Question : The side $BC$ of a triangle $ABC$ is extended to $D$. If $\angle ACD = 120^{\circ}$ and $\angle ABC = \frac{1}{2} \angle CAB$, then the value of $\angle ABC$ is:
Option 1: $80^{\circ}$
Option 2: $40^{\circ}$
Option 3: $60^{\circ}$
Option 4: $20^{\circ}$
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Correct Answer: $40^{\circ}$
Solution :
Given that $\angle ACD = 120^\circ$ and $\angle ABC = \frac{1}{2} \angle CAB$
Let $\angle ABC = x$, Then $\angle CAB = 2x$
An exterior angle of a triangle is equal to the sum of opposite interior angles.
So, $\angle ACD = \angle ABC + \angle CAB$
⇒ $120^\circ = x + 2x$
⇒ $3x = 120^\circ$
$\therefore x = 40^\circ$
Hence, the correct answer is $40^\circ$.
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