Gradient P to Q from common point R

$K=100$, $C=0$, staff vertical: $D=Ks\cos^2\theta$, $V=\tfrac12Ks\sin2\theta$. The zenith angles give the vertical angles $\theta=90^\circ-z$: from P, $90^\circ-81^\circ=+9^\circ$; from Q, $90^\circ-85^\circ=+5^\circ$. The middle hair reading is the axial reading $h$.

Step 1: Distances and vertical components

Point R from P (bearing 60°30', angle of elevation $9^\circ 00'$):

$$\begin{aligned} s&=3.000-1.000=2.000\ \text{m}\\ D&=100\times2.000\times\cos^2 9^\circ 00'=195.106\ \text{m}\\ V&=\tfrac12\times100\times2.000\times\sin 2(9^\circ 00')=30.902\ \text{m} \end{aligned}$$

Point R from Q (bearing 335°45', angle of elevation $5^\circ 00'$):

$$\begin{aligned} s&=3.600-1.300=2.300\ \text{m}\\ D&=100\times2.300\times\cos^2 5^\circ 00'=228.253\ \text{m}\\ V&=\tfrac12\times100\times2.300\times\sin 2(5^\circ 00')=19.970\ \text{m} \end{aligned}$$

Step 2: RL of P and Q (RL of R = 1200.000 m)

$$\text{RL}_R=\text{RL}_{station}+HI+V-h\ \Rightarrow\ \text{RL}_{station}=\text{RL}_R-HI-V+h$$ $$\begin{aligned} \text{RL}_P&=1200.000-1.380-30.902+2.000=1169.718\ \text{m}\\ \text{RL}_Q&=1200.000-1.420-19.970+2.450=1181.060\ \text{m} \end{aligned}$$

Step 3: Horizontal distance PQ

Take P as the origin, with coordinates written as (N, E). R is $D_P=195.106$ m from P on a bearing of $60^\circ30'$. The bearing $335^\circ45'$ is from Q to R, so Q is found by going back from R a distance $D_Q=228.253$ m along that bearing:

$$\begin{aligned} R&=(96.075,\ 169.811)\\ Q&=R-D_Q(\cos335^\circ45',\ \sin335^\circ45')=(96.075-208.112,\ 169.811+93.748)=(-112.038,\ 263.559)\\ PQ&=\sqrt{(-112.038)^2+263.559^2}=286.384\ \text{m},\qquad \text{bearing of PQ}=113^\circ 01'48'' \end{aligned}$$

Step 4: Gradient from P to Q

$$\text{Gradient}=\frac{\text{RL}_Q-\text{RL}_P}{PQ}=\frac{+11.342}{286.384}=0.03960$$

Answer: RL of P = 1169.718 m, RL of Q = 1181.060 m, PQ = 286.384 m; the gradient from P to Q is rising 1 in 25.2 (3.96 %).