tan(6°15') below axis A, so axis A = 1282.500 + 125 tan 6°15' = 1296.190 m. (Key point: Axis of A) 4. What is the RL of P and the elevation angles used from B and A? A. 1708.376 m; 25°04' from B and 27°39' from A B. 1708.376 m; 64°56' from B and 62°21' from A C. 1296.190 m; 25°04' from B and 27°39' from A D. 911.767 m; 62°21' from B Answer: A. Elevation = 90° - mean zenith, giving 25°04' and 27°39'; the solution gives x = 911.767 m and RL P = 1708.376 m. (Key point: Result) ## Close variants set in other exams - [RL of hill station P (A-B 120 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2074b-ab-120) — Surveying II, chapter 3, asked 2× ## Related in Surveying II - [RL of hill top P (A to B 112 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2073m-ab-112) — Surveying II, chapter 3, asked 4× - [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 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 - [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× - [RL of hill top P (A to B 120 m, vane 2.5 m)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2072m-vane-25) — Surveying II, chapter 3 - [RL of hill station P, A-B 150 m, target 3.5 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-p-vane-15-ab) — Surveying II, chapter 3, asked 2× ## Related in other subjects - [Level book, 20 m interval, RL 1280 to 1279.924](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-20m-interval-1280-1279-924) — Surveying I, chapter 5 - [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×