SURVY1 202 (FUNDAMENTALS OF SURVEYING)
END-TERM QUIZ NO. 1 (1st Semester AY 2023-2024)
“The eyes of the Lord are in every place, Keeping watch on the evil and the good.” - Proverbs 15:3
I. MULTIPLE CHOICE. Choose the letter of the correct answer. Erasures not allowed. (2 points each)
_____1. It is the symbol used to represent the grid north.
a. Star
b. small blackened circle
c. Full arrow
d. Half arrow
_____2. A fixed line of reference parallel to the central meridian of a system of plane rectangular coordinates is
known as
a. True Meridian
b. Grid Meridian
c. Magnetic Meridian
d. True Meridian
_____ 3. The Angle between a line and the prolongation of the preceding line is called.
a. Interior angle
b. Angle to the left
c. Angle to the right
d. Deflection angle
_____ 4. What unit of angular measurement is used when the circumference is divided into 6,400 parts.
a. Angular
b. Circular
c. Grad
d. Mil
_____ 5. It is the Angle between a line and the prolongation of the preceding line
a. Deflection angle
b. Interior angle
c. Angle to the right
d. Azimuth
_____ 6. The characteristic phenomenon of the compass needle to be attracted downward from the horizontal plane
due to the earth’s magnetic lines of force is called ________________.
a. Local Attraction
b. Magnetic Declination
c. Magnetic Dip
d. Magnetic Pull
_____ 7. The angle and direction from the true meridian to the magnetic meridian is called
a. Declination
b. Azimuth
c. Angle to the right
d. Bearing
_____ 8. What is the acute horizontal angle between the reference meridian and the line?
a. Azimuth
b. Bearing
c. Deflection angle
d. Interior angle
_____9. The deflection angle 83.4486 grad is equal to
a. 75°10’ 37”
b. 83g 26c 55cc
c. 75°06’13”
d. 1,335.18 Mils
_____10. A bearing of S 88°15’ W if expressed in azimuth from north is equivalent to
a. 88° 15’
b. 268° 15’
c. 271°45’
d. 91°45’
II. PROBLEM SOLVING. Show your complete solutions in separate sheets of paper. (5 points each)
_____1. Compute angle RST from the given magnetic bearings: RS, S 42° 15’ E and ST, S 22° 05’ E.
a. 64° 20’
b. 20° 10’
c. 137°45’
d. None of these
Solution:
R
𝜃RST = 180⁰ - 42⁰ 15’ + 22⁰ 05’
S 42° 15’ E
𝜃RST = 159⁰ 50’
S
𝜃RST
S 22° 05’ E
_____2. Determine the bearing of line XY If its back azimuth is 53⁰ 40’.
a. S 53° 40’ W
b. S 46° 20’ E
c. N 53° 40’ E
Solution:
T
d. N 46° 20’ E
Y
Bearing of XY = 𝜃XY = N 53⁰ 40’ E
𝜃XY
X
53° 40’
_____3. The magnetic azimuth from north of line AB is 288° 41’. If the magnetic declination in the area is 3° 15’ W,
compute the true bearing of line AB.
a. N 15° 26’ W
b. N 78° 15’ W
c. S 78°15’ E
d. None of these
Solution:
TN
MN
𝜃AB = 360⁰ - 288⁰ 41’ + 3⁰ 15’ = 74⁰ 34’
𝜃AB
B
True Bearing of AB = N 74⁰ 34’ W
‫٭‬
3° 15’
A 288° 41’
____4. The interior angles of a five sided closed figure are ϴA = 85° 35’, ϴB = 98° 12’, ϴC = 133° 53’, ϴD = 126° 44’,
and ϴE = 94⁰46’. If the bearing of line CD = S 42° 15’ E, determine the bearing of line EA. Traverse in
clockwise route.
a. N 0° 45’ W
b. N 0° 25’ W
c. S 84°05’ E
d. N 84° 05’ W
𝜃BC
Solution:
B
C
∑ 𝑖𝑛𝑡. 𝑎𝑛𝑔𝑙𝑒𝑠 = 360⁰
𝜃
B
𝜃C
Int. Angle M-Value Error
Corr.
C-Value
42° 15’ 42° 15’
𝜃BA
𝜃A
85° 35’
+ 0⁰ 10’ 85° 45’
D
𝜃B
98° 12’
+ 0⁰ 10’ 98° 22’
𝜃D
𝜃AB
𝜃C
133° 53’
+ 0⁰ 10’ 134° 03’
𝜃ED 𝜃DE
𝜃D
126° 44’
+ 0⁰ 10’ 126° 54’
𝜃
EA
𝜃A
𝜃E
94⁰46’
+ 0⁰ 10’ 94⁰56’
A
𝜃E
Sum
539⁰10’ 0⁰50’
540⁰00’
E
𝜃AE
𝜃DE = 180⁰ - 126° 54’ - 42° 15’ = 10° 51’
𝜃EA = 94° 56’ - 10° 51’ = 84° 05’
Bearing of EA = N 84⁰ 05’ W
____5. Still in problem no. 4, determine the azimuth of line BC. (ans. 271⁰ 48’)
a. 3° 42’
b. 268° 12
c. 88° 12’
Solution:
𝜃AB = 180⁰ - 𝜃A - 𝜃AE = 180⁰ - 85° 45’ - 84⁰ 05’ = 10⁰ 10’
𝜃BC = 360⁰ - (𝜃B - 𝜃BA) = 360⁰ - (98° 22’ - 10⁰ 10’) = 271⁰ 48’
d. None of these
Azimuth of line BC = 271⁰ 48’
____6. The declination of the magnetic meridian in the year 1950 was 5° 15’ W and in 1990 it was 3° 12’ E. If the
magnetic bearing of line MN in 1950 was S 3° 08’ E, compute the true azimuth of line MN. (ans. 351⁰ 37’)
a. 171° 37’
b. 3° 07’
c. 8°23’
d. None of these
1950
MN
TN
Solution:
The True azimuth of a line never changes with time and remains
the same.
𝜃MN = 360⁰ - 5⁰ 15’ - 3⁰ 08’ = 351⁰ 37’
True Azimuth of MN = 351⁰ 37’
‫٭‬
5° 15’
𝜃MN
M
5° 15’
3° 08’
N