Question : If in $\triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}$ and $\angle \mathrm{ACD}=125^{\circ}$, then $\angle \mathrm{BAC}$ is:
Option 1: $75^\circ$
Option 2: $55^\circ$
Option 3: $60^\circ$
Option 4: $70^\circ$
Correct Answer: $70^\circ$
Solution : Given: $\angle ACD= 125^\circ$ $\therefore \angle ACB+\angle ACD=180^\circ$ ⇒ $\angle ACB=180^\circ-\angle ACD=180^\circ-125^\circ=55^\circ$ Also, $AB = AC$ $\therefore \angle ACB = \angle ABC=55^\circ$ In $\triangle ABC$, $\angle BAC+\angle ABC +\angle ACB=180^\circ$ ⇒ $\angle BAC +55^\circ+55^\circ =180^\circ$ $\therefore\angle BAC =70^\circ$ Hence, the correct answer is $70^\circ$.
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Option 2:
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