RL of hill station P (A-B 125 m, RL B 1280.50)

The face right reading of B to P is printed as 239°00', which does not match the face left reading (64°52'); it is taken as 295°00' (a misprint).

Mean vertical angles

A to B is a depression, so A is higher than B.

RL of instrument axes

• Axis of B = RL of B + HI = 1280.500 + 1.42 = 1281.920 m
• Sight A to B: angle -6°15' (depression). The vane at B (2 m above the foot) is at 1280.500 + 2 = 1282.500 m, so the axis of A = 1282.500 − 125 tan(-6°15') = 1296.190 m.

Elevation angles to P: from B = 25°04'00", from A = 27°39'00". Take D = 125 m as the horizontal distance BA.

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 $D = 125$ 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 $\alpha_B = 25°04'00"$ ($\tan = 0.46772$), $\alpha_A = 27°39'00"$ ($\tan = 0.52390$), axis RL of B = 1281.920 m and of A = 1296.190 m:

$$x = \frac{(1296.190 - 1281.920) - 125\times 0.52390}{0.46772 - 0.52390} = 911.767\ \text{m}$$ $$RL_{P} = 1281.920 + 911.767\times 0.46772 = 1708.376\ \text{m}$$

Check from A: $1296.190 + (911.767 - 125)\times 0.52390 = 1708.376$ m (same).

Answer: RL of the hill station P = 1708.376 m.