# Theodolites, Compass and Traverse Surveying

 Question 1
If the magnetic bearing of the Sun at a place at noon is $S2^{\circ}E$, the magnetic declination (in degrees) at that place is
 A $2^{\circ}E$ B $2^{\circ}W$ C $4^{\circ}E$ D $4^{\circ}W$
GATE CE 2022 SET-2   Geometics Engineering
Question 1 Explanation:
$MB=S2^{\circ}E=180^{\circ}-2^{\circ}=178^{\circ}$
$TB= 180^{\circ}$
Declination, $\delta =TB-MB=180-178=2^{\circ} \; or \; 2^{\circ}E$
 Question 2
The bearing of a survey line is $N31^{\circ}17'W$. Its azimuth observed from north is ______ deg. (round off to two decimal places)
 A 328.71 B 458.25 C 124.65 D 625.25
GATE CE 2022 SET-1   Geometics Engineering
Question 2 Explanation: $WCB=360-\left ( 31+\frac{17}{60} \right )^{\circ}=329.716^{\circ}$
 Question 3
For a given traverse, latitudes and departures are calculated and it is found that sum of latitudes is equal to +2.1 m and the sum of departures is equal to -2.8 m. The length and bearing of the closing error, respectively, are
 A $3.50 \mathrm{~m} \text { and } 53^{\circ} 7^{\prime} 48^{\prime \prime} \mathrm{NW}$ B $2.45 \mathrm{~m} \text { and } 53^{\circ} 7^{\prime} 48^{\prime \prime} \text { NW }$ C $0.35 \mathrm{~m} \text { and } 53.13^{\circ} \mathrm{SE}$ D $3.50 \mathrm{~m} \text { and } 53.13^{\circ} \mathrm{SE}$
GATE CE 2021 SET-2   Geometics Engineering
Question 3 Explanation:
\begin{aligned} e_{L} &=+2.1 \mathrm{~m} \\ e_{D} &=-2.8 \mathrm{~m} \\ e &=\sqrt{e_{L}^{2}+e_{D}^{2}} \\ &=\sqrt{(2.1)^{2}+(2.8)^{2}}=3.5 \mathrm{~m} \\ \text { Bearing of closing error } &=\tan ^{-1}\left(\frac{e_{D}}{e_{L}}\right) \\ &=\tan ^{-1}\left(\frac{-2.8}{2.1}\right)=-53.13^{\circ} \\ &=53^{\circ} 7^{\prime} 48^{\prime \prime} \mathrm{NW} \end{aligned}
 Question 4
A horizontal angle $\theta$ is measured by four different surveyors multiple times and the values reported are given below.
$\begin{array}{|c|c|c|} \hline \text { Surveyor } & \text { Angle } \theta & \text { Number of observations } \\ \hline 1 & 36^{\circ} 30^{\prime} & 4 \\ \hline 2 & 36^{\circ} 00^{\prime} & 3 \\ \hline 3 & 35^{\circ} 30^{\prime} & 8 \\ \hline 4 & 36^{\circ} 30^{\prime} & 4 \\ \hline \end{array}$
he most probable value of the angle $\theta$ ( in degree, round off to two decimal placesis ________
 A 12 B 28 C 36 D 44
GATE CE 2021 SET-2   Geometics Engineering
Question 4 Explanation:
\begin{aligned} \mathrm{MPV} &=\frac{\left(36^{\circ} 30^{\prime} \times 4\right)+\left(36^{\circ} \times 3\right)+\left(35^{\circ} 30^{\prime} \times 8\right)+\left(36^{\circ} 30^{\prime} \times 4\right)}{4+3+8+4} \\ &=36^{\circ} \end{aligned}
 Question 5
Traversing is carried out for a closed traverse PQRS. The internal angles at vertices P, Q, R and S are measured as $92^{\circ}$,$62^{\circ}$,$123^{\circ}$ and $77^{\circ}$, respectively. If fore bearing of line PQ is $27^{\circ}$, fore bearing of line RS (in degrees, in integer) is _________
 A 258 B 753 C 159 D 218
GATE CE 2021 SET-1   Geometics Engineering
