RL of tower top Q from stations P and R

Take 120 m as the horizontal distance PR. R is the lower station (the sight from R to the staff at P is an angle of elevation). The angle of elevation to Q is larger from P, so P is nearer to the tower and lies between R and Q.

RL of instrument axes

• Axis of R = RL of R + HI = 112.780 + 1.64 = 114.420 m
• Sight R to 2 m above the foot of the staff at P: elevation $15^\circ11'$. Point sighted = 114.420 + 120 tan 15°11' = 146.986 m, so the ground at P = 146.986 − 2.00 = 144.986 m.
• Axis of P = 144.986 + 1.87 = 146.856 m

Elevation of Q: from P $= 90^\circ - 51^\circ39' = 38°21'$, from R $= 90^\circ - 68^\circ42' = 21°18'$.

Distance and RL of Q

Both stations lie in the same vertical plane as Q, with P between R and Q. Let $x$ = horizontal distance from R to Q, so the distance from P to Q is $x - D$ with $D = 120$ m.

                                  Q
                            _.--'|
                      _.--'      |
            P  _.--'  aP        |
   R _.--'  aR  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of Q above axis of R = $x\tan\alpha_R$ and above axis of P = $(x-D)\tan\alpha_P$. The RL of Q is the same from both stations:

$$\begin{aligned} RL_{axis\,R} + x\tan\alpha_R &= RL_{axis\,P} + (x-D)\tan\alpha_P \\ x &= \frac{(RL_{axis\,P} - RL_{axis\,R}) - D\tan\alpha_P}{\tan\alpha_R - \tan\alpha_P} \end{aligned}$$

With $\alpha_R = 21°18'00"$ ($\tan = 0.38988$), $\alpha_P = 38°21'00"$ ($\tan = 0.79117$), axis RL of R = 114.420 m and of P = 146.856 m:

$$x = \frac{(146.856 - 114.420) - 120\times 0.79117}{0.38988 - 0.79117} = 155.761\ \text{m}$$ $$RL_{Q} = 114.420 + 155.761\times 0.38988 = 175.149\ \text{m}$$

Check from P: $146.856 + (155.761 - 120)\times 0.79117 = 175.149$ m (same).

Answer: RL of the top of the tower Q = 175.149 m.