Missing bearings of BC and DE in traverse

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. The lengths of BC and DE are known but their bearings are not, so the two lines must together cancel the resultant of the four lines AB, CD, EF and FA. The lines BC, DE and that resultant form a triangle, which is solved by the cosine rule.

Step 1: Resultant of the known lines

$$\begin{aligned} r&=\sqrt{915.391^2+407.637^2}=1002.05\ \text{m}\\ \phi&=\tan^{-1}\frac{-407.637}{-915.391}=204^\circ 00'\ \ (\text{direction of the vector BC + DE}) \end{aligned}$$

Step 2: Triangle test

BC and DE must together make up a vector of length $r$, which is possible only if

$$|DE-BC|=84\ \text{m}\ \le\ r\ \le\ BC+DE=728\ \text{m}$$

Here $r=1002.05$ m is greater than $BC+DE=728$ m. A triangle with sides 322 m, 406 m and 1002.05 m does not exist, so no real bearings of BC and DE satisfy closure with the lengths and bearings as recorded. The cosine rule gives

$$\cos\alpha=\frac{r^2+l_{BC}^2-l_{DE}^2}{2\,r\,l_{BC}}=\frac{1002.05^2+322^2-406^2}{2\times1002.05\times322}=1.461$$

A cosine cannot exceed 1, so the triangle is impossible. With the printed figures the closure check fails by a large margin, so one or more of the recorded lengths or bearings (AB, CD, EF, FA, BC, DE) must be wrong and need to be re-observed.

Method to apply once the data are correct

1. Compute $r$ and $\phi$ as in Step 1.
2. Compute $\alpha$ from the cosine rule above; then bearing of BC $=\phi\pm\alpha$.
3. Find the bearing of DE from the remaining vector: $\tan\theta_{DE}=\dfrac{-(\Sigma\text{Dep}+\text{Dep}_{BC})}{-(\Sigma\text{Lat}+\text{Lat}_{BC})}$, placing it in the correct quadrant.
4. Check that $\Sigma\text{Lat}=\Sigma\text{Dep}=0$.

Answer: with the data as given, the bearings of BC and DE cannot be computed, because the resultant of the known lines (1002.05 m) exceeds the sum of the two unknown lengths (728 m); the field notes need to be rechecked.