Subtense bar and tacheometer, A-B distance and level

Reading of the data: the distance AB comes from the subtense bar (angle measured at A, bar at B); the levels come from the theodolite at C ($K=100$, $C=0$, staff vertical). The distance CB is not read directly (only the middle hair is read on B), so it is found from triangle CAB with the stated angle $ACB=60^\circ00'$. (The circle reading $80^\circ24'20''$ for B in the table does not agree with this; the stated angle is used.) The height of instrument (1.55 m) is the same for both sights and cancels in the difference of level.

Step 1: Distance AB from the subtense bar

Bar length $l=2$ m, angle $\beta=2^\circ40'20''$:

$$AB=\frac{l}{2}\cot\frac{\beta}{2}=\cot(1^\circ20'10'')=42.875\ \text{m}$$

Step 2: Distance CA and vertical component from the stadia readings

Vertical angle $-5^\circ10'$ (depression):

Point A (angle of depression $5^\circ 10'$):

$$\begin{aligned} s&=1.839-1.459=0.380\ \text{m}\\ D&=100\times0.380\times\cos^2 5^\circ 10'=37.692\ \text{m}\\ V&=\tfrac12\times100\times0.380\times\sin 2(5^\circ 10')=-3.408\ \text{m} \end{aligned}$$

Step 3: Distance CB from triangle ACB

$$AB^2=CA^2+CB^2-2\,CA\,CB\cos60^\circ$$ $$CB^2-37.692\,CB+(37.692^2-42.875^2)=0\ \Rightarrow\ CB=46.644\ \text{m}$$

(the other root is negative).

Step 4: Vertical component for B and difference in level

Vertical angle $+10^\circ23'30''$ and the middle hair reading 1.235 m:

$$V_B=CB\tan10^\circ23'30''=46.644\times0.18338=+8.554\ \text{m}$$ $$\begin{aligned} \text{RL}_B-\text{RL}_A&=(V_B-1.235)-(V_A-1.649)\\ &=(+8.554-1.235)-(-3.408-1.649)=+12.376\ \text{m} \end{aligned}$$

Answer: horizontal distance AB = 42.875 m (from the bar); B is 12.376 m higher than A.