Question : If in a $\triangle ABC$, as drawn in the figure, $AB = AC$ and $\angle ACD = 120^{\circ}$, then angle A is equal to:
Option 1: $50^{\circ}$
Option 2: $60^{\circ}$
Option 3: $70^{\circ}$
Option 4: $80^{\circ}$
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Correct Answer: $60^{\circ}$
Solution : Given that $AB = AC$ in $\triangle ABC$. This is an isosceles triangle, such that $\angle BAC = \angle ABC$. We have, $\angle ACD = 120^\circ$ $\therefore \angle ACB = 180^\circ - 120^\circ = 60^\circ$ Since the sum of angles in a triangle is $180^\circ$. ⇒ $\angle BAC + \angle ABC + \angle ACB = 180^\circ$ ⇒ $2\angle BAC + 60^\circ = 180^\circ$ $\therefore\angle BAC = \frac{180^\circ - 60^\circ}{2} = 60^\circ$ Hence, the correct answer is $60^\circ$.
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