--- title: "RL of tower top Q from stations P and R" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:tower-top-q-from-p-and" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tower-top-q-from-p-and site: BCT Engineering --- # RL of tower top Q from stations P and R Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2072 Magh, 6 marks - Video: [Trigonometric Levelling: Case-II Object & Instrument lies in Same vertical plane Board Exam Solution](https://www.youtube.com/watch?v=cag1wI8Qd0o) — OPengineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/tower-top-q-from-p-and - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question The top (Q) of a tower was sighted from two stations at very different level and in same vertical plane with Q. Find R.L. of the top of tower from the following observed data: | Inst. St. | H.I | R.L | Target | Zenithal angle | |---|---|---|---|---| | P | 1.87 m | - | Q | 51°39' | | R | 1.64 m | 112.78 m | Q | 68°42' | The distance between instrument stations P and R is 120 m. The angle of elevation from R to 2 m above the root of the staff held at P was 15°11'. ## Short answer From zenith angles, elevations to Q are 21°18' from R and 38°21' from P. Axis R = 114.420 m and axis P = 146.856 m; solving equal-RL equations gives x = 155.761 m from R and RL Q = 175.149 m. ## Answer Take 120 m as the horizontal distance PR. R is the lower station (the sight from R to the staff at P is an angle of elevation). The angle of elevation to Q is larger from P, so P is nearer to the tower and lies between R and Q. #### RL of instrument axes - Axis of R = RL of R + HI = 112.780 + 1.64 = 114.420 m - Sight R to 2 m above the foot of the staff at P: elevation $15^\circ11'$. Point sighted = 114.420 + 120 tan 15°11' = 146.986 m, so the ground at P = 146.986 − 2.00 = 144.986 m. - Axis of P = 144.986 + 1.87 = 146.856 m Elevation of Q: from P $= 90^\circ - 51^\circ39' = 38°21'$, from R $= 90^\circ - 68^\circ42' = 21°18'$. #### Distance and RL of Q Both stations lie in the same vertical plane as Q, with P between R and Q. Let $x$ = horizontal distance from R to Q, so the distance from P to Q is $x - D$ with $D = 120$ m. ```text Q _.--'| _.--' | P _.--' aP | R _.--' aR - - - - - - - - | |<--- D --->|<--- (x - D) --->| |<------------ x ------------>| ``` Height of Q above axis of R = $x\tan\alpha_R$ and above axis of P = $(x-D)\tan\alpha_P$. The RL of Q is the same from both stations: $$ \begin{aligned} RL_{axis\,R} + x\tan\alpha_R &= RL_{axis\,P} + (x-D)\tan\alpha_P \\ x &= \frac{(RL_{axis\,P} - RL_{axis\,R}) - D\tan\alpha_P}{\tan\alpha_R - \tan\alpha_P} \end{aligned} $$ With $\alpha_R = 21°18'00"$ ($\tan = 0.38988$), $\alpha_P = 38°21'00"$ ($\tan = 0.79117$), axis RL of R = 114.420 m and of P = 146.856 m: $$ x = \frac{(146.856 - 114.420) - 120\times 0.79117}{0.38988 - 0.79117} = 155.761\ \text{m} $$ $$ RL_{Q} = 114.420 + 155.761\times 0.38988 = 175.149\ \text{m} $$ Check from P: $146.856 + (155.761 - 120)\times 0.79117 = 175.149$ m (same). **Answer: RL of the top of the tower Q = 175.149 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. R has RL 112.78 m and HI 1.64 m. What is the axis RL of R? A. 112.780 m B. 111.140 m C. 114.420 m D. 146.856 m Answer: C. Axis = ground RL + HI = 112.780 + 1.64 = 114.420 m. (Key point: Axis RL of R) 2. R sights a point 2 m above the foot of the staff at P at an elevation of 15°11'. How is the axis RL of P found? A. Sighted point = 114.420 + 120 tan 15°11'; subtract 2.00 for the ground at P; add HI 1.87 B. 114.420 + 120 tan 15°11' + 2.00 C. 114.420 + 1.87 D. 114.420 + 120 tan 15°11' - 1.87 Answer: A. The sighted point is 2 m above ground. So ground = 146.986 - 2.00 = 144.986 m and axis P = 144.986 + 1.87 = 146.856 m. (Key point: Axis RL of P) 3. The elevation of Q is 38°21' from P and 21°18' from R. What follows? A. P is farther, so x is measured from P B. Both stations are equally distant from Q C. Q lies between R and P D. R is farther (smaller angle), so P lies between R and Q Answer: D. The smaller elevation angle means the greater distance, so x is R to Q and P to Q is x - 120. (Key point: Which station is farther) 4. If x = 155.761 m is the horizontal distance from R to Q, what is the distance from P to Q? A. 155.761 m B. 35.761 m C. 275.761 m D. 120.000 m Answer: B. P is between R and Q, so P to Q = x - 120 = 35.761 m. 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