RL of hill station P, A-B 150 m, target 3.5 m

Zenith angles are converted to elevation angles ($\alpha = 90^\circ - z$). The sight from A to B (z = 105°30') is a depression, so A is above B.

RL of instrument axes

• Axis of B = RL of B + HI = 120.000 + 1.35 = 121.350 m
• Sight A to B: angle -15°30' (depression). The vane at B (1.5 m above the foot) is at 120.000 + 1.5 = 121.500 m, so the axis of A = 121.500 − 150 tan(-15°30') = 163.099 m.

Elevation angles to P: from B = 20°15'00", from A = 30°45'00". Take D = 150 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 $D = 150$ 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 $\alpha_B = 20°15'00"$ ($\tan = 0.36892$), $\alpha_A = 30°45'00"$ ($\tan = 0.59494$), axis RL of B = 121.350 m and of A = 163.099 m:

$$x = \frac{(163.099 - 121.350) - 150\times 0.59494}{0.36892 - 0.59494} = 210.125\ \text{m}$$ $$RL_{P} = 121.350 + 210.125\times 0.36892 = 198.869\ \text{m}$$

Check from A: $163.099 + (210.125 - 150)\times 0.59494 = 198.869$ m (same).

The RL found is for the target on top of the station. As the target is 3.5 m above the ground:

$$RL_{\text{ground at P}} = 198.869 - 3.5 = 195.369\ \text{m}$$

Answer: RL of the target at P = 198.869 m; RL of the hill station (ground) = 195.369 m.