# Aptitude Question ID : 93489

The side BC of a triangle ABC is extended to D. If ∠ACD = 120° and ∠ABC $= \frac{1}{2}$ ∠CAB, then the value of ∠ABC is:
[A]20°
[B]80°
[C]40°
[D]60°

40°
∠CAB = 2 ∠ABC
∠ACB + ∠ACD = 180°
= ∠ACB + 120° = 180°
= ∠ACB = 180° – 120° = 60°
∴ ∠A + ∠B = 180° – 60° = 120°
= 2 ∠B + ∠B = 120°
$= \angle B = \frac{120^{\circ}}{3} = 40^{\circ}$
Hence option [C] is the right answer.