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 2 of 4.
T.Aashish said:
5 years ago
Thanks for explaining @Meghanada.
(1)
Sumit said:
5 years ago
In finding true angle (satellite station) reduction to center, then which formula is used? Please tell me.
Tarun Halder said:
6 years ago
Include angle b/w ABC = 108 - 146°30" + 68°30" = 102°.
(1)
Mohmmed said:
6 years ago
The Azimuth of a line AB is 114 and the angle ABC 126°30'20". Calculate the Bearing of BC?
Please answer this.
Please answer this.
KAN said:
7 years ago
BB of AB + FB of BC.
BB of AB is 180-146.30 = 33.30,
FB of BC is 68.30.
So, 33.30 + 68.30 = 102.
BB of AB is 180-146.30 = 33.30,
FB of BC is 68.30.
So, 33.30 + 68.30 = 102.
Shramik shanti said:
7 years ago
Applying formula..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.
(2)
Rakesh Kumar Baral said:
7 years ago
AB = 107*15'.
BC= 22*00'.
AB-BC= 107*15-22*00= 85*15'.
Included angle ABC= 180*-85*15'= 94*45'.
BC= 22*00'.
AB-BC= 107*15-22*00= 85*15'.
Included angle ABC= 180*-85*15'= 94*45'.
Aadarsh Devkate said:
7 years ago
180°-146°30°= 33°30°00°.
So,
<ABC = 68°30°+ 33°30°00° = 102°.
So,
<ABC = 68°30°+ 33°30°00° = 102°.
Saurabh kumar said:
8 years ago
As per my knowledge, Interior angle = BB of AB - FB of BC.
Are interior angle and included angle same.
Are interior angle and included angle same.
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!
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