--- title: "RL of hill top P (A to B = 53 m)" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:hill-top-rl-2078p-ab-53" source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-top-rl-2078p-ab-53 site: BCT Engineering --- # RL of hill top P (A to B = 53 m) Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2078 Poush, 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-top-rl-2078p-ab-53 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question Compute the RL of top of the hill point P from the following data. | Inst Stn | Target Stn | Zenith angle | Height of Instrument (m) | |---|---|---|---| | A | P | 62°30' | 1.48 | | B | P | 71°15' | 1.42 | | A | B | 80°00' | 1.48 | | A | BM | 90°00' | 1.48 | While sighting from A to B and BM corresponding staff readings are 1.75 m and 1.00 m. Distance between A and B is 53.00 m and RL of BM is 1260.00 m. ## Short answer RL BM 1260.00, staff reading 1.00 m gives axis A = 1261.000 m. Sight A to B (vane 1.75 m) gives RL_B = 1268.595 m, axis B = 1270.015 m. With αA=27°30', αB=18°45', D=53 m, solving x = 202.114 m yields RL_P = 1338.624 m. ## Answer The data are read as: staff reading on the BM from A = 1.00 m (horizontal sight), staff (vane) reading at B = 1.75 m, horizontal distance AB = 53 m. #### RL of instrument axes - Axis of A = 1260.000 + 1.00 = 1261.000 m - Sight A to B: elevation $= 90^\circ - 80^\circ = 10°00'$. Reading point at B = 1261.000 + 53 tan(10°00') = 1270.345 m. Ground at B = 1270.345 − 1.75 = 1268.595 m. - Axis of B = 1268.595 + 1.42 = 1270.015 m Elevation angles to P: from A = $90^\circ - 62^\circ30' = 27°30'$, from B = $90^\circ - 71^\circ15' = 18°45'$. B sees P at the smaller angle, so B is taken as the farther station; the result $x > D$ below confirms this. #### 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 $D = 53$ m. ```text P _.--'| _.--' | A _.--' aA | B _.--' aB - - - - - - - - | |<--- D --->|<--- (x - D) --->| |<------------ x ------------>| ``` The height of P above the axis of B is $x\tan\alpha_B$, and above the axis of A it is $(x-D)\tan\alpha_A$. 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 $\alpha_B = 18°45'00"$ ($\tan = 0.33945$), $\alpha_A = 27°30'00"$ ($\tan = 0.52057$), axis RL of B = 1270.015 m and axis RL of A = 1261.000 m: $$ x = \frac{(1261.000 - 1270.015) - 53\times 0.52057}{0.33945 - 0.52057} = 202.114\ \text{m} $$ $$ RL_{P} = 1270.015 + 202.114\times 0.33945 = 1338.624\ \text{m} $$ Check from A: $1261.000 + (202.114 - 53)\times 0.52057 = 1338.624$ m (same). **Answer: RL of the top of the hill P = 1338.624 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the staff reading on BM (1.00 m) added to the RL of BM to get the axis of A? A. The sight is inclined, so a trigonometric correction is needed B. Because HI of A (1.48 m) is added to the BM RL C. The sight is horizontal (zenith 90°00'), so axis A = 1260 + 1.00 = 1261.000 m D. Because BM is at the same level as the ground at A Answer: C. A level sight to a BM gives the axis RL directly as BM RL + staff reading. (Key point: Axis from a horizontal sight on BM) 2. The reading point at B is 1270.345 m (axis A + 53 tan 10°00') and the reading is 1.75 m. With HI of B = 1.42 m, what is the axis RL of B? A. 1270.345 - 1.75 + 1.42 = 1270.015 m B. 1270.345 + 1.75 + 1.42 = 1273.515 m C. 1270.345 - 1.42 = 1268.925 m D. 1270.345 + 1.42 = 1271.765 m Answer: A. Ground at B = 1270.345 - 1.75 = 1268.595 m. Add HI 1.42 m to get the axis at 1270.015 m. (Key point: Axis RL of B) 3. The elevation of P is 27°30' from A and 18°45' from B. Which statement is correct? A. A is farther from P than B, so x is measured from A B. The angles show A and B are at the same level C. x - 53 is the horizontal distance from B to P D. B is farther (smaller elevation angle), so x is B to P and A to P is x - 53 Answer: D. The smaller angle belongs to the farther station. So B is farther, and A lies between B and P. (Key point: Which station is farther) 4. Which equation is solved for x (B to P)? A. axis B + (x - 53) tan 18°45' = axis A + x tan 27°30' B. axis B + x tan 18°45' = axis A + (x - 53) tan 27°30' C. axis B + x tan 27°30' = axis A + x tan 18°45' D. x tan 18°45' = (x - 53) tan 27°30' Answer: B. The RL of P from B (x tan aB) equals the RL of P from A ((x - 53) tan aA), each added to its own axis RL. 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