Question : In $\triangle A B C $, AB and AC are produced to points D and E, respectively. If the bisectors of $\angle C B D$ and $\angle B C E$ meet at the point O, and $\angle B O C=57^{\circ}$, then $\angle A$ is equal to:
Option 1: 93°
Option 2: 66°
Option 3: 114°
Option 4: 57°
Correct Answer: 66°
Solution :
Given: $\angle BOC = 57^\circ$
⇒ $\angle BOC = 90^\circ – \frac{\angle BAC}{2}$
⇒ $\frac{\angle BAC}{2} = 90^\circ – 57^\circ$
⇒ $\frac{\angle BAC}{2} = 33^\circ$
⇒ $\angle BAC = 66^\circ$
Hence, the correct answer is $66^\circ$.
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