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
102°
78°
45°
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
40 comments Page 2 of 4.

Apurba said:   8 years ago
360 - (180 + 146°30) + 68°30 = 102.

Indrajit Maiti said:   8 years ago
The bb of AB - fb of BC = {(180°+146°30')- 68°30'}= 258°,
Angle = (360°-258°) =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.

Pritam said:   8 years ago
But in book, Interior angle = BB of AB - FB of BC.

Are interior angle and included angle same.

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°.

Shashi said:   8 years ago
Yes, right @Chandan.

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°.

T vijay Bhaskar said:   8 years ago
Included angle =180-146°30'+68°30' = 102.

Leo said:   8 years ago
Option d is right.

Becuase when bearing of two line measured not from their point of intersection. The rule of finding angle as different.

Leo said:   8 years ago
Great explanation @ Meghanada. Thanks.


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