2. It is required to provide horizontal control in the area covered by points t1, t12, t3,14, t5 and t6 as shownin 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%3DDistances :A - tl = 69.97 mtl - t2 =75.52 mTo Pt2 - t3 = 80.11 mt3 - 14 = 100.18 m14- t5 = 95.59 mt2At5 - t6 = 65.74 mt4t6 - B = 90.80 mt1t3The datum points have the following coordinates:Figure 2t6t5N(m)EA1000.00200.00B693.50316.50BP1160.10392.00To Q518.80560.40Adjust the bearing sheet and compute the traverse to determine the coordinates of the new points.

Bearing is the horizontal angle measured from the north end of the reference meridian in the…

Answered: Adjust the bearing sheet and compute…

Engineering

Civil Engineering

Adjust the bearing sheet and compute the traverse to determine the coordinates of the new points.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

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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