--- title: "RL of hill station P, A-B 150 m, target 3.5 m" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:hill-station-p-vane-15-ab" source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-station-p-vane-15-ab site: BCT Engineering --- # RL of hill station P, A-B 150 m, target 3.5 m Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2066 Magh (old course), 8 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-station-p-vane-15-ab - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question The top of a hill station P was sighted from two stations A and B at a different level and at the same vertical plane with the target. The zenith angle from A to P and B to P were 59°15' and 69°45' respectively. The zenith angle from A to B to a vane 1.5 m above the foot of the vane was 105°30'. If the height of instrument of A and B were 1.45 m and 1.35 m respectively and distance between A and B was 150 m, and RL of B was 120.00 m. Find the RL of hill station (if target is 3.5 m above the ground). ## Short answer Axis RLs from vane sight: A = 163.099 m, B = 121.350 m. Using elevation angles 30°45′ and 20°15′, solving simultaneous height equations gives RL of target = 198.869 m and ground RL of hill station = 195.369 m. ## Answer Zenith angles are converted to elevation angles ($\alpha = 90^\circ - z$). The sight from A to B (z = 105°30') is a depression, so A is above B. #### RL of instrument axes - Axis of B = RL of B + HI = 120.000 + 1.35 = 121.350 m - Sight A to B: angle -15°30' (depression). The vane at B (1.5 m above the foot) is at 120.000 + 1.5 = 121.500 m, so the axis of A = 121.500 − 150 tan(-15°30') = 163.099 m. Elevation angles to P: from B = 20°15'00", from A = 30°45'00". Take D = 150 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 $D = 150$ 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 $\alpha_B = 20°15'00"$ ($\tan = 0.36892$), $\alpha_A = 30°45'00"$ ($\tan = 0.59494$), axis RL of B = 121.350 m and of A = 163.099 m: $$ x = \frac{(163.099 - 121.350) - 150\times 0.59494}{0.36892 - 0.59494} = 210.125\ \text{m} $$ $$ RL_{P} = 121.350 + 210.125\times 0.36892 = 198.869\ \text{m} $$ Check from A: $163.099 + (210.125 - 150)\times 0.59494 = 198.869$ m (same). The RL found is for the target on top of the station. As the target is 3.5 m above the ground: $$ RL_{\text{ground at P}} = 198.869 - 3.5 = 195.369\ \text{m} $$ **Answer: RL of the target at P = 198.869 m; RL of the hill station (ground) = 195.369 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A sights the vane at B (level 121.500 m) at a depression of 15°30' over 150 m. How is axis A found? A. 121.500 + 150 tan 15°30' = 163.099 m B. 121.500 - 150 tan 15°30' = 79.90 m C. 121.350 + 150 tan 15°30' = 162.95 m D. 121.500 + 1.45 = 122.95 m Answer: A. A looks downward to the vane, so A's axis is higher than the vane by D tan(depression). (Key point: Axis RL of A) 2. RL of B is 120.00 m and HI of B is 1.35 m. What is the axis RL of B? A. 120.000 m B. 121.500 m C. 121.350 m D. 121.450 m Answer: C. Axis B = 120.000 + 1.35 = 121.350 m. The 1.5 m vane height belongs to the A-to-B sight only. (Key point: Axis RL of B) 3. The zenith angles to P are 59°15' from A and 69°45' from B. What are the elevations? A. 20°15' from A and 30°45' from B B. 59°15' and 69°45' C. 31°15' from A and 21°45' from B D. 30°45' from A and 20°15' from B Answer: D. Elevation = 90° - z, so A gives 30°45' and B gives 20°15'. (Key point: Elevation angles) 4. The RL found for the target is 198.869 m, and the target is 3.5 m above the ground. What is the ground RL? A. 198.869 - 3.5 = 195.369 m B. 198.869 + 3.5 = 202.369 m C. 198.869 m D. 198.869 - 1.5 = 197.369 m Answer: A. The sighted target is above the ground, so the ground RL is the target RL minus 3.5 m. (Key point: Ground level at P) ## Close variants set in other exams - [RL of hill station P, A-B 150 m, RL B 1500](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-p-vane-175-ab) — Surveying II, chapter 3, asked 2× ## Related in Surveying II - [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 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 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 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 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 P (A to B 112 m, RL 1290)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2077c-ab-112) — Surveying II, chapter 3, asked 4× ## Related in other subjects - [Mean angles and VCR, station O (reading 179°59′40″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-first-reversed-179-59-40) — Surveying I, chapter 7, asked 3×