Question 5 Explanation: $\begin{array}{l} Q=\left[\begin{array}{l} B B \text { of } P Q=27^{\circ}+180^{\circ}=207^{\circ} \\ F B \text { of } Q R=207^{\circ}-68^{\circ}=139^{\circ} \end{array}\right. \\ R=\left[\begin{array}{l} B B \text { of } Q R=139^{\circ}+180^{\circ}=319^{\circ} \\ F B \text { of } R S=319^{\circ}-123^{\circ}=196^{\circ} \end{array}\right. \end{array}$ $\begin{array}{l} Q=\left[\begin{array}{l} B B \text { of } P Q=27^{\circ}+180^{\circ}=207^{\circ} \\ F B \text { of } Q R=207^{\circ}+68^{\circ}=275^{\circ} \end{array}\right. \\ R=\left[\begin{array}{l} B B \text { of } Q R=275^{\circ}-180^{\circ}=95^{\circ} \\ F B \text { of } R S=95^{\circ}+123^{\circ}=218^{\circ} \end{array}\right. \end{array}$
 Question 6
A theodolite is set up at station A. The RL of instrument axis is 212.250 m. The angle of elevation to the top of a 4 m long staff, held vertical at station B, is $7^{\circ}$. The horizontal distance between station A and B is 400 m. Neglecting the errors due to curvature of earth and refraction, the RL (in m, round off to three decimal places) of station B is __________
 A 257.363 B 145.126 C 472.156 D 324.422
GATE CE 2020 SET-2   Geomatics Engineering
Question 6 Explanation: \begin{aligned} V&=400 \tan 7^{\circ}\\ &=49.113\\ x&=(49.113-4)=45.113\\ RL_B&=212.25+45.113\\ &=257.363m \end{aligned}
 Question 7
A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was $45^{\circ}$. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), is
 A 413.05 B 413.955 C 431.05 D 435.955
GATE CE 2020 SET-2   Geomatics Engineering
Question 7 Explanation: \begin{aligned} \frac{x}{20}&=\tan 45^{\circ}\\ x&=20m\\ RL \; of \; Q&=433.05+2.905-x-2\\ &=433.05+2.905-20-2\\ &=413.955m \end{aligned}
 Question 8
The length and bearings of a traverse PQRS are: The length of line segment SP (in m, round off to two decimal places), is ________.
 A 33.07 B 25.36 C 47.78 D 44.79
GATE CE 2020 SET-1   Geomatics Engineering
Question 8 Explanation:
\begin{aligned} \Delta L&= 40 \cos 80^{\circ}+50 \cos 10^{\circ}+30 \cos 210^{\circ}\\ &=30.20 \\ \Delta D&=40 \sin 80^{\circ}+50 \sin 10^{\circ}+30 \sin 210^{\circ} \\ &=33.07 \\ \text{Length, \; SP}&=\sqrt{\Delta L^2+\Delta D^2}\\ &=44.79m \end{aligned}
 Question 9
An open traverse PQRST is surveyed using theodolite and the consecutive coordinates obtained are given in the table If the independent coordinates (Northing, Easting) of station P are (400 m, 200 m) the independent coordinates (in m) of station T, are
 A 194.7, 370.1 B 205.3, 429.9 C 394.7, 170.1 D 405.3, 229.9
GATE CE 2020 SET-1   Geomatics Engineering
Question 9 Explanation: \begin{aligned} \Delta L &=-5.3 \\ \Delta D&=-29.9 \\ T, \; \text{Northing}&\{400+(-5.3)\}=394.7 \\ T, \; \text{Easting}&\{200+(-29.9)\}=170.1\\ &T[394.7m, 170.1m] \end{aligned}
 Question 10
The data from a closed traverse survey PQRS (run in the clockwise direction) are given in the table The closing error for the traverse PQRS (in degrees) is ___
 A 2 B 3 C 4 D 5
GATE CE 2019 SET-2   Geomatics Engineering
Question 10 Explanation:
Assuming it as anticlockwise traverse.
Mathematically sum of interior angle for a closed traverse $=(2n-4)\times 90 = (2 \times 4 -4) \times 90 = 360^{\circ}$
Given sum of interior angles, $= 88 + 92 + 94 + 89 = 363^{\circ}$
Then error in interior angle $= 363 - 360 = 3^{\circ}$
Note: In this question as per clockwise traverse included angle should be taken as exterior angle. But if we take exterior angle then we get all interior angles more than $180^{\circ}$.
There are 10 questions to complete.

### 1 thought on “Theodolites, Compass and Traverse Surveying”

1. Qno. 10 diagramn is wrong
Thanks for ur effort