--- title: "RL of hill station P, A-B 150 m, RL B 1500" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:hill-station-p-vane-175-ab" source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-station-p-vane-175-ab site: BCT Engineering --- # RL of hill station P, A-B 150 m, RL B 1500 Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2081 Chaitra, 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-station-p-vane-175-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 instrument stations A and B at a very 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 1.75 m above the foot of the vane was 105°30'. If the height of the instrument of A and B were 1.56 m and 1.48 m respectively and distance between A and B was 150 m, RL of B was 1500 m. Find the RL of hill station (The target is 2.5 m above the ground). ## Short answer Axis RLs: B = 1501.480 m, A = 1543.349 m. Solving with elevation angles 30°45′ and 20°15′ yields target RL = 1578.803 m, ground RL of hill station = 1576.303 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 = 1500.000 + 1.48 = 1501.480 m - Sight A to B: angle -15°30' (depression). The vane at B (1.75 m above the foot) is at 1500.000 + 1.75 = 1501.750 m, so the axis of A = 1501.750 − 150 tan(-15°30') = 1543.349 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 = 1501.480 m and of A = 1543.349 m: $$ x = \frac{(1543.349 - 1501.480) - 150\times 0.59494}{0.36892 - 0.59494} = 209.594\ \text{m} $$ $$ RL_{P} = 1501.480 + 209.594\times 0.36892 = 1578.803\ \text{m} $$ Check from A: $1543.349 + (209.594 - 150)\times 0.59494 = 1578.803$ m (same). The target is 2.5 m above the ground at P, so $$ RL_{\text{ground at P}} = 1578.803 - 2.5 = 1576.303\ \text{m} $$ **Answer: RL of the target = 1578.803 m; RL of the hill station (ground) = 1576.303 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The zenith angle from A to P is 59°15'. What is the angle of elevation used in the calculation? A. 59°15' B. 30°45' C. 20°15' D. 149°15' Answer: B. Elevation = 90° - zenith angle = 90° - 59°15' = 30°45'. (Key point: Zenith to elevation) 2. The sight from A to B has z = 105°30', so it is a depression of 15°30'. How is the axis RL of A found? A. Vane level at B (1501.75) plus 150 tan 15°30' B. Axis of B minus 150 tan 15°30' C. RL of B plus 1.56 D. Vane level at B minus 150 tan 15°30' Answer: A. A looks down at the vane, so the vane level equals axis A minus 150 tan(15°30'); hence axis A = 1501.75 + 150 tan 15°30' = 1543.349 m. (Key point: Axis of A) 3. With A between B and P, what condition gives the horizontal distance x from B to P? A. x tan(alpha_B) = (x - D) tan(alpha_A) B. RL axis B + x tan(alpha_B) = RL axis A + (x - D) tan(alpha_A) C. RL axis B - x tan(alpha_B) = RL axis A - x tan(alpha_A) D. x sin(alpha_B) = D sin(alpha_A) Answer: B. P has one RL, so the height reached from B (distance x) must equal the height reached from A (distance x - D). (Key point: Equation for x) 4. RL of the target at P is 1578.803 m and the target is 2.5 m above the ground. What is the RL of the hill station (ground)? A. 1581.303 m B. 1578.803 m C. 1576.303 m D. 1543.349 m Answer: C. The ground is 2.5 m below the target, so 1578.803 - 2.5 = 1576.303 m. (Key point: Ground RL of the hill) ## Close variants set in other exams - [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 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 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 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 1290)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2077c-ab-112) — 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 ## 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×