Gradient between X and Y from R

Anallatic lens: $C=0$, $K=100$. The middle hair reading is the axial reading $h$. The height of instrument is not given, so only the difference in level between X and Y is found (it cancels).

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 X (bearing 345°00', angle of elevation $15^\circ 00'$):

$$\begin{aligned} s&=2.585-0.915=1.670\ \text{m}\\ D&=100\times1.670\times\cos^2 15^\circ 00'=155.813\ \text{m}\\ V&=\tfrac12\times100\times1.670\times\sin 2(15^\circ 00')=41.750\ \text{m} \end{aligned}$$

Point Y (bearing 75°00', angle of elevation $10^\circ 00'$):

$$\begin{aligned} s&=3.715-0.760=2.955\ \text{m}\\ D&=100\times2.955\times\cos^2 10^\circ 00'=286.590\ \text{m}\\ V&=\tfrac12\times100\times2.955\times\sin 2(10^\circ 00')=50.533\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{Y}-h_{Y})-(V_{X}-h_{X})\\ &=(+50.533-2.240)-(+41.750-1.750)=+8.293\ \text{m} \end{aligned}$$

Step 3: Horizontal distance XY

The angle at R between the two lines is the difference of the bearings, $90^\circ 00'$ (345°00' and 75°00').

$$\begin{aligned} XY&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{155.813^2+286.590^2-2\times155.813\times286.590\cos 90^\circ 00'}=326.207\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{XY}=\frac{8.293}{326.207}=0.02542$$

The line from X to Y is rising: 1 in 39.3, i.e. 2.54 %, an angle of $1^\circ 27'23''$ with the horizontal.

Answer: XY = 326.21 m; difference in level = +8.293 m; gradient rising 1 in 39.3 (2.54 %, 1°27').