Gradient P to Q from station A

Reading of the data: the staff is held at right angles (normal) to the line of sight at both points, and no instrument constants are given, so $K=100$ and $C=0$ are taken. No heights of instrument are given, so only the difference in level between P and Q is found; the height of instrument cancels.

Staff normal to the line of sight, angle of elevation $\theta$, axial hair reading $h$ ($C=0$). The axial reading lies along the staff, which is tilted by $\theta$ from the vertical, so it adds $h\sin\theta$ to the horizontal distance and $h\cos\theta$ vertically:

$$D=Ks\cos\theta+h\sin\theta,\qquad V=Ks\sin\theta,\qquad \text{RL of staff point}=\text{RL of axis}+V-h\cos\theta$$

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

Step 1: Horizontal distance and vertical component of each sight

Point P (bearing 84°36', angle of elevation $3^\circ 30'$, $h=2.100$ m):

$$\begin{aligned} s&=2.850-1.350=1.500\ \text{m}\\ D&=(100\times1.500)\cos 3^\circ 30'+2.100\sin3^\circ30'=149.720+0.128=149.848\ \text{m}\\ V&=(100\times1.500)\sin 3^\circ 30'=9.157\ \text{m},\qquad h\cos\theta=2.096\ \text{m} \end{aligned}$$

Point Q (bearing 142°24', angle of elevation $2^\circ 45'$, $h=2.860$ m):

$$\begin{aligned} s&=3.765-1.955=1.810\ \text{m}\\ D&=(100\times1.810)\cos 2^\circ 45'+2.860\sin2^\circ45'=180.792+0.137=180.929\ \text{m}\\ V&=(100\times1.810)\sin 2^\circ 45'=8.684\ \text{m},\qquad h\cos\theta=2.857\ \text{m} \end{aligned}$$

Step 2: Difference in level

The height of instrument is the same for both sights, so it cancels:

$$\begin{aligned} \Delta H&=(V_{Q}-h_{Q}\cos\theta_Q)-(V_{P}-h_{P}\cos\theta_P)\\ &=(8.684-2.857)-(9.157-2.096)=5.827-7.061=-1.234\ \text{m} \end{aligned}$$

Step 3: Horizontal distance PQ

The angle at A between the two lines is the difference of the bearings: $142^\circ24'-84^\circ36'=57^\circ 48'$.

$$\begin{aligned} PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{149.848^2+180.929^2-2\times149.848\times180.929\cos 57^\circ 48'}=162.158\ \text{m} \end{aligned}$$

Step 4: Gradient

$$\text{Gradient}=\frac{\Delta H}{PQ}=\frac{1.234}{162.158}=0.00761$$

The line from P to Q is falling: 1 in 131.4, i.e. 0.76 %, an angle of $0^\circ 26'09''$ with the horizontal.

Answer: PQ = 162.16 m; difference in level = -1.234 m; gradient falling 1 in 131.4 (0.76 %, 0°26').

If the staff is instead taken as vertical, the corresponding result is $PQ=161.81$ m and gradient 1 in 132.0 (difference in level $-1.226$ m), which agrees closely because the vertical angles are small.