--- title: "Two point resection, point C (A, B at 7928.474 N)" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:two-point-resection-c-a-b" source: https://www.bctengineer.com/q/semester_4/surveying_ii/two-point-resection-c-a-b site: BCT Engineering --- # Two point resection, point C (A, B at 7928.474 N) Surveying II · Chapter 5: Orientation - Asked in: 2076 Baisakh, 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/two-point-resection-c-a-b - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question Two points A and B have coordinates (7928.474 mN, 8464.418 mE) and (6363.275 mN, 10602.054 mE). From two points C and D located south and west of AB angles observation are ∠ACB = 70°35'50", ∠DCA = 52°25'40", ∠ADB = 65°27'35" & ∠BDC = 32°16'40". Determine the coordinates of resection point C. ## Short answer Hansen's method: solve quadrilateral ADCB with sine rule and tangent relation, compute distances AD and BC, then coordinates from bearings. Resection point C is (9209.871 mE, 5365.771 mN). Point D is (7961.869 mE, 5853.684 mN). ## Answer This is the two-point resection (Hansen's problem), solved by the sine rule and the tangent relation. Coordinates are (Easting, Northing): A = (8464.418 E, 7928.474 N), B = (10602.054 E, 6363.275 N). C and D are on the south-west side of AB (C south, D west). Observed: $\angle ACB = 70°35'50"$, $\angle DCA = 52°25'40"$ at C and $\angle ADB = 65°27'35"$, $\angle BDC = 32°16'40"$ at D. ```text A ___________________ B \ \ __---' \ \ __--' D --- C ``` #### Step 1: Base AB $$ c = AB = \sqrt{(10602.054-8464.418)^2 + (6363.275-7928.474)^2} = 2649.403\ \text{m},\qquad \text{bearing of AB} = 126°12'43" $$ #### Step 2: Angles of the quadrilateral ADCB $$ \angle ADC = \angle ADB + \angle BDC = 97°44'15",\qquad \angle DCB = \angle ACB + \angle DCA = 123°01'30" $$ $$ S = x + y = \angle DAB + \angle CBA = 360^\circ - 97°44'15" - 123°01'30" = 139°14'15" $$ #### Step 3: Solve for x = ∠DAB $P_0 = 180^\circ - \angle ADB = 114°32'25"$ and $$ K = \frac{\sin\angle ADB\,\sin\angle DCA}{\sin\angle ACB\,\sin\angle ADC} = 0.771427 $$ $$ \tan x = \frac{\sin P_0 - K\sin S}{\cos P_0 - K\cos S} \quad\Rightarrow\quad x = \angle DAB = 67°24'14" $$ $$ y = \angle CBA = S - x = 71°50'01" $$ #### Step 4: Distances AD and BC $\angle ABD = 180^\circ - \angle ADB - x = 47°08'11"$ and $\angle BAC = 180^\circ - \angle ACB - y = 37°34'09"$ $$ AD = \frac{c\sin\angle ABD}{\sin\angle ADB} = 2134.785\ \text{m},\qquad BC = \frac{c\sin\angle BAC}{\sin\angle ACB} = 1712.655\ \text{m} $$ #### Step 5: Coordinates Bearing of AD = bearing of AB + x $= 193°36'57"$; bearing of BC = bearing of BA − y $= 234°22'42"$ (C and D are on the right-hand side of the direction A to B). $$ E_D = 8464.418 + 2134.785\sin 193°36'57" = 7961.869,\qquad N_D = 7928.474 + 2134.785\cos 193°36'57" = 5853.684 $$ $$ E_C = 10602.054 + 1712.655\sin 234°22'42" = 9209.871,\qquad N_C = 6363.275 + 1712.655\cos 234°22'42" = 5365.771 $$ **Check:** the angles recomputed from these coordinates are $\angle ACB = 70°35'50"$, $\angle DCA = 52°25'40"$, $\angle ADB = 65°27'35"$, $\angle BDC = 32°16'40"$, equal to the observed values. The length CD = 1339.988 m from the coordinates agrees with triangle ADC. **Answer: C = (9209.871 E, 5365.771 N); D = (7961.869 E, 5853.684 N).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is S = ∠DAB + ∠CBA in the worked problem? A. 360° - 97°44'15" - 123°01'30" = 139°14'15" B. 180° - 97°44'15" = 82°15'45" C. 97°44'15" + 123°01'30" = 220°45'45" D. 360° - 70°35'50" - 65°27'35" = 223°56'35" Answer: A. S = 360° minus the interior angles ∠ADC and ∠DCB of quadrilateral ADCB, where ∠ADC = 65°27'35"+32°16'40" and ∠DCB = 70°35'50"+52°25'40". (Key point: Sum S of the two unknown angles) 2. How is the interior angle ∠DCB obtained from the observations? A. ∠ACB - ∠DCA B. ∠ACB + ∠DCA = 123°01'30" C. 180° - ∠ACB D. ∠BDC + ∠ADB Answer: B. Ray CA lies between CD and CB, so ∠DCB = ∠ACB + ∠DCA = 70°35'50" + 52°25'40". (Key point: Angle at C) 3. Which equation gives x = ∠DAB? A. tan x = (sin S - K sin P0)/(cos S - K cos P0) B. tan x = K tan S C. tan x = (sin P0 - K sin S)/(cos P0 - K cos S), with P0 = 180° - ∠ADB D. sin x = K sin S Answer: C. From sin(P0 - x) = K sin(S - x) one gets tan x = (sin P0 - K sin S)/(cos P0 - K cos S); here x = 67°24'14". (Key point: Solving for x) 4. How is the bearing of AD obtained, and what is the final check? A. Bearing of AB + x = 193°36'57"; CD from coordinates must agree with CD from triangle ADC B. Bearing of AB - x; the check is that AB equals CD C. Bearing of BA + x; the check is that angle sum is 180° D. Bearing of AB + y; the check is that AD = BC Answer: A. C and D are on the right of A to B, so bearing AD = bearing AB + x. The distance CD computed from coordinates must equal that from triangle ADC (1339.988 m). 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