--- title: "RL of hill top P (DAB = 58 m)" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:hill-top-rl-2078c-ab-58" source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-top-rl-2078c-ab-58 site: BCT Engineering --- # RL of hill top P (DAB = 58 m) Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2078 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-top-rl-2078c-ab-58 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question Compute the RL of top of a hill station P from the following data. | Inst St'n | Target | Zenith Angle | Height of Instrument | Target ht & Distance | |---|---|---|---|---| | A | P | 64°30' | 1.42 | DAB = 58 m | | B | P | 74°15' | 1.48 | B = 1.75 m | | A | B | 78°00' | 1.42 | BM = 1.25 m | | A | BM | 90°00' | 1.42 | | RL of BM is 1280.00 m. ## Short answer RL BM 1280.00, staff reading 1.00 m gives axis A = 1281.250 m. Sight A to B gives RL_B = 1291.828 m, axis B = 1293.308 m. With αA=25°30', αB=15°45', D=58 m, solving x = 203.763 m yields RL_P = 1350.775 m. ## Answer The data are read as: staff reading on the BM from A = 1.25 m (horizontal sight, 90°), vane height at B = 1.75 m, horizontal distance AB = 58 m. #### RL of instrument axes - Axis of A = RL of BM + staff reading = 1280.000 + 1.25 = 1281.250 m - Sight A to B: elevation angle $= 90^\circ - 78^\circ00' = 12°00'$. Vane at B = 1281.250 + 58 tan(12°00') = 1293.578 m. Ground at B = 1293.578 − 1.75 = 1291.828 m. - Axis of B = 1291.828 + 1.48 = 1293.308 m Elevation angles to P: from A = $90^\circ - 64^\circ30' = 25°30'$, from B = $90^\circ - 74^\circ15' = 15°45'$. 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 $D = 58$ 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 $\alpha_B = 15°45'00"$ ($\tan = 0.28203$), $\alpha_A = 25°30'00"$ ($\tan = 0.47698$), axis RL of B = 1293.308 m and axis RL of A = 1281.250 m: $$ x = \frac{(1281.250 - 1293.308) - 58\times 0.47698}{0.28203 - 0.47698} = 203.763\ \text{m} $$ $$ RL_{P} = 1293.308 + 203.763\times 0.28203 = 1350.775\ \text{m} $$ Check from A: $1281.250 + (203.763 - 58)\times 0.47698 = 1350.775$ m (same). **Answer: RL of the top of the hill P = 1350.775 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the 90°00' sight from A to the BM useful? A. It gives the horizontal distance AB B. It gives the angle of elevation of P C. The sight is horizontal, so axis RL of A = RL of BM + staff reading = 1281.250 m D. The sight is horizontal, so axis RL of A = RL of BM + HI = 1281.420 m Answer: C. A zenith angle of 90° is a level sight, so the axis RL is the BM RL plus the staff reading (1.25 m). (Key point: Axis from a horizontal sight on BM) 2. The vane level at B is 1293.578 m and the vane is 1.75 m above the foot. What is the axis RL of B (HI of B = 1.48 m)? A. 1293.308 m B. 1291.828 m C. 1295.058 m D. 1293.578 m Answer: A. Ground at B = 1293.578 - 1.75 = 1291.828 m. Axis of B = ground + HI = 1291.828 + 1.48 = 1293.308 m. (Key point: Axis RL of B) 3. The elevation of P is 25°30' from A and 15°45' from B. What does this show? A. A is farther from P, so B lies between A and P B. Both stations are the same distance from P C. P is lower than both instrument axes D. B is farther from P, so A lies between B and P Answer: D. A smaller elevation angle to the same top means a longer horizontal distance. So x is measured from B and A to P is x - 58. (Key point: Which station is farther) 4. How is the RL of P = 1350.775 m checked? A. Compute from A using x (not x - 58) with tan 25°30' B. Compute from A: 1281.250 + (x - 58) tan 25°30' gives the same value C. Repeat the zenith readings D. Average the axis RLs of A and B Answer: B. P has one RL, so the value from A, 1281.250 + (203.763 - 58) x 0.47698, must equal the value from B. 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