Bearings of AB and EA from latitudes, departures

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

$$\begin{aligned} \Sigma\text{Lat}&=+1017.690\ \text{m},\qquad \Sigma\text{Dep}=-702.910\ \text{m}\\ \text{Closing vector}&:\ \text{Lat}=-1017.690,\ \text{Dep}=+702.910\\ r&=\sqrt{1017.690^2+702.910^2}=1236.841\ \text{m}\\ \phi&=\tan^{-1}\frac{+702.910}{-1017.690}=145^\circ 22' \end{aligned}$$

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'.)