Question : I is the incenter of a triangle ABC. If $\angle$ ABC = 65$^\circ$ and $\angle$ ACB = 55$^\circ$, then the value of $\angle$ BIC is:
Option 1: 130$^\circ$
Option 2: 120$^\circ$
Option 3: 140$^\circ$
Option 4: 110$^\circ$
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Correct Answer: 120$^\circ$
Solution :
Given, triangle ABC with I as an incenter of a triangle.
And, $\angle$ ABC = 65$^\circ$ and $\angle$ ACB = 55$^\circ$
We know that the incenter is the intersection point of all angle bisectors.
So, $\angle$ IBC = $\frac{\angle ABC}{2}$ = $\frac{65^\circ}{2}$
Also, $\angle$ ICB = $\frac{\angle ACB}{2}$ = $\frac{55^\circ}{2}$
Applying angle sum property in $\triangle$ BIC,
$\angle$ IBC +$\angle$ ICB + $\angle$ BIC = 180$^\circ$
or, $\frac{65^\circ}{2}+ \frac{55^\circ}{2}+\angle$ BIC = 180$^\circ$
or, $\angle$ BIC = 120$^\circ$
Hence, the correct answer is 120$^\circ$.
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