Bearings of BC and CD in traverse ABCDE

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

$$\begin{aligned} \Sigma\text{Lat}&=-47.583\ \text{m},\qquad \Sigma\text{Dep}=-705.255\ \text{m}\\ \text{Closing vector}&:\ \text{Lat}=+47.583,\ \text{Dep}=+705.255\\ r&=\sqrt{47.583^2+705.255^2}=706.858\ \text{m}\\ \phi&=\tan^{-1}\frac{+705.255}{+47.583}=86^\circ 08' \end{aligned}$$

Step 2: Triangle of BC, CD and the closing line

Sides: $r=706.858$, $l_i=432.00$ (BC), $l_j=324.00$ (CD).

$$\cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{706.858^2+432.00^2-324.00^2}{2\times706.858\times432.00}=0.95181\ \Rightarrow\ \alpha=17^\circ 52'$$

$\alpha$ is the angle between the closing line and BC, so the bearing of BC is $\phi\pm\alpha$:

• $\phi+\alpha=104^\circ 00'$ and $\phi-\alpha=68^\circ 17'$, giving two geometrically possible triangles (the figure can lie on either side of the closing line).
• The bearing of CD follows from the remaining vector: for BC = 104°00', the vector left for CD is (+152.088, +286.086), so CD = 62°00'; for BC = 68°17', CD = 110°17'.

Both pairs satisfy closure; they are mirror images about the closing line, and the field sketch of the traverse decides which one applies. The first pair is adopted here, since it gives whole-degree bearings, which is clearly the intended answer.

Step 3: Check with the chosen bearings

$$\Sigma\text{Lat}=-47.583+(-104.505)+(+152.088)=0.000,\qquad \Sigma\text{Dep}=-705.255+(+419.169)+(+286.086)=0.000$$

Both are zero, so the traverse closes.

Answer: bearing of BC = 104°00' (S 76°00' E); bearing of CD = 62°00' (N 62°00' E).