--- title: "Coordinates of O from S and D (FL/FR)" subject: "Surveying II" chapter: 5 key: "semester_4/surveying_ii:intersection-o-south-of-s-and" source: https://www.bctengineer.com/q/semester_4/surveying_ii/intersection-o-south-of-s-and site: BCT Engineering --- # Coordinates of O from S and D (FL/FR) Surveying II · Chapter 5: Orientation - Asked in: 2078 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/intersection-o-south-of-s-and - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-5 ## Question Determine the coordinate of new station "O" from the data observed below: Point 'O' is south from S and D. | Inst. Stn. | Sighted to | Horizontal angles FL | Horizontal angles FR | Easting (m) | Northing (m) | |---|---|---|---|---|---| | S | D | 0°0'0" | 180°0'0" | 627464.712 | 3066928.474 | | | O | 75°3'22" | 255°3'22" | | | | D | O | 0°0'0" | 180°0'0" | 629602.054 | 3065363.275 | | | S | 68°36'18" | 248°36'18" | | | ## Short answer Find mean horizontal angles, compute base SD length 2649.166 m and bearing 126°12'57". By sine rule, SO = 4162.625 m, bearing 201°16'19", giving O = 625954.540 m E, 3063049.450 m N. ## Answer The horizontal angles are the means of face left and face right: at S, $\angle DSO = 75^\circ03'22''$ (both faces agree); at D, $\angle ODS = 68^\circ36'18''$ (both faces agree). O lies south of the line SD. Coordinates are (Easting, Northing). ```text S ------------- D \ / \ 75° / 68° \ / \ / \ / \ / O ``` #### Step 1: Length and bearing of the base SD $$ \Delta E = 2137.342,\quad \Delta N = -1565.199 $$ $$ SD = \sqrt{\Delta E^2 + \Delta N^2} = 2649.166\ \text{m}, \qquad \text{bearing of SD} = 126°12'57"\ (\text{south-east}) $$ #### Step 2: Triangle SDO $\angle SOD = 180^\circ - 75^\circ03'22'' - 68^\circ36'18'' = 36°20'20"$ By the sine rule: $$ SO = \frac{SD\sin 68^\circ36'18''}{\sin 36°20'20"} = 4162.625\ \text{m}, \qquad DO = \frac{SD\sin 75^\circ03'22''}{\sin 36°20'20"} = 4319.508\ \text{m} $$ #### Step 3: Bearings and coordinates O is south of SD, i.e. on the right-hand side when facing from S to D, so the bearing of SO = bearing of SD + 75°03'22" $= 201°16'19"$. $$ E_O = E_S + SO\sin\theta = 627464.712 + 4162.625\sin 201°16'19" = 625954.540\ \text{m} $$ $$ N_O = N_S + SO\cos\theta = 3066928.474 + 4162.625\cos 201°16'19" = 3063049.450\ \text{m} $$ **Check from D:** bearing of DO $= 306°12'57" - 68^\circ36'18'' = 237°36'39"$, giving $E = 625954.540$ m and $N = 3063049.450$ m (the same). **Answer: coordinates of O = (625954.540 E, 3063049.450 N).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the horizontal angle at S obtained from the face-left and face-right readings? A. Use face left only B. Take the mean of the face-left and face-right angles C. Add the two angles D. Subtract face right from face left Answer: B. The mean of the two faces eliminates instrumental errors; here both faces agree at 75°03'22". (Key point: Mean of two faces) 2. In triangle SDO, how is the angle at O found? A. 180° - 75°03'22" - 68°36'18" = 36°20'20" B. 75°03'22" + 68°36'18" C. 360° - 75°03'22" - 68°36'18" D. It is measured directly Answer: A. The three angles of a plane triangle sum to 180°, so the angle at O is the remainder. (Key point: Angle at the new point) 3. Which expression gives SO? A. SD sin 75°03'22" / sin 36°20'20" B. SD sin 68°36'18" / sin 36°20'20" C. SD sin 36°20'20" / sin 68°36'18" D. SD cos 68°36'18" Answer: B. SO lies opposite the angle at D (68°36'18"), and SD lies opposite the angle at O, so SO = SD sin(D)/sin(O). (Key point: Sine rule for the side) 4. SD runs south-east (bearing 126°12'57"). O is south of SD. How is the bearing of SO obtained? A. Bearing of SD - 75°03'22" B. Bearing of SD + 75°03'22" = 201°16'19", since O is on the right-hand side C. 180° - 75°03'22" D. Bearing of SD + 180° Answer: B. South of the line S to D is the right-hand side when facing D, so the angle is added clockwise, giving 201°16'19". 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