Bearings of DE and EA in traverse ABCDE

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Here the bearings of DE and EA are unknown, but all lengths are known. The two unknown lines together must cancel the resultant of the other lines, so DE, 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

$$\begin{aligned} \Sigma\text{Lat}&=-54.105\ \text{m},\qquad \Sigma\text{Dep}=+423.651\ \text{m}\\ \text{Closing vector (D to A)}&:\ \text{Lat}=+54.105,\ \text{Dep}=-423.651\\ r&=\sqrt{54.105^2+423.651^2}=427.092\ \text{m}\\ \phi&=\tan^{-1}\frac{-423.651}{+54.105}=277^\circ 17' \end{aligned}$$

Step 2: Triangle of DE, EA and the closing line

Sides: $r=427.092$, $l_i=240.30$ (DE), $l_j=265.40$ (EA).

$$\cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{427.092^2+240.30^2-265.40^2}{2\times427.092\times240.30}=0.82683\ \Rightarrow\ \alpha=34^\circ 14'$$

$\alpha$ is the angle between the closing line and DE, so the bearing of DE is $\phi\pm\alpha$. This gives two geometrically possible triangles, mirror images about the closing line:

• DE $=\phi-\alpha=243^\circ 03'$: the vector left for EA is $(+163.005,\ -209.444)$, so EA $=307^\circ 54'$.
• DE $=\phi+\alpha=311^\circ 30'$: the vector left for EA is $(-105.135,\ -243.688)$, so EA $=246^\circ 40'$.

Both pairs satisfy closure, and a field sketch would settle which one applies. With the first pair the bearings increase steadily round the traverse (36°45', 97°12', 169°45', 243°03', 307°54'), so every turn is to the right and ABCDE is a convex pentagon. The second pair makes a re-entrant angle at E. The convex figure, which is the usual case, is adopted.

Step 3: Check with the chosen bearings

DE: Lat $=240.30\cos243^\circ03'=-108.900$, Dep $=-214.208$; EA: Lat $=+163.005$, Dep $=-209.444$.

$$\Sigma\text{Lat}=-54.105+(-108.900)+(+163.005)=0.000,\qquad \Sigma\text{Dep}=+423.651+(-214.208)+(-209.444)\approx0.000$$

Both are zero, so the traverse closes.

Answer: bearing of DE = 243°03' (S 63°03' W); bearing of EA = 307°54' (N 52°06' W). (The other mathematically possible solution, giving a re-entrant figure, is DE = 311°30', EA = 246°40'.)