The sight from A to B has z = 105°30', so it is a depression of 15°30'. How is the axis RL of A found?

Vane level at B (1501.75) plus 150 tan 15°30'

Axis of B minus 150 tan 15°30'

Vane level at B minus 150 tan 15°30'

With A between B and P, what condition gives the horizontal distance x from B to P?

x tan(alpha_B) = (x - D) tan(alpha_A)

RL axis B - x tan(alpha_B) = RL axis A - x tan(alpha_A)

x sin(alpha_B) = D sin(alpha_A)

Asked 2/26
2081 Chaitra · 6 marks

RL of hill station P, A-B 150 m, RL B 1500

Short answer

Axis RLs: B = 1501.480 m, A = 1543.349 m. Solving with elevation angles 30°45′ and 20°15′ yields target RL = 1578.803 m, ground RL of hill station = 1576.303 m.

The top of a hill station P was sighted from two instrument stations A and B at a very different level and at the same vertical plane with the target. The zenith angle from A to P and B to P were 59°15' and 69°45' respectively. The zenith angle from A to B to a vane 1.75 m above the foot of the vane was 105°30'. If the height of the instrument of A and B were 1.56 m and 1.48 m respectively and distance between A and B was 150 m, RL of B was 1500 m. Find the RL of hill station (The target is 2.5 m above the ground).

Zenith angles are converted to elevation angles ( $\alpha = 90^\circ - z$ ). The sight from A to B (z = 105°30') is a depression, so A is above B.

RL of instrument axes

Axis of B = RL of B + HI = 1500.000 + 1.48 = 1501.480 m
Sight A to B: angle -15°30' (depression). The vane at B (1.75 m above the foot) is at 1500.000 + 1.75 = 1501.750 m, so the axis of A = 1501.750 − 150 tan(-15°30') = 1543.349 m.

Elevation angles to P: from B = 20°15'00", from A = 30°45'00". Take D = 150 m as the horizontal distance BA.

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 $D = 150$ 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 $\alpha_B = 20°15'00"$ ( $\tan = 0.36892$ ), $\alpha_A = 30°45'00"$ ( $\tan = 0.59494$ ), axis RL of B = 1501.480 m and of A = 1543.349 m:

x = \frac{(1543.349 - 1501.480) - 150\times 0.59494}{0.36892 - 0.59494} = 209.594\ \text{m}

RL_{P} = 1501.480 + 209.594\times 0.36892 = 1578.803\ \text{m}

Check from A: $1543.349 + (209.594 - 150)\times 0.59494 = 1578.803$ m (same).

The target is 2.5 m above the ground at P, so

RL_{\text{ground at P}} = 1578.803 - 2.5 = 1576.303\ \text{m}

Answer: RL of the target = 1578.803 m; RL of the hill station (ground) = 1576.303 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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