RL of radio tower top from hill tops A, B

A, B and T are in one vertical plane. B sees A at an angle of elevation, so A is the higher station and nearer the tower. Distance AB = 75 m is taken as horizontal. All the angles are angles of elevation.

RL of instrument axes

• Axis of B = RL of B + HI = 1121.500 + 1.5 = 1123.000 m
• Sight B to A: elevation 14°56'. The vane (3 m above the foot of A) is at 1123.000 + 75 tan(14°56') = 1143.003 m, so the ground at A = 1140.003 m and axis of A = 1140.003 + 1.4 = 1141.403 m.

Elevation angles to T: from B = 21°18'00", from A = 26°21'00". Take D = 75 m as the horizontal distance BA.

Distance and RL of T

Both stations lie in the same vertical plane as T, with A between B and T. Let $x$ = horizontal distance from B to T, so the distance from A to T is $x - D$ with $D = 75$ m.

                                  T
                            _.--'|
                      _.--'      |
            A  _.--'  aA        |
   B _.--'  aB  - - - - - - - - |
   |<--- D --->|<--- (x - D) --->|
   |<------------ x ------------>|

Height of T above axis of B = $x\tan\alpha_B$ and above axis of A = $(x-D)\tan\alpha_A$. The RL of T 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 $\alpha_B = 21°18'00"$ ($\tan = 0.38988$), $\alpha_A = 26°21'00"$ ($\tan = 0.49532$), axis RL of B = 1123.000 m and of A = 1141.403 m:

$$x = \frac{(1141.403 - 1123.000) - 75\times 0.49532}{0.38988 - 0.49532} = 177.800\ \text{m}$$ $$RL_{T} = 1123.000 + 177.800\times 0.38988 = 1192.321\ \text{m}$$

Check from A: $1141.403 + (177.800 - 75)\times 0.49532 = 1192.321$ m (same).

Horizontal distance from A to the tower $= x - D = 177.800 - 75 = 102.800$ m (and from B: 177.800 m).

Answer: RL of the top of the tower T = 1192.321 m; horizontal distance from A to the tower = 102.800 m.