Gradient and bearing of PQ from R

$K=100$, $C=0$. The zenith angles are converted to vertical angles by $\theta=90^\circ-z$: for P, $90^\circ-79^\circ28'=+10^\circ32'$ (elevation); for Q, $90^\circ-95^\circ06'=-5^\circ06'$ (depression). The middle reading is the axial hair reading $h$. The height of instrument is not given and cancels in the difference of levels.

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 S60°E = 120°00', angle of elevation $10^\circ 32'$):

$$\begin{aligned} s&=2.360-1.250=1.110\ \text{m}\\ D&=100\times1.110\times\cos^2 10^\circ 32'=107.291\ \text{m}\\ V&=\tfrac12\times100\times1.110\times\sin 2(10^\circ 32')=19.950\ \text{m} \end{aligned}$$

Point Q (bearing S30°W = 210°00', angle of depression $5^\circ 06'$):

$$\begin{aligned} s&=2.940-1.300=1.640\ \text{m}\\ D&=100\times1.640\times\cos^2 5^\circ 06'=162.704\ \text{m}\\ V&=\tfrac12\times100\times1.640\times\sin 2(5^\circ 06')=-14.521\ \text{m} \end{aligned}$$

Step 2: Difference in level

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

$$\begin{aligned} \Delta H&=(V_{Q}-h_{Q})-(V_{P}-h_{P})\\ &=(-14.521-2.120)-(+19.950-1.810)=-34.781\ \text{m} \end{aligned}$$

Step 3: Horizontal distance PQ

The angle at R between the two lines is the difference of the bearings, $90^\circ 00'$ (120°00' and 210°00').

$$\begin{aligned} PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{107.291^2+162.704^2-2\times107.291\times162.704\cos 90^\circ 00'}=194.894\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{PQ}=\frac{34.781}{194.894}=0.17846$$

The line from P to Q is falling: 1 in 5.6, i.e. 17.85 %, an angle of $10^\circ 07'06''$ with the horizontal.

Step 5: Bearing of PQ

Taking the instrument station as the origin:

$$\begin{aligned} P&=(-53.645,\ +92.916),\qquad Q=(-140.906,\ -81.352)\\ \tan\theta&=\frac{\Delta E}{\Delta N}=\frac{-174.268}{-87.261}\ \Rightarrow\ \theta_{PQ}=243^\circ 24'06'' \end{aligned}$$

Bearing of PQ = 243°24' (S 63°24' W).

Answer: PQ = 194.89 m; difference in level = -34.781 m; gradient falling 1 in 5.6 (17.85 %, 10°07'); bearing of PQ = 243°24'.