Closed traverse ABCDA, B at 26500 N 22400 E

The deflection angles are to the left, so the traverse runs anticlockwise and bearing of a line = bearing of previous line − left deflection. For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$, which gives the length and bearing of the last line.

Step 1: Bearings of the known lines

• AB: S40°W = 220°00'
• BC: 220°00' − 116°00' = 104°00'
• CD: 104°00' − 60°00' = 44°00'

Step 2: Latitudes and departures of the known lines

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

Step 3: Length and bearing of DA

For closure, the missing line must cancel the sums:

$$\begin{aligned} \text{Lat}_{DA}&=-(-53.759)=+53.759\ \text{m}\\ \text{Dep}_{DA}&=-(+372.933)=-372.933\ \text{m}\\ L_{DA}&=\sqrt{(+53.759)^2+(-372.933)^2}=376.788\ \text{m}\\ \tan\theta&=\frac{-372.933}{+53.759}\ \Rightarrow\ \theta_{DA}=278^\circ 12' \end{aligned}$$

(The signs of the latitude and departure put the line in the north-west quadrant.)

Step 4: Missing deflection angles

Left deflection at a station = bearing of previous line − bearing of next line (add 360° if the result is negative).

• At A (DA to AB): 278°12' − 220°00' = 58°12'
• At D (CD to DA): 44°00' − 278°12' + 360° = 125°48'

Check: sum of left deflections = 116°00' + 60°00' + 58°12' + 125°48' = 360°00' (anticlockwise traverse).

Step 5: Coordinates (B = 26500 N, 22400 E)

Independent coordinates = previous coordinates + latitude/departure of the line.

Check: A plus line AB gives (26500.000 N, 22400.000 E), which is the given point, so the traverse closes.

Summary of the table