RL of hill top P (A to B 120 m, vane 2.5 m)

The face right reading of A to P is printed as 304°36', which would not agree with the face left reading (65°18'); it is taken as 294°36' (a misprint).

Mean vertical angles

A to B is a depression (12°48'00" below the horizontal), so A is higher than B. The angle of elevation to P is also larger from A, so A is nearer to P and B is the farther station.

RL of instrument axes

• Axis of B = 1280.000 + 1.47 = 1281.470 m
• Sight A to B: the vane at B is at (axis A) $- 120\tan(12°48'00")$, so ground B + 2.5 m = axis A $+ 120\tan(-12°48'00")$. Axis A = 1280.000 + 2.5 + 120 tan(12°48'00") = 1309.763 m (the HI at A is not needed, because the axis level of A comes directly from this sight)

Distance and RL of P

Both stations lie in the same vertical plane as P, with A between B and P. Let $x$ = horizontal distance from B to P, so the distance from A to P is $x - D$ with $D = 120$ m.

                                  P
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of P above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of P is the same from both stations:

$$\begin{aligned} RL_{axis\,B} + x\tan\alpha_B &= RL_{axis\,A} + (x-D)\tan\alpha_A \\ x &= \frac{(RL_{axis\,A} - RL_{axis\,B}) - D\tan\alpha_A}{\tan\alpha_B - \tan\alpha_A} \end{aligned}$$

With $\alpha_B = 20°04'00"$ ($\tan = 0.36529$), $\alpha_A = 24°39'00"$ ($\tan = 0.45889$), axis RL of B = 1281.470 m and of A = 1309.763 m:

$$x = \frac{(1309.763 - 1281.470) - 120\times 0.45889}{0.36529 - 0.45889} = 286.034\ \text{m}$$ $$RL_{P} = 1281.470 + 286.034\times 0.36529 = 1385.955\ \text{m}$$

Check from A: $1309.763 + (286.034 - 120)\times 0.45889 = 1385.955$ m (same).

Answer: RL of the hill top P = 1385.955 m.