--- title: "Bearings of AB and EA from latitudes, departures" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:latitude-departure-bearings-ab-ea" source: https://www.bctengineer.com/q/semester_4/surveying_ii/latitude-departure-bearings-ab-ea site: BCT Engineering --- # Bearings of AB and EA from latitudes, departures Surveying II · Chapter 1: Traversing - Asked in: 2065 Kartik (old course), 10 marks - Video: [Calculating Length and Bearing of a line between two given coordinates, (with solved example)](https://www.youtube.com/watch?v=M1oiq2D3FLY) — CIVIL Solved - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/latitude-departure-bearings-ab-ea - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question Given the following latitudes and departures of traverse ABCDEA, the bearings of AB and EA having been omitted. | Line | Latitude (m) | Departure (m) | Length (m) | |---|---|---|---| | AB | --- | --- | 1970 | | BC | +841.11 | +336.71 | | | CD | +877.18 | -311.74 | | | DE | -700.60 | -727.88 | | | EA | --- | --- | 1181 | Determine the bearings of AB and EA. ## Short answer Use the resultant of the known lines as the closing side, then solve the triangle formed with AB and EA by the cosine rule. Adopted bearings: AB = 110°52' (S 69°08' E), EA = 254°29' (S 74°29' W); mirror alternative AB = 179°52', EA = 36°16'. ## Answer For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Here the bearings of AB and EA are unknown, but their lengths are known. The two unknown lines together must cancel the resultant of the other lines, so AB, EA and that resultant (the closing line) form a triangle whose three sides are known. The angles of this triangle are found by the cosine rule. #### Step 1: Resultant of the known lines | Line | Length (m) | Latitude | Departure | |---|---|---|---| | AB | 1970 | ? | ? | | BC | | +841.11 | +336.71 | | CD | | +877.18 | -311.74 | | DE | | -700.60 | -727.88 | | EA | 1181 | ? | ? | | **Sum of given** | | +1017.69 | -702.91 | $$ \begin{aligned} \Sigma\text{Lat}&=+1017.690\ \text{m},\qquad \Sigma\text{Dep}=-702.910\ \text{m}\\ \text{Closing vector}&:\ \text{Lat}=-1017.690,\ \text{Dep}=+702.910\\ r&=\sqrt{1017.690^2+702.910^2}=1236.841\ \text{m}\\ \phi&=\tan^{-1}\frac{+702.910}{-1017.690}=145^\circ 22' \end{aligned} $$ #### Step 2: Triangle of AB, EA and the closing line Sides: $r=1236.841$, $l_i=1970.00$ (AB), $l_j=1181.00$ (EA). $$ \cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{1236.841^2+1970.00^2-1181.00^2}{2\times1236.841\times1970.00}=0.82409\ \Rightarrow\ \alpha=34^\circ 30' $$ $\alpha$ is the angle between the closing line and AB, so the bearing of AB is $\phi\pm\alpha$: - $\phi+\alpha=179^\circ 52'$ and $\phi-\alpha=110^\circ 52'$, giving two geometrically possible triangles (the figure can lie on either side of the closing line). - The bearing of EA follows from the remaining vector: for AB = 110°52', the vector left for EA is (-316.080, -1137.917), so EA = 254°29'; for AB = 179°52', the vector left is (+952.305, +698.481), so EA = 36°16'. Both pairs satisfy closure; they are mirror images about the closing line, and only a field sketch of the traverse can decide which one applies. The pair AB = 110°52', EA = 254°29' is adopted here. #### Step 3: Check with the chosen bearings $$ \Sigma\text{Lat}=+1017.690+(-701.610)+(-316.080)=0.000,\qquad \Sigma\text{Dep}=-702.910+(+1840.827)+(-1137.917)=0.000 $$ Both are zero, so the traverse closes. **Answer:** bearing of AB = 110°52' (S 69°08' E); bearing of EA = 254°29' (S 74°29' W). (The other possible solution is AB = 179°52', EA = 36°16'.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. AB and EA have unknown bearings but known lengths. What principle is used? A. Sum of interior angles is 540° B. AB and EA must cancel the resultant of the known lines, forming a triangle of three known sides C. AB must equal EA D. Bowditch's rule Answer: B. Sum Lat = 0 and sum Dep = 0 mean AB, EA and the closing vector of the other lines form a closed triangle with known sides. (Key point: Closed traverse condition) 2. Known lines give sum Lat = +1017.69 and sum Dep = -702.91. The closing vector (AB + EA) has: A. Lat +1017.69, Dep -702.91 B. Lat -1017.69, Dep +702.91 C. Lat +702.91, Dep -1017.69 D. Lat -702.91, Dep +1017.69 Answer: B. The unknown lines must cancel the known sums, so their combined lat/dep are the negatives: -1017.69 and +702.91, r = 1236.841 m. (Key point: Resultant vector) 3. The angle alpha between the closing line r and AB (length l_i) with EA (l_j) is found from: A. cos alpha = (r^2 + l_i^2 - l_j^2) / (2 r l_i) B. cos alpha = (l_i^2 + l_j^2 - r^2) / (2 l_i l_j) C. cos alpha = r / l_i D. tan alpha = l_j / l_i Answer: A. The angle at the end of r and l_i is opposite l_j in the triangle, which gives this cosine rule form. (Key point: Cosine rule) 4. Why are two bearing pairs possible? A. The data contain a blunder B. Bearing is ambiguous by 180° C. Two lengths are equal D. The triangle can lie on either side of the closing line, so bearing = phi + alpha or phi - alpha Answer: D. The two triangles are mirror images about the closing line; a field sketch decides. The answer adopts AB = 110°52', EA = 254°29'. 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