Question : In $\Delta ABC$, the external bisector of the angles, $\angle B$ and $\angle C$ meet at the point $O$. If $\angle A = 70^\circ$, then the measure of $\angle BOC$:
Option 1: $55^\circ$
Option 2: $75^\circ$
Option 3: $60^\circ$
Option 4: $50^\circ$
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Correct Answer: $55^\circ$
Solution :
Given: $\angle A = 70^\circ$
To find: $\angle BOC$
We know that,
Angle formed by external angle bisectors = $90^\circ-\frac{1}{2}\angle A$
⇒ $\angle BOC$ = $90^\circ-\frac{1}{2}\angle A$
= $90^\circ-\frac{1}{2}×70^\circ$
= $90^\circ-35^\circ$
= $55^\circ$
Hence, the correct answer is $55^\circ$.
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