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Q1: The bearing of AB is 190° and that of CB is 260° 30'. The included angle ABC, is

A none of these

B 70° 30'

C 80° 30'

D 99° 30'

ANS:C - 70° 30'

BB of line AB = 190-180 = 10.
BB of line CB = 360-260 30' = 100 30'.
= 100 30'+10 = 110 30'.
= 180-110 30' = 70 30'. (190*-180*)+(270*-260*30')+90*.
=10*+9*30'+90*.
109*30'. The included angle ABC is 260°30'-190°=70°30'00''. As per your calculation, BB of line CB = 360-260 30' = 99 30' (check properly). AB=270-190=80.
BC=270-260.30=10.30.
Then 80-10.30=70.30.



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