Traverse missing length CD, bearing EA

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Two quantities are unknown, the length of CD and the bearing of EA, so the two closure equations are enough to find them.

Step 1: Latitudes and departures of the known lines

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

The given bearings in whole circle form are AB = S52°30'E = 127°30', BC = N48°45'E = 48°45', CD = N18°15'W = 341°45', DE = S78°30'W = 258°30'. Let the unknown length of CD be $x$ (bearing 341°45'). Its latitude is $x\cos\theta$ and departure $x\sin\theta$. The line EA has length 238.80 m and must close the traverse, so

$$(16.610+x\cos\theta)^2+(189.257+x\sin\theta)^2=238.80^2$$

with $\cos\theta=0.94970$ and $\sin\theta=(-0.31316)$ for 341°45'.

Step 2: Solve for the length

$$x^2+2(16.610\times0.94970+189.257\times(-0.31316))x+(16.610^2+189.257^2-238.80^2)=0$$ $$x^2+(-86.988)x+(-20931.47)=0$$ $$x=\frac{-(-86.988)\pm\sqrt{91292.8}}{2}=194.567\ \text{or}\ -107.580$$

The negative root has no meaning, so length of CD = 194.567 m.

Step 3: Bearing of EA

Latitude of EA $=-(16.610+194.567\times0.94970)=(-201.390)$ m and departure $=-(189.257+194.567\times(-0.31316))=(-128.325)$ m.

$$\tan\theta=\frac{(-128.325)}{(-201.390)}\ \Rightarrow\ \theta=212^\circ 30'$$

The signs of latitude and departure show the line lies in the south-west quadrant, so the whole circle bearing is 212°30' (S 32°30' W).

Step 4: Check

Sums are nil, so the traverse closes.

Answer: length of CD = 194.567 m; bearing of EA = 212°30'.