SATHEE: Quantitative Aptitude Ques 1821

Quantitative Aptitude Ques 1821

Question: The bisectors BI and CI of $ \angle B $ and $ \angle C $ of a $ \Delta ABC $ meet in I. What is $ \angle BIC $ equal to?

Options:

A) $ 90{}^\circ -\frac{A}{4} $

B) $ 90{}^\circ +\frac{A}{4} $

C) $ 90{}^\circ -\frac{A}{2} $

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

Show Answer

Answer:

Correct Answer: D

Solution:

Given that, $ BI $ and $ CI $ are angle bisectors of $ \angle B $ and $ \angle C $ respectively. Now, in $ \Delta BIC $ $ x{}^\circ +\frac{B}{2}+\frac{C}{2}=180{}^\circ $ [let $ \angle BIC=x{}^\circ $ ]

$ \Rightarrow $ $ x{}^\circ =180{}^\circ -\frac{1}{2}(B+C) $

$ \Rightarrow $ $ x{}^\circ =180{}^\circ -\frac{1}{2}(180{}^\circ -A) $ $ [\because \text{in }\Delta ABC,A+B+C=180{}^\circ ] $

$ \Rightarrow $ $ x{}^\circ =180{}^\circ -90{}^\circ +\frac{A}{2} $

$ \therefore $ $ \angle BIC=x{}^\circ =90{}^\circ +\frac{A}{2} $

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