Subtense bar, level difference and gradient A-B

Reading of the data: the tacheometer is taken as $K=100$, $C=0$ (not stated). The bar at B is horizontal and normal to the line of sight. The theodolite measures the angle it subtends in the horizontal plane, so the distance found from this angle is the horizontal distance, whatever the inclination of the sight. The height of instrument is the same for both sights and cancels in the difference of level. The bar height 1.180 m is the height of the bar's sighting target above B.

Step 1: Staff point A (stadia)

Zenith angle $96^\circ30'$ gives $\theta=-6^\circ30'$ (depression).

Point A (bearing 345°00', angle of depression $6^\circ 30'$):

$$\begin{aligned} s&=1.850-0.650=1.200\ \text{m}\\ D&=100\times1.200\times\cos^2 6^\circ 30'=118.462\ \text{m}\\ V&=\tfrac12\times100\times1.200\times\sin 2(6^\circ 30')=-13.497\ \text{m} \end{aligned}$$

Step 2: Bar point B (subtense)

Horizontal distance from the bar angle $\beta=0^\circ42'15''$ and bar length $l=2$ m:

$$D_B=\frac{l}{2}\cot\frac{\beta}{2}=\cot\frac{0^\circ42'15''}{2}=162.732\ \text{m}$$

Zenith angle $85^\circ00'$ gives $\theta=+5^\circ00'$ (elevation):

$$V_B=D_B\tan5^\circ=162.732\times0.08749=+14.237\ \text{m}$$

Step 3: Difference in level

$$\begin{aligned} \text{RL}_B-\text{RL}_A&=(V_B-1.180)-(V_A-1.250)\\ &=(+14.237-1.180)-(-13.497-1.250)=+27.804\ \text{m} \end{aligned}$$

Step 4: Horizontal distance AB

Bearings from R: A $=345^\circ$, B $=225^\circ$, so angle ARB $=120^\circ$.

$$AB=\sqrt{118.462^2+162.732^2-2\times118.462\times162.732\cos120^\circ}=244.525\ \text{m}$$

Step 5: Gradient

$$\text{Gradient}=\frac{+27.804}{244.525}=0.11371$$

Answer: B is 27.804 m higher than A; AB = 244.525 m; gradient from A to B rising 1 in 8.8 (11.37 %).