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.
Karthikranga said:
1 decade ago
How you get this answer can you explain the procedure for 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)
Mr. Salim Mulla said:
9 years ago
152° 30'+ 180° = 332°30'.
360° - 332° 30' = 27° 30°.
Then do 124° 28' - 27° 30' = 96° 58'.
Hope you understand.
360° - 332° 30' = 27° 30°.
Then do 124° 28' - 27° 30' = 96° 58'.
Hope you understand.
Sameer said:
7 years ago
Thanks a lot for explaining @Mr Salim.
(1)
Alok said:
7 years ago
I am not getting the solution. Please, anyone explain me.
Vikas Dubey said:
6 years ago
Bearing of any line = BB.of line + angle.
If angle is more than 180' than, -360'.
You will find the answer.
If angle is more than 180' than, -360'.
You will find the answer.
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'.
Adeel Rehman said:
5 years ago
Best Explanation, Thanks @Julliet Osunde.
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°).
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