RL of hill top P (A to B 112 m, RL 1290)

Station A is higher than B (the sight from B to A is an elevation). The angle of elevation to P is also larger from A, so A is nearer to P and B is the farther station. Take $D = 112$ m as the horizontal distance AB.

Mean vertical angles

RL of instrument axes

• Axis of B: $RL_{axis\,B} = 1290.000 + 1.42 = 1291.420$ m
• Sight B to A: the vane (1.75 m above the foot) at A is at 1291.420 + 112$\tan$(4°27'00") = 1300.136 m, so the ground at A is 1298.386 m and $RL_{axis\,A} = 1298.386 + 1.47 = 1299.856$ 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 $D = 112$ 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 $\alpha_B = 20°08'30"$ ($\tan = 0.36677$), $\alpha_A = 30°39'00"$ ($\tan = 0.59258$), axis RL of B = 1291.420 m and of A = 1299.856 m:

$$x = \frac{(1299.856 - 1291.420) - 112\times 0.59258}{0.36677 - 0.59258} = 256.560\ \text{m}$$ $$RL_{P} = 1291.420 + 256.560\times 0.36677 = 1385.519\ \text{m}$$

Check from A: $1299.856 + (256.560 - 112)\times 0.59258 = 1385.519$ m (same).

Answer: RL of the top of the hill P = 1385.519 m.