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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Question : $\triangle ABC$ is an isosceles triangle with AB = AC. If $\angle BAC=50^\circ$, then the degree measure of $\angle ABC$ is equal to:
Option 1: $70^\circ$
Option 2: $55^\circ$
Option 3: $60^\circ$
Option 4: $65^\circ$
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: $90^{\circ}$
Question : In figure, $\angle \mathrm{ADB}$ is equal to:
Option 1: $40^\circ$
Option 2: $60^\circ$
Option 3: $80^\circ$
Option 4: $50^\circ$
Question : $ABC$ is an isosceles triangle with $AB = AC$, The side $BA$ is produced to $D$ such that $AB = AD$. If $\angle ABC = 30^{\circ}$, then $\angle BCD$ is equal to:
Option 1: $45^{\circ}$
Option 2: $90^{\circ}$
Option 3: $30^{\circ}$
Option 4: $60^{\circ}$
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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