Question : The internal bisectors of the angles B and C of a triangle ABC meet at I. If $\angle$BIC = $\frac{\angle A}{2}$ + X, then X is equal to:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $90^{\circ}$
Option 4: $45^{\circ}$
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Correct Answer: $90^{\circ}$
Solution :
$\angle$BIC = $\frac{\angle A}{2}$ + X
In $\triangle$ABC,
⇒ $\angle$A + $\angle$B + $\angle$C = $180^{\circ}$
⇒ $\angle$B + $\angle$C = $180^{\circ}$ – $\angle$A
⇒ $\frac{1}{2}$($\angle$B + $\angle$C) = $90^{\circ}$ – $\frac{\angle A}{2}$
In $\triangle$BIC,
⇒ $\frac{\angle B}{2}$ + $\frac{\angle C}{2}$ + $\angle$BIC = $180^{\circ}$
⇒ $90^{\circ}$ – $\frac{\angle A}{2}$ + $\angle$BIC = $180^{\circ}$
⇒ $\angle$BIC = $180^{\circ}$ – $90^{\circ}$ + $\frac{\angle A}{2}$ = $90^{\circ}$ + $\frac{\angle A}{2} = \frac{\angle A}{2}$ + X
⇒ X = $90^{\circ}$
Hence, the correct answer is $90^{\circ}$.
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