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The bisectors BI and CI of \( \Large \angle B\) and \( \Large \angle C\) of \( \Large \triangle ABC\) meet in I. What is \( \Large \angle BIC\) equal to?

A) \( \Large 90 ^{\circ} - \frac{A}{4} \)
B) \( \Large 90 ^{\circ} + \frac{A}{4} \)
C) \( \Large 90 ^{\circ} - \frac{A}{2} \)
D) \( \Large 90 ^{\circ} + \frac{A}{2} \)

Correct Answer:




D) \( \Large 90 ^{\circ} + \frac{A}{2} \)

Description for Correct answer:

Given that, BI and CI are angle bisectors of \( \Large \angle B \) and \( \Large \angle C\), respectively.

Now, in \( \Large \triangle BIC, \)

\( \Large x ^{\circ} +\frac{B}{2}+\frac{C}{2}=180 ^{\circ} \) [let \( \Large \angle BIC = x ^{\circ}\)]

=> \( \Large x ^{\circ} = 180 ^{\circ} - \frac{1}{2} \left(B+C\right) \)

=> \( \Large x ^{\circ} = 180 ^{\circ} - \frac{1}{2} \left(180 ^{\circ} -A\right) \)

[\( \Large \because In \triangle ABC, A+B+C=180 ^{\circ} \)]

\( \Large x ^{\circ} = 180 ^{\circ} - 90 ^{\circ} +\frac{A}{2} \)

\( \Large \therefore \angle BIC = x ^{\circ} = 90 ^{\circ} +\frac{A}{2} \)

Part of solved Geometry questions and answers : >> Elementary Mathematics >> Geometry

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