Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 1 (Q.No. 40)
40.
The bearings of the lines AB and BC are 146° 30' and 68° 30'. The included angle ABC is
Discussion:
40 comments Page 1 of 4.
Apcivilian said:
8 years ago
Guys take as simple.
∠ABC = F.B of BC - B.B of AB
= 68°30'- (146°30'+180°)
= -258°.
= -258° + 360°.
= 102°.
Hope it will help!
∠ABC = F.B of BC - B.B of AB
= 68°30'- (146°30'+180°)
= -258°.
= -258° + 360°.
= 102°.
Hope it will help!
Dipankar Barman said:
4 years ago
BB of AB - FB of BC
(180° + 146°30') - 68°30'
= 326°30'- 68°30'.
= 258°(exterior angle),
So included angle = 360°- 258°,
= 102°.
(180° + 146°30') - 68°30'
= 326°30'- 68°30'.
= 258°(exterior angle),
So included angle = 360°- 258°,
= 102°.
(11)
Er Meghanada said:
8 years ago
Force bearing of AB 146°30'.
Back Bearing of AB 146°30'+180°=326°30'.
Exterior angle ABC=B.B-BC.
=326°30'-68°30'=258°.
Interior angle ABC=360°-258°=102°.
Back Bearing of AB 146°30'+180°=326°30'.
Exterior angle ABC=B.B-BC.
=326°30'-68°30'=258°.
Interior angle ABC=360°-258°=102°.
Darshan H A said:
10 years ago
Included angle = Bearing of previous line - Bearing of next line
= bearing of ba - bearing of bc
= (180+146.5)-(68.5) = 258 which is exterior angle.
Hence interior angle = 360-258 = 102.
= bearing of ba - bearing of bc
= (180+146.5)-(68.5) = 258 which is exterior angle.
Hence interior angle = 360-258 = 102.
Mukesh Shah said:
5 years ago
Includes angle = fb of next line - bb of the previous line.
= 68.30 - (146.30 + 180).
= 63.38 - 326.30.
= -258.
Angel can not be negative so it will be added by 360.
= -258 + 360.
=102.
= 68.30 - (146.30 + 180).
= 63.38 - 326.30.
= -258.
Angel can not be negative so it will be added by 360.
= -258 + 360.
=102.
(4)
Danesh said:
1 decade ago
Included angle = f.b of next line - b.b of previous line.
= f.b of BC - b.b of AB.
= 68.30 - 146.30.
= 78.
+180 = 102.
= f.b of BC - b.b of AB.
= 68.30 - 146.30.
= 78.
+180 = 102.
Sandip mondal said:
8 years ago
FB of AB=146°30'.
BB of AB=180°+146°30'=326°30.
Ext<ABC=326°30'-BB of BC=326°30'-68*30'=258°.
Int<ABC=360°-258°=102° ans.
BB of AB=180°+146°30'=326°30.
Ext<ABC=326°30'-BB of BC=326°30'-68*30'=258°.
Int<ABC=360°-258°=102° ans.
Sura said:
8 years ago
FB of AB 146°30'.
B. B. of AB 146°30'+180°=326°30'.
Exterior angle ABC=326°30'-68°30'=258°.
Interior angle ABC=360°-258°=102°.
B. B. of AB 146°30'+180°=326°30'.
Exterior angle ABC=326°30'-68°30'=258°.
Interior angle ABC=360°-258°=102°.
Ramjan said:
3 years ago
Angle ABC= 360*(326°30'-68°30')
=102°00'00"
since this will be anti-clockwise so;
Interior angle =360*[(BB)AB-(FB)BC].
=102°00'00"
since this will be anti-clockwise so;
Interior angle =360*[(BB)AB-(FB)BC].
(3)
Diganta Bikash Maji said:
8 months ago
FB of AB=146°30',
So BB of AB or FB of BA = 180° + 146°30' =326°30',
Exterior Angle ABC = (326°30' - 68°30') =258°,
Interior Angle = 360°- 258°) = 102°.
So BB of AB or FB of BA = 180° + 146°30' =326°30',
Exterior Angle ABC = (326°30' - 68°30') =258°,
Interior Angle = 360°- 258°) = 102°.
(3)
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