Gradient between P and Q (constants 100, 0)

$K=100$, $C=0$. The height of instrument is not given and cancels in the difference of levels. The middle reading is the axial hair reading $h$.

Staff held vertical, line of sight inclined at angle $\theta$ (additive constant zero, which is the case for an anallatic lens or $C=0$):

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

where $s$ is the staff intercept, $\theta$ the vertical angle and $K=100$, $C=0$.

Step 1: Horizontal distance and vertical component of each sight

Point P (bearing 100°00', angle of elevation $8^\circ 20'$):

$$\begin{aligned} s&=2.600-1.100=1.500\ \text{m}\\ D&=100\times1.500\times\cos^2 8^\circ 20'=146.849\ \text{m}\\ V&=\tfrac12\times100\times1.500\times\sin 2(8^\circ 20')=21.510\ \text{m} \end{aligned}$$

Point Q (bearing 200°00', angle of depression $2^\circ 30'$):

$$\begin{aligned} s&=2.500-1.320=1.180\ \text{m}\\ D&=100\times1.180\times\cos^2 2^\circ 30'=117.775\ \text{m}\\ V&=\tfrac12\times100\times1.180\times\sin 2(2^\circ 30')=-5.142\ \text{m} \end{aligned}$$

Step 2: Difference in level

The height of instrument is the same for both sights, so it cancels (RL of point $=$ RL of axis $+V-h$):

$$\begin{aligned} \Delta H&=(V_{Q}-h_{Q})-(V_{P}-h_{P})\\ &=(-5.142-1.910)-(+21.510-1.850)=-26.712\ \text{m} \end{aligned}$$

Step 3: Horizontal distance PQ

The angle at R between the two lines is the difference of the bearings, $100^\circ 00'$ (100°00' and 200°00').

$$\begin{aligned} PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{146.849^2+117.775^2-2\times146.849\times117.775\cos 100^\circ 00'}=203.574\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{PQ}=\frac{26.712}{203.574}=0.13122$$

The line from P to Q is falling: 1 in 7.6, i.e. 13.12 %, an angle of $7^\circ 28'32''$ with the horizontal.

Answer: PQ = 203.57 m; difference in level = -26.712 m; gradient falling 1 in 7.6 (13.12 %, 7°29').