Question : In $\triangle A B C, O$ is the point of intersection of the bisectors of $\angle B$ and $\angle A$. If $\angle B O C=108^{\circ}$, then $\angle B A O=$?
Option 1: 27°
Option 2: 40°
Option 3: 18°
Option 4: 36°
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Correct Answer: 18°
Solution :
In $\triangle$ABC, O is the point of intersection of the bisectors of $\angle$B and $\angle$A
$\angle$BOC = 108°
If the bisector of $\angle$ABC and $\angle$ACB of a triangle ABC meet at a point O then, $\angle$BOC = $90^\circ + \frac{1}{2}$$\angle$A
⇒ $108^\circ = 90^\circ + \frac{1}{2}$$\angle$A
⇒ $\frac{1}{2} \angle \text{A} = 108^\circ - 90^\circ$
⇒ $\frac{1}{2} \angle\text{ A} = 18^\circ$
⇒ $\angle$A = 36$^\circ$
We know that an angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.
⇒ $\angle$ BAO = $\frac{\angle \text{A}}{2}$
⇒ $\angle$ BAO = $\frac{36}{2}$
⇒ $\angle$ BAO = 18$^\circ$
$\therefore$ The value of $\angle$BAO = 18$^\circ$
Hence, the correct answer is 18$^\circ$.
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