Omitted quantities in closed traverse ABCDE

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

Step 1: Latitudes and departures of the known lines

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

Let the unknown length of BC be $x$ (bearing 151°24'). Its latitude is $x\cos\theta$ and departure $x\sin\theta$. The line EA has length 359.60 m and must close the traverse, so

$$((-100.618)+x\cos\theta)^2+((-244.066)+x\sin\theta)^2=359.60^2$$

with $\cos\theta=(-0.87798)$ and $\sin\theta=0.47869$ for 151°24'.

Step 2: Solve for the length

$$x^2+2((-100.618)\times(-0.87798)+(-244.066)\times0.47869)x+((-100.618)^2+(-244.066)^2-359.60^2)=0$$ $$x^2+(-56.984)x+(-59620.10)=0$$ $$x=\frac{-(-56.984)\pm\sqrt{241727.5}}{2}=274.321\ \text{or}\ -217.337$$

The negative root has no meaning, so length of BC = 274.321 m.

Step 3: Bearing of EA

Latitude of EA $=-((-100.618)+274.321\times(-0.87798))=341.467$ m and departure $=-((-244.066)+274.321\times0.47869)=112.751$ m.

$$\tan\theta=\frac{112.751}{341.467}\ \Rightarrow\ \theta=18^\circ 16'$$

The signs of latitude and departure show the line lies in the north-east quadrant, so the whole circle bearing is 18°16' (N 18°16' E).

Step 4: Check

Sums are nil, so the traverse closes.

Answer: length of BC = 274.321 m; bearing of EA = 18°16'.