--- title: "RL of hill top P (A to B 120 m, vane 2.5 m)" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:hill-top-rl-2072m-vane-25" source: https://www.bctengineer.com/q/semester_4/surveying_ii/hill-top-rl-2072m-vane-25 site: BCT Engineering --- # RL of hill top P (A to B 120 m, vane 2.5 m) Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2072 Magh, 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-2072m-vane-25 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question In a trigonometrical levelling a hill station "P" was sighted from two instrument station A and B which were at very different level but with same line of sight to that of target and following information were noted. | Inst. st'n | HI (m) | Target | Zenith FL | Zenith FR | Distance | |---|---|---|---|---|---| | A | 1.42 | P | 65°18' | 304°36' | 120.00 m (st'n A to st'n B) | | B | 1.47 | P | 69°52' | 290°00' | | | A | 1.42 | B | 102°52' | 257°16' | | Determine the RL of P, if RL of ground point B was 1280.00 m and vane height while sighting from A to B was 2.50 m above the foot of the vane. ## Short answer Correct the printed FR to 294°36'. Mean elevations are A to P 24°39', B to P 20°04', A to B -12°48'; axes are B = 1281.470 m and A = 1309.763 m. Solving gives x = 286.034 m from B, so RL P = 1385.955 m. ## Answer The face right reading of A to P is printed as 304°36', which would not agree with the face left reading (65°18'); it is taken as 294°36' (a misprint). #### Mean vertical angles | Station | Face left | Face right (360° − FR) | Mean zenith $z$ | Elevation $\alpha = 90^\circ - z$ | |---|---|---|---|---| | A to P | 65°18' | 65°24' | 65°21'00" | 24°39'00" | | B to P | 69°52' | 70°00' | 69°56'00" | 20°04'00" | | A to B | 102°52' | 102°44' | 102°48'00" | -12°48'00" | A to B is a depression (12°48'00" below the horizontal), so A is higher than B. The angle of elevation to P is also larger from A, so A is nearer to P and B is the farther station. #### RL of instrument axes - Axis of B = 1280.000 + 1.47 = 1281.470 m - Sight A to B: the vane at B is at (axis A) $- 120\tan(12°48'00")$, so ground B + 2.5 m = axis A $+ 120\tan(-12°48'00")$. Axis A = 1280.000 + 2.5 + 120 tan(12°48'00") = 1309.763 m (the HI at A is not needed, because the axis level of A comes directly from this sight) #### 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 $D = 120$ 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 $\alpha_B = 20°04'00"$ ($\tan = 0.36529$), $\alpha_A = 24°39'00"$ ($\tan = 0.45889$), axis RL of B = 1281.470 m and of A = 1309.763 m: $$ x = \frac{(1309.763 - 1281.470) - 120\times 0.45889}{0.36529 - 0.45889} = 286.034\ \text{m} $$ $$ RL_{P} = 1281.470 + 286.034\times 0.36529 = 1385.955\ \text{m} $$ Check from A: $1309.763 + (286.034 - 120)\times 0.45889 = 1385.955$ m (same). **Answer: RL of the hill top P = 1385.955 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. RL 1280.00 m is of the ground at B and HI of B is 1.47 m. What is the axis RL of B? A. 1278.530 m B. 1280.000 m C. 1281.470 m D. 1282.500 m Answer: C. Axis = ground RL + HI = 1280.000 + 1.47 = 1281.470 m. (Key point: Axis RL of B) 2. A sights a vane 2.5 m above the foot at B, at a depression of 12°48' (D = 120 m). How is axis A found? A. 1282.5 - 120 tan 12°48' = 1255.237 m B. 1280 + 2.5 + 120 tan 12°48' = 1309.763 m C. 1280 + 1.47 + 120 tan 12°48' = 1308.733 m D. 1281.470 m Answer: B. The vane level is 1282.5 m. As A looks down to it, A's axis is higher by D tan(depression). (Key point: Axis RL of A) 3. What is wrong with FR = 304°36' for A to P (FL 65°18')? A. Nothing is wrong B. FR should be 245° C. FR must always be below 270° D. 360° - 304°36' = 55°24' disagrees with 65°18'; it is taken as 294°36' Answer: D. Face right should be about 360° - FL = 294°42'. So 294°36' is the intended reading. (Key point: Misprint in the FR reading) 4. With x = 286.034 m from B to P, which expression gives the RL of P? A. 1281.470 + 286.034 tan 24°39' = 1412.73 m B. 1281.470 + 286.034 tan 20°04' = 1385.955 m C. 1309.763 + 286.034 tan 24°39' = 1441.02 m D. 1280.000 + 286.034 tan 20°04' = 1384.485 m Answer: B. From B use the axis of B and the angle at B with distance x. Using A requires x - 120 and tan 24°39'. 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