Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > In a close traverse, the fore bearing of line...
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In a close traverse, the fore bearing of lines AB and BC are 45°30' and 120°15' respectively. The angle ∠B will be equal to,
- a)254°45'
- b)200°15'
- c)210°45'
- d)215°15'
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In a close traverse, the fore bearing of lines AB and BC are 45°30...
Included angle = fore bearing of the forward line-back bearing of the previous line
Forward line = BC, previous line = AB
∠B = Fore bearing of BC -back bearing of AB
Back bearing of AB = 45°30' + 180° = 225°30'
∠B = Fore bearing of BC -back bearing of AB = 120°15'-225°30' = -105°15'
∠B = -105°15'+360° = 254°45'
Forward line = BC, previous line = AB
∠B = Fore bearing of BC -back bearing of AB
Back bearing of AB = 45°30' + 180° = 225°30'
∠B = Fore bearing of BC -back bearing of AB = 120°15'-225°30' = -105°15'
∠B = -105°15'+360° = 254°45'
Most Upvoted Answer
In a close traverse, the fore bearing of lines AB and BC are 45°30...
To solve this problem, we need to determine the relationship between the fore bearings of lines AB and BC.
In a close traverse, the sum of the interior angles is always equal to (n-2) * 180 degrees, where n is the number of sides or lines in the traverse.
Since we are given the fore bearing of lines AB and BC as 45 degrees, we can conclude that the sum of the interior angles of the traverse is 45 + 45 = 90 degrees.
Using the formula for the sum of the interior angles, we can set up the equation:
(n-2) * 180 = 90
Simplifying the equation, we have:
n - 2 = 0.5
n = 2.5
Since n cannot be a fractional value, we conclude that there is no solution to this problem.
In a close traverse, the sum of the interior angles is always equal to (n-2) * 180 degrees, where n is the number of sides or lines in the traverse.
Since we are given the fore bearing of lines AB and BC as 45 degrees, we can conclude that the sum of the interior angles of the traverse is 45 + 45 = 90 degrees.
Using the formula for the sum of the interior angles, we can set up the equation:
(n-2) * 180 = 90
Simplifying the equation, we have:
n - 2 = 0.5
n = 2.5
Since n cannot be a fractional value, we conclude that there is no solution to this problem.
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In a close traverse, the fore bearing of lines AB and BC are 45°30 and 120°15 respectively. The angle ∠B will be equal to,a)254°45b)200°15c)210°45d)215°15Correct answer is option 'A'. Can you explain this answer?
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