Adjust the bearing sheet and compute the traverse to determine the coordinates of the new points.
Adjust the bearing sheet and compute the traverse to determine the coordinates of the new points.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question

Transcribed Image Text:2. It is required to provide horizontal control in the area covered by points t1, t12, t3,14, t5 and t6 as shown
in Figure 2 below. The extract of the reduced data from the field notebook is as follows:
Bearings:
A - P = 50° 10' 15",
tl – t2 = 84 59' 58",
t3 – 12 = 00° 00'16",
14 – 15 = 205° 00' 11",
t6 – 15 = 80°00' 15",
B- Q = 125° 36' 30"
A – tl = 135° 00' 07",
t2 – tl = 265° 00’ 00",
t3 – 14 = 86 59' 59",
t5 – 14 = 25°00' 03",
16 – B = 182° 00' 24",
tl – A = 314 59' 52",
12 – 13 = 180° 00' 13",
14 – 13 = 266° 59' 53",
t5 – t6 = 259° 59' 59",
B – t6 = 02° 00' 24",
%3D
%3D
%3D
Distances :
A - tl = 69.97 m
tl - t2 =
75.52 m
To P
t2 - t3 = 80.11 m
t3 - 14 = 100.18 m
14- t5 = 95.59 m
t2
A
t5 - t6 = 65.74 m
t4
t6 - B = 90.80 m
t1
t3
The datum points have the following coordinates:
Figure 2
t6
t5
N
(m)
E
A
1000.00
200.00
B
693.50
316.50
B
P
1160.10
392.00
To Q
518.80
560.40
Adjust the bearing sheet and compute the traverse to determine the coordinates of the new points.
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