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DEPARTMENT OF CIVIL ENGINEERING

TOPIC-DETERMINATION OF INCLUDED ANGLE

SEMISTER-4^TH

PREPARED BY – ER SAROJ KUMAR NAYAK

(Lect. In Civil Engg Dept.)

AY:2021-2022

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Line	Fore bearing
AB	112°30 '
BC	27°15'
CD	276° 0'
DE	187°45'
EA	118°30'

The following bearings were noted down with a compass in an anticlockwise traverse. Calculate the interior angles of a closed traverse

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1. Angle A:

∠A = [FB of line AB - BB of line EA]

As fore bearing of EA < 180°, we have to add 180° to the FB.

BB of line EA

= [118°30' + 180° ]

= 298°30'

∠A = [112°30' - 298°30']

= -186°

�

In the whole circle bearing system, the angle cannot be -ve. In such cases, we have to add 360° to the -ve angle.

∠A = [-186° + 360°]

= 174°

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2. Angle B:

∠B = [FB of line BC - BB of line AB]

As fore bearing of AB < 180°, we have to add 180° to the FB.

BB of line A B

= [112°30' + 180° ]

= 292°30'

∠B = [27°15' - 292°30']

= -265°15'

∠B = [-265°15' + 360°]

= 94°45

3. Angle C:

∠C = [FB of line CD - BB of line BC]

As fore bearing of BC < 180°, we have to add 180° to the FB.

BB of line BC

= [27°15' + 180° ]

= 207°15'

∠C = [276° - 207°15']

= 68°45

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4. Angle D:

∠D = [FB of line DE - BB of line CD]

As fore bearing of CD > 180°, we have to deduct 180° from the FB.

BB of line CD

= [276° - 180° ]

= 96°

∠D = [187°45' - 96°]

= 91°45

5. Angle E:

∠E = [FB of line EA - BB of line DE]

As fore bearing of DE > 180°, we have to deduct 180° from the FB.

BB of line DE

= [187°45' - 180° ]

= 7°45'

∠E = [118°30' - 7°45']

= 110°45'

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Check:

The total sum of interior angles

= [∠A +∠B + ∠C +∠D + ∠E]

= [174° + 94°45' + 68°45' + 91°45' + 110°45']

= 540°

The total value of the angle should be equal to [(2n -4) ✕90°], where n= no. of sides.

[(2 ✕ 5 - 4) ✕ 90°] = [6 ✕ 90°] = 540° ✔