Question : In a $\triangle$ ABC, BC is extended to D and $\angle$ ACD = $120^{\circ}$. $\angle$ B = $\frac{1}{2}\angle$ A. Then $\angle$ A is:
Option 1: $60^{\circ}$
Option 2: $75^{\circ}$
Option 3: $80^{\circ}$
Option 4: $90^{\circ}$
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Correct Answer: $80^{\circ}$
Solution : $\angle$ ACD = $120^{\circ}$ ⇒ $\angle$ B = $\frac{1}{2}\angle$ A ⇒ $\angle$ACD = $\angle$A + $\angle$B ⇒ $120^{\circ}$ = $\angle$A + $\frac{1}{2}\angle$ A = $\frac{3}{2}\angle$ A ⇒ $\angle$A = $\frac{120×2}{3}$ = $80^{\circ}$ Hence, the correct answer is $80^{\circ}$.
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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:
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