For P to Q, face left is 59°18' and face right is 300°36'. How is the mean zenith found?

Average of 59°18' and (360° - 300°36') = 59°24', giving 59°21'

Average of 59°18' and 300°36'

With P between R and Q, which expression gives x, the horizontal distance R to Q?

(axis P - axis R - D tan(alpha_P)) / (tan(alpha_R) - tan(alpha_P))

(axis R - axis P) / (tan(alpha_R) + tan(alpha_P))

D / (tan(alpha_R) - tan(alpha_P))

(axis P + axis R) / tan(alpha_R)

What is the RL of Q, and how can it be checked?

1375.519 m; check from P using (x - 112) tan(alpha_P)

1281.420 m; check from R using the zenith angle

1289.856 m; no check is possible

1375.519 m; check by adding 1.47 m

Asked 4/26
2076 Baisakh · 6 marks

RL of hill top Q from stations P and R

Short answer

Mean zenith angles reduced to elevation; axis RLs: R = 1281.420 m, P = 1289.856 m. Solving simultaneous height equations gives RL of hill top Q = 1375.519 m.

The top of the hill station Q was sighted from two instrument stations P and R located at very different level and in the same vertical plane to the target and following observations were noted.

Inst st'n	HI (m)	Target	Zenith Angle FL	Zenith Angle FR	Distance	Target ht.
P	1.47	Q	59°18'	300°36'	PR = 112.00 m
R	1.42	Q	69°52'	290°09'
R	1.42	P	85°32'	274°26'		1.75 m (vane)

If RL of station R was 1280.00 m, find the RL of top of hill.

Mean vertical angles

Sight	Face left	360° − FR	Mean zenith	Elevation
P to Q	59°18'	59°24'	59°21'00"	30°39'00"
R to Q	69°52'	69°51'	69°51'30"	20°08'30"
R to P	85°32'	85°34'	85°33'00"	4°27'00"

RL of instrument axes

Axis of R = RL of R + HI = 1280.000 + 1.42 = 1281.420 m
Sight R to P: elevation 4°27'. The vane (1.75 m above the foot of P) is at 1281.420 + 112 tan(4°27') = 1290.136 m, so the ground at P = 1288.386 m and axis of P = 1288.386 + 1.47 = 1289.856 m.

Elevation angles to Q: from R = 20°08'30", from P = 30°39'00". Take D = 112 m as the horizontal distance RP.

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 $D = 112$ 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 $\alpha_R = 20°08'30"$ ( $\tan = 0.36677$ ), $\alpha_P = 30°39'00"$ ( $\tan = 0.59258$ ), axis RL of R = 1281.420 m and of P = 1289.856 m:

x = \frac{(1289.856 - 1281.420) - 112\times 0.59258}{0.36677 - 0.59258} = 256.560\ \text{m}

RL_{Q} = 1281.420 + 256.560\times 0.36677 = 1375.519\ \text{m}

Check from P: $1289.856 + (256.560 - 112)\times 0.59258 = 1375.519$ m (same).

Answer: RL of the top of the hill Q = 1375.519 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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