Gradient P to Q, constants 100 and 0

Reading of the data: only one RL is given (RL of X = 1197.339 m), so the staff point sighted from Q is taken to be the same point X as sighted from P (the common-point layout). With $K=100$, $C=0$ and the staff vertical:

$$D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta,\qquad \theta=90^\circ-z$$

Vertical angles: from P, $90^\circ-80^\circ10'=+9^\circ50'$ (elevation); from Q, $90^\circ-95^\circ20'=-5^\circ20'$ (depression).

Step 1: Distances and vertical components

Point X from P (bearing 70°20', angle of elevation $9^\circ 50'$):

$$\begin{aligned} s&=2.900-2.230=0.670\ \text{m}\\ D&=100\times0.670\times\cos^2 9^\circ 50'=65.046\ \text{m}\\ V&=\tfrac12\times100\times0.670\times\sin 2(9^\circ 50')=11.274\ \text{m} \end{aligned}$$

Point X from Q (bearing 335°00', angle of depression $5^\circ 20'$):

$$\begin{aligned} s&=1.440-1.000=0.440\ \text{m}\\ D&=100\times0.440\times\cos^2 5^\circ 20'=43.620\ \text{m}\\ V&=\tfrac12\times100\times0.440\times\sin 2(5^\circ 20')=-4.072\ \text{m} \end{aligned}$$

Step 2: RL of P and Q

$$\text{RL}_X=\text{RL of axis}+V-h\ \Rightarrow\ \text{RL of axis}=\text{RL}_X-V+h$$ $$\begin{aligned} \text{Axis at P}&=1197.339-11.274+2.565=1188.630\ \text{m}\\ \text{RL}_P&=1188.630-1.46=1187.170\ \text{m}\\ \text{Axis at Q}&=1197.339-(-4.072)+1.220=1202.631\ \text{m}\\ \text{RL}_Q&=1202.631-1.38=1201.251\ \text{m} \end{aligned}$$

Step 3: Horizontal distance and bearing of PQ

Take P as the origin. X is at $D_P=65.046$ m on bearing $70^\circ20'$, and Q is $D_Q=43.620$ m from X in the direction opposite to the bearing $335^\circ00'$ of the line from Q to X:

$$\begin{aligned} X&=(21.891,\ 61.251)\ \text{(N, E)}\\ Q&=X-D_Q(\cos335^\circ,\ \sin335^\circ)=(21.891-39.533,\ 61.251+18.435)=(-17.642,\ +79.686)\\ PQ&=\sqrt{17.642^2+79.686^2}=81.616\ \text{m},\qquad \theta_{PQ}=\tan^{-1}\frac{+79.686}{-17.642}=102^\circ 29'01'' \end{aligned}$$

Step 4: Gradient from P to Q

$$\text{Gradient}=\frac{\text{RL}_Q-\text{RL}_P}{PQ}=\frac{+14.081}{81.616}=0.17253$$

Answer: RL of P = 1187.170 m, RL of Q = 1201.251 m, PQ = 81.616 m (bearing 102°29'). The gradient from P to Q is rising 1 in 5.8 (17.25 %).