--- title: "Resection, two point problem, intersection" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:resection-two-point-problem-ab-p" source: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-two-point-problem-ab-p site: BCT Engineering --- # Resection, two point problem, intersection Surveying II · Chapter 5: Orientation - Asked in: 2073 Bhadra, 4+4 marks - Video: [Two Point Problem| Surveying|Devesh Saini](https://www.youtube.com/watch?v=UpUnqyfF3u0) — Civil Beats - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/resection-two-point-problem-ab-p - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question What is resection? Explain the two point problem. What is intersection? The coordinates of known stations A (7492 mN, 3932 mE) and station B (7487 mN, 2960 mE). Calculate the coordinate of unknown point P, where the observed horizontal angles taken to P from A is 44°52'36" and to P from B is 75°33'22" respectively. ## Short answer Resection determines an unknown station from angles measured at that point; the two-point problem uses two known stations and two auxiliary unknown points. Intersection finds P from angles at known stations. Here AB=972.013 m, bearing AB=269°42'19". By sine rule, AP=1091.700 m, bearing AP=314°34'55", giving P=(3154.440 mE, 8258.295 mN). ## Answer #### Resection **Resection** is the method of finding the coordinates of an unknown station P by measuring the horizontal angles at P to three (or two) stations of known coordinates. The instrument is set up only at the unknown point. #### Two-point problem The **two-point problem** is used when only two stations of known coordinates (A and B) are available. Two points alone cannot fix one unknown station by angles at that station, so two unknown stations (C and D, which are intervisible) are set up. At each of them the angles to A and B and to the other unknown station are observed. The angles of the quadrilateral give the angles at A and B (Hansen's method), and the sine rule then gives the distances and the coordinates of both stations. #### Intersection **Intersection** is the method of finding the coordinates of an unknown point P by measuring the horizontal angles to P from two (or more) stations of known coordinates. The instrument is set up at the known stations, not at P. #### Numerical (coordinates of P from the angles at A and B) The observed angles are at the known stations A and B, so P is found by **intersection**. Coordinates are (Easting, Northing): A = (3932 E, 7492 N), B = (2960 E, 7487 N). The angles are $\angle BAP = 44^\circ52'36''$ at A and $\angle ABP = 75^\circ33'22''$ at B. P is taken on the north side of AB (the south side would give the mirror image). $$ AB = \sqrt{(2960-3932)^2 + (7487-7492)^2} = 972.013\ \text{m}, \quad \text{bearing of AB} = 269°42'19" $$ $\angle APB = 180^\circ - 44^\circ52'36'' - 75^\circ33'22'' = 59°34'02"$ $$ AP = \frac{AB\sin 75^\circ33'22''}{\sin 59°34'02"} = 1091.700\ \text{m},\qquad BP = \frac{AB\sin 44^\circ52'36''}{\sin 59°34'02"} = 795.426\ \text{m} $$ Bearing of AP = bearing of AB + 44°52'36" $= 314°34'55"$ (AB runs almost due west, so turning clockwise puts P to the north). $$ E_P = 3932 + 1091.700\sin 314°34'55" = 3154.440\ \text{m},\qquad N_P = 7492 + 1091.700\cos 314°34'55" = 8258.295\ \text{m} $$ Check from B: bearing of BP $= 89°42'19" - 75^\circ33'22'' = 14°08'57"$ gives $E = 3154.440$, $N = 8258.295$ (same). **Answer: P = (3154.440 E, 8258.295 N).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the two-point problem with only A and B known, what is needed to fix the unknown stations? A. One unknown station with angles to A and B only B. Two intervisible unknown stations C and D, with angles observed to A, B and each other C. Only the distance AB D. A third known station is mandatory Answer: B. Two known points alone cannot fix one station by angles, so two intervisible unknown stations are set up and Hansen's method is used. (Key point: Two-point problem setup) 2. In the numerical part, angles are observed at A and B to P. This is therefore: A. Resection B. Intersection C. Traversing D. Levelling Answer: B. The instrument is on the known stations A and B, so P is found by intersection. (Key point: Which method in the numerical) 3. Which expression gives AP? A. AB sin 44°52'36" / sin 59°34'02" B. AB sin 75°33'22" / sin 59°34'02", since AP is opposite the angle at B C. AB sin 59°34'02" D. AB cos 75°33'22" Answer: B. AP is opposite the angle at B (75°33'22"), and AB is opposite angle APB = 59°34'02", so AP = AB sin B / sin P. (Key point: Sine rule for AP) 4. Why is the bearing of AP taken as bearing of AB + 44°52'36" = 314°34'55"? A. Because AB runs nearly due west, a clockwise turn puts P to the north B. Because angles are always added C. Because P is south of AB D. Because AB runs east Answer: A. AB has bearing 269°42'19" (almost west); adding the angle clockwise points toward the north-west, the stated side of P. (Key point: Bearing of AP) ## Related in Surveying II - [Resection point O from stations A, B, C](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-abc-2075b) — Surveying II, chapter 5 - [Two point resection, point C from A, B, D](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atwo-point-resection-acb-dca-adb) — Surveying II, chapter 5 - [Three point analytical resection, average coordinates of O](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Athree-point-resection-p-w-d) — Surveying II, chapter 5 - [Two point resection, point C (A, B at 7928.474 N)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atwo-point-resection-c-a-b) — Surveying II, chapter 5 - [Intersection vs resection, three point method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aintersection-vs-resection-three-point) — Surveying II, chapter 5 - [Resection point P from L, M, N](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-lmn-p-2079j) — Surveying II, chapter 5 ## Related in other subjects - [Resection by three point problem](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Athree-point-resection-sketches) — Surveying I, chapter 6 - [Mean horizontal angles, station O (P, Q, R)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-o-pqrp) — Surveying I, chapter 7 - [Two-point resection in plane tabling](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-point-resection-plane-table) — Surveying I, chapter 6 - [Two point problem; advantages, disadvantages](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-point-problem-adv-disadv) — Surveying I, chapter 6