Basic Civil Engineering (3110004)

BE | Semester-1   Winter-2019 | 08-01-2020

Q5) (c)

Given below are the bearings of lines of a closed traverse ABCD. Calculate the interior angles.
Line F.B.
AB N45°E
BC N75°E
CD S35°W
DA N65°W

∠A = BB of line EA – FB of line AB = 180° – 45° – 65° = 70°
 
∠B = BB of line AB – FB of line BC = 180° – 75° + 45° = 150°
 
∠C = 360° – FB of line CD + BB of line BC = 75° – 35° = 40°
 
∠D= 360° – FB of line DE + BB of line CD = 65° + 35° = 100°
 
Sum of Included Angles = ∠A + ∠B + ∠C + ∠D = 70° + 150° + 40° + 100° = 360°
 
Check = (2N - 4) X 90° = 360°
 
Here sum of included angles = check   (hence O.K.)