RL of hill top from two stations (A-B 118 m)

The observation recorded against "C" in the table is taken as the sight from B to A (instrument station C is a misprint for B; the target height 1.65 m is the vane at A).

The sight from B to A is an elevation, so A is higher than B. B sees P at the smaller angle of elevation, so B is the station farther from P. Take $D = 118$ m as the horizontal distance AB.

Mean vertical angles

RL of instrument axes

• Axis of B: $RL_{axis\,B} = 1590.000 + 1.47 = 1591.470$ m
• Sight B to A: the vane at A (1.65 m above the foot) is at 1591.470 + 118$\tan$(4°27'00") = 1600.653 m, so the ground at A is at 1599.003 m and $RL_{axis\,A} = 1599.003 + 1.45 = 1600.453$ m

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 $D = 118$ m.

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

The height of P above the axis of B is $x\tan\alpha_B$, and above the axis of A it is $(x-D)\tan\alpha_A$. 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 $\alpha_B = 20°08'30"$ ($\tan = 0.36677$), $\alpha_A = 30°39'00"$ ($\tan = 0.59258$), axis RL of B = 1591.470 m and axis RL of A = 1600.453 m:

$$x = \frac{(1600.453 - 1591.470) - 118\times 0.59258}{0.36677 - 0.59258} = 269.884\ \text{m}$$ $$RL_{P} = 1591.470 + 269.884\times 0.36677 = 1690.456\ \text{m}$$

Check from A: $1600.453 + (269.884 - 118)\times 0.59258 = 1690.456$ m (same).

Answer: RL of the top of the hill P = 1690.456 m.