RL of hill top P (DAB = 58 m)

The data are read as: staff reading on the BM from A = 1.25 m (horizontal sight, 90°), vane height at B = 1.75 m, horizontal distance AB = 58 m.

RL of instrument axes

• Axis of A = RL of BM + staff reading = 1280.000 + 1.25 = 1281.250 m
• Sight A to B: elevation angle $= 90^\circ - 78^\circ00' = 12°00'$. Vane at B = 1281.250 + 58 tan(12°00') = 1293.578 m. Ground at B = 1293.578 − 1.75 = 1291.828 m.
• Axis of B = 1291.828 + 1.48 = 1293.308 m

Elevation angles to P: from A = $90^\circ - 64^\circ30' = 25°30'$, from B = $90^\circ - 74^\circ15' = 15°45'$. 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 $D = 58$ 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 $\alpha_B = 15°45'00"$ ($\tan = 0.28203$), $\alpha_A = 25°30'00"$ ($\tan = 0.47698$), axis RL of B = 1293.308 m and axis RL of A = 1281.250 m:

$$x = \frac{(1281.250 - 1293.308) - 58\times 0.47698}{0.28203 - 0.47698} = 203.763\ \text{m}$$ $$RL_{P} = 1293.308 + 203.763\times 0.28203 = 1350.775\ \text{m}$$

Check from A: $1281.250 + (203.763 - 58)\times 0.47698 = 1350.775$ m (same).

Answer: RL of the top of the hill P = 1350.775 m.