Closed traverse ABCDA missing figures (C=2500)

All the given deflections 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: S50°W = 230°00'
• BC: 230°00' − 85°00' = 145°00'
• CD: 145°00' − 135°00' = 10°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:

$$\begin{aligned} \text{Lat}_{DA}&=-(-159.737)=+159.737\ \text{m}\\ \text{Dep}_{DA}&=-(+47.562)=-47.562\ \text{m}\\ L_{DA}&=\sqrt{(+159.737)^2+(-47.562)^2}=166.667\ \text{m}\\ \tan\theta&=\frac{-47.562}{+159.737}\ \Rightarrow\ \theta_{DA}=343^\circ 25' \end{aligned}$$

(Positive latitude and negative departure put DA in the north-west quadrant, N 16°35' W.)

Step 4: Missing deflection angles

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

• At A (DA to AB): 343°25' − 230°00' = 113°25'
• At D (CD to DA): 10°00' − 343°25' + 360° = 26°35'

Check: sum of left deflections = 85°00' + 135°00' + 113°25' + 26°35' = 360°00' (anticlockwise traverse).

Step 5: Coordinates (C = 2500 mN, 2500 mE)

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

Check: B plus line BC gives (2500.000 N, 2500.000 E), which is the given point, so the traverse closes.

Summary of the table