Closed traverse ABCDA, missing data

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$, so the missing line DA (length and bearing) is the line that closes the figure formed by AB, BC and CD.

Step 1: Whole circle bearings and latitudes, departures

AB = N45°30'W = 314°30', BC = N5°30'E = 5°30', CD = N88°20'E = 88°20'.

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Step 2: Length of DA

The latitude and departure of DA must cancel these sums:

$$\begin{aligned} \text{Lat}_{DA}&=-(+675.069)=-675.069\ \text{m}\\ \text{Dep}_{DA}&=-(+81.621)=-81.621\ \text{m}\\ L_{DA}&=\sqrt{(-675.069)^2+(-81.621)^2}=679.99\ \text{m} \end{aligned}$$

Step 3: Bearing of DA

$$\tan\theta=\frac{\text{Dep}}{\text{Lat}}=\frac{-81.621}{-675.069}\ \Rightarrow\ \theta=186^\circ 54'$$

Latitude is negative and departure is negative, so DA lies in the south-west quadrant: whole circle bearing = 186°54', i.e. S 6°54' W.

Check

Lat: +675.069 + (-675.069) = 0.000; Dep: +81.621 + (-81.621) = 0.000, so the traverse closes.

Answer: DA = 679.99 m, bearing 186°54' (S 6°54' W).