This question was also avoided by most of the candidates and the few that attempted it
did poorly in it. From the given position vectors A, B and C, AB = -2i – 3j, BC = 4i-j. AB. BC = (-2i-3j). (4i-j) = -5. Thus, -5 = (√13) (√12) cos θ which makes cos θ = 0.3363.
This implies that θ = 109.65° ~ 109.7°. Hence, the required angle = 180° – 109.7° =70.3°.
In the part (b), the resolution of the forces gives.
(-T1 cos 30 ) ( T2 = cos 45) (50 cos 90 ) (0)
( ) + ( ) + ( ) =
( T1 sin 30) ( T2 sin 45) (50 sin 90) (0)
i.e. -√3T1 + √2 T2 + 0 = 0
2 2
1/2 T1 + √2 T2 + 50 = 0
2
Solving the equations simultaneously,
T1 = 36.603N and T2 = 44.829 N
Lami’s theorem could also be applied here. 
