RL of hill top P (A to B = 53 m)

The data are read as: staff reading on the BM from A = 1.00 m (horizontal sight), staff (vane) reading at B = 1.75 m, horizontal distance AB = 53 m.

RL of instrument axes

• Axis of A = 1260.000 + 1.00 = 1261.000 m
• Sight A to B: elevation $= 90^\circ - 80^\circ = 10°00'$. Reading point at B = 1261.000 + 53 tan(10°00') = 1270.345 m. Ground at B = 1270.345 − 1.75 = 1268.595 m.
• Axis of B = 1268.595 + 1.42 = 1270.015 m

Elevation angles to P: from A = $90^\circ - 62^\circ30' = 27°30'$, from B = $90^\circ - 71^\circ15' = 18°45'$. B sees P at the smaller angle, so B is taken as the farther station; the result $x > D$ below confirms this.

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 $D = 53$ 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 $\alpha_B = 18°45'00"$ ($\tan = 0.33945$), $\alpha_A = 27°30'00"$ ($\tan = 0.52057$), axis RL of B = 1270.015 m and axis RL of A = 1261.000 m:

$$x = \frac{(1261.000 - 1270.015) - 53\times 0.52057}{0.33945 - 0.52057} = 202.114\ \text{m}$$ $$RL_{P} = 1270.015 + 202.114\times 0.33945 = 1338.624\ \text{m}$$

Check from A: $1261.000 + (202.114 - 53)\times 0.52057 = 1338.624$ m (same).

Answer: RL of the top of the hill P = 1338.624 m.