Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 8 (Q.No. 14)
14.
The bearing of line AB is 152° 30' and angle ABC measured clockwise is 124° 28'. The bearing of BC is
Discussion:
21 comments Page 1 of 3.
Kanchan thakur said:
5 years ago
Bearing of AB=152°30'
<ABC=124°28'
Bearing of BC=?
Back bearing of AB = BA = F.B + 180 = 152°30' + 180° = 332°30'
Bearing of bc = b.b of AB+<ABC = 332°30' + 124°28' = 456°58'
This is more than 360 so;
Bearing of BC = 456°58' - 360° = 96°58'.
<ABC=124°28'
Bearing of BC=?
Back bearing of AB = BA = F.B + 180 = 152°30' + 180° = 332°30'
Bearing of bc = b.b of AB+<ABC = 332°30' + 124°28' = 456°58'
This is more than 360 so;
Bearing of BC = 456°58' - 360° = 96°58'.
Nada said:
4 years ago
The reduced bearing of the centerline of two roads AB and AC are N 30° W and N 30° E respectively. These two roads connected by a third road BC. The length and the bearing of BC is 200 m and N 75° E. Find the length of AB and AC. Can anyone answer this?
DHEERAJ said:
1 decade ago
Back bearing of ab line is 332 degree 30 min. Therefore angle reduced bearing with north is 360 minus above bearing which is 27 degrees 30 min. Then subtract this from above angle that is 124 degree 28 min minus 27 degree 30 min gives you answer.
Juliet osunde said:
1 decade ago
Bearing of AB = 152° 30'
Angle at B = 124° 28'
The bearing of BC=?
Back bearing= forward bearing + included angle,
CB=152 30' + 124 28'= 276 58`.
Since we asked to find the forward bearing i.e BC,
BC= 276 58` - 180 = 96 58'.
Angle at B = 124° 28'
The bearing of BC=?
Back bearing= forward bearing + included angle,
CB=152 30' + 124 28'= 276 58`.
Since we asked to find the forward bearing i.e BC,
BC= 276 58` - 180 = 96 58'.
(1)
Philologue said:
4 years ago
Bearing of AB=152°30'.
Angle ABC = 124°28'.
Bearing BC=?
Bearing BA = AB + 180 = 152°30'+180°= 332°30'.
Bearing BC= AB + angle ABC = 332°30' + 124°28' = 456°58'.
Then 456°58'-360°= 96°58'.
Angle ABC = 124°28'.
Bearing BC=?
Bearing BA = AB + 180 = 152°30'+180°= 332°30'.
Bearing BC= AB + angle ABC = 332°30' + 124°28' = 456°58'.
Then 456°58'-360°= 96°58'.
(10)
Burhan Ali said:
5 years ago
FB of AB=152°30'
BB of AB=152°30 ' +180° = 332°30'
FB of BC = BB of AB at B=332 °30' + 124°28' = 456°58'.
= 456°58' - 360° = 96°58'(remember if angle is more then 360° Subtract 360°).
BB of AB=152°30 ' +180° = 332°30'
FB of BC = BB of AB at B=332 °30' + 124°28' = 456°58'.
= 456°58' - 360° = 96°58'(remember if angle is more then 360° Subtract 360°).
Prakash Singh Thakuri said:
4 years ago
Fore bearing BC = back bearing AB + clockwise Angle ±180° ±540°.
BC = 152°30'+124°28'±180°±540°.
BC = 276°58'-180°.
BC = 96°58'.
Fore bearing BC=96°58' answer.
BC = 152°30'+124°28'±180°±540°.
BC = 276°58'-180°.
BC = 96°58'.
Fore bearing BC=96°58' answer.
(3)
Rajan said:
1 year ago
Bearing of AB =152°30'.
Angle ABC = 124°28'.
Bearing BC = ?
Bearing BA = AB + 180 = 152°30' + 180°= 332°30'.
Bearing BC = AB + angle ABC = 332°30' + 124°28' = 456°58'.
Then 456°58' - 360° = 96°58'.
Angle ABC = 124°28'.
Bearing BC = ?
Bearing BA = AB + 180 = 152°30' + 180°= 332°30'.
Bearing BC = AB + angle ABC = 332°30' + 124°28' = 456°58'.
Then 456°58' - 360° = 96°58'.
(2)
Aswathy said:
6 years ago
Forward Bearing of AB =152°30'
Back Bearing of AB= 180°-152°30=27°30'
FB of BC= Included angle - BB of AB
=124°28'-27°30'
=96°58'.
Back Bearing of AB= 180°-152°30=27°30'
FB of BC= Included angle - BB of AB
=124°28'-27°30'
=96°58'.
Niranjan said:
4 years ago
FB of BC = FB of AB+<ABC+-180.
= 152 * 30' + 124 * 28'-180° (first two values sum is more than 180 so use -ve sign).
= 276 * 58'-180° = 96° 58'.
= 152 * 30' + 124 * 28'-180° (first two values sum is more than 180 so use -ve sign).
= 276 * 58'-180° = 96° 58'.
(6)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers