Q1: The bearings of the lines AB and BC are 146° 30' and 68° 30'. The included angle ABC is

ANS:A - 102°

Includes angle = fb of next line - bb of the previous line.
= 68.30 - (146.30 + 180).
= 63.38 - 326.30.
= -258.

Angel can not be negative so it will be added by 360.
= -258 + 360.
=102. 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°. 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. Included angle = Bearing of previous line - Bearing of next line
= bearing of ba - bearing of bc
= (180+146.5)-(68.5) = 258 which is exterior angle.
Hence interior angle = 360-258 = 102. Both f.b.
So BC of B.B =68°30'+180° = 248°30'.
AB of f.b. = 146°30'.
/ABC = 248°30' - 146°30' = 102° answer.