Tacheometer depression angles, distance A to B and RL

Reading of the data: the tacheometer at A is first sighted on the staff held on the benchmark (BM, RL 1000.000 m) and then on the staff at B. Both sights are angles of depression. The constants are $K=100$ and $C=0.3$ m, and the staff is vertical, so

$$D=Ks\cos^2\theta+C\cos\theta,\qquad V=\tfrac12Ks\sin2\theta+C\sin\theta$$

with $\theta$ negative for depression. The horizontal distance "from A to B" is the distance from the instrument at A to the staff at B.

Step 1: Sight on the BM (readings 0.640, 2.200, 3.760; angle $-6^\circ19'$)

Point BM (angle of depression $6^\circ 19'$):

$$\begin{aligned} s&=3.760-0.640=3.120\ \text{m}\\ D&=100\times3.120\cos^2 6^\circ 19'+0.30\cos 6^\circ 19'=308.521\ \text{m}\\ V&=\tfrac12\times100\times3.120\sin 2(6^\circ 19')+0.30\sin 6^\circ 19'=-34.152\ \text{m} \end{aligned}$$

Since the BM is below the axis, RL of axis $=$ RL$_{BM}$ $+|V|+h$:

$$\text{RL}_{axis}=1000.000+34.152+2.200=1036.352\ \text{m}$$

Step 2: Sight on B (readings 0.010, 2.120, 4.230; angle $-7^\circ42'$)

Point B (angle of depression $7^\circ 42'$):

$$\begin{aligned} s&=4.230-0.010=4.220\ \text{m}\\ D&=100\times4.220\cos^2 7^\circ 42'+0.30\cos 7^\circ 42'=414.721\ \text{m}\\ V&=\tfrac12\times100\times4.220\sin 2(7^\circ 42')+0.30\sin 7^\circ 42'=-56.073\ \text{m} \end{aligned}$$ $$\text{RL}_B=\text{RL}_{axis}-|V|-h=1036.352-56.073-2.120=978.159\ \text{m}$$

Answer: horizontal distance from A to B = 414.721 m; R.L. of B = 978.159 m.