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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