# Theodolites, Compass and Traverse Surveying

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 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 8
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 8 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 $180^{\circ}$.
 Question 9
The interior angles of four triangles are given below:

Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?
 A Both P and R B Both Q and R C Both P and S D Both Q and S
GATE CE 2019 SET-1   Geomatics Engineering
Question 9 Explanation:
For an ill conditioned traingle in triangulation survey, any angle can be less than $38^{\circ}$, and can be greater than $38^{\circ}$.
For traingles Q and S, the above condition is valid.
 Question 10
The following details refer to a closed traverse:

The length and direction (whole circle bearing) of closure, respectively are
 A 1 m and $90^{\circ}$ B 2 m and $90^{\circ}$ C 1 m and $270^{\circ}$ D 2 m and $270^{\circ}$
GATE CE 2018 SET-1   Geomatics Engineering
Question 10 Explanation:
\begin{aligned} \text { Latitude }_{P Q} &=L_{P Q}=-437 \mathrm{m} \\ L_{S P} &=-83 \mathrm{m} \\ L_{O R} &=+101 \mathrm{m}\\ L_{R S} &=+419 \mathrm{m} \\ \Sigma \text { Latitudes } &=-437-83+101+419 \\ &=0 \mathrm{m} \\ \text { Departure }_{P O} &=D_{P O}=+173 \mathrm{m} \\ D_{O R} &=+558 \mathrm{m} \\ D_{8 S} &=-96 \mathrm{m} \\ D_{S P} &=-634 \mathrm{m} \end{aligned}
$\Sigma$ Departures =1 m For closure of traverse
$\Sigma$ Latitude =0
$\Sigma$ Departure =0
Then,
Departure of closure =-1 m,
Latitude of closure =0

So the length and direction (whole circle bearing of closure is 1m and $270^{\circ}$ respectively.
There are 10 questions to complete.