Length and bearing of CD from stations A, B

Anallatic lens: $C=0$, $K=100$, staff vertical. The coordinates are taken as (Northing, Easting): A (800, 1800) and B (950, 2500). Only the horizontal distances $D$ are needed (the heights of instrument do not enter), so each sight gives the position of the staff point from its station.

$$D=Ks\cos^2\theta$$

Step 1: Distances AC and BD

Point C (bearing 15°14', angle of elevation $8^\circ 09'$):

$$\begin{aligned} s&=2.600-1.100=1.500\ \text{m}\\ D&=100\times1.500\times\cos^2 8^\circ 09'=146.985\ \text{m}\\ V&=\tfrac12\times100\times1.500\times\sin 2(8^\circ 09')=21.050\ \text{m} \end{aligned}$$

Point D (bearing 340°18', angle of elevation $2^\circ 03'$):

$$\begin{aligned} s&=2.500-1.320=1.180\ \text{m}\\ D&=100\times1.180\times\cos^2 2^\circ 03'=117.849\ \text{m}\\ V&=\tfrac12\times100\times1.180\times\sin 2(2^\circ 03')=4.218\ \text{m} \end{aligned}$$

Step 2: Coordinates of C and D

$$\begin{aligned} N_C&=800+146.985\cos15^\circ14'=941.821\\ E_C&=1800+146.985\sin15^\circ14'=1838.620\\ N_D&=950+117.849\cos340^\circ18'=1060.951\\ E_D&=2500+117.849\sin340^\circ18'=2460.274 \end{aligned}$$

Step 3: Length and bearing of CD

$$\begin{aligned} \Delta N&=N_D-N_C=+119.130,\qquad \Delta E=E_D-E_C=+621.653\\ CD&=\sqrt{119.130^2+621.653^2}=632.965\ \text{m}\\ \theta_{CD}&=\tan^{-1}\frac{+621.653}{+119.130}=79^\circ 09'06'' \end{aligned}$$

Answer: length of CD = 632.965 m, bearing of CD = 79°09' (N 79°09' E).

If the coordinates are instead read as (Easting, Northing), the same method gives CD = 672.956 m with a bearing of 6°06'44